Class 12 Medical Physics - Electromagnetic Induction

Define coefficient of mutual induction. If in the primary coil of a transformer, the current decreases from $0.8A$ to $0.2A$ in $4$ milliseconds, calculate the induced e.m.f in the secondary coil. Mutual inductance is $1.76H$ .

Coefficient of mutual induction:-

It is a measure of the induction between two circuits, It is the ratio of the electromotive force in a circuit to the corresponding change of current in neighboring circuit; usually measured in henries.

In given circuit,

$I_1=0.8 , I_2=0.2 , \Delta t=4\times 10^{-3}, M=1.76$

But,

Induced emf

$=V_{ind}=M\times \dfrac{I_1-I_2}{\Delta t}$

$\therefore V_{ind}=1.76\times \dfrac{0.8-0.2}{4\times 10^{-3}}$

$\therefore V_{ind}=264V$

A rectangular frame of wire $abcd$ has dimensions $32\ cm\times 8.0\ cm$ and a total resistance of $2.0\Omega$ . It is pulled out of a magnetic field $B = 0.020\ T$ by applying a force of $3.2\times 10^{-5}N$ (figure). It is found that the frame moves with constant speed. Find the emf induced in the loop.

In the given figure, emf induced

$e$ =

$VBl$

If loop move with constant speed therefore electromagnetic force=external

$F$ .

$F=\dfrac{VB^{2}l^{2}}{R}=F$

$V=25 ms^{-1}$

Now,emf

$e=25 \times 0.02 \times 0.08$

$e=0.04V$

Answer the following :
A metallic rod $PQ$ of length $l$ is rotated with an angular velocity $\omega$ about an axis passing through its mid-point (O) and perpendicular to the plane of the paper, in uniform magnetic field $\vec{B}$ , as shown in the figure. What is thee potential difference developed between the two ends of the rod, P and Q ?

Given,

Length

$= l$

Magnetic field = B

Emf induce in

$O Q$ will be equal to

$OP$ .

And O will be at high potential and P, Q will at low potential

$V_{OQ} = V_{OQ} = \dfrac{1}{2} \omega^2 \left(\dfrac{l}{2} \right)$

$\therefore V_P = V_Q$

$\therefore V_{PQ} = O$

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A wire length $10cm$ translates in a direction making an angle of ${60}^{o}$ with its length. The plane of motion is perpendicular to a uniform magnetic field of $1.0T$ that exists in the space. Find the emf induced between the ends of the rod if the speed of translation is $20cm$ ${s}^{-1}$

$l=10cm=0.1m;\theta ={60}^{o};B=1T;V=20m/s=0.2m/s$

$E=Bvl\sin{60}^{o}$
As we know to take that component of length vector which is perpendicular to the velocity vector

$=1\times 0.2\times 0.1\times \sqrt{3}/2$

$=17.32\times {10}^{-3}V$
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Figure shows a metallic wire of resistance $0.20\Omega$ sliding on a horizontal, U-shaped metallic rail. The separation between the parallel arms is $20cm$ . An electric current of $2.0\mu A$ passes through the wire when it is slid at a rate of $20cm$ ${s}^{-1}$ . If the horizontal component of the earth's magnetic field is $3.0\times { 10 }^{ -5 }T$ . Calculate the dip at the place.

$l=20cm=20\times { 10 }^{ -2 }m;v=20m/s=20\times { 10 }^{ -2 }m/s;{ B }_{ H }=3\times { 10 }^{ -5 }T;i=2\mu A=2\times { 10 }^{ -6 }A;R=0.2\Omega$

$i=\cfrac { { B }_{ v }lv }{ R } \Rightarrow { B }_{ v }=\cfrac { iR }{ lv } =\cfrac { 2\times { 10 }^{ -6 }\times 2\times { 10 }^{ -1 } }{ 20\times { 10 }^{ -2 }\times 20\times { 10 }^{ -2 } } =1\times { 10 }^{ -5 }Tesla$

$\tan { \delta } =\cfrac { { B }_{ v } }{ { B }_{ H } } =\cfrac { 1\times { 10 }^{ -5 } }{ 3\times { 10 }^{ -5 } } =\cfrac { 1 }{ 3 } \Rightarrow \delta (dip)\tan ^{ -1 }{ 1/3 }$

A current of $1.0A$ is established in a tightly wound solenoid of radius $2cm$ having $1000$ turns/metre. Find the magnetic energy stored in each metre of the solenoid.

$i=1.0A,r=2cm,n=1000turn/m$
magnetic energy stored

$=\cfrac { { B }^{ 2 }V }{ 2{ \mu }_{ 0 } }$
where

$B$ -magnetic field,

$V$ - Volume of solenoid

$=\cfrac { { \mu }_{ 0 }{ n }^{ 2 }{ i }^{ 2 } }{ 2{ \mu }_{ 0 } } \times \pi { r }^{ 2 }h=\cfrac { 4\pi \times { 10 }^{ -7 }\times { 10 }^{ 6 }\times 1\times \pi \times 4\times { 10 }^{ -4 }\times 1 }{ 2 }$ (

$h=1m$ )

$=8{ \pi }^{ 2 }\times { 10 }^{ -5 }=7.9\times { 10 }^{ -4 }J$

An average emf of $20V$ is induced in an inductor when the current in it is changed from $2.5A$ in one direction to the same value in the opposite direction in $0.1s$ . Find the self-inductance of the inductor.

$V=20V;dI={ I }_{ 2 }-{ I }_{ 1 }=2.5-(-2.5)=5A;dt=0.1s\quad$

$V=L\cfrac { dI }{ dt } \Rightarrow 20=L(5/0.1)\Rightarrow 20=L\times 50\Rightarrow L=20/50=4/10=0.4Henr$

How many times, the current produced by $AC$ Generator reverses its direction in one complete revolution of its coil?

$AC$ generator supplies

$AC$ current. To determine how many times the direction of an Alternating Current changes we have to find out about its frequency. Frequency of an

$AC$ current implies number of complete waves present in a second. If the coil completes one revolution in one second then frequency of the

$AC$ generator is said to be

$1\ hertz$ . The given picture will clear the concept.
In the picture coil completes a revolution in a second, hence its frequency is

$1$ hertz. Hence we can say that in one revolution the current change its direction twice i.e first positive to negative then negative to positive.
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How many times $AC$ supply in India reversed its direction in one second.

Frequency of an

$AC$ current implies number of complete waves present in a second. In the picture there is one wave in a second, hence its frequency is

$1$ hertz. Hence we can say that in one revolution the current change its direction twice i.e first positive to negative then negative to positive.

In India, the

$Ac$ current has a frequency of

$50\ Hertz$ .

$50\ hertz$ frequency means that there are

$50$ waves present in the interval of one second. And we know that in one wave current changes its direction twice, so in

$50$ waves it will change its direction

$100$ times.

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A plot of magnetic flux $(\varphi )$ versus current $(I)$ is shown in the figure for two inductors $A$ and $B$ . Which of the two has larger value of self inductance?

$\varphi =LI$
For same current,

$\varphi _A > \varphi_B$ , so

$L_A > L_B$
i.e., Inductor

$A$ has larger value of self- inductance.

Solve a problem differing from the foregoing one by a magnetic field with induction $B = 0.8\ T$ replacing the electric field.

Choose

$\vec {B}$ in the

$z$ direction, and the velocity

$\vec {v} = (v\sin \alpha, 0, v\cos \alpha)$ in the

$x - z$ plane, then in the

$K'$ frame,

$\vec {E'}_{\parallel} = \vec {E}_{\parallel} = 0 | \vec {B'}_{\parallel} = \vec {B}_{\parallel}$

$\vec {E'}_{\perp} = \dfrac {\vec {v}\times \vec {B}}{\sqrt {1 - v^{2}/c^{2}}} | \vec {B'}_{\perp} = \dfrac {\vec {B}_{\perp}}{\sqrt {1 - v^{2}/ c^{2}}}$
We find similarly,

$E' = \dfrac {c\beta B\sin \alpha}{\sqrt {1 - \beta^{2}}}$

$B' = B \sqrt {\dfrac {1 - \beta^{2} \cos^{2} \alpha}{1 - \beta^{2}}} \tan \alpha' = \dfrac {\tan \alpha}{\sqrt {1 - \beta^{2}}}$ .

An inductor with an inductance of $3.00H$ and a resistance of $7.00\Omega$ is connected to the terminals of a battery with an emf of $12.0V$ and negligible internal resistance. Find the initial rate of increase of current in the circuit.

In the initial stage the inductor will acts like a open circuit and all the voltage drop across the inductor will be equal to the e.m.f of battery

$\quad E=L{ \left( \cfrac { di }{ dt } \right) }_{ 0 }$

${ \left( \cfrac { di }{ dt } \right) }_{ 0 }\quad =\cfrac { E }{ L } =\cfrac { 12 }{ 3 } =4A/s$

Which end of the inductor, $a$ or $b$ , is at a higher potential?

Current is decreasing from b to a.

According to Lenz's law emf is induced such as to maintain current constant.

i.e,induced voltage is such a way

$v_a >v_b$

What is the current when $t=2.0ms$ ?

${V}_{L}=L\cfrac { di }{ dt }$

$\therefore \quad di=\cfrac { 1 }{ L } \left( { V }_{ L }dt \right)$

$\therefore \quad \int { di=i= } \cfrac { 1 }{ L } \int { { V }_{ L }dt }$
At

$t=2ms$

$i={ \left( 150\times { 10 }^{ -3 } \right) }^{ -1 }\left( \cfrac { 1 }{ 2 } \times 2\times { 10 }^{ -3 }\times 5 \right) =3.33\times { 10 }^{ -2 }A\quad$

According to which law current $I$ flowing in the rod must vary for the rod to rotate at a constant angular speed. Begin to measure the time from the instant when the rod is in its right-hand horizontal position. Consider the current to be positive when it flows from the axis of rotation toward the ring.

Lenz law:

Lenz's law staes that the direction of induced current or voltage will be such as to oppose the cause producing it.

Given situation illustrates that when magnet move towards loop current induced anticlockwise it means mechanical energy converted into electrical energy.

Interpret K' - K.

Some possible interpretations are
i. It is the magnetic energy stored in shell.
magnetic energy = magnetic energy density (B

$^2$ /2

$\mu_0$ )

$\times$ Volume (

$\pi R^2 l$ ) and / or
ii. It is the self inductance energy (1/2Li

$^2$ ) of the system.
iii. Poynting vector argument can also show that it is magnetic energy.
(2

$\pi$ )Rl

$\displaystyle{\left(\frac{1}{\mu_0}\int{\bar{E} \times \bar{B} dt} \right)}$ = K' - K =

$\displaystyle{\frac{-B\pi R^2 l}{2\mu_0}}$

A frame $ABCD$ is rotating with an angular velocity $\omega$ about an axis passing through point $O$ perpendicular to the plane of paper as shown in the figure. A uniform magnetic field $\vec { B }$ is applied into the plane of the paper in the region as in the figure. Match the following.

Potential difference between the ends of a rod of length

$l$ rotating about one of its end with angular speed

$w$ is given by

$\mathcal{E} = \dfrac{Bwl^2}{2}$ .............(1)

(A) : Join A and O

The rod AO rotates with

$w$ about O

$\therefore$

$\mathcal{E_{AO}} = \dfrac{Bw (AO)^2}{2} = \dfrac{Bw (\sqrt{2}L)^2}{2}$

$\implies \mathcal{E_{AO}} = BwL^2 = constant$

(B) : Join O and D

Similarly

$\mathcal{E_{DO}} = BwL^2 = constant$

$OC$ rotates with

$w$ about O.

$\therefore$

$\mathcal{E_{CO}} = \dfrac{Bw (CO)^2}{2}$

$\implies \mathcal{E_{CO}} = \dfrac{BwL^2}{2}$

Potential between C and D

$\mathcal{E_{DC}} = \mathcal{E_{DO}} - \mathcal{E_{CO}}$

$\therefore \mathcal{E_{DC}} = BwL^2 - \dfrac{BwL^2}{2} = \dfrac{BwL^2}{2} = constant$

(D) : Potential between A and D

$\mathcal{E_{DA}} = \mathcal{E_{DO}} - \mathcal{E_{AO}}$

$\therefore \mathcal{E_{DA}} = BwL^2 - {BwL^2} = 0$

A conducting rod $AB$ moves parallel to the x-axis in a uniform magnetic field pointing in the positive z direction. The end $A$ of the rod gets positively charged. explain.

Let

$l$ be the length of rod AB.

Applying right hand rule,emf induced in direction of B to A.

i.e, A is higher potential induced emf

$e=vBl$

Electric field

$\overrightarrow { E }=\dfrac{e}{l}=vB$ from A to B.

The magnetic flux linked with the armature coil changed in a generator = $\Phi (t) = xNBA\cos (\omega t)$ then x=

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A solenoid has a cross sectional area of $6.0 \times 10^{-4} m^{2}$ , consists of 400 turns per meter, and carries a current 0.4 A and are connected to a circumference of the solenoid. The ends of the coil are connected to a 1.5 $\Omega$ resistor. Suddenly, a switch is opened, and the current in the solenoid dies to zero in a time 0.050 s. Find the average current passing through the coil during this time.

$\varepsilon = \dfrac{-d \phi}{dt}$

for

$N$ no. of turns,

$\varepsilon = -N \dfrac{d \phi}{dt}$

$= -N \dfrac{d(BA \cos (0))}{dt} =-NA \dfrac{dB}{dt}$

$I =- \dfrac{N \dfrac{dB}{dt} A}{R} =- \dfrac{NA \dfrac{\mu_0 n \Delta I}{dt}}{R}$

$= \dfrac{(10) (6 \times 10^{-4}) \dfrac{(4\pi \times 10^{-7}) (400) (0.4)}{0.050}}{1.5 \Omega}$

$I = 1.6 \times 10^{-5} A$

The voltage applied to a purely inductive coil of self inductance $15.9mH$ is given by the equation $V=100sin 314t$ + $75sin942t$ + $450sin1570t.$
Find the equation of current wave.

$\begin{array}{l}V = L\frac{{di}}{{dt}}\\or\,i = \frac{1}{L}\int {Vdt} \end{array}$

$\begin{array}{l}V = 100\sin 314t + 75\sin 942t + 50\sin 1570t\\i = \frac{1}{L}\int {\left( {100\sin 314t + 75\sin 942t + 50\sin 1570t} \right)dt} \\\,\,\, = \frac{1}{{15.9 \times {{10}^{ - 3}}}}\left[ { - \frac{{100}}{{314}}\cos 314t - \frac{{75}}{{942}}\cos 942t - \frac{{50}}{{1570}}\cos 1570t} \right]\\\,\,\, = \frac{1}{{15.9 \times {{10}^{ - 3}}}} \times \frac{1}{{1570}}\left[ {500\cos 314t + 125\cos 924t + 50\cos 1570t} \right]\\\,\,\, = \frac{{ - 1}}{{24.963}} \times 25\left[ {20\cos 314t + 5\cos 924t + 2\cos 1570t} \right]\\\,\,\, \approx - \left( {20\cos 314t + 5\cos 924t + 2\cos 1570} \right)\end{array}$
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Additional Problems(67)
(a) A flat, circular coil does not actually produce a uniform magnetic field in the area it encloses. Nevertheless, estimate the inductance of a flat, compact, circular coil with radius $R$ and $N$ turns by assuming the field at its center is uniform over its area. (b) A circuit on a laboratory table consists of a $1.50-volt$ battery, a $270-\Omega$ resistor, a switch, and three $30.0-cm$ -long patch cords connecting them. Suppose the circuit is arranged to be circular. Think of it as a flat coil with one turn. Compute the order of magnitude of its inductance and (c) of the time constant describing how fast the current increases when you close the switch.

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Energy Carried by Electromagnetic Waves(31)
Review. An AM radio station broadcasts isotropically (equally in all directions) with an average power of $4.00 \,kW$ . A receiving antenna $65.0 \,cm$ long is at a location $4.00 \,mi$ from the transmitter. Compute the amplitude of the emf that is induced by this signal between the ends of the receiving antenna.

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A solenoid of length $1 m$ and $0.05 m$ diameter has $500$ turns. If a current of $2 A$ passes through the coil, calculate the co-efficient of self induction of the coil.

Self-inductance

$=L=\dfrac { { \mu }_{ 0 }{ N }^{ 2 }A }{ l }$

$=\dfrac { 4\pi \times { 10 }^{ -7 }\times 500\times 500\times 3.14{ \left( 0.025 \right) }^{ 2 } }{ 1 }$

$=0.616\times { 10 }^{ -3 }H$

On what factors does the induced electromotive force depend?

Induced e.m.f for a coil of wire depends on :
A) Magnetic strength of the core in the coil of wire.
B) Number of turns of wire in the coil
C) The cross-sectional area of the coil.
D) Rate at which magnet is moved into/out of the coil.

Give two definitions of mutual inductance and write its unit.

$(i)$ Mutual inductance of two coils is equal to the

$e.m.f$ induced in one coil when when rate of changes of current through the other coil is unity

$.$

$(ii)$ Mutual inductance of two coils is defined as the magnetic flux linked with the secondary coil when the current in primary coil is

$1$ ampere

$.$

$S.I.$ unit of mutual induction is

$henry.$

What is an AC generator? Obtain an expression for the sinusoidal emf induced in the coil of ac generator, rotating with a uniform angular speed in a uniform magnetic field.

An AC generator is an electric generator that converts mechanical energy into electrical energy in form of alternative emf or alternating current. AC generator works on the principle of ”Electromagnetic Induction”

When the number of turn in a coil is doubled without, any change in the length of the coil , its self inductance becomes

We know that,

Self-inductance of a solenoid $=\dfrac{\mu A{{n}^{2}}}{l}$

So, self-induction proportional to ${{n}^{2}}$

So, induction becomes $4$ times when $n$ is doubled.

Hence, this is the required solution

A conducting circular loop having a radius of $5.0$ cm, is placed perpendicular to a magnetic field of $0.50 T$ . It is removed from the field in $0.50 s$ . Find the average emf produced in the loop during this time.

${ \phi }_{ 1 }=B.A=0.5\times { \left( 5\times { 10 }^{ -2 } \right) }^{ 2 }=5\pi .25\times { 10 }^{ -5 }=125\times { 10 }^{ -5 };{ \phi }_{ 2 }=0\quad$

$E=\cfrac { { \phi }_{ 1 }-{ \phi }_{ 2 } }{ t } =\cfrac { 125\pi \times { 10 }^{ -5 } }{ 5\times { 10 }^{ -1 } } =7.8\times { 10 }^{ -3 }\quad \quad$

A circular coil of $200$ turns and of radius $20 cm$ carries a current of $5A$ . Calculate the magnetic induction at a point along its axis, at a distance three times the radius of the coil from its centre.

Given:

$n=200\\a=20\ cm=.2\ m\\I=5\ A\\x=3a$

$B = \dfrac{n \mu_0 I a^2}{2 (a^2 + x^2)^{3/2}}$

Substitution and simplification

$B = 9.9 \times 10^{-5} T$

A stiff semi-circular wire of radius $R$ is rotated in a uniform magnetic field $B$ about an axis passing through its ends. If the frequency of rotation of wire is $f$ , calculate the amplitude of alternating emf induced in the wire.

Induced motion emf at the semicircle endpoints will be

$=\frac{1}{2}Bwl^{2}$

Here.

$l=2r$

Induced motion emf at the semicircle endpoints will be

$=\frac{1}{2}Bw(2r)^{2}$

Induced motion emf at semicircle endpoints will be

$=\pi ^{2}BR^{2}f$ .

A coil of insulated copper wire is connected to a galvanometer. What will happen if a bar magnet is pushed into the coil?

There is a momentary deflection in the needle of the galvanometer.

A pair of adjacent coils has a mutual inductance of $1.5 H$ . If the current in one coil changes from $0$ to $10 A$ in $0.2 s$ , what is the change of flux linkage with the other coil?

Given, mutual incidence of coil

$M=1.5H$ ,

Current change in coil

$dl=10-0=10A$ ,

Time taken in change

$dt=0.2s$ ,

Induced emf in the coil,

$e=M\frac{dl}{dt}=\frac{d\phi}{dt}$

or,

$d\phi =M.dl$

$=1.5\left ( \frac{10}{0.2} \right )$

$d\phi =75Wb$

Thus, the change in flux linkage is

$75Wb$ .

Figure shows a square loop of resistance 1 $\Omega$ of side 1 m being moved towards right at a constant speed of 1 m/s. The front edge enters the 3 m wide magnetic field (B = 1T) at t = 0 Draw the graph of current induced in the loop as time passes ( Take anticlockwise direction of current as positive)

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In a fluorescent lamp choke (a small transformer) $100$ V of reverse voltage is produced when the choke current changes uniformly from $0.25$ A to $0$ in a duration of $0.025$ ms. The self-inductance of the choke (in mH) is estimated to be _______.

Given

$\epsilon=100v$

$\Delta t=0.025ms=2.5\times 10^{-5}s$

$\Delta i=0.25-0=0.25A$

$\epsilon=L.\dfrac{\Delta i}{\Delta t}$

$100=L\times \dfrac{0.25}{2.5\times 10^{-5}}$

$100=L\times 10^4$

$L=0.01H$

$L=10\,mH$

$\boxed{Answer=10}$

What is a solenoid? Draw a diagram with a solenoid connected in a circuit. How can you increase the strength of a solenoid?

The solenoid is a long cylindrical coil of wire consisting of a large number of turns bound together very tightly.

The length of the coil should be longer than its diameter.

Magnetic field around a current carrying solenoid is shown in figure. These appear to be similar to that of a bar magnet.

When soft iron rod is placed inside the solenoid, it behaves like an electromagnet. The use of soft iron as core in the solenoid produces the strong magnetism.

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A circular coil of wire consist of exactly 100 turns with a total resistance 0.20 $\omega$ . The area of the coil is $100 cm^2$ . The coil is kept in a uniform magnetic field B as shown in fig 4.The magnetic field is increased at a constant rate of 2 T/s. Find the induced current in the coil in A

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Two discharge tubes have identical material structure and the same gas is filled in them. The length of one tube is $10 cm$ and that of the other tube is $20 cm$ . Sparking starts in both the tubes when the potential difference between the cathode and the anode is $100 V$ . If the pressure in the shorter tube is $1.0 mm$ of mercury, what is the pressure in the longer tube ?

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The current in a long solenoid of radius $R$ and having $n$ turns per unit length is given by $i={i}_{0}\sin \omega t$ . A coil having $N$ turns is wound around it near the centre. Find (a) the induced emf in the coil and (b) the mutual inductance between the solenoid and the coil.

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Name one device which converts electrical energy into mechanical energy.

Electrical motor is a device that converts electrical energy into mechanical energy.

Name the phenomenon which is made use of in an electric generator.

Electromagnetic Induction is the phenomenon used for the electric generator.

If you hold a coil of wire next to a magnet, no current will flow in the coil. What else in needed to induce a current?

In order to induce the current in the wire, relative motion between the coil and the magnet is required.

The inductance of a closely wound coil is such that an emf of 3.00 mV is induced when the current changes at the rate of 5.00 A/s. A steady current of 8.00 A produces a magnetic flux of $40.0\mu Wb$ through each turn. (a) Calculate the inductance of the coil. (b) How many turns does the coil have?

(a) From Eq. $\mathscr{E} _L=-L\dfrac{di}{dt}\text{ (self-induced current) }$ , we find $L=(3.00 \mathrm{mV}) /(5.00 \mathrm{A} / \mathrm{s})=0.600 \mathrm{mH}$

(b) since $N \Phi=i L \text { (where } \Phi=40.0 \mu \mathrm{Wb} \text { and } i=8.00 \mathrm{A}),$ we obtain $N=120$

How can the magnitude of the induced current be increased?

The magnitude of the induced current can be increased by:
(i) Taking the conductor in the form of a coil of many turns of insulated wire.
(ii) Increasing the strength of the magnetic field used.
(iii) Increasing the rate of change of magnetic flux associated with the coil.

Suggest two ways in an a.c. generator to produce a higher e.m.f.

Two ways in an a.c generator to produce a higher e.m.f. are:

(1) By increasing the speed of rotation of the coil.

(2) By increasing the number of turns of coil.

A magnetic substance of volume $30 cm^{3}$ is placed in a magnetising field of $5$ oersted. It produces a magnetic moment of $6 A/m^{2}$ . Calculate the magnetic induction.

Volume of magnetic substance

$V = 30cm^{3} = 30\times 10^{-6}m^{3}3$

Magnetising field

$H = 5$ oersted

Magnetic moment

$M = 6 A/m^{2}$

$B = \mu_{0}(H+I)$

$B = \mu_{0}\left[H+\dfrac{M}{V}\right]$

$B = 4\pi \times 10^{-7}\left[5\times 10^{3}+\dfrac{6}{30\times 10^{-6}}\right]$

$B = 4\times 3.14\times 10^{-7}[5\times 10^{3}+200\times 10^{3}]$

$B = 4\times 3.14\times 10^{-7}\times 205 \times 10^{3}$

$B = 2574.8\times 10^{-4}$

$B = 0.257T$

Two identical loops, one of copper and another of aluminium are rotated with the same speed in the same magnetic field. In which case, the
(a) The induced emf and
(b) induced current will be more and why?

The induced emf will be same in both but the induced current will be more in copper loop as its resistance will be lesser as compared to that of aluminium loop.

A series RLC circuit is driven by a generator at a frequency of $2000Hz$ and an emf amplitude of $170 V$ . The inductance is $60.0 mH$ , the capacitance is $0.400\mu F$ , and the resistance is $200\Omega$ . (a) What is the phase constant in radians? (b) What is the current amplitude?

(a) We observe that

$\omega _{d}=12566rad/s$ . Consequently,

$X_{L}=754\Omega$ and

$X_{C}=199\Omega$ . Similarly,

$\phi=\tan^{-1}\left ( \frac{X_{L}-X_{C}}{R} \right )=1.22rad$ .

(b) We find the current amplitude from equation

$I=\frac{\varepsilon _{m}}{\sqrt{R^{2}+\left ( X_{L}-X_{C} \right )^{2}}}$ :

$I=\frac{\varepsilon _{m}}{\sqrt{R^{2}+\left ( X_{L}-X_{C} \right )^{2}}}=0.288A$ .

A horizontal wire $20 m$ long extending from east to west is falling with a velocity of $10 m/s$ normal to the Earths magnetic field of $0.5*10^{-4} T$ . What is the value of induced emf in the wire?

Here,

Length of the wire,

$l=20m$ ,

Speed,

$v=10m/s$

Earth's magnetic field,

$B=0.5*10^{-4}T$

Value of e.m.f. induced in the wire,

$=Blv$ .

$=0.5*10^{-4}*20*10$

$=100*10^{-4}V$

$=10mV$ .

A horizontal straight wire $10m$ long extending from east to west is falling with a speed of $5.0ms^{-1}$ , at right angles to the horizontal component of the earth's magnetic filed, $0.30*10^{4}Wbm^{2}$ .
(a) What is the instantaneous value of the emf induced in the wire?
(b) What is the direction of the emf?
(c) Which end of the wire is at the higher electrical potential?

Here, length of the wire,

$l=10m$ ;
Velocity of the wire,

$V=5.0ms^{-1}$
Horizontal component of earth's magnetic field,

$B_{H}=0.30*10^{4}WBm^{-2}$
(a) Now,

$e=B_{H}l_{v}=0.30*10^{-4}*10*5.0$

$=1.5*10^{-3}V$ .
(b) The induced e.m.f. will be set up from west to east end.
(c) The eastern end will be at higher potential.

Maximum voltage produced in an AC generator completing $60$ cycles in $30$ seconds is $250\,V.$
What is the maximum emf produced when the armature completes $1800$ rotation ?

The maximum emf produced by the AC generator is the peak voltage of the generator. It will be

$250\ \text{V}$ .

Conceptual Questions
Consider this thesis: "Joseph Henry, America's first professional physicist, caused a basic change in the human view of the Universe when he discovered self-induction during a school vacation at the Albany Academy aboutBefore that time, one could think of the Universe as composed of only one thing: matter. The energy that temporarily maintains the current after a battery is removed from a coil, on the other hand, is not energy that belongs to any chunk of matter. It is energy in the massless magnetic field surrounding the coil. With Henry's discovery, Nature forced us to admit that the Universe consists of fields as well as matter."
(a) Argue for or against the statement.
(b) In your view, what makes up the Universe?

We can think of Henry's discovery of self-inductance as fundamentally new. Before a certain school vacation at the Albany Academy about 1830, one could visualize the universe as consisting of only one thing, matter. All the forms of energy then known (kinetic, gravitational, elastic, internal, electrical) belonged to chunks of matter. But the energy that temporarily maintains a current in a coil after the battery is removed is not energy that belongs to any bit of matter. This energy is vastly larger than the kinetic energy of the drifting electrons in the wires. This energy belongs to the magnetic field around the coil. Beginning in 1830, Nature has forced us to admit that the universe consists of matter and also of fields, massless and invisible, known only by their effects.
The idea of a field was not due to Henry, but rather to Faraday, to whom Henry personally demonstrated self-induction. Still the thesis stated in the question has an important germ of truth. Henry precipitated a basic change if he did not cause it.
(b) A list today of what makes up the Universe might include quarks, electrons, muons, tauons, and neutrinos of matter; photons of electric and magnetic fields; W and Z particles; gluons; energy; charge; baryon number; three different lepton numbers; upness; downness; strangeness; charm; topness; and bottomness. Alternatively, the relativistic interconvertibility of mass and energy, and of electric and magnetic fields, can be used to make the list look shorter. Some might think of the conserved quantities energy, charge, ... bottomness as properties of matter, rather than as things with their own existence.

A 1.0 m long metallic rod is rotated with an angular frequency of 400 rad $s^{-1}$ about an axis normal to the rod passing through its one end. The other end of the rod is in contact with a circular metallic ring. A constant and uniform magnetic field of 0.5 T parallel to the axis exists everywhere. Calculate the emf developed between the centre and the ring.

The emf is given by

$e=\dfrac{Bl^{2}\omega }{2}=\dfrac{0.5\times 1^{2} \times 400}{2}=100V$

A pair of adjacent coil has a mutual inductance of $1.5 \,H$ . If the current in one coil changes from $0$ to $20 \,A$ in $0.5$ sec, what is the change of flux linkage with the other coil?

We know that

$e=\dfrac{d\phi }{dt}=M\dfrac{di}{dt}$

Therefore,

$d\phi =M{di}=1.5\times (20-0)=30Wb$

A spectral line caused by the transition $^{3} D1 \rightarrow ^{3}P_{0}$ experiences the Zeeman splitting in a weak magnetic field. When observed at right angles to the magnetic field direction, the interval between the neighbouring components of the split line is $\Delta \omega = 1.32 \times 10^{10}\,s^{-1}$ . Find the magnetic field induction B at the point where the source is located.

1918708_1855584_ans_2c31702e711e421ca49def90406e9307.jpg

Derive the expression for the self inductance of a long solenoid of cross sectional area $A$ and length $l$ , having $n$ turns per unit length.

Total number of turns in solenoid

$N = nl$

Magnetic field inside the long solenoid

$B = \mu_o ni$

Flux through one turn

$\phi_1 = BA = \mu_o ni A$

Thus total flux through N turns

$\phi_t = N \phi_1 = nl \times \mu_o niA =\mu_o n^2 lAi$

Using

$\phi_t = Li$ where

$L$ is the self inductance of the coil

$\therefore$

$\mu_o n^2 lA i = Li$

$\implies L = \mu_o n^2 Al$

Define self-inductance of a coil. Show that magnetic energy required to build up the current I in coil of coil of self inductance L is given by $\displaystyle \frac{1}{2}LI^{2}$

Self Inductance of a coil is the magnetic flux linked with the coil when the current through coil is

$IA$

$\phi=li$ where

$L$ is the constant of proportionality .

Let a source of emf be connected to an inductor

$L$

With increase in current ,the opposite emf ,

$e=-L\dfrac{di}{dt}$

Work done

$dw=\left| e \right| idt=Li\dfrac { di }{ dt } dt=Lidi$

$W=\int _{ 0 }^{ i }{ Lidi=\dfrac { 1 }{ 2 } L{ i }^{ 2 } }$

Energy stored in magnetic inductor,

$U=\dfrac { 1 }{ 2 } L{ i }^{ 2 }$

Two concentric circular coils of radii r and R are placed coaxially with centres coiciding. If R >> r then calculate the mutual inductance between the coils.

Let current

$I_2$ flow in the outer coil.
Hence magnetic field at its center=

$\dfrac{\mu_0 I_2}{2R}$

Since

$r<<R$ , this magnetic field can be considered to be constant over the entire face of smaller coil.

Hence flux through the coil=

$\phi=\dfrac{\mu_0I_2}{2R}\pi r^2$

By definition of mutual inductance,

$\phi=M_{12}I_2=\dfrac{\mu_0 I_2}{2R}\pi r^2$

$\implies M_{12}=\dfrac{\mu_0 \pi r^2}{2R}$

An aircraft having a wingspan of $20.48 m$ flies due north at a speed of $40{ ms }^{ -1 }$ . If the vertical component of earth's magnetic field at the place is $2\times { 10 }^{ -5 }T$ , calculate the emf induced between the ends of the wings.

Given data:

$I=20.48m$ ;

$V=40{ ms }^{ -1 }$ ;

$B=2\times { 10 }^{ -5 }T$ ;

$e=$ ?

Solution:

$e=-BlV$

$=-2\times { 10 }^{ -5 }\times 20.48\times 40$

$e=-0.0164$ Volt.

The unit of self-inductance in SI system is ________ .

The unit of self-inductance in

$SI$ is Henry.

It is denoted as

$H$ .

$1H=1 weber/amperce$

Name and state the principle of a simple a.c. generator. What is its use?

A Generator is a device which converts mechanical energy into electrical energy. A.C Generator works on the principle of electromagnetic induction. In generator an induced emf is produced by rotating a coil in a magnetic field.
Electromagnetic induction:
Electromagnetic induction is the production of an electromotive force across a conductor when it is exposed to a varying magnetic field. It is a process where a conductor placed in a changing magnetic field (or a conductor moving through a stationary magnetic field) causes the production of a voltage across the conductor. This process of electromagnetic induction, in turn, causes an electrical current.
The ac generator is used to produce alternating current which has high transmission power and power generation capability.

State one factor that determines the magnitude of induced e.m.f.

The Faraday's laws of electromagnetic induction says that the E.M.F. induced in a coil 'e' = -(rate of change of magnetic flux linkage)

where,
the Flux linkage =number of turns'N' x magnetic field'B' x area'A' x cos $\theta$
where,
theta is angle between magnetic field B and area A.
Theta at any instant 't'= $(angular \ velocity'w')\times (time \ instant't')$ .

That is, Theta = $w\times t$ .

E.M.F. induced in a coil 'e'= $N\times B\times A\times w\times sinw\times t$ .
The factors involved in the induced emf of a coil are:

The induced e.m.f. is directly proportional to N, the total number of turns in the coil.
The induced e.m.f. is directly proportional to A, the area of cross-section of the coil.
The induced e.m.f. is directly proportional to B, the strength of the magnetic field in which the coil is rotating.
The induced e.m.f. is directly proportional to 'w', the angular velocity of coil.
The induced e.m.f. also varies with time and depends on instant 't'.
The induced e.m.f. is maximum when plane of coil is parallel to magnetic field B and e.m.f. is zero when plane of coil is perpendicular to magnetic field B.

What is the source of energy associated with the current obtained in part when a magnet is moved towards a coil having a galvanometer at its ends?

When a magnet is moved closer to the current carrying coil it will generate electricity as the coil moves through the magnetic field. As the magnet is moved, there will be an induced electro-motive force (EMF) which can cause a current in the coil. Once the magnet stops moving, the current will go to zero.
Hence, when a galvanometer is connected to the circuit, there will be deflection due to the flow of electricity. As the magnet is moved toward the coil of wire, the needle of the galvanometer moves one direction. As the magnet is moved away from the coil of wire, the needle of the galvanometer moves the opposite direction. If the magnet is moved faster, the magnitude of the deflection increases.
The mechanical movement/energy is converted into electrical energy.

$Match \: the \: statements \: in \: Column \: A, \: with \: those \: in \: Column \: B.$

A-Faraday law of Electromagnetic induction statement.

B-step up transformer used to increase the voltage from primary to secondary.

C-Generator used to convert Mechanical energy to electrical energy.

D-inverse of B

E-induced current.

Define the term 'self-inductance' of a coil. Write its S.I. unit.

Self-inductance of a coil is defined as the phenomenon due to which an emf is induced in a coil when the magnetic flux of coil , linked with the coil changes or current in coil changes . Its S.I. unit is

$Henry (H)$ .

Fill in the blanks.
The type of electric current that changes its direction twice during one cycle of the dynamo is called _____________ .

Alternating current changes its direction twice during one cycle of the dynamo.

State Faraday's laws of electromagnetic induction.

What is motional emf? State any two factors on which it depends.

Motional emf is the emf induced in a conducting rod when it moves in a region of magnetic field.
Motional emf is calculated by

$\mathcal{E} = v Bl$
where

$B$ is the magnetic field and

$l$ is the length of the rod and

$v$ is the velocity of rod perpendicular to the length of the rod.
Motional emf depends on

1. The magnitude of length of rod and

2. Velocity of rod and

3. Magnetic field.

Two coils A and B have mutual inductance $2\times { 10 }^{ -2 }$ henry. If the current in the primary is $i=5sin(10\pi t)$ , then the maximum value of e.m.f. induced in coil B is :

Given,

$M=2\times 10^{-2}H$

Current in primary coil A,

$i=5sin(10\pi t)$

The emf induced in coil B,

$e=-M\dfrac{di}{dt}$

$e=-M\times 5\times 10\pi cos(10\pi t)$

$e=-2\times 10^{-2}\times 50\pi cos(10\pi t)$

The maximum value of emf induced in coil B is

$e_{max}=\pi$

What is a generator state the principle on which generators work

Generators are useful appliances that supply electrical power during a power outage and prevent discontinuity of daily activities or disruption of business operations. Generators are available in different electrical and physical configurations for use in different applications.

An electric generator is a device that converts mechanical energy obtained from an external source into electrical energy as the output.

The generator works on the principle of electromagnetic induction discovered by Michael Faraday in 1831-32. Faraday discovered that the above flow of electric charges could be induced by moving an electrical conductor, such as a wire that contains electric charges, in a magnetic field. This movement creates a voltage difference between the two ends of the wire or electrical conductor, which in turn causes the electric charges to flow, thus generating electric current.

Describe the activity that shows that a current-carrying conductor experiences a force perpendicular to its length and the external magnetic field. How does Flemings' left-hand rule help us to find the direction of the force acting on the current-carrying conductor?

The direction of force in a current carrying conductor can be understood clearly From Fleming's left hand rule.
Fleming's left hand rule states that When an electric current passes through a straight wire, an external magnetic field is applied across that flow, the wire experiences a force perpendicular both to the field and to the direction of the current flow. This rule can be used to determine the direction of the force, magnetic field and current.
A left hand can be held, so that the thumb, first finger and second finger are held mutually perpendicular to each other. The thumb represents the direction of Motion resulting from the force on the conductor. The first finger represents the direction of the magnetic Field. The second finger represents the direction of the Current. Please refer the figure given below.

The current through an indicator of $1H$ is given by $i=3t\sin { t }$ . Find the voltage across the inductor.

$\left| e \right| =\left| L.\cfrac { \Delta i }{ \Delta t } \right| \quad or\quad \left| L\cfrac { di }{ dt } \right|$
Here,

$L=1H$
And,

$\cfrac { di }{ dt } =3\left[ \sin { t } +t\cos { t } \right]$

$=3(t\cos { t } +\sin { t } )$

the self inductance of the primary coil.

${ L }_{ 1 }=\cfrac { { N }_{ 2 }{ \phi }_{ 2 } }{ { i }_{ 1 } } =\cfrac { \left( 600 \right) \left( 5\times { 10 }^{ -3 } \right) }{ 3 }$

$=1H$

At the instant when the current in an inductor is increasing at a rate of $0.0640A/s$ , the magnitude of the self-induced emf is $0.0160V$ . If the inductor is a solenoid with $400$ turns, what is the average magnetic flux through each turn when the current is $0.720A$ ?

$Inductance, L=\cfrac { N\phi }{ i }$

$\therefore Flux, \quad \phi =\cfrac { Li }{ N } =\cfrac { \left( 0.25 \right) \left( 0.72 \right) }{ 400 } =4.5\times { 10 }^{ -4 }Wb$

When a current of $4A$ between two coils changes to $12A$ in $0.5s$ in primary and induces an emf of $50mV$ in the secondary. Calculate the mutual inductance between the two coils.

$M=\cfrac { { e }_{ 2 } }{ d{ i }_{ 2 }/dt } =\cfrac { \left( 50\times { 10 }^{ -3 } \right) }{ \left( 8/0.5 \right) }$

$=3.125\times { 10 }^{ -3 }H=3.125mH$

At the instant when the current in an inductor is increasing at a rate of $0.0640A/s$ , the magnitude of the self-induced emf is $0.0160V$ . What is the inductance of the inductor?

$e=\left| L\cfrac { di }{ dt } \right| \quad \Rightarrow L=\left| \cfrac { e }{ di/dt } \right| =\cfrac { 0.016 }{ 0.064 }$

$=0.25H$

The current (in Ampere) in an inductor is given by $I=5+16t$ , where $t$ is in seconds. The self-induced emf in it is $10mV$ . Find the self-inductance.

$\dfrac{dI}{dt}=16A/s$

$\therefore \quad L=\left| \cfrac { e }{ \dfrac{dI}{dt} } \right| =\cfrac { 10\times { 10 }^{ -3 } }{ 16 } =0.625\times { 10 }^{ -3 }H$

$=0.625mH$

Two bar magnets are placed side by side by side that the north pole of one magnet is next to the south pole of the other magnet. If these magnets are then pushed toward a coil of wire, would you expect an emf to be induced in the coil. Explain your answer.

Since both the magnets are aligned opposite to each other, so both the magnets will cancel out the effect due to each other while being pushed towards a coil. Let say, one magnet induces a clockwise emf in the coil, then the other magnet will induce anticlockwise emf in the coil and thus net emf induced in the coil is zero.

emf induced in the arm PQ

The equivalent circuit is shown in the figure. Here

$PQ=l=10 cm=0.1$
The induced emf

$\varepsilon=Blv=(0.1)(0.1)v=0.01v$

Current in a circuit falls from $5.0 A$ to $0.0 A$ in $0.1\ s$ . If an average emf of $200\ V$ induced, find an estimate of the self-inductance of the circuit.

The emf is related to inductance by the equation,

$e=L\dfrac{di}{dt}$ .

Now, $L=\dfrac{e}{\dfrac{di}{dt}}=\dfrac{200}{\dfrac{5}{0.1}}=4H$ .

A rectangular wire loop of sides 8 cm and 2 cm with a small cut is moving out of a region of uniform magnetic field of magnitude 0.3 T directed normal to the loop. What is the emf developed across the cut if the velocity of the loop is 1 cm $s^{-1}$ in a direction normal to the (a) longer side, (b) shorter side of the loop? For how long does the induced voltage last in each case?

Length of the rectangular wire, $l=8cm=0.08m$

Width of the rectangular wire, $b=2cm=0.02m$

Hence, area of the rectangular loop,

$A=lb$

$=0.08\times 0.02$

$=16\times 10^{-4}m^2$

Magnetic field strength, $B=0.3T$

Velocity of the loop, $v=1cm/s=0.01m/s$

(a)

Emf developed in the loop is given as:

$e=Blv$

$=0.3\times 0.08\times 0.01=2.4\times 10^{-4}V$

Time taken to travel along the width, $t\displaystyle =\frac{Distance\, travelled}{Velocity}=\frac{b}{v}$

$\displaystyle =\frac{0.02}{0.01}=2s$

Hence, the induced voltage is $2.4\times 10^{-4}V$ which lasts for 2 s.

(b)

Emf developed, $e=Bbv$

$=0.3\times 0.02\times 0.01=0.6\times 10^{-4}V$

Time taken to travel along the length, $\displaystyle t=\frac{Distance\, traveled}{Velocity}=\frac{l}{v}$

$=\displaystyle \frac{0.08}{0.01}=8s$

Hence, the induced voltage is $0.6\times 10^{-4}V$ which lasts for 8 s.

(a) A rod of length l is moved horizontally with a uniform velocity 'v' in a direction perpendicular to its length through a region in which a uniform magnetic field is acting vertically downward. Derive the expression for the emf induced across the ends of the rod.
(b) How does one understand this motional emf by invoking the Lorentz force acting on the free charge carriers of the conductor? Explain.

(a)Let a straight conductor of length

$l$ be moving in u shaped conductor in perpendicular magnetic field.

$d \phi=B(vldt)=Bvldt$

$\dfrac{d\phi}{dt}=Bvl$

Induced emf

$\varepsilon =\dfrac { d\phi }{ dt } =Bvl$

(b)Due to motion of the conductor,free electrons move from one end to other end.

Due to this,both ends generate positive and negative charge and electric force acts on it.

Now according to Lorentz law,

$F_{net}=F_e+F_m$

At equilibrium,

$F_{net}=0$

$F_e+F_m=0$

$qE+q\left( \overrightarrow { v } \times \overrightarrow { B } \right) =0$

$E=-\left( \overrightarrow { v } \times \overrightarrow { B } \right)$

$\\ \left| E \right| =Bv\sin { \theta }$ ,when velocity perpendicular when magnetic field

$\theta=90$

$\left| E \right| =Bv$

Also,

$\\ \dfrac { d\xi }{ dl } =\left| E \right| =Bv$

$\\ d\xi =Bvdl$

Induced emf=

$\\ Bvl=\xi$

The property of a conductor which enables to induce an EMF due to change of current in the same coil is ____________.

Self inductance is defined as the property of a conductor which enables it to induce EMF due to change of current in the same coil.

Derive Faradays law of induction from law of conservation of energy.

Consider a situation shown in figure a rod of length

$(l)$ slide over two rail S R rod P Q is moving with speed V towards right. The electromagnetic force act on the rod is

${ f }_{ m }=\dfrac { { B }^{ 2 }{ l }^{ 2 }V }{ R }$ towards R is the resistance of the circuit then the work done by the mechanical movement is

${W}_{m} = {f}_{m}.dx$

Mechanical power develop

${ F }_{ m }=\dfrac { { B }^{ 2 }{ l }^{ 2 }V }{ R }$

${ p }_{ mech }=-{ f }_{ m }\dfrac { dx }{ dt } ={ -f }_{ m }V\\ =\dfrac { { -B }^{ 2 }{ l }^{ 2 }{ V }^{ 2 } }{ R } \rightarrow (1)$

If current induced in the circuit due to induced emf. e then electrical power developed is

${ p }_{ e }=\dfrac { { \varepsilon }^{ 2 } }{ R } \rightarrow (ii)$

If energy is conserve then

${p}_{e} = {p}_{mech}$ using equation (1)

$-\dfrac { { B }^{ 2 }{ l }^{ 2 }{ V }^{ 2 } }{ R } =\dfrac { { \varepsilon }^{ 2 } }{ R } \\ \varepsilon =-BlV\rightarrow (iii)$

We know that flux

$\phi =\int { B.ds } =B.lx$

Area of PQRS =

$l x$

Now equation (iii) written as

$\varepsilon =-Bl\dfrac { dx }{ dt } =\dfrac { -d }{ dt } Blx=\dfrac { -d\phi }{ dt }$

Here we can see that induced emf

$\varepsilon =\dfrac { -d\phi }{ dt }$ that is Farady's law.

An emf of 2V is induced in a coil when current in it is changed from 0A to 10A in 0.40 sec. Find the coefficient of self-inductance of the coil.

Let self-inductance of the coil be

$L$ , emf of the cell be

$e$ and current in the circuit be

$i$

$L=\dfrac{e}{di/dt}$

$=\dfrac{2}{10}\times 0.40$

$= 0\cdot 08 H$

A metal rod $\cfrac { 1 }{ \sqrt { \pi } } m$ long rotates about one of its ends perpendicular to a plane whose magnetic induction is $4\times { 10 }^{ -3 }T$ . Calculate the number of revolutions made by the rod per second if the e.m.f induced between the ends of the rod is $16mV$ .

$E=\cfrac { 1 }{ 2 } B\omega { l }^{ 2 }$

$\Rightarrow 16\times { 10 }^{ -3 }=\cfrac { 1 }{ 2 } \times 4\times { 10 }^{ -3 }\times \omega \times { \left( \cfrac { 1 }{ \sqrt { \pi } } \right) }^{ 2 }$

$\Rightarrow 16\times { 10 }^{ -3 }=\cfrac { 1 }{ 2 } \times 4\times { 10 }^{ -3 }\times \omega \times { \left( \cfrac { 1 }{ \pi } \right) }$

$\Rightarrow n=\cfrac { 16\times { 10 }^{ -3 }\times 2 }{ 4\times { 10 }^{ -3 }\times 2 }$

$n=4$

A coil of self inductance $2\cdot 5 H$ and resistance $20\Omega$ is connected to a battery of emf 120V having internal resistance of $5 \Omega$ . Find the current in the circuit in steady state.

Current in the circuit

$i_0=\dfrac{EMF}{\text{internal resistance}}=\dfrac{E}{R}=\dfrac{120}{20}$

$=6A$

Write Faraday's law of electromagnetic induction.

Faraday's law states that any change in the magnetic field of a coil of wire will cause an emf to be induced in the coil. This emf induced is called induced emf and if the conductor circuit is closed, the current will also circulate through the circuit and this current is called induced current.

Write the Faraday's law of electromagnetic induction.

Faraday's First Law states that any change in the magnetic field of a coil of wire will cause an emf to be induced in the coil. This emf induced is called induced emf and if the conductor circuit is closed, the current will also circulate through the circuit and this current is called induced current.

Faraday's Second Law states that the magnitude of emf induced in the coil is equal to the rate of change of flux that linkages with the coil. The flux linkage of the coil is the product of number of turns in the coil and flux associated with the coil.

Current in a circuit fall from $5.0$ A to zero in $0.1$ S. If an average emf of $100$ Volt is induced then calculate self-inductance of a inductor in the circuit.

Induced emf in an inductor is given as

$EMF_{ind}=L\dfrac{di}{dt}$

$\implies 100=L(\dfrac{5}{0.1})$

$\implies L=2H$

State Faraday's laws of electromagnetic induction and Lenz's law.

Faraday's laws of electromagnetic induction.
(a) First law : Whenever there is a change in the magnetic flux associated with a circuit, an e.m.f. is induced in the circuit.
(b) Second law : The magnitude of the induced e.m.f. is directly proportional to the time rate of change of magnetic flux through the circuit.
Lenz law : The direction of induced current is such as to oppose the change that produces it.

Explain self-induction of a coil. Arrive at an expression for the induced emf in a coil and the rate of change of current in it.

Inductance is the property of a component that opposes the change of current flowing through it and even a straight piece of wire will have some inductance.
Inductors do this by generating a self-induced emf within itself as a result of their changing magnetic field. In an electrical circuit, when the emf is induced in the same circuit in which the current is changing this effect is called Self-induction.
Magnetic field gives rise to magnetic flux

$\phi$ passing through each turn in the coil. Coil having N turns will have flux

$N\phi$ .
Magnetic field is proportional to the current i flowing through the coil and so is the flux.

$N\phi \propto i$

$N\phi =L i$

$L$ is the self inductance.
Now lets consider the current to change in the coil with time which changes the flux as well.
So this rate of change of flux induces an emf (e) by Farady's law of electromagnetic induction.

$\begin{equation} e=-\frac{d (N\phi)}{dt}=-\frac{d (Li)}{dt}=-L\frac{di}{dt} \end{equation}$
If

$\dfrac{di}{dt}=1 A/s$ we have,

$e=-L$
Thus self inductance is numerically equal to the emf induced in coil when current changes at a rate of 1 Ampere per second.

Current in a circuit falls from $5.0 A$ to $0.0 A$ in $0.1 sec$ . If an average emf of $200 V$ is induced, give an estimate of the self-inductance of the circuit.

Change in current,

$\Delta I = 0.0 - 5.0 = -5.0$

$A$

Rate of change of current,

$\dfrac{\Delta I}{\Delta t } = \dfrac{-5.0}{0.1} = -50.0$

$A/s$

Average induced emf,

$\mathcal{E } = 200$

$V$

Using

$\mathcal{E} = -L\dfrac{\Delta I}{\Delta t}$ where

$L$ is the self inductance of circuit

$\therefore$

$200 = -L\times (-50.0)$

$\implies$

$L = 4.0$

$H$

What are the methods of producing induced emf?

The induced emf can be produced by changing :
(i) the magnetic induction

$\left( B \right)$ ,
(ii) area enclosed by the coil

$\left( A \right)$ and
(iii) the orientation of the coil

$\left( \theta \right)$ with respect to the magnetic field.

What is the self inductance of a straight conductor?

Since there are no enclosed loops in a straight conductor, the self inductance is zero.

a) Switch in the primary is kept in the ON position. Will the bulb in the secondary glow? Justify your answer.
b) When the current in the primary is switched OFF. What happens to the magnetic field in the secondary coil? How will this change influence the current in the secondary and the glow in the electric bulb?

a) If the primary circuit is kept ON, there is no change in magnetic flux. Hence, no induced emf and so the bulb in the secondary does not glow.
b) At the time of turning the switch OFF, there is a sudden change in magnetic flux which causes an induced emf. Hence, the bulb in the secondary glows momentarily.

Explain seld induction and mutual induction.

Self inductance is defined as the property of the coil due to which it opposes the change of current flowing through it. Inductance is attained by a coil due to the self-induced emf produced in the coil itself by changing the current flowing through it. It's given by the following:

$e= L \frac{dI}{dt}$

Where e is self induced emf, L is self inductance, I is current and t is time passed. The unit of self inductance, L is Henry.

Mutual Inductance between the two coils is defined as the property of the coil due to which it opposes the change of current in the other coil, or you can say in the neighboring coil. When the current in the neighboring coil is changing, the flux sets up in the coil and because of this changing flux emf is induced in the coil called Mutually Induced emf and the phenomenon is known as Mutual Inductance.

In short flux through second coil is directly proportional to current flowing in first coil, which is given by the following expression:

$\phi_2 = M_{21}I_1$ and

$\phi_1= M_{12}I_2$

Where

$\phi_1$ and

$\phi_2$ is flux through coil 1 and coil 2 respectively,

$I_1$ and

$I_2$ is current in coil 1 and coil 2 respectively and

$M_{12}=M_{21}=M$ is mutual induction.

Draw a labelled diagram of an ac generator. Obtain the expression for the emf induced in the rotating coil of N turns each of cross-sectional area A, in the presence of a magnetic field $\overrightarrow{B}$ .

Let say at the instant

$t=0$ , the plane of the coil is perpendicular to the direction of magnetic field i.e. area vector of coil points in same direction as that of magnetic field. So maximum flux passes through the coil.

Flux passing through the coil at

$t=0$ ,

$\phi_{max} = NBA$

The coil rotates about an axis with constant angular velocity

$w$ as shown in the figure.

Angle rotated by coil in time

$t$ is

$wt$ .

Flux passing through the coil at that instant,

$\phi = NBA\cos wt$

Rate of change of flux

$\dfrac{d\phi}{dt} = NBA \dfrac{d}{dt} (\cos wt)$

$\therefore$

$\dfrac{d\phi}{dt} = NBA \times (-w\sin wt)$

We get

$\dfrac{d\phi}{dt} = -NBA w\sin wt$

Induced emf in the coil

$\mathcal{E} = -\dfrac{d\phi}{dt}$

$\implies$

$\mathcal{E} =NBAw \sin wt$

Current in a circuit falls from 5.0 A to 0.0 A in 0.1 s. If an average emf of 200V induced, calculate the self-inductance of the circuit.

Initial current,

$I_1=5.0 A$

Final current,

$I_2=0A$

Change in current,

$dl=I_1-I_2=5A$

Time taken for the charge,

$t=0.1s$

Average emf,

$e=200V$

For self inductance(L) of the coil, we have the relation for average emf as:

$e=L\dfrac{di}{dt}$

$L=\dfrac{e}{\dfrac{di}{dt}}$

$= \dfrac{200}{\dfrac{5}{0.1}=4H}$

Hence the self inductance of the coil is

$4H$ .

What is the effect on self inductance of a solenoid, if a core of soft iron is placed in it?

A core of soft iron is placed in a solenoid will enhance the self-induction as the magnetic flux linkage between the coils is improved due to high magnetic permeability of soft iron.

Derive expression for the self-induction of solenoid. What factors affect its value and how?

Self Inductance of long solenoid: Consider a solenoid of length l, cross-section area A. Having total number of turns N. If I current flow through the solenoid there e.m.f. at the centre of solenoid.

$B=\mu_ont$
But

$n=\displaystyle\frac{N}{1}$

$\Rightarrow B=\displaystyle\frac{\mu_oNI}{1}$

$\Phi =NBAcos\theta$

$\Rightarrow L=\displaystyle\frac{\mu_oN^2IA}{II}A cos\theta$

$\Phi =\displaystyle\frac{\mu_oN^2IA}{l}$

$L=\displaystyle\frac{\mu_oN^2IA}{l}$

If the relative permeability of medium inside the solenoid is

$\mu_r$ , then,

$L=\displaystyle\frac{\mu_o\mu_rN^2IA}{l}$

It is clear that self inductance of solenoid depends upon the following factors:
(1) On relative permeability of material inside the solenoid: If a soft iron core placed inside the solenoid, the magnetic flux linked with the solenoid increased hence, self inductance of the solenoid will also increases.
(2) On total number of turns in solenoid: If total number of turns in the solenoid increases then self inductance of the solenoid will also increases.
(3) On cross-section area of solenoid: If cross-section area of solenoid increase then self inductance of solenoid will also increases.
(4) On length of solenoid: If length of solenoid increases then its self-inductance decrease.

A rectangular coil has $60$ turns and its length and width is $20\ cm$ . and $10\ cm$ respectively. The coil rotates at a speed of $1800$ rotation per minute in a uniform magnetic field of $0.5\ T$ about its one of the diameter. Calculate maximum induced emf will be

Given that,

Number of turns $N=60$

Length $l=20\,cm$

Width $b=10\,cm$

Magnetic field $B=0.5\,T$

Rotation per min $f=1800$

We know that,

$\phi =\left( AB\cos \theta \right)N$

$\phi =NAB\cos \omega t$

$\dfrac{-d\phi }{dt}=\varepsilon$

$\varepsilon =-\dfrac{d\left( NAB\cos \omega t \right)}{dt}$

$\varepsilon =-AB\omega \left( -\sin \omega t \right)N$

$\varepsilon =NAB\omega \sin \omega t$

$\omega =2\pi f$

$\varepsilon =2\pi fNAB\sin \omega t$

${{\varepsilon }_{\max }}=2\pi fNAB$

${{\varepsilon }_{\max }}=2\times 3.14\times 60\times \left( 20\times 10\times {{10}^{-4}} \right)\times 0.5\times 30$

${{\varepsilon }_{\max }}=113.04\,volt$

Hence, the maximum induced emf is $113.04\,volt$

A charged particle oscillates about its equilibrium position with an frequency of $100\ MHz$ . What is the frequency of electromagnetic waves produced by the oscillator?

The frequency of an electromagnetic wave produced by the oscillator is the same as that of a charged particle oscillating about its mean position i.e.

$100\ MHz$ .

A circular coil of cross-sectional area $200cm^2$ and 20 turns is rotated about the vertical diameter with angular speed of $50 rad {s}^{-1}$ in a uniform magnetic field of magnetic $3.0 \times10^{-2}T$ . Calculate the maximum value of the current in the coil.

Here,

$A=200 cm^{2}$

$N=20$

$\omega=50 \dfrac{rad}{s}$

$B=3\times { 10 }^{ -2 }T$

For maximum current emf should be maximum.

So for maximum emf

$e=NAB\omega\sin\omega t$

$\sin\omega t=1$

$e_{max}=NAB\omega$

$e_{max}=20\times 200\times { 10 }^{ -4 }\times 50\times 3\times { 10 }^{ -2 }$

$e_{max}$ =

$0.6 V$

Since resistance is not given,let us consider

$R$

$i_{max}$ =

$\dfrac { e_{max} }{ R } =\dfrac { 0.6 }{ R } \\$

A circular coil of cross-sectional area $20 cm^2$ and $20$ turns is rotated about the diameter with angular speed of 50 rad $s^{-1}$ in a uniform magnetic field of magnetic $3.0 \times 10^{-2}T$ . Calculate the maximum value of the current in the coil.

Here,

$A=200 cm^2$

$N=20$

$w=50\dfrac{rad}{s}$

$B=3 \times 10^{-2} T$

For maximum current emf should be maximum.

So for maximum emf

$e=NABwsinwt$

$sinwt=1$

$e_{max} = NABw$

$e_{max} = 20 \times 200 \times 10*{-4} \times 50 \times3 \times10^{-2}$

$e_{max} = 0.6 V$

Since resistance is not given,let us consider

$R$

$i_{max} = \dfrac{e_{max}}{R} = \dfrac{0.6}{R}$

Draw a labeled diagram of AC generator.Derive the expression for the instantaneous value of the emf induced in the coil.

Construction of AC generator:

Main parts of an AC generator include,

1. Armature-Rectangular coil ABCD.

2. Field magnet

3. Slip Rings

4. Brushes $B_1,B_2$

Working: As the armature coil is rotated in magnetic field,angle $\theta$ between the field and normal to the coil changes continuously ,therefore magnetic flux linked with the coil changes.An emf is induced in the coil.According to Fleming’s right hand rule,to calculate emf,Suppose,

$A$ -Area of each turn the coil.

$N$ -Number of turn in the coil.

$\overrightarrow { B }$ -strength of magnetic field.

$\theta$ -angle which normal to the coil make with $\overrightarrow{B}$ at any instant t.

Magnetic Flux linked with coil in this position.

$\phi=N\left( \overrightarrow { B } .\overrightarrow { A } \right) =NBA\cos { \theta }$

$=NBA\cos { \omega t }$

$e=-\dfrac { d\phi }{ dt } =-NBA(-\left( \sin { \omega t } \right) \omega)$

$e=+NBA\omega \sin { \omega t }$

Why self induction is called inertia of electricity?

Self-induction of coil is the property by virtue of which it tends to maintain the magnetic flux linked with it and opposes any change in the flux by inducing current in it. This property of a coil is analogous to mechanical inertia. That is why self-induction is called the inertia of electricity.

A solenoid of $500$ turns, diameter 2 $0 cm$ and resistance $2$ $\Omega$ is rotated about its vertical diameter through $\pi$ radian in 1/4 s, when a horizontal field of $3 \times 10^{-5} T$ acts normal to its plane. Find the emf induced and current thereof.

emf induced is

$e$ .

$e=NBA\omega \sin\theta$ where ,

$N$ =number of turns.

$B$ =magnetic field.

$\omega$ =angular velocity.

$\omega=\dfrac{\pi}{\dfrac{1}{4}} \dfrac{rad}{s}$

$=4\pi \dfrac{rad}{s}$

$A$ =area of cross section=

$\dfrac{\pi d^{2}}{4}$

(

$\theta=30° , \sin\theta=1$ )

$e=500\times 3\times { 10 }^{ -5 }\times 4\pi\times \dfrac { \pi{ \left( 0.2 \right) }^{ 2 } }{ 4 } \times 1$

$=500\times 3\times 4\times { 10 }^{ -7 }\times \pi$

$=18840\times { 10 }^{ -7 } V$

$=1.884mV$

Current

$i=\dfrac{e}{R}$ =

$\dfrac{1.884}{2} A$

$i=0.945A$

A rectangular frame of wire $abcd$ has dimensions $32\ cm\times 8.0\ cm$ and a total resistance of $2.0\Omega$ . It is pulled out of a magnetic field $B = 0.020\ T$ by applying a force of $3.2\times 10^{-5}N$ (figure). It is found that the frame moves with constant speed. What is constant speed of the frame?

If loop abcd moves with constant speed ,therefore electromagnetic force

$f_m$ is equal to external force

$f$ .

$f_m=f=3.2 \times 10^{-5}$

$ilB=3.2\times 10^{-5}$

$\dfrac{e}{R}lB=\dfrac{VBl}{R}lB=\dfrac{VB^{2}l^{2}}{R}$ where,

$V$ =speed of loop

$R$ =resistance of abcd.

$i$ =current in the loop.

$B$ =magnetic field

$\Longrightarrow \dfrac { V\times { \left( 0.2 \right) }^{ 2 }\times { { \left( 8\times { 10 }^{ -2 } \right) }^{ 2 } } }{ 2 } =3.2\times { 10 }^{ -5 }$

$\Longrightarrow \dfrac { V\times 4\times { 10 }^{ -4 }\times 64\times { 10 }^{ -4 } }{ 2 } =3.2\times { 10 }^{ -5 }$

$\Longrightarrow V=\dfrac { 2\times 3.2\times { 10 }^{ 3 } }{ 4\times 64 } =25\dfrac { m }{ s }$

Velocity required =

$25 ms^{-1}$

Define coefficient of self inductance and write its unit.

Definition:-

Coefficient of self inductance is a ratio of electromotive force produced in a circuit by self induction to the rate of change of current producing it.

Unit of coefficient of self inductance is

$Henry$ .

Derive the expression for the motional emf induced in a conductor moving in a uniform magnetic field.

Consider a metallic frame

$MSRN$ placed in a constant uniform magnetic field. Let the magnetic field

$B$ be perpendicular to the plane of the coil. Let a metal rod

$PQ$ of length '

$l$ ' placed on it be moving towards left with a velocity

$v$ as shown in the figure.

Let the distance of

$PQ$ from

$SR$ be

$x$

$\phi_B=BA \cos \theta$

$=Blx\cos 0 = Blx$

As the rod

$PQ$ is moving towards left with a velocity

$v$ ,

$X$ is changing and

$\overrightarrow{v} = \dfrac{-dx}{dt}$

Hence Induced emf

$\in=\dfrac{-d\phi_B}{dt}=\dfrac{-d(Blx)}{dt} = -Bl\dfrac{dx}{dt}$

$\Rightarrow \in = Blv$

This emf is induced in the road because of the motion of the rod in the magnetic field.

Therefore this emf is called motional emf.

Match the followings.

A. Angular momentum= moment of inertia

$\times$ angular velocity

$= r \times p$

$=Kgm^2s^{-1}=[ML^2T^{-1}]$

B. Latent Heat=

$Joule/Kg= [ML^2Q^{-2}]$

C. Torque= Newton metre=

$[ML^2T^{-2}]$

D. Capacitance= Charge/Voltage=

$[M^{-1}L^{-2}T^{2}Q^{2}]$

E. Inductance= Ohm sec=

$[L^2T^{-2}]$

F. Resistivity= Ohm/Length=

$[ML^3T^{-1}Q^{-2}]$

What is self inductance? Name the factors on which self inductance depends.

Self-inductance definition is an inductance in which an electromotive force is produced by self-induction. It is the as the property of the coil due to which it opposes the change of current flowing through it.

Self inductance depends on-

1-Size of coil

2- Shape of the coil

3- Material of the coil

4-Medim

$e.g$ Self inductance of solenoid is

$\dfrac{\mu N^2A}{l}$

What will be the magnitude and polarity of induced emf across the rod, if it is rotating about an axis at a distance $\cfrac{L}{4}$ from one of its ends?

Consider, The rotating with angular velocity

$\omega$

$\begin{array}{l} e=\int _{ -\frac { L }{ 4 } }^{ \frac { { 3L } }{ 4 } }{ \omega Bxdx } \\ =\omega B{ \left[ { \frac { { { x^{ 2 } } } }{ 2 } } \right] _{ \frac { { -L } }{ 4 } } }^{ \frac { { 3L } }{ 4 } } \\ =\frac { { \omega B{ L^{ 2 } } } }{ 4 } \end{array}$

And polarity longer end will be positive

$wrt$ origin.

A semicircle loop $PQ$ of radius $'R'$ is moved with velocity $'v'$ in transverse magnetic field as shown in figure. The value of induced emf. at the end of loop is :-

Induced

$\begin{array}{l} emf=\frac { { -d\theta } }{ { dt } } \\ E=\frac { { -d } }{ { dt } } \left( { BA\cos \theta } \right) \\ E=-B\frac { { dA } }{ { dt } } \cos \theta \end{array}$ because the ring falls with its vertical plane in horizontal mehnetic induction

$\begin{array}{l} \vec { B } .\theta =0\, and\cos \theta =1 \\ or\, E=-B.\frac { d }{ { dt } } \left( { 2Rdx } \right) \\ E=-2RB\left( { \frac { { dx } }{ { dt } } } \right) \\ \left| E \right| =2RBV \end{array}$

This is equal to the emf developed in an imaginary rod joining M &Q

According to fleming right hand rule, M is at low potential and Q is at high potential and induced emf is from M to Q.

What is eddy current? Mention two applications of eddy current.

Eddy currents are the currents induced in a metallic plate when it is kept in a time varying magnetic field. Magnetic flux linked with the plate changes and so the induced current is set up. Eddy currents are sometimes so strong, that metallic plate become red hot.

Application:-

(1)-In induction furnace, the metal to be heated is placed in rapidly varying magnetic field produced by high frequency alternating current. Strong eddy currents are set up in the metal produce so much heat that the metal melts. This process is used in extracting a metal from its ore.The arrangement of heating the metal by means of strong induced current is called the induction furnace.

(2)-Induction motor, the eddy currents may be used to rotate the rotor. When a metallic cylinder(or rotor) is placed in a rotating magnetic field , eddy currents are produced in it. According to Lenz's law, these currents tends to reduce to relative motion between the cylinder and the field. The cylinder, therefore, begins to rotate in the direction in the field. This is the principle of induction motion.

A $200km$ long telegraph wire has a capacitance of $0.014\mu F/km$ . If it carries an ac of $5kHz$ , what should be the inductance required to be connected in series, so that the impedance is minimum?
Take $\pi=\sqrt {10}$

For minimum impedance

$\begin{array}{l} { X_{ L } }={ X_{ C } } \\ \omega L=\dfrac { 1 }{ { \omega C } } \\ L=\dfrac { 1 }{ { { \omega ^{ 2 } }C } } =\dfrac { 1 }{ { { { \left( { 2\pi f } \right) }^{ 2 } }C } } \\ For\, 1km\, C=0.14\mu F \\ For\, 200km\, C=0.14\times 200\mu F \\ L=\dfrac { 1 }{ { \left( { 2\; \times 3.14\times 5\times { { 10 }^{ 3 } } } \right) \times 0.014\times 200 } } =0.36mH \end{array}$

A direct current $I$ flows along a lengthy straight wire. From the point $O$ (Fig.) the current spreads radially all over an infinite conducting plane perpendicular to the wire. Find the magnetic induction at all points of space.

It is easy to convince oneself that both in the regions.

$1$ and

$2$ there can only be a circuital magnetic field (i.e. the component

$B_{\varphi})$ . Any radial field in region

$1$ or any

$B_{Z}$ away from the current plane will imply a violation of Gauss' law of magnetostatics,

$B_{\varphi}$ must obviously by symmetrical about the straight wire. Then in

$1$ ,

$B_{\varphi} 2\pi r = \mu_{0} I$
or,

$B_{\varphi} = \dfrac {\mu_{0}}{2\pi} \dfrac {I}{r}$

$In\ 2, B_{\varphi} \cdot 2\pi r = 0$ , or

$B_{\varphi} = 0$ .
1787897_1056091_ans_d763a9abfe92423db4f6c82fde90ccd6.jpg

A rectangular coil of area $A$ , having the number of turns $N$ is rotated at $f$ revolutions per second in a uniform magnetic field $B$ the field is perpendicular to the coil. Prove that the maximum emf induced in the coil is $2 \pi fNBA$ .

$\begin{array}{l}E = BAN\omega \sin \theta \\\omega = 2\pi f\,and\,at\,\theta = {90^ \circ }\\E = BAN\left( {2\pi f} \right)\sin 90\\ = 2\pi fBAN\end{array}$

A coil having resistance $20\Omega$ and inductance $2H$ is connected to a battery of emf $4.0V$ . Find (a) the current at $0.20s$ after the connection is made and (b) the magnetic field energy in the coil at the instant

$L=2H\quad R=20\Omega \quad emf=4V\quad t=0.20s$

$I_0=\dfrac{e}{R}=\dfrac{4}{20}=0.2$

$t=\dfrac{L}{R}=\dfrac{2}{20}=0.1$

$(a)$

$I=I_0(1-e^{-0.2/0.1})=0.17A$

$(b)$

$\dfrac{1}{2}LI^2=\dfrac{1}{2}\times2\times0.17^2=0.03J$

A conducting circular loop of area $1$ nm is placed compulsary with a long, straight wire at a distance of current which changes from $10$ A to zero in $0.1$ s. Find the average emf induced in the loop in $0.1$ s.

A = 1mm² ; i = 10A, d = 20cm ; dt = 0.1s

A coil draws a current of 1ampere and a power of 100 watt from an AC.Source of 100 volt and $\frac { 5\sqrt { 22 } }{ \pi }$ Hertz.Find the inductance and resistance of the coil

1328331_1114780_ans_cd9d17ab276942a4ad3d7e2ce242f1c8.jpg

A copper disc of radius $0.1$ m is rotated about its center with $20$ revolution per second in a uniform magnetic field of $0.1$ T with its plane perpendicular to the field. The emf induced across the radius of the disc is?

Emf induced is given by

$=\dfrac { 1 }{ 2 } B\omega { R }^{ 2 }$ ---(1) where B=magnetic field=0.1T;

$\omega$ =angular velocity=20 revolution per second and R=radius of the disc=0.1 m

Now putting the above mentioned values in equation 1

emf

$=\dfrac { 1 }{ 2 } 0.1\times 2\pi \times 20\times (0.1)^{ 2 }\\ =0.0628$

On what principle $AC$ generator works?

AC generator operates on the principle of electromagnetic induction, discovered (1831) by Michael Faraday. According to the principle when a conductor passes through a magnetic field, a voltage is induced across the ends of the conductor. In the simplest form of generator the conductor is an open coil of wire rotating between the poles of a permanent magnet, thus voltage is induced.

A conducting disc of radius r rotates with a small constant angular velocity $\omega$ about its axis. A uniform magnetic field B exists parallel to the axis of rotations. Find the motional emf between the center and the periphery of the disc.

Rotating dics can be consider as a rotating rod of same radius.

v at a distance $r/2$ From the centre $= r \dfrac{w}{2}$

$E=Blv$

$E= B \times r \times \dfrac{rw}{2}$

$E= \dfrac{1}{2} B r^2 w$

1081239_1129516_ans_4e6300fb321d42828e6bb9ee3aab68af.PNG

Figure shows a straight, long wire carrying a current $i$ and a rod of length $l$ coplanar with the wire and perpendicular to it. The rod moves with a constant velocity $v$ in direction parallel to the wire. The distance of the wire from the centre of the rod is $x$ . Find the motional emf induced in the rod.

In this case

$\vec { B }$ varies
hence considering a small element at centre of rod of length

$dx$ at a distance

$x$ from the wire

$\vec { B } =\cfrac { { \mu }_{ 0 }i }{ 2\pi x }$
so

$de=\cfrac { { \mu }_{ 0 }i }{ 2\pi x } \times vdx$

$e=\int _{ 0 }^{ e }{ de } =\cfrac { { \mu }_{ 0 }iv }{ 2\pi x } =\int _{ x-l/2 }^{ x+l/2 }{ \cfrac { dx }{ x } } =\cfrac { { \mu }_{ 0 }iv }{ 2\pi } \left[ \ln { (x+l/2) } -\ln { (x-l/2) } \right] =\cfrac { { \mu }_{ 0 }iv }{ 2\pi } \ln { \left[ \cfrac { x+l/2 }{ x-l/2 } \right] = } \cfrac { { \mu }_{ 0 }iv }{ 2\pi } \ln { \left( \cfrac { 2x+l }{ 2x-l } \right) }$

A constant current exists in a inductor-coil connected to a battery. The coil is short-circuited and the battery is removed. Show that the charge flown through the coil after the short-circuiting is the same as that which flows in one time constant before the short-circuiting.

In this case there is no resistor in the circuit
So, the energy stored due to the inductor before and after removal of battery remains same

${V}_{1}={V}_{2}=\cfrac{1}{2}L{i}^{2}$
So, the current will also remain same
Thus charge flowing through the conductor is the same

An LR circuit has $L=1.0H$ and $R=20\Omega$ . It is connected across an emf of $2.0V$ at $t=0$ What are the values of the self-induced emf in the circuit at the times indicated there in?

1604154_1723260_ans_fc82e3bacef742e78009ca1872030534.jpg

Consider a small cube of volume $1{mm}^{2}$ at the centre of a circular loop of radius $10cm$ carrying a current of $4A$ . Find the magnetic energy stored inside the cube.

Energy density

$\cfrac { { B }^{ 2 } }{ 2{ \mu }_{ 0 } }$
total energy stored

$\cfrac { { B }^{ 2 }V }{ 2{ \mu }_{ 0 } } =\cfrac { { \left( { \mu }_{ 0 }i/2r \right) }^{ 2 } }{ 2{ \mu }_{ 0 } } V=\cfrac { { \mu }_{ 0 }{ i }^{ 2 } }{ 4{ r }^{ 2 }\times 2 } =\cfrac { 4\pi \times { 10 }^{ -7 }\times { 4 }^{ 2 }\times 1\times { 10 }^{ -9 } }{ 4\times { \left( { 10 }^{ -1 } \right) }^{ 2 }\times 2 } =8\pi \times { 10 }^{ -4 }J$

The magnetic field at a point inside a $2.0mH$ inductor-coil becomes $0.80$ of its maximum value in $20\mu s$ when the inductor is joined to a battery. Find the resistance of the circuit.

${ i }_{ }={ i }_{ 0 }\left( 1-{ e }^{ -t/\tau } \right)$

$\Rightarrow { \mu }_{ 0 }ni={ \mu }_{ 0 }n{ i }_{ 0 }\left( 1-{ e }^{ -t/\tau } \right) \Rightarrow B={ B }_{ 0 }\left( 1-{ e }^{ -IR/L } \right)$

$\Rightarrow 0.8{ B }_{ 0 }={ B }_{ 0 }\left( 1-{ e }^{ -20\times { 10 }^{ -3 }\times R/2\times { 10 }^{ -3 } } \right) \Rightarrow 0.8\left( 1-{ e }^{ -R/100 } \right)$

$\Rightarrow { e }^{ -R/100 }=0.2\Rightarrow \ln { \left( { e }^{ -R/100 } \right) } =\ln { \left( 0.2 \right) }$

$\Rightarrow -R/100=-1.609\Rightarrow R=16.9=160\Omega \quad$

Consider the situation shown in figure. The wires ${ P }_{ 1 }{ Q }_{ 1 }$ and ${ P }_{ 2 }{ Q }_{ 2 }$ are made to slide on the rails with the same speed $5cm$ ${s}^{-1}$ . Find the electric current in the $19\Omega$ resistor if (a) both the wires move towards right and (b) if ${ P }_{ 1 }{ Q }_{ 1 }$ moves towards left but ${ P }_{ 2 }{ Q }_{ 2 }$ moves towards right

(a) The wires constitute 2 parallel emf
Net emf

$=Blv=1\times 4\times { 10 }^{ -2 }\times 5\times { 10 }^{ -2 }=20\times { 10 }^{ -4 }\quad$
Net resistance

$=\cfrac { 2\times 2 }{ 2\times 2 } +19=20\Omega$
Net current

$=\cfrac { 20\times { 10 }^{ -4 } }{ 20 } =0.1mA\quad$
(b) When both the wires move towards opposite directions then net emf=0
net current

$=0$

1565009_1723185_ans_4a9f3b34e7094f17b5bb9c85f63ee80c.png

A long wire carries a current of $4.00A$ . Find the energy stored in the magnetic field inside a volume of $1.00{mm}^{2}$ at a distance of $10.0cm$ from the wire.

$I=4.00A,V=1{ mm }^{ 3 },d=10cm=0.1m$

$\vec { B } =\cfrac { { B }^{ 2 } }{ 2{ \mu }_{ 0 } }$
Now magnetic energy stored

$=\cfrac { { B }^{ 2 } }{ 2{ \mu }_{ 0 } } V=\cfrac { { \mu }_{ 0 }^{ 2 }{ i }^{ 2 } }{ 4{ \pi r }^{ 2 } } \times \cfrac { 1 }{ 2{ \mu }_{ 0 } } \times V=\cfrac { 4\pi \times { 10 }^{ -7 }\times 16\times 1\times 1\times { 10 }^{ -9 } }{ 4\times 1\times { 10 }^{ -2 }\times 2 } =\cfrac { 8 }{ \pi } \times { 10 }^{ -14 }=2.55\times { 10 }^{ -14 }J\quad$

An inductor of inductance $2.00H$ is joined in series with a resistor of resistance $200\Omega$ and a battery of emf $2.00V$ . At $t=10ms$ , find (a) the current in the circuit, (b) the power delivered by the battery, (c) the power dissipated in heating the resistor and (d) the rate at which energy is being stored in magnetic field.

$L=2H,R=200\Omega, E=2V,t=10ms$
(a)

$l={ l }_{ 0 }\left( 1-{ e }^{ -t/\tau } \right) =\cfrac { 2 }{ 200 } \left( 1-{ e }^{ -10\times { 10 }^{ -3 }\times 200/2 } \right) =0.01(1-{ e }^{ -1 })=0.01(1-0.3678)=6.3A$
(b) Power delivered by the battery

$=VI=E{ I }_{ 0 }\left( 1-{ e }^{ -t/\tau } \right) =\cfrac { { E }^{ 2 } }{ R } { \left( 1-{ e }^{ -tR/L } \right) }=\cfrac { 2\times 2 }{ 200 } \left( 1-{ e }^{ -10\times { 10 }^{ -3 }\times 200/2 } \right) =0.02(1-{ e }^{ -1 })=0.1264=12mw$
(c) Power dissipated in heating the resistor

${ \left[ { i }_{ 0 }\left( 1-{ e }^{ -t/\tau } \right) \right] }^{ 2 }R={ (6.3mA) }^{ 2 }\times 200=6.3\times 6.3\times 200\times { 10 }^{ -6 }=7.938\times { 10 }^{ -3 }=8mA$
(d) Rate at which energy is stored in the magnetic field

$d/dt(1/2 L{I}^{2})$

$==\cfrac { L{ I }_{ 0 }^{ 2 } }{ \tau } \left( { e }^{ -t/\tau }-{ e }^{ -2t/\tau } \right) =\cfrac { 2\times { 10 }^{ -4 } }{ { 10 }^{ -2 } } \left( { e }^{ -1 }-{ e }^{ -2 } \right) =2\times { 10 }^{ -2 }\left( 0.2325 \right) =4.6\times { 10 }^{ -3 }=4.6mW$

An ac generator with emf $\xi=\xi _{m}\sin \omega _{d}t$ , where $\xi _{m}=25.0V$ and $\omega _{d}=377rad/s$ , is connected to a $4.15\mu F$ capacitor. (a) What is the maximum value of the current? (b) When the current is a maximum, what is the emf of the generator? (c) When the emf of the generator is $-12.5 V$ and increasing in magnitude, what is the current?

(a) The circuit consists of one generator across one capacitor; therefore,

$\xi _{m}=V_{C}$ . Consequently, the current amplitude is,

$I=\dfrac{\varepsilon _{m}}{X_{C}}=\omega C\varepsilon _{m}=\left ( 377\ rad/s \right )\left ( 4.15 \times 10^{-6}F \right )\left ( 25.0V \right )=3.91 \times 10^{-2}A$ .

(b) When the current is at a maximum, the charge on the capacitor is changing at its largest rate. This happens not when it is fully charged (

$\pm q_{max}$ ), but rather as it passes through the (momentary) states of being uncharged (

$q = 0$ ). Since

$q = CV$ , then the voltage across the capacitor (and at the generator, by the loop rule) is zero when the current is at a maximum. Stated more precisely, the time-dependent emf

$ε(t)$ and current

$i(t)$ have a

$\phi =-90^{o}$ phase relation, implying

$ε(t) = 0$ when

$i(t) = I$ . The fact that

$\phi =-90^{o}=-\dfrac{\pi}{2} \ rad$ is used in part (c).

(c) Consider equation

$\varepsilon =\varepsilon _{m}\sin \omega _{d}t$ with

$\varepsilon =-\dfrac{1}{2}\varepsilon _{m}$ . In order to satisfy this equation, we require

$sin( \omega _{d}t)=-1/2$ . Now we note that the problem states that

$ε$ is increasing in magnitude, which (since it is already negative) means that it is becoming more negative. Thus, differentiating equation with respect to time (and demanding the result be negative) we must also require

$\cos \left ( \omega _{d}t \right )<0$ . These conditions imply that

$ωt$ must equal

$\left ( 2n\pi -5\pi /6 \right )$ [

$n$ = integer]. Consequently, equation

$i=I\sin \left ( \omega _{d}t+\phi \right )$ yields (for all values of

$n$ )

$i=I\sin \left ( 2n\pi -\dfrac{5\pi}{6}+\dfrac{\pi}{2} \right )=\left ( 3.91 \times 10^{-3}A \right )\left (- \dfrac{\sqrt{3}}{2} \right )=-3.38 \times 10^{-2}A$ .
or,

$\left | i \right |=3.38 \times 10^{-2}A$

Explain the various part of $AC$ generator.

$AC$ generator is a device which produces an alternating current when a changing magnetic field is applied on a conductor coil. The frequency of the current is same as the frequency of rotation of the conductor coil. There are various parts of an

$AC$ generator which are shown in the image:-
The parts of

$AC$ generator are:-

Rectangular coil: Rectangular Coil is the conductor coil in which current is produced. When the coil rotates in an applied magnetic field due to electromagnetic induction a current is produced.
Permanent magnet: permanent magnets are another most important part. They provide a steady magnetic field in which the coil rotates. Without them there is no magnetic field and there will be no electromagnetic induction.
Slip rings: When the coil rotated, the wire attached to them through which we will get electricity will also rotate and will create a problem by as the wire will get a curl up. So the solution to this problem is slip rings. The current which is produced through coil is transported to the rings. Inner part of the ring is allowed to have rotation but the exterior part is not allowed and inner and exterior part are both connected. So the current passes to exterior part from inner part and then to the wire.
Carbon brushes: Carbon brushes are the fixed part which is connected to slip rings. They are fixed and current from slip rings passes to carbon brush and then to the wire.
Shaft: Shaft is like a handle attached to the coil, which is used for moving the coil. It basically makes rotation easier and faster.

1633233_1774036_ans_c1d5673230e04e4bacae0f09103dcd75.png

On what factors does the magnitude of the emf induced in the circuit due to magnetic flux depend?

It depends on the rate of change in magnetic flux (or simply change in magnetic flux)

$|\omega|=\dfrac {\Delta \varphi}{\Delta t}$

How does one understand this motional emf by invoking the Lorentz force acting on the free charge carriers of the conductor? Explain.

Suppose any arbitrary charge

$q$ in the conductor of length

$l$ moving inward in the field as shown in figure, the charge

$q$ also moves with velocity

$V$ in the magnetic field

$B$
The Lorentz force on the charge

$q$ is

$F=qvB$ and its direction is downwards.
So, work done in moving the charge

$q$ along the conductor of length

$l$

$W=F.l$

$W=qvBl$
Since emf is the work done per unit charge

$\therefore \varepsilon =\dfrac{W}{q}=Blv$
This equation gives emf induced across the rod.

State the rule to determine the direction of a current induced in coil due to its rotation in a magnetic field.

Fleming's right hand rule is used to find out the direction of the current induced in coil due to its rotation in a magnetic field.
According to Fleming's right-hand rule:
If we put our thumb, index finger and middle finger mutually perpendicular to each other, such that thumb points in the direction of the motion of conductor and index finger point in the direction of the applied magnetic field.
Then middle/ centre finger will show the direction of the induced or generated current within the conductor.

At a certain location in the Philippines Earth's magnetic field of $39\hspace{0.05cm} \mu T$ is horizontal and directed due north. Suppose the net filed is zero exactly $8.0\hspace{0.05cm} cm$ above a long, straight, horizontal wire that carries a constant current. What are the (a) magnitude and (b) direction of the current?

(a) The field due to the wire, at a point

$8.0 cm$ from the wire, must be

$39\hspace{0.05cm}\mu T$ and must be directed due south. Since

$B=\mu_0 i/2\pi r$ ,

$i=\dfrac{2\pi rB}{\mu_{0}}=\dfrac{2\pi(0.080\hspace{0.05cm})(39\times 10^{-6}\hspace{0.05cm}T)}{4\pi \times 10^{-7}\hspace{0.05cm}T.m/A}=16 A$

(b)The current must be from west to east to produce a field that is directed southward at points below it.

A bar magnet $M$ is dropped so that it falls vertically through the coil $C$ . The graph obtained for voltage produced across the coil versus time is showing in figure(b)
Explain the shape of the graph.

When the bar magnet falls through the coil, the magnetic flux linked with the coil changes, so an emf (or

$pd$ ) is developed across the coil.
Initially the rate of increases of flux increases, becomes maximum and then it decreases, becomes zero. Now magnetic flux begins to decrease, the rate of decrease increases becomes maximum and then it decreases and when the magnet is sufficiently far on the other side, the flux becomes zero and so pd induced becomes zero.

Answer the following questions
Define mutual inductance.

Mutual Inductance is the interaction of one coils magnetic field on another coil as it induces a voltage in the adjacent coil

The figure shows a loop model (loop L) for a paramagnetic
material. (a) Sketch the field lines through and about the material
due to the magnet. What is the direction of (b) the loops net magnetic dipole moment $\vec \mu$ , (c) the conventional current i in the loop
(clockwise or counterclockwise in the figure), and (d) the magnetic
force acting on the loop?

(a) A sketch of the field lines (due to the presence of the bar magnet) in the vicinity of the loop is shown.

(b) For paramagnetic materials, the dipole moment

$\vec μ$ is in the same direction as

$\vec B$ . From
the above figure,

$\vec μ$ points in the –x-direction.

(c) Form the right-hand rule, since

$\vec μ$ points in the –x-direction, the current flows
counterclockwise, from the perspective of the bar magnet.

(d) The effect of

$\vec F$ is to move the material toward regions of larger

$|\vec B|$ values. Since the
size of

$|\vec B|$ relates to the “crowdedness” of the field lines, we see that

$\vec F$ is toward the left,
or –x.
1662202_1779003_ans_fafbb8328a704b3d92c60299e92a9692.jpg

$t=0$ , the current in the long straight wire is 5.0 A rightward.
Thereafter, the current changes according to $i=5.0A-(2.0A/s^2)t^2$ . (The straight wire is insulated; so there is no electrical contact between it and the wire of the loop.) What is the magnitude of the current induced in the loop
at times $t>0$ ?

The field (due to the current in the straight wire) is out of the page in the upper half of the circle and is into the page in the lower half of the circle, producing zero net flux, at any time. There is no induced current in the circle.

A 12 H inductor carries a current of 2.0 A. At what rate must the current be changed to produce a 60 V emf in the inductor?

Since
$\varepsilon =-L(di/dt)$ , we may obtain the desired induced emf by setting

$\dfrac{d i}{d t}=-\dfrac{\varepsilon}{L}=-\dfrac{60 \mathrm{V}}{12 \mathrm{H}}=-5.0 \mathrm{A} / \mathrm{s}$

or $|d i / d t|=5.0 \mathrm{A} / \mathrm{s} .$ We might, for example, uniformly reduce
the current from $2.0 \mathrm{A}$ to zero in $40 \mathrm{ms}$

A cylindrical bar magnet is kept along the axis of a circular coil and neat it as shown in figure. Will there be any induced emf at the terminals of the coil, when the magnet is rotated About its own axis.

When the magnet is rotated about its own axis, then due to symmetry of magnet the magnetic flux linked with circular coil remains unchanged, hence no emf is induced at terminals of coil.

In the given diagram, a coil $B$ is connected to low voltage bulb $L$ and placed parallel to another coil $A$ as shown. Explain bulb gets dimmer if the coil $B$ moves upwards.

When coil

$B$ moves upwards, the magnetic flux linked with

$B$ decreases and hence lesser current is induced in

$B$ .

Define $1$ henry.

$1$ henry is self-inductance of that coil in which

$1$ volt emf is produced when the rate of change of current in that coil is

$1\ A/s$ .

The magnetic field of a cylindrical magnet that has a pole-face diameter of 3.3 cm can be varied sinusoidally between 29.6 T and 30.0 T at a frequency of 15 Hz. (The current in a wire wrapped around a permanent magnet is varied to give this variation in the net field.)
At a radial distance of 1.6 cm, what is the amplitude of the electric field induced by the variation?

The magnetic field $B$ can be expressed as

$B(t)=B_{0}+B_{1} \sin \left(\omega t+\phi_{0}\right)$

where $B_{0}=(30.0 \mathrm{T}+29.6 \mathrm{T}) / 2=29.8 \mathrm{T}$ and $B_{1}=(30.0 \mathrm{T}-29.6 \mathrm{T}) / 2=0.200 \mathrm{T} .$ Then

from Eq. $E=\dfrac{r}{2}\dfrac{dB}{dt}$

$E=\dfrac{1}{2}\left(\dfrac{d B}{d t}\right) r=\dfrac{r}{2} \dfrac{d}{d t}\left[B_{0}+B_{1} \sin \left(\omega t+\phi_{0}\right)\right]=\dfrac{1}{2} B_{1} \omega r \cos \left(\omega t+\phi_{0}\right)$

We note
that $\omega=2 \pi f$ and that the factor in front of the cosine is the
maximum value of the field. Consequently,

$E_{\max }=\dfrac{1}{2} B_{1}(2 \pi f) r=\dfrac{1}{2}(0.200 \mathrm{T})(2 \pi)(15 \mathrm{Hz})\left(1.6 \times 10^{-2} \mathrm{m}\right)=0.15 \mathrm{V} / \mathrm{m}$

A long cylindrical solenoid with 100 turns/cm has a radius of 1.6 cm. Assume that the magnetic field it produces is parallel to its axis and is uniform in its interior. (a) What is its inductance per meter of length? (b) If the current changes at the rate of 13 A/s, what emf is induced per meter?

(a) The self-inductance per meter is

$\dfrac{L}{\ell}=\mu_{0} n^{2} A=\left(4 \pi \times 10^{-7} \mathrm{H} / \mathrm{m}\right)(100 \mathrm{turns} / \mathrm{cm})^{2}(\pi)(1.6 \mathrm{cm})^{2}=0.10 \mathrm{H} / \mathrm{m}$

(b) The induced emf per meter is

$\dfrac{\varepsilon}{\ell}=\dfrac{L}{\ell} \dfrac{d i}{d t}=(0.10 \mathrm{H} / \mathrm{m})(13 \mathrm{A} / \mathrm{s})=1.3 \mathrm{V} / \mathrm{m}$

In the given diagram, a coil $B$ is connected to low voltage bulb $L$ and placed parallel to another coil $A$ as shown. Explain bulb lights if the coil $B$ moves upwards.

Bulb lights up due to induced current in

$B$ because of change in flux linked with it as a consequence of continuous variation of magnitude of alternative current flowing in

$A$

Consider an infinitely long wire carrying a current $I(t)$ , with $\dfrac {dI}{dt} = \lambda = constant$ .
Find the current produced in the rectangular loop of wire $ABCD$ if its resistance is $R$ (Fig.).

Consider a strip of width

$dr$ and length

$l$ inside a rectangle at distance

$r$ from the surface of current carrying conductor. The magnetic field across strip of length

$l = B(r) = \dfrac {\mu_{0}I}{2\pi r} l.B(r)$ is perpendicular to the paper upward.

$\therefore$ Flux in strip

$\phi = \dfrac {\mu_{0}I}{2\pi} l \int_{x_{0}}^{x} \dfrac {dr}{r}$

$\phi = \dfrac {\mu_{0}Il}{2\pi} [\log_{e}r]_{x_{0}}^{x} = \dfrac {\mu_{0}Il}{2\pi}\log_{e} \dfrac {x}{x_{0}}$

$\epsilon = \dfrac {-d\pi}{dt}$
So

$IR = \dfrac {d\phi}{dt}$

$I = \dfrac {1}{R} \dfrac {d}{dt}\left [\dfrac {\mu_{0}Il}{2\pi} \log_{e} \dfrac {x}{x_{0}}\right ] = \dfrac {\mu_{0}l}{2\pi R} . \log_{e} \dfrac {x}{x_{0}} \dfrac {dI}{dt}$

$I = \dfrac {\mu_{0}\lambda l}{2\pi R} \log_{e} \dfrac {x}{x_{0}} \left [\because \dfrac {dI}{dt} = \lambda (given)\right ]$ .

State two factors on which the magnitude of induced e.m.f. depend.

Magnitude of induced e.m.f depend upon:

(i) The change in the magnetic flux.

(ii) The time in which the magnetic flux changes.

Identify the figures and explain their use.

Figure (c) represents a DC generator. It is a device that generates electricity by rotating its rotor in a magnetic field. Thus, it converts mechanical energy into electrical energy.

On what factor do induced emf of a rotating coil (rectangular loop) in a magnetic field depend?

The induced e.m.f. is given by coil:

$\varepsilon =\varepsilon _0 sin\omega t=NAB \omega sin \omega t$

Where-

$N$ =Number of turn,

$A$ =Area of cross section,

$B$ =Magnetic field,

$\omega$ = Angular velocity.

How would we wrap two coils so that maximum value of induced emf is obtained?

The two coils should be wrapped opposite to one another. This makes the current flow in the reverse direction between the coils. This leads to the magnetic field to be in the same direction as given by the right-hand screw rule. Due to this, the induced emf due to the two coils will be maximum.

Define the coefficient of mutual inductance. Give its unit and dimensional formula.

Coefficient of Mutual Inductance

If orientation, size and shape of perimary coil

$C_1$ and secondary coil

$C_2$ , remains same and in coil

$C_1$ current is

$l_1$ , then second coil

$C_2$ is related to magnetic-flux which is proportional to the flow of current

$i_2.$ i.e.,

$\phi_2 \propto l_1$

$\phi_2=Ml_ 1 .........(1)$

Here,

$M$ is proportionality constant, which is called mutual induction. Its value depend upon number of turns,area of secondary coil and medium.

If the current in coil

$C_1$ changes with time, then flux linked with

$C_2, i.e., \phi_2$ changes .

Thus, induced emf in coil

$C_2,$

$\varepsilon _2=-\dfrac{d \phi_2}{dt}$

Thus,

$\varepsilon _2=-\dfrac{MdI_1}{dt}...............(2)$

The negative sign in equation (2), indicate that the direction of induced emf in secondary coil opposes the growth or decay of current flow in a primary coil, from equation(1),

$\phi_2=Ml_1$

$l_1=1 Amp$

then

$M=\phi_2...........(2)$

Thus the coefficient of mutual inductance is equal to the magnetic flux linked with secondary coil when the current flow in a primary coil is unity, from equation(2),

$\varepsilon _2= M\dfrac{dI_1}{dt }$

$\therefore M=\dfrac{\varepsilon _2}{-\dfrac{dI_1}{dt}}, if -\dfrac{dI_1}{dt}=1$

then

$m= \varepsilon _2$

Hence the coefficient of mutual inductance is equal to the induced emf when decay rate of current in primary coil is unity.

The unit of mutual inductance

$M=\dfrac{Weber}{Amp}$ or henry

$(H)$

$\dfrac{Volt \times sec}{Ampere}=Volt s^{-1}A^{-1}$

The dimensional Formula

$\dfrac{Dimension \quad of \quad magnetic \quad flux}{Dimention \quad of \quad current}$

$\dfrac{[MLT^{-2}A^{-1}]}{[A]}=[MLT^{-2}A^{-2}]$

Mutual inductance depends upon the number of turns in a coil, area of cross section and medium.

The

$S.I.$ unit of

$M$ is

$Wb/A$ or

$VA/A$ or henry

$(H)$ and its dimensions are

$[M^1L^2T^{-2}A^{-2}]$

A conducting wire is in North-South direction. It is freely dropped towards Earth. Is an emf induced between its ends? Why?

No there is no emf induced because there is no change in magnetic flux.

What is the value of self inductance, keeping the number of turns in the coil same and doubling the cross-sectional area?

Coefficient of self induction of coil is given by the relation:

$L= \dfrac{N^2\mu_0A}{l}$

$\therefore L \propto A$ So when the area is doubled, the self induction is

$2$ times.

A coil in a magnetic field is removed with:
(i) fast seed,
(ii) slow speed.
In which case, is the induced emf and work done more?

By the relation:

The induced emf is given as:

$\varepsilon =-\frac{d\phi}{dt},$

for fast speed

$dt$ is minimum so the induced emf is more.

The self inductance of a coil is $1 H$ . What do you understand?

Self inductance of coil

$L=1H$

We know that

Induced emf

$( \varepsilon )=-L\dfrac{dI}{dt} \therefore \varepsilon =-\dfrac{dI}{dt}$

Hence, when the coefficient of self inductance of coil is unit then the decay rate of current is equal to the induced emf in coil.

A conducting wire of length $20 cm$ is moving normally in a magnetic field of $5 \times 10^4 Wb/m.$ If the conducting wire covers $1m$ distance in $4s$ , then determine induced emf at the ends of the conducting wire.

Given , length of conductor

$(I) = 20 cm = 0.2 m$

Magnetic filed

$(B) = 5 \times 10^{-4} Wb/m^2$

Time

$= 4 s$

distance covered (s)

$= 1$ meter ,

Velocity of conductor

$(v) = \dfrac {s}{t} = \dfrac {1}{5} m/s$

$\therefore$ Induced

$emf ( \epsilon) = Blv$

$= 5 \times 10^{-4} \times 0.2 \times \dfrac {1}{4}$

$= 2.5 \times 10^{-5} volt$

A metallic rod of length $2m$ is moved (i) vertically ,(ii) horizontally with a speed of $15 km/h$ from West to East. If the horizontal component of Earth's magnetic field is $0.5 \times 10^{-5} Wb/m^2$ , calculate induced emf between the ends of the rod in each case.

Given,
Length of rod

$= 2 m$
Velocity of rod

$= 15 km /h = \dfrac { 15 \times 5}{18} = \dfrac {25}{6} m/s W to E$

The horizontal component of earth's magnetic filed

$(B) = 0.5 \times 10^{-5} Wb/m^2$

(i) When the rod is moving perpendicular then

$B_H , \overrightarrow {v}$ and

$\overrightarrow {l}$ are mutually perpendicular to each other
induced

$emf(e) = B_Hvl$

$= 0.5 \times 10^{-5} \times \dfrac {25}{6} \times 2$

$= 4 16 \times 10^{-5} volt$

(ii) When the rod is moving horizontal then :

$\epsilon = BLv sin 0$

$\epsilon = 0$

A conducting rod of length $L$ is rotated in a magnetic field $B$ with an angular velocity $\omega$ , so that the rotational plane of rod is perpendicular to the magnetic field Determine the induced emf between the ends of the rod.

Induced emf in a Metal Rod Rotating in a Uniform Magnetic Field

In figure

$9.10,$ a uniform magnetic field is shown by cross (x) whose direction is normally inwards the surface. A conducting rod

$OA$ of length

$L$ rotates in the anticlockwise direction in magnetic field

$B$ with angular velocity

$\omega$ . The plane of rotation of rod is perpendicular to the magnetic field. When an element

$dI$ of rod moves with velocity

$v$ normally with the magnetic field, the induced emf of the element

$d \varepsilon =Bvdl$

If the distance between small element and centre is

$l$ , then

$v=\omega l$

Thus, to determine induced emf in the rod, we intergrate the above equation from zero to

$L$ ,

$\int d\varepsilon =\int _{ 0 }^{ L } B\omega l dl$

$\varepsilon =\dfrac{1}{2} B\omega L^2 ...........(1)$

Using Fleming's right hand rule, the direction of induced current in rod is from

$A$ to

$O$ . Thus, the end

$O$ of the rod is positive and

$A$ is negative.

If the frequency of rotation of rod is

$f$ , then to

$\omega=2 \pi f$

Thus,

$\varepsilon =\dfrac{1}{2} B \times 2\pi f\times L^2$

$B\times \pi L^2 \times f$

If the area covered by the rod in magnetic field is represented by

$A$ then

$\pi L^2=A$

then

$\varepsilon =BAf ..........(2)$

Let the resistance of rod

$'R'$ then

$\therefore$ Induced current

$(I)=\dfrac{\varepsilon }{R}$

$\therefore$ form the equation (1) and (2),

$I=\dfrac{1}{2} \dfrac{B\omega L^2}{R}=\dfrac{1B\omega L^2}{2R}.....(3)$

and

$I=\dfrac{1}{2} \dfrac{BAf}{R}=\dfrac{BAf}{2R} .....(4)$

The heat produced due to the induced current:

$H=I^2Rt$

From the equation (3)

$H=\left( \dfrac { 1B\omega L^2 }{ 2R } \right)^2 Rt$

$=\dfrac{B^2\omega^2L^4}{R^2}Rt=\dfrac{B^2\omega^2L^4t}{R}.............(5)$

The fore produced in a conductor:

$F=IBl \quad sin 90=IBl$

From equation (3)

$F=\dfrac{1}{2} \dfrac{B\omega L^2}{R} BL=\dfrac{1}{2}\dfrac{B^2\omega L^3}{R}......(6)$

1758796_1836567_ans_d06859304c2e4bb2a5438edb751ec1ba.png

A coil is made from soft iron of length $0.1 m$ and radius $0.01m.$ If the relative magnetisation of soft iron is $1200$ , then calculate the number of turns in the coil.
[Self inductance of coil is $0.25 H$ ].

Given length of coil

$= I = 0.1 m$

Radius

$(R) = 0.01 m$

Relative permeability of the space =

$\mu_r = 1200$

coefficient of self induction

$(L) = 0.25 H$

The self induction of the coil

$(L) = \dfrac { \phi}{I}$

$\therefore \quad L = \dfrac {NBA}{I} = \dfrac {N( \dfrac { \mu NI}{2R}) \times \pi R^2 }{I}$

$L = \dfrac { \mu N^2 \pi R}{2} \quad \quad [ but \mu = \mu_r \mu_0]$

$\therefore N = \sqrt { \dfrac {2L}{ \mu_r \mu_0 \pi R } } = \sqrt { \dfrac {2 \times 0.25 }{ 4 \times 3.14 \times 10^{-7} \times 1200 \times 3.14 \times 0.1} }$

$= 1.027 \times 10^2$

$N = 10^2 m$

The magnetic flux linked with the coil of $50 turns$ is $\phi_B=0.02$ cos $100 \pi t$ Wb. Determine:
(a)Mximum induced volatage.
(b)Induced emf at $t=0.01s.$
(c) Induced electric current at $t=0.005 s.$
(if external resistance is $100 \Omega$ )

Number of turns

$- N = 50$
Magnetic flux

$( \phi_B) = 0.02 cos (100 \pi t) Wb$
(a) The induced emf

$(\epsilon)$ =

$- \dfrac { Nd \phi}{dt}$

$= -50 \dfrac {d}{dt} [ 0.02 cos ( 100 \pi t) ]$

$= 50 \times 0.02 \times 100 \pi sin ( 100 \pi t)$

$= 314 sin ( 100 \pi t)$
Compare by the relation

$\epsilon = \epsilon_0 sin wt$

$\therefore \quad \epsilon_0 = 314 V$ and

$W = 100 \pi$

$(b) t = 0.01 s$

$\epsilon = 314 sin ( 100 \pi \times 0.01)$

$= 314 sin \pi = 0$

(c) The external resistance ( R) =

$100 \Omega , t = 0.005 s$

$\therefore$ induced current (I)

$= \dfrac {\epsilon}{R} = 314 sin \dfrac {(100 \pi r) }{R}$

$= \dfrac { 314 sin ( 100 \times \pi \times 0.005)}{ 100 }$

$= \dfrac {314 sin ( \pi /2) }{ 100} = 3.14$

If a current of $5 A$ flowing in a primary coil nullifies in $2$ min, then induced emf is $25 kV$ . Calculate the coefficient of mutual inductance.

Given , change in current

$dl) = 0-5 = -5A$

Time

$(dt)= 2 ms = 2 \times 10^{-3} s$

Induced emf in coil

$\epsilon = kV = 2.5 \times 10^4 volt$

$\therefore \quad \epsilon = -M \dfrac {dI}{dt}$

$\therefore \quad M = - \dfrac { \epsilon}{dI} \times dt$

$= - \dfrac { 2.5 \times 10^4 \times 2 \times 10^{-3} }{-5A} = 10 H$

What is self induction? Explain the experiments of self induction and calculate self inductance for a solenoid.

Self induction

when current in a coil changes the magnetic flux linked with the coil also changes and hence emf is induced in the coil . this phenomenon is known as self induction.

According to len'z law the induced emf opposes the change in magnetic flux

Experimental representation in fig a conducting coil C is connected with a battery and key in series. when current is passed through the key hence magnetic flux is linked with the coil changes. when key is pressed , on increasing current, the change in magnetic flux also increase which produces emf and the induced current oppose the current that flows through the circuit . Similarly when key is opened the current in the circuit is zero . Then , magnetic flux is linked with the coil decreases and hence the induced current flows in the direction of the current supplied. this is shown in figure .

Experiment to demonstrate self induction :

take a solenoid having a large number of insulated wire wound over over a soft iron core . such a solenoid is called a choke coil. connect the solenoid in series with battery . a eheostat and a tapping key . connect a 6 v bulb in parallel with the solenoid . press the tapping key and adjust the current with the help pf rheostat so that the bulb just glows faintly . As the tapping key is released , the bulb glows brightly for a moment and then goes out . this is because as the circuit is broken suddenly . vanishes is the rate of charge of magnetic flux linked with the coil is very large. hence large self induced emf and current are produced in the coil which make the bulb glow brightly for a moment.

1758998_1836582_ans_a23fb91e380448adb32532aab1fee48a.png

A conducting rod of length $1m$ rotates with an angular velocity at the rate $50$ revolutions per second normally to a magnetic field of $0.001 T$ from one end. Calculate the induced emf between the ends of the rod.

Length of rod =

$1 m$

Rotational frequency

$(f) = 50 rps$

$\therefore \quad angular \quad velocity (w) = \dfrac { 2 \pi f}{t}$

$= \dfrac { 2 \times 3.14 \times 50}{ 1} = 314 rad /sec$

Magnetic filed

$(B) = 0.001 T$

The induced emf across the ends of rod

$\rightarrow$

$\epsilon = \dfrac {1}{2} B w l^2$

$= \dfrac {1}{2} \times 0.001 \times 314 \times 1 \times 1$

$= \dfrac { 0.314}{2} = 0.157 volt$

A solenoid of radius $2 cm$ and $100$ turns has a length of $50 cm$ . If vacuum is inside the solenoid, then calculate the self inductance of solenoid.

radius of solenoid

$r) = 2 cm = 2 \times 10^{-2} ,$

Number of turns

$(N) = 100$

Length of solenoid

$(I) = 50 cm = 50 \times 10^{-2} m$

$\therefore$ Self inductance of solenoid

$L) = \dfrac { \mu_0 N^2A}{l}$

$= \dfrac { \mu_0 N^2 \pi r^2 }{ l }$

$= \dfrac { 4 \pi \times 10^{-7} \times 100 \times 100 \times 3.14 \times ( 2 \times 10^{-2})^2 }{ 50 \times 10^{-2}}$

$= 31.55 \times 10^{-6} H$

$= 31.55 \mu H$

What do you mean by electromagnetic induction? State Faradays law of induction.

Faraday's law of electromagnetic induction, also known as Faraday's law is the basic law of electromagnetism which helps us predict how a magnetic field would interact with an electric circuit to produce an electromotive force (EMF). This phenomenon is known as electromagnetic induction.

Faraday's laws of Electromagnetic Induction:

Faraday's laws of Electromagnetic Induction consists of two laws. The first law describes the induction of emf in a conductor and the second law quantifies the emf produced in the conductor.

Faraday's First law of Electromagnetic Induction:

The discovery and understanding of electromagnetic induction are based on a long series of experiments carried out by Faraday and Henry. From the experimental observations, Faraday arrived at the conclusion that an emf is induced in the coil when the magnetic flux across the coil changes with time. With this in mind, Faraday formulated his first law of electromagnetic induction as, whenever a conductor is placed in varying magnetic field, an electromotive force is induced. If the conductor circuit is closed, a current is induced which is called induced current.

Changing the magnetic field intensity in a closed loop. Shown in below figure.

Faraday's second law of Electromagnetic Induction:

Faraday's second law of electromagnetic induction states that, The induced emf in a coil is equal to the rate of change of flux linkage.

The flux is the product of the number of turns in the coil and the flux associated with the coil. The formula of Faraday's law is given below:

$\varepsilon =-N\frac{\Delta \phi}{\Delta t}$

Where,

$\varepsilon$ is the electromotive force.

$\phi$ is the magnetic flux.

$N$ is the number of turns.

The negative sign indicates that the direction of the induced emf and change in direction of magnetic fields have opposite signs.

Additionally, there is another key law known as Lenz's law that describes electromagnetic induction as well.

1763728_1841896_ans_94e84bdf7ee0410da201a4f4ed616352.png

Current is induced when the armature of a generator rotates. Slip rings and brushes are the ways and means by which this current is brought to the outer circuit. Is the arrangement necessary if the magnet in generator is made to rotate ?

Current is induced when the armature of a generator rotates. Slip rings and brushes are the ways and means by which this current is brought to the outer circuit. No, this arrangement is not necessary if the magnet in generator is made to rotate.

There is only one type of generator AC generator. Write down your responses about this statement.

According to the nature of output, generators can be classified into 2 types AC generator and DC generator. Though ac current is produced in a DC generator with the help of split-ring commutator ac is converted into dc current.

The diagram shown above is an arrangement for producing $10\,V$ AC using electromagnetic induction. Observe the diagram carefully and answer the question.
Find out the frequency by drawing a time emf graph if the armature completes $10$ cycles in $5$ seconds.

$f = \dfrac{10}{5} = 2\,Hz$
1784992_1848936_ans_ee772b4fc04842d2afb0519ba54b34a6.png

Write down the similarities and differences in the structure of an AC generator and a DC generator.

The Similarities between AC and DC generator are:
(i) Field magnet and Armature are present.
(ii) Works on the principle of electromagnetic inductance.
(iii) AC produced on armature.

The differences between the AC and DC generators are:

AC generator	DC generator
Slip ring AC	Split ring commutator
produced in outer circuit	DC produced in outer circuit
Armature or magnet can be rotated	Only armature can rotated

Parts of an AC generator are given below.
Field magnet, Armature, Slip rings, Brush.
a. Explain the position of these parts in an AC generator.
b. Write down the functions of any two.

a. Armature rotates in the magnetic field of the field magnet. Springs are Soldered to the ends of armature coil, Graphite Brush is connected in such a way as to have constant contact with springs.
b. Armature - Generate emf.
Field magnet - Creates magnetic flux.

When $50 \,Hz$ AC is used, how many times will the direction of current change in the circuit ?

50 Hertz (Hz) means the rotor of the generator turns 50 cycles per second, the current changes 50 times per second back and forth, direction changes 100 times. That means the voltage changes from positive to negative, and from negative to positive voltage, this process converts 50 times/second.

Line diagrams of a generator are given.
What is the specialty of the electricity reaching the galvanometer if the armatures of both the generators are made to rotate ?

DC in first and AC in second.

Maximum voltage produced in an AC generator completing $60$ cycles in $30$ seconds is $250\,V.$
What is the period of the armature.

Period = Time taken to complete one full rotation.
Period

$= \dfrac{Time \,taken \,for \,rotation}{Number \,of \,fractions} = \dfrac{30}{60} = \dfrac{1}{2} = 0.5\,s$
When the Armature ABCD rotates, change in the magnetic flux depends on its direction and intensity.

A plane loop shown in Fig. is shaped as two squares with sides $a = 20\ cm$ and $b = 10\ cm$ and is introduced into a uniform magnetic field at right angles to the loop's plane. The magnetic induction varies with time as $B = B_{0}\sin \omega t$ , where $B_{0} = 10\ mT$ the current induced in the loop if its resistance per unit length is equal to $\rho = 50\ m\Omega/ m$ . The inductance of the loop is to be neglected.

The loops are connected in such a way that if the current is clockwise in one, it is anticlockwise in the order. Hence the e.m.f. in loop

bb.

Lenz's Law (26)
Consider the arrangement shown in Figure. Assume that $R = 6.00 \,\Omega, \,$ , and a uniform $2.50-T$ magnetic field is directed into the page. At what speed should the bar be moved to produce a current of $0.500 \,A$ in the resistor?

the induced current formula is given by:

I = ε/R =(−LvB)/R

where the − sign indicates the current is counterclockwise (out of the page), so current flows upward through the bar.

We can solve this equation for v, using only the magnitude of I

v = (IR)/(LB)= 1.00 m/s

A wire loop enclosing a semi-circle of radius $a$ is located on the boundary of a uniform magnetic field of induction $B$ (Fig.). At the moment $t = 0$ the loop is set into rotation with a constant angular acceleration $\beta$ about an axis $O$ coinciding with a line of vector $B$ on the boundary. Find the emf induced in the loop as a function of time $t$ . Draw the approximate plot of this function. The arrow in the figure shows the emf direction taken to be positive.

Flux at any moment of time,

$|\Phi_{t}| = |\vec {B} \cdot d\vec {S}| = B\left (\dfrac {1}{2}R^{2} \varphi \right )$
where

$\varphi$ is the sector angle, enclosed by the field.
Now, magnitude of induced e.m.f. is given by,

$\xi_{in} = \left |\dfrac {d\Phi_{t}}{dt}\right | = \left |\dfrac {BR^{2}}{2} \dfrac {d\varphi}{dt}\right | = \dfrac {BR^{2}}{2}\omega$ ,
where

$\omega$ is the angular velocity of the disc. But as it starts rotating from rest at

$t = 0$ with an angular acceleration

$\beta$ its angular velocity

$\omega (t) = \beta t$ . So,

$\xi_{in} = \dfrac {BR^{2}}{2}\beta t$ .
According to Lenz law the first half cycle current in the loop is in anticlockwise sense, and in subsequent half cycle it is in clockwise sense.
Thus is general,

$\xi_{in} = (-1)^{n} \dfrac {BR^{2}}{2}\beta t$ , where

$n$ in number of half revolutions.
The plot

$\xi_{in}(t)$ , where

$t_{n} = \sqrt {2\pi n/\beta}$ is shown in the answer sheet.

Calculate the inductance of a unit length of a double taped line (Fig.) if the tapes are separated by a distance $h$ which is considerably less than their width $b$ , namely, $b/h = 50$ .

Neglecting end effects the magnetic field

$B$ , between the plates, which is mainly parallel to the plates, is

$B = \mu_{0} \dfrac {I}{b}$
(For a derivation)
Thus, the associated flux per unit length of the plates is,

$\Phi = \mu_{0} \dfrac {I}{b}\times h\times 1 = \left (\mu_{0} \dfrac {h}{b}\right ) \times I$ .
So,

$L_{1} =$ inductance per unit length

$= \mu_{0} \dfrac {h}{b} = 24\ nH/ m$ .
1798276_1855099_ans_a941d5c3d888476c95058f4456df99d5.jpg

A current $I_{0} = 1.9\ A$ flows in a long closed solenoid. The wire it is wound of is in a superconducting state. Find the current flowing in the solenoid when the length of the solenoid is increased by $\eta = 5\%$ .

In a solenoid, the induction

$L = \mu \mu_{0} n^{2}V = \mu \mu_{0} \dfrac {N^{2}S}{l}$ ,
where

$S =$ area of cross section of the solenoid,

$l =$ its length,

$V = Sl, N = nl =$ total number of turns.
When the length of the solenoid is increased, for example, by pulling it, its inductance will decrease. If the current remains unchanged, the flux, linked to the solenoid, will also decrease. An induced e.m.f. will then come into play, which by Lenz's law will try to oppose the decrease of flux, for example, by increasing the current. In the superconducting state the flux will not change and so,

$\dfrac {I}{l} = constant$
Hence,

$\dfrac {I}{l} = \dfrac {I_{0}}{l_{0}}$ , or,

$I = I_{0} \dfrac {l}{l_{0}} = I_{0} (1 + \eta)$ .

In a certain region of the inertial reference frame there is magnetic field with induction $B$ rotating with angular velocity $\omega$ . Find $\bigtriangledown \times E$ in this region as a function of vectors $\omega$ and $B$ .

A rotating magnetic field can be represented by,

$B_{x} = B_{0}\cos \omega t; B_{y} = B_{0}\sin \omega t$ and

$B_{z} = B_{zo}$
Then curl,

$\vec {E} = \dfrac {\partial \vec {B}}{\partial t}$ .
So,

$-(Curl\ \vec {E})_{x} = -\omega B_{0}\sin \omega t = -\omega B_{y}$

$- (Curl\ vec {E})_{y} = \omega B_{0} \cos \omega t = \omega B_{x}$ and

$-(Curl\ \vec {E})_{z} = 0$
Hence,

$Curl\ \vec {E} = -\vec {\omega} \times \vec {B}$ ,
where,

$\vec {\omega} = \vec {e_{3}} \omega$ .

Calculate the mutual inductance of a long straight wire and a rectangular frame with sides $a$ and $b$ . The frame and the wire lie in the same plane, with the side $L$ being closest to the wire, separated by a distance $l$ from it and oriented parallel to it.

Here,

$B = \dfrac {\mu_{0}I}{2\pi r}$ at a distance

$r$ from the wire. The flux through the frame is obtained as,

$\Phi_{12} = \int_{1}^{a + 1} \dfrac {\mu_{0}I}{2\pi r} bdr = \dfrac {\mu_{0}b}{2\pi} I\ ln \left (1 + \dfrac {a}{l}\right )$
Thus,

$L_{12} = \dfrac {\Phi_{12}}{I} = \dfrac {\mu_{0}b}{2\pi} ln \left (1 + \dfrac {a}{l}\right )$
Thus,

$L_{12} = \dfrac {\Phi_{12}}{I} = \dfrac {\mu_{0}b}{2\pi} ln \left (1 + \dfrac {a}{l}\right )$ .
1798315_1855108_ans_b73cf2c665e444f3a4790a11eaa2f310.jpg

If a rotor in a power generator completes $3000$ rotations in a minute.
Which type of AC is produced here, single-phase or three-phase ?

Three phase AC

Three-phase electric power is a common method of alternating current electric power generation, transmission, and distribution. It is a type of polyphase system and is the most common method used by electrical grids worldwide to transfer power. It is also used to power large motors and other heavy loads.

A non-relativistic point charge $q$ moves with a constant velocity $v$ . Using the field transformation formulas, find the magnetic induction $B$ produced by this charge at the point whose position relative to the charge is determined by the radius vector $r$ .

In the reference frame

$K$ , moving with the particle,

$\vec {E'} \cong \vec {E} + \vec {v_{0}} \times \vec {B} = \dfrac {\vec {q}}{4\pi \epsilon_{0} r^{3}}$

$\vec {B'} \cong \vec {B} - \vec {v_{0}} \times \vec {E}/ c^{2} = 0$ .
Here,

$\vec {v_{0}} =$ velocity of

$K'$ , relative to the

$K$ frame, in which the particle has velocity

$\vec {v}$ .
Clearly,

$\vec {v_{0}} = \vec {v}$ . From the second equation,

$\vec {B} \cong \dfrac {\vec {v} \times \vec {E}}{c^{2}} = \epsilon_{0}\mu_{0} \times \dfrac {q}{4\pi \epsilon_{0}} \dfrac {\vec {v}\times \vec {r}}{r^{3}} = \dfrac {\mu_{0}}{4\pi} \dfrac {q(\vec {v}\times \vec {r})}{r^{3}}$ .

Maximum voltage produced in an AC generator completing $60$ cycles in $30$ seconds is $250\,V.$
How many cycles are completed in $T / 2$ seconds ?

Given: time taken to produce 60 cycles

$=30$ sec.

$\therefore$ time period

$T = \dfrac{30}{60} = \dfrac{1}{2}$

Now

$T' = \dfrac{T}{2} = \dfrac{1}{4}$

We know,

$v = \dfrac{1}{T}$

$v = \dfrac{1}{1/4} = 4$ cycles / sec

Hence 4 cycle are completed in

$\dfrac{T}{2}$ seconds.

(a) Obtain an expression for the mutual inductance between a long straight wire and a square loop of side a as shown in Fig.
(b) Now assume that the straight wire carries a current of 50 A and the loop is moved to the right with a constant velocity, v = 10 m/s. Calculate the induced emf in the loop at the instant when x = 0.2 m. Take a = 0.1 m and assume that the loop has a large resistance.

Take a small element

$dy$ in the loop at a distance

$y$ from the long straight wire.

Magnetic flux associated with element,

$dy$ is

$d\phi=BdA$

Where,

$dA=a\times dy$ is Area of element

$dy$

$B=\mu_oI/2\pi y$ is magnetic field at

$y$ .

$I$ is current in the wire.

$\displaystyle \therefore d\phi=\mu_oIa\times dy/2\pi y$

$\displaystyle \phi=(\mu_oIa/2\pi)\int_x^{a+x}{dy/y}$

$\Rightarrow \phi=(\mu_oIa/2\pi)ln(\dfrac{a+x}{x})$

Mutual Inductance

$M=\phi/I=(\mu_oa/2\pi)ln(\dfrac{a+x}{x})$

Emf induced in the loop,

$e=Bav=(\mu_oI/2\pi x)av$

From the given values,

$e=5\times10^{-5}\ V$

A line charge $\lambda$ per unit length is lodged uniformly onto the rim of a wheel of mass $M$ and radius $R$ . The wheel has light non-conducting spokes and is free to rotate without friction about its axis as shown in above figure. A uniform magnetic field extends over a circular region within the rim. It is given by,
$B\, =\, -B_0\, k\, \quad\, (r\, \leq \, a;\, a\, <\, R)$ , or $B\,=\,0$ (otherwise).
What is the angular velocity of the wheel after the field is suddenly switched off?

Line charge per unit length,

$\lambda=Q/2\pi r$

at distance r, the magnetic force is balanced by the centripetal force.

So,

$BQv=Mv^2/r$

also the angular velocity,

$\omega=v/R=B2\pi \lambda r^2/MR$

So for

$r\leq a;a\lt R$ , we get:

$\omega=-2B_o a^2 \lambda k/MR$

It is desired to measure the magnitude of field between the poles of a powerful loud speaker magnet. A small flat search coil of area $2\, cm^2$ with 25 closely wound turns, is positioned normal to the field direction, and then quickly snatched out of the field region. Equivalently, one can give it a quick $90^{\circ}$ turn to bring its plane parallel to the field direction). The total charge flown in the coil (measured by a ballistic galvanometer connected to coil) is 7.5 mC. The combined resistance of the coil and the galvanometer is 0.50 $\Omega$ . Estimate the field strength of magnet.

$\displaystyle Q\, =\, \int_{t_i}^{t_f}Idt$

$\displaystyle =\, \displaystyle \frac{1}{R}\, \int_{t_i}^{t_f}\varepsilon dt$

$=\, -\, \displaystyle \frac{N}{R}\, \int_{\Phi_i}^{\Phi_f}d\Phi$

$=\, \displaystyle \frac{N}{R}\, \left ( \Phi _i\, -\, \Phi _f \right )$

$N = 25, R = 0.50\Omega, Q = 7.5\, \times\, 10^{-3}\, C$

$\Phi _f\, =\, 0, \quad A\, =\, 2.0\, \times\, 10^{-4}\, m^2$

$\implies \Phi _i\, =\, 1.5\, \times\, 10^{-4}\, Wb$

$B\, =\, \Phi_i\, /\, A\, =\, 0.75\, T$

Explain in detail the principle, construction and working of a single phase AC generator.

AC Generator (Dynamo) - Single Phase: The ac generator is a device used for converting mechanical energy into electrical energy.

Principle: It is based on the principle of electromagnetic induction according to which an emf is induced in a coil when it is rotated in a uniform magnetic field.

Essential parts of an AC generator

i) Armature: Armature is a rectangular coil consisting of a large number of loops or turns of insulated copper wire wound over a laminated soft iron core or ring. The soft iron core not only increases the magnetic flux but also serves as a support for the coil.

ii) Field magnets: The necessary magnetic field is provided by permanent magnets in the case of low power dynamos. For high power dynamos, field is provided by electron magnet. Armature rotates between the magnetic poles such that the axis of rotation is perpendicular to the magnetic field.

iii) Slip rings: The ends of the armature coil are connected to two hollow metallic rings

${ R }_{ 1 }$ and

${ R }_{ 2 }$ called slip rings. These rings are fixed to a shaft, to which the armature is also fixed. When the shaft rotates, the slip rings along with the armature also rotate.

iv) Brushes:

${ B }_{ 1 }$ and

${ B }_{ 2 }$ are two flexible metallic plates or carbon brushes. They provide contact with the slip rings by keeping themselves pressed against the rig. They are used to pass on the current from the armature to the external power line through the slip rings.

Working: Whenever, there is a change in orientation of the coil, the magnetic flux linked with the coil changes, producing and induced emf in the coil. The direction of the induced current is given by Fleming's right hand rule.

Suppose the armature

$ABCD$ is initially in the vertical position. It is rotated in the anticlockwise direction. The side

$AB$ of the coil moves downwards and the side

$DC$ moves upwards Figure (a). Then according to Flemings right hand rule the current induced in arm

$AB$ flows from

$B$ to

$A$ and in

$CD$ it flows from

$D$ to

$C$ . Thus the current flows along

$DCBA$ in the coil. In the external circuit the current flows from

${ B }_{ 1 }$ to

${ B }_{ 2 }$ .

On further rotation, the arm

$AB$ of the coil moves upwards and

$DC$ moves downwards. Now the current in the coil flows along

$ABCD$ . In the external circuit the current flows from

${ B }_{ 2 }$ to

${ B }_{ 1 }$ . As the rotation of the coil continues, the induced current in the external circuit keeps changing its direction for every half a rotation of the coil. Hence the induced current is alternating in nature figure (b).

As the armature completes

$v$ -rotations in one second, alternating current of frequency

$v$ cycles per second is produced. The induced emf at any instant is given by

$e={ E }_{ 0 }\sin { \omega t }$ .

The peak value of the emf,

${ E }_{ 0 }=NBA\omega$ where

$N$ is the number of turns of the coil,

$A$ is the area enclosed by the coil,

$B$ is the magnetic field and

$\omega$ is the angular velocity of the coil.

What is self inductance? Establish expression for self inductance of a long solenoid.

Self inductance: When the current is passed through a coil, the magnetic flux linked with the coil is directly proportional is the current.
If the magnetic flux is

$\Phi$ for the current I, then

$\Phi \propto I$
or

$\Phi =L.I$
Where L is constant, its value depends upon shape and size of the coil meidum and number of turns. It is called self inductance or coefficient of self induction of the coil.
Expression- Consider a solenoid whose length l is very large in comparison to its radius r. A uniform magnetic filed is induced within it at every point when a current I is passed through it. The intensity of this induced magnetic field.

$B=\displaystyle\frac{\mu BI}{l}$ ..........(1)
Here

$\mu$ is the relative permeability of the material of core and N is the number of turns of the conducting wire on the solenoid. The magnetic flux linked with the solenoid is.

$\Phi = B\times$ total effective area of solenoid.
or

$\Phi =B\times N\pi r^2$ ........(2)
If self inductance is L, then

$\Phi =LI$ ........(3)
Comparing eqns. (2) and (3),

$LI=B\times N\pi r^2$
or

$L=\displaystyle\frac{B\times N\pi r^2}{I}$ ........(4)
Substituting B from eqn. (1) in eqn. (4),

$L=\displaystyle\frac{\mu NI}{l}\cdot \frac{N\pi r^2}{I}$
or

$L=\displaystyle\frac{\mu \pi N^2r^2}{l}$
This is the required expression.
Factors affecting the self inductance of the solenoid-
(i) On N the number of turns in the solenoid:

$L\propto N^2$ i.e., the self inductance increases on increasing the number of turns in the solenoid.
(ii) On the radius(r) of the solenoid:

$L\propto r^2$ i.e., the self inductance increases on increasing the radius.
(iii) On the length (l) of the solenoid:

$L\propto \displaystyle\frac{1}{l}$ i.e., self inductance decreases on increasing the length of solenoid.
(iv) On the relative permeability

$(\mu)$ of the material placed inside the solenoid:

$L\propto \mu$ i.e., if a soft iron rod is placed inside the solenoid its self inductance increases.

Figure shows a square frame of wire having a total resistance r placed coplaner with a long, straight wire. The wire carries a current i given by $i = {i_0}\sin \omega t.$ . Find (a) the flux of the magnetic field through the square frame; (b) the emf induced in the frame and (c) the heat developed in the frame in the time interval 0 to ${{20\pi } \over \omega }$ .

1596803_1034126_ans_0b949a200e1e49ffa9394b7f70cb00f3.jpg

Prove theoretically, the relation between e.m.f. induced and rate of change of magnetic flux in a coil moving in a uniform magnetic field.

Let us consider a rectangular wire loop

$PQRS$ of width

$l$ , and its plane perpendicular to a uniform magnetic field. The loop is being pulled out of magnetic field at a constant speed

$v$ . At any instant, let '

$x$ ' be the length of loop in the magnetic field. As the loop moves towards right, the area of the loop inside the field changes by

$dA=ldx=lvdt$ . So the change in magnetic flux is given by

$d\phi =BdA=Blvdt$
so

$\dfrac { d\phi }{ dt } =\dfrac { Blvdt }{ dt } =Bvl$
Now a straight current-carrying conductor of length

$L$ experiences a magnetic force given by

$\vec { F } =\vec { IL } \times \vec { B }$
Its direction is given by Fleming's left and rule.
So, the forces

${F}_{1}$ and

${F}_{2}$ on wires

$PR$ and

$QS$ respectively are equal in magnitude and opposite in direction and have the same line of action. Hence they balance each other. There is no force on the wire

$RS$ as it lies outside the field. The force

$\vec{{F}_{3}}$ on the wire

$PQ$ has magnitude

${F}_{3}=IBL$ directed towards left. To move the loop with constant velocity

$\vec{v}$ an external force must be applied. So work done by the external agent is given by

$dw=Fdx=-IlBdx=-IBdA$

$=-Id{ \phi }_{ m }$
So rate of doing work

$P=\dfrac { dw }{ dt }$

$=I\left( -\dfrac { d{ \phi }_{ m } }{ dt } \right)$
but

$P=EI$
so

$\boxed { E=-\frac { d{ \phi }_{ m } }{ dt } }$

Define self-inductance of a coil.
(a) Obtain an expression for the energy stored in a solenoid of self-inductance $'L'$ when the current though it grows from zero to $'I'$ .
(b) A square loop MNOP of side 20 cm is placed horizontally in a uniform magnetic field acting vertically downwards as shown in the figure. The loop is pulled with a constant velocity of 20 cm $s^{-1}$ till it goes out of the field.
(i) Depict the direction of the induced current in the loop as it goes out of the field. For how long would the current in the loop persist?
(ii) Plot a graph showing the variation of magnetic flux and induced emf as a function of time.

1) Self-inductance- It is defined as the ration of total flux linked with the coil to the current flowing through the coil. It is denoted as

$L$ .

$L=\dfrac{\phi _B}{I}$ . . . . . . .(1)

Expression for the energy stored in a solenoid-

The induced emf in the coil is given by

$e=-\dfrac{d\phi_B}{dt}$ . . . . .(2)

From equation (1) and (2), we get

$e=-L\dfrac{dI}{dt}$ . . . . . . . . .(3)

The self induced emf is also called back emf, as it opposes any change in the current. So, work needs to be done against back emf in the establishing current and this work done is stored as magnetic potential energy.

The rate of doing work is,

$\dfrac{dW}{dt}=|e|I=LI\dfrac{dI}{dt}$

The total work done,

$W=\int dW=\int _0 ^ILIdI$

$W=\dfrac{1}{2}LI^2$

(i) The direction of induced current in the loop is clockwise.

Given,

$d=20cm$

$v=20cm/s$

The current will persist till the entire loop comes out of the field,

$t=\dfrac{d}{v}=\dfrac{20}{20}=1sec$

(ii) The maximum flux is,

$\phi= Bla$ , a=side of the square.

The magnetic flux remains constant inside the magnetic field. This flux will starts dropping once the loop comes out of the magnetic field and it is zero when it comes completely out from the magnetic field as shown in the above figure.

The maximum Induced emf in coil is,

$e=-\dfrac{d\phi}{dt}=-Blv$

The

$e$ remains constant till the entire comes out from the magnetic field, when it comes out then emf drops to zero as shown in the above figure.

1421375_492955_ans_6a08c2b7e29a46439b102ef20ee844fc.png

A closed coil having 100 turns is rotated in a uniform magnetic field $B=4.0\ \times { 10 }^{ -4 }T$ about a diameter which is perpendicular to the field. The angular velocity of rotation is 300 revolution per minute. The area of the coil is $25\ { cm }^{ 2 }$ and its resistance is $4.0\ \Omega$ . Find (a) the average emf developed in half a turn from a position where the coil is perpendicular to the magnetic field, (b)the average emf in full turn and (c) the net charge displaced in part (a).

Given,

$N=100$ turns

$\\ B=4\times { 10 }^{ -4 }\quad T$

$\\ A=25\quad { cm }^{ 2 }=25\times { 10 }^{ -4 }{ \quad m }^{ 2 }$

(a) When the coil is perpendicular to field.

$\phi=NBA$

Average emf=

$\dfrac { 2NBA }{ t }$

$\dfrac { 2\times 100\times 4\times { 10 }^{ -4 }\times 25\times { 10 }^{ -4 } }{ \dfrac { 1 }{ 600 } \times 60 } =2\times { 10 }^{ -3 }V$

(b)In full turn average emf will be zero.

(c)Charged induced=

$\dfrac { 2NBA }{ R } =\dfrac { 2\times 100\times 4\times { 10 }^{ -4 }\times 25\times { 10 }^{ -4 } }{ 4 } \approx 5\times { 10 }^{ -5 }C$

In a motor, a rotor is fitted with the armature that has current of 10 A. The rotor rotates with angular speed of 3 rad/s.Magnetic field of magnitude 2 T varies in direction is such a way that it is always perpendicular to the loop area. If the rotor coil has N number of turns and area of each loop is $0.45 m^2$ then find the value N. Given that motor consumes 2106 W power and there are no losses.

Given that motor consume

$2106$ W power.

Current value=

$10A$

Voltage across motor armature=

$\dfrac{2106}{10}=210.6V$

emf induced=

$NBA\omega$

$N=\dfrac{210.6}{BA\omega}$

$N=\dfrac{210.6}{2\times0.45\times3}=\dfrac{210.6}{2.7}$

$N=78$

Deduce an equation $U=\cfrac { 1 }{ 2 } L{ I }^{ 2 }$ for an inductor.

$\varepsilon =-L\dfrac{di}{dt}$

The work done by voltage source during a time interval

$dt$ is

$dw=Pdt$

$dw=-\varepsilon idt$

$dw=-iL\dfrac{di}{dt}dt$

$dw=-Lidi$ .......(i)

Integrate both sides in eq (i)

$w=-\displaystyle \int_{0}^{I}{Lidi}$

$w=-1/2LI^2$

$\boxed{U=-W=1/2LI^2}$

1441411_961675_ans_1a8bdd4cfc004c5e808327c7beb254da.png

An infinitesimally small bat magnet $\bar M$ is pointing and moving with the speed $v$ in the $\bar x-$ direction. A small closed circular conducting loop of radius $a$ and negligible self inductance lies in the $y-z$ plane with its centre at $x=0$ and its axes coinciding with the $x-$ axis. Find the force opposing the motion of the magnet, if the resistance of the loop is $R$ . Assume that the distance $x$ of the magnet from the centre of the loop is much greater than $a$ .

1093263_1011409_ans_6ddb7b50493542c395cc88eee4971fbf.jpg

A wire forming one cycle of sine curve is moved in $x-y$ plane with velocity $\vec V={V}_{x}\hat i+{V}_{y}\hat j$ . There exist a magnetic field $\vec B=-{B}_{0}\hat k$ . Find the motional emf developed across the ends $PQ$ of wire.

1944439_1022166_ans_79440693c5ed45bba63d5567cfa915a2.jpg

The magnetic field in a region is given by $\overrightarrow { B } =\cfrac { { B }_{ 0 } }{ L } xk$ , where $L$ is a fixed length. A conducting rod of length $L$ lies along the X-axis between the origin and the point $(L,0,0)$ . If the rod moves with a velocity $\overrightarrow { v } ={ v }_{ 0 }\hat { j }$ , find the emf induced between the ends of the rod

1596776_1023335_ans_ae0b538dcbad4562be85a07427c2e905.jpg

Crosses represent uniform magnetic field directed into the paper. A conductor XY moves in the field towards right side. Find the direction of induced current in the conductor. Name the rule you applied. What will be the direction of current if the direction of field and the direction of motion of the conductor both are reversed?

Velocity

$\vec V$ is rightward and field

$\vec B$ is into the paper so

$\vec V \times \vec B$ will be

$upward$ so

$electron$ being negatively charged will shift downward so current can be thought of in

$upward$ direction. As force

$\vec F = q \vec V \times \vec B$ for electron

$q=-e$

if direction of field and velocity both are reversed the direction of induced current will remain same as

$-\vec V \times -\vec B= \vec V \times \vec B$

The figure shows a small circular coil of area A suspended from a point O by a string of length l in a uniform magnetic induction B in the horizontal direction. If the coil is set into oscillations likes simple pendulum by displacing it a small angle ${\theta _0}$ as shown, find emf induced in the coil as a function of time. Assume the plane of the coil is always in the plane of string.

The instantaneous induced emf to the coil of area

$A$ suspended by string of length

$l$ ,

$e=-\dfrac{d\phi}{dt}$

magnetic flux,

$\phi=BAcos\theta_0$ put in the above equation.

$e=-\dfrac{(dBAcos\theta_0)}{dt}$

$e=BAsin\theta_0\dfrac{d\theta_0}{dt}$

when

$\theta_0$ is very small.

$e=BA\theta_0\dfrac{d\theta_0}{dt}$

1421377_1034651_ans_bfed7de7648042f8ac0533c50702b34f.png

A conducting circular loop is placed in a uniform magnetic field $B=0.020T$ with its plane perpendicular to the field. Somehow, the radius of the loop starts shrinking at a constant rate of $1 mm/s.$ Find the induced current in the loop at an instant when the radius is $2 cm.$

Area

$A = \pi {r^2}$

$\dfrac{{dA}}{{dt}} = 2\pi r\dfrac{{dr}}{{dt}}$

when,

$r = 2\,cm\, = 0.02\,m$

$\dfrac{{dr}}{{dt}} = 1\,mm/s\,\, = {10^{ - 3}}m/s$

emf

$= N.B \times 2\pi r.\dfrac{{dr}}{{dt}}$

$=1 \times 0.02 \times 2\pi \times 0.02 \times {10^{ - 3}}V$

$=2.51 \times {10^{ - 6}}\,V$

$= 2.51\,\mu V$

Find the induced emf about ends of the rod in each case.

Refer image (i).

$|E|\vec{B}=\dfrac{d\vec{A}}{dt}$

here

$\dfrac{dA}{dt}=(2R)\times v$

$\therefore E=B(2R)v$

Refer image (ii)-

$XY=2l\sin\theta$

$LN=(2l\sin\theta)+2l(1+\sin\theta)$

$\therefore E=B2l(1+\sin\theta)v$

Refer image (ii)

(a) Refer image (iii-a)

(b) same as (iii) a Noe emf induced across LN since

$d\vec{A}=\vec{v}dt\times d\vec{l}$

$=0$

Refer image (iv)

No emf induced in the rod

$PQ$ and

$RS$

$\therefore E=B(3l)V$

Refer image (v)

$E=Blu$

1485667_1113887_ans_25ac19f6ccf1481fa8a3c9233f20063b.png

Consider the situation shown in the figure. The wire PQ has mass m, resistance r and can slide on the smooth horizontal parallel rails of separated by a distance $l$ . The resistance of the rails is negligible . A uniform magnetic field B exist in the rectangular region and a resistance R connects the rails outside the field region. At t=0, the wire PQ is pushed towards right with a speed of $v_{0}$ . Find
(a) the current in the loop in an instant when the speed of the wire PQ is v.
(b) the acceleration of the wire at this instant
(c) Velocity v as a function

(a) When the speed is

$V$

$Emf=Blv$
Resistance

$=r+R$
Current

$=\cfrac { Blv }{ r+R }$

(b) Force acting on the wire

$F=ilB$

$F=\cfrac { BlvlB }{ R+r } =\cfrac { { { l }^{ 2 }B }^{ 2 }v }{ r+R }$
Acceleration on the wire

$=\dfrac Fm =\cfrac { { { l }^{ 2 }B }^{ 2 }v }{ m(r+R) }$

(c)

$v={ v }_{ 0 }+at={ v }_{ 0 }-\cfrac { { { l }^{ 2 }B }^{ 2 }vt }{ m(r+R) } ={ v }_{ 0 }-\cfrac { { { l }^{ 1 }B }^{ 2 }x }{ m(r+R) }$

A coil has an inductance of 5H and resistance $20\Omega$ . An emf of $100$ v is applied to it. What is the energy stored in the magnetic field, when the current has reached its final value .

given

$L=5$ H

$R=20\Omega$

${E}_{v}=100$ v

When current reaches its final value

$L=0$

${I}_{0}=\dfrac{{E}_{0}}{R}$

$=\dfrac{\sqrt{2}{E}_{v}}{R}$

$=\dfrac{100\sqrt{2}}{20}$

$=5\sqrt{2}$ A

Energy stored in magnetic field

$=\dfrac{1}{2}L{I}_{\circ}^{2}$

$=\dfrac{1}{2}\times 5\times{\left(5\sqrt{2}\right)}^{2}$

$=\dfrac{1}{2}\times 5\times 25\times 2$

$=125$ J

A coil of resistance $40\Omega$ is connected across a $4.0$ V battery. $0.10$ s after the battery is connected, the current in the coil is $63$ mA. Find the inductance of the coil.

$R=40\Omega,\,E=4$ V,

$\,t=0.1,\,\,i=63$ mA

$i={i}_{\circ}\left(1-{e}^{\frac{tR}{2}}\right)$

$\Rightarrow 63\times{10}^{-3}=\dfrac{4}{40}\left(1-{e}^{-0.1\times\frac{40}{L}}\right)$

$\Rightarrow 63\times {10}^{-2}=1-{e}^{\frac{-4}{L}}$

$\Rightarrow 1-0.63={e}^{\frac{-4}{L}}$

$\Rightarrow {e}^{\frac{-4}{L}}=0.37$

$\Rightarrow \dfrac{-4}{L}=\ln{\left(0.37\right)}=-0.994$

$\Rightarrow L=\dfrac{-4}{-0.994}=4.024$ H

$=4$ H

Two protons are projected simultaneously from a fixed point with the same velocity v into a region, where there exists a uniform magnetic field. The magnetic field strength at B and it is prependicular to the initial direction of v. One proton starts at time t=o and another proton at t= $\frac { \pi m }{ 2qB }$ . The separation between then at time $\frac { \pi m }{ 2qB }$ (where, m and q are the mass and charge of proton), will be approximately.

2045919_1387684_ans_0bcb489879b641bbab70d7291cf9eb51.jpg

The closed loop $(P Q R S)$ of wire is moved out of a uniform magnetic field at right angles to the plane of the paper as shown. Predict the direction of induced current in the loop.

So far the loop remains in the magnetic field, there is no change in magnetic flux linked with the loop and so no current will be induced in it, but when the loop comes out of the magnetic field, the flux linked with it will decrease and so the current will be induced so as to oppose the decrease in magnetic flux, i.e. It will cause magnetic field downwards; so the direction of current will be clockwise.

Figure shows a smooth pair of thick metallic rails connected across a battery of emf $\varepsilon$ having a negligible internal resistance. A wire $ab$ of length $l$ and resistance $r$ can slide smoothly on the rails. The entire system lies in a horizontal plane and is immersed in a uniform vertical magnetic field $B$ . At an instant $t$ , the wire is given a small velocity $v$ towards right. (A) Find the current in it at this instant. What is the direction of the current? (b) What is the force acting on the wire at this instant? (c) Show that after some time the wire $ab$ will slide with a constant velocity. Find this velocity.

Net emf = E-Bvl

$I=\dfrac{E-Bvl}{r}$ from b to a

$F=IlB$

$=\left(\dfrac{E-Bvl}{r}\right)lb=\dfrac{lB}{r}(E-Bvl)$ towards right

After some time when

$E=Bvl$

Then the wire moves constant velocity v

Hence

$v=E/Bl$

2038211_1723211_ans_362a5fb7f3f549d9a41a57391efde281.png

Find the mutual inductance between the circular coil of radius $a^{'}$ and the loop of radius $a$ as shown in figure when the slider on the rheostat is moved. The total resistance of rheostat is R and the distance between the center of two coils is x.

Refer image,

The magnetic field at centre of coil 2, due to coil 1 is given by:

$B=\dfrac{\mu_0\,ia^2}{2(a^2+x^2)^{3/2}}$

The fluxed linked with coil 2 is given by

$\Phi=BA'$

$=\dfrac{\mu_0\,ia^2}{2(a^2+x^2)^{3/2}}\pi a^{1^2}$

Now, let y be the distance of the sliding contact from its left end.

Given:

$v=\dfrac{dy}{dt}$

Total Resistance of rheostat = R

When the distance of the sliding contact from the left end is y, the resistance of rheostat is given by

$r'=\dfrac{R}{L}y$

The current in the coil is the function of distance y travelled by the sliding contact of the rheostat. It is given by

$i=\dfrac{E}{\left(\dfrac{Ry}{L}+r\right)}$

The magnitude of emf induced can be calculated as

$e=\dfrac{d\Phi}{dt}=\dfrac{\mu_0a^2a^{1^2}\pi}{2(a^2+x^2)^{3/2}}\dfrac{di}{dt}$

$e=\dfrac{\mu_0\,\pi a^2a^{1^2}}{2(a^2+x^2)^{3/2}}\dfrac{d}{dt}\dfrac{E}{\left(\dfrac{R}{L}y+r\right)}$

$e=\dfrac{\mu_0\,a^2a^{1^2}}{2(a^2+x^2)^{3/2}}E\left[\dfrac{\left(-\dfrac{R}{L}v\right)}{\left(\dfrac{R}{L}y+r\right)^2}\right]$

emf induced,

$e=\dfrac{\mu_0\pi \,a^2a^{1^2}}{2(a^2+x^2)^{3/2}}E\left[\dfrac{-\dfrac{R}{L}v}{\left(\dfrac{Ry}{L}y\right)^2}\right]$

Now, emf induced in coil can be also given as

$\dfrac{di}{dt}=\left[\dfrac{E\left(-\dfrac{R}{L}v\right)}{\left(\dfrac{Ry}{L}y\right)^2}\right]$

$e=\dfrac{Mdi}{dt}$

$\dfrac{di}{dt}=\left[\dfrac{E\left(-\dfrac{Rv}{L}\right)}{\left(\dfrac{Ry}{L}+r\right)^2}\right]$

$\therefore M=\dfrac{e}{\dfrac{di}{dt}}=\dfrac{\mu_0\pi a^2a^{1^2}}{2(a^2+x^2)^{3/2}}$

2069993_1723628_ans_c9e31d09f3a643df9961781aecb6ad96.png

Figure (a) shows, in cross section, three current-carrying wires that are long, straight, and parallel to one another. Wires 1 and 2 are fixed in place on an $x$ axis, with separation d.Wire 1 has a current of 0.750 A, but the direction of the current is not given. Wire 3, with a current of 0.250 A out of the page, can be moved along the $x$ axis to the right of wireAs wire 3 is moved, the magnitude of the net magnetic force $\vec{F_{2}}$ on wire 2 due to the currents in wires 1 and 3 changes. The $x$ component of that force is $F_{2x}$ and the value per unit length of wire 2 is $F_{2x}/L_{2}$ . Figure (b) gives $F_{2x}/L_{2}$ versus the position $x$ of wireThe plot has an asymptote $F_{2x}/L_{2}=-0.627\hspace{0.05cm} \mu N/m$ as $x \rightarrow \infty$ . The horizontal scale is set by $x_{s}=12\hspace{0.05cm}cm$ . cm. What are the (a) size and (b) direction (into or out of the page) of the current in wire 2?

(a) The fact that the curve in Figure (b) passes through zero implies that the currents in wires 1 and 3 exert forces in opposite directions on wire 2. Thus, current

$i_1$ points out of the page. When wire 3 is a great distance from wire 2, the only field that affects wire 2 is that caused by the current in wire 1; in this case the force is negative according to Figure (b). This means wire 2 is attracted to wire 1, which implies that wire 2’s current is in the same direction as wire 1’s current: out of the page. With wire 3 infinitely far away, the force per unit length is given (in magnitude) as

$6.27 × 10^{−7} N/m$ .We set this equal to

$F_{12}=\mu_{0}i_1i_2/2\pi d$ . When wire 3 is at x=0.004m the curve passes through zero point previously mentioned, so the force between 2 and 3 must be equal

$F_{12}$ there. This allows us to solve for the distance between wire 1 and wire 2:

$d=(0.04m)(0.750A)/(0.250A)=0.12m$
Then we solve

$6.27\times 10^{-7} N/m=\mu_{0} i_1 i_2/2\pi d$ and obtain

$i_2=0.50A$

(b)The direction of

$i_2$ is out of the page.

An ac generator has emf $\xi=\xi _{m}\sin\left ( \omega _{d}t-\pi/4 \right )$ , where $\xi _{m}=30.0V$ and $\omega _{d}=350rad/s$ . The current produced in a connected circuit is $i\left ( t \right )=I\sin \left ( \omega _{d}t-3\pi /4 \right )$ , where $I=620 mA$ . At what time after $t=0$ does (a) the generator emf first reach a maximum and (b) the current first reach a maximum? (c) The circuit contains a single element other than the generator. Is it a capacitor, an inductor, or a resistor? Justify your answer. (d) What is the value of the capacitance, inductance, or resistance, as the case may be?

(a) The generator emf and the current are given by,

$\varepsilon =\varepsilon _{m}\sin \left ( \omega _{d}-\pi /4 \right )$ ,

$i\left ( t \right )=I\sin \left ( \omega _{d}-3\pi /4 \right )$ .
The expressions show that the emf is maximum when,

$\sin \left ( \omega _{d}t-\pi /4 \right )=1$ or

$\omega _{d}t-\pi/4=\left ( \pi /2 \right )\pm 2n\pi$ . [

$n$ = integer].

The first time this occurs after

$t = 0$ is when

$\omega _{d}t-\pi/4=\pi /2$ (that is,

$n = 0$ ). Therefore,

$t=\frac{3\pi}{4\omega _{d}}=\frac{3\pi}{4\left ( 350rad/s \right )}=6.73*10^{-3}s$ .

(b) The current is maximum when

$\sin \left ( \omega _{d}t-3\pi/4 \right )=1$ or,

$\omega _{d}t-3\pi /4=\left ( \pi /2 \right )\pm 2n\pi$ , [

$n$ = integer]
The first time this occurs after

$t = 0$ is when

$\omega _{d}-3\pi /4=\pi /2$ (as in part (a),

$n = 0$ ). Therefore,

$t=\frac{5\pi}{4\omega _{d}}=\frac{5\pi }{4\left ( 350rad/s \right )}=1.12-10^{-2}s$ .

(c) The current lags the emf by

$+π / 2 rad$ , so the circuit element must be an inductor.

(d) The current amplitude

$I$ is related to the voltage amplitude

$V_{L}$ by

$V_{L}=IX_{L}$ , where

$X_{L}$ is the inductive reactance, given by

$X_{L}=\omega _{d}L$ . Furthermore, since there is only one element in the circuit, the amplitude of the potential difference across the element must be the same as the amplitude of the generator emf:

$V_{L}=\varepsilon _{m}$ . Thus,

$\varepsilon _{m}=I\omega _{d}L$ and

$L=\frac{\varepsilon _{m}}{I\omega _{d}}=\frac{30.0V}{\left ( 620*10^{-3}A \right )\left ( 350rad/s \right )}=0.138H$ .

Note: The current in the circuit can be rewritten as,

$i\left ( t \right )=I\sin \left ( \omega _{d}-\frac{3\pi}{4} \right )=I\sin \left ( \omega _{d}-\frac{\pi}{4}-\phi \right )$ .

where

$\phi =+\pi /2$ . In a purely inductive circuit, the current lags the voltage by

$90°$ .

For above figure, show that the average rate at which energy is dissipated in resistance $R$ is a maximum when $R$ is equal to the internal resistance $r$ of the $ac$ generator. (we tacitly assumed that $r=0$ .)

This circuit contains no reactances, so

$\varepsilon _{rms}=I_{rms}R_{total}$ . Using equation

$P_{avg}=I_{rms}^{2}R$ , we find the average dissipated power in resistor

$R$ is,

$P_{R}=I_{rms}^{2}R=\left ( \frac{\varepsilon _{m}}{r+R} \right )^{2}R$ .

In order to maximize

$P_{R}$ we set the derivative equal to zero:

$\frac{dP_{R}}{dR}=\frac{\varepsilon _{m}^{2}\left [ \left ( r+R \right )^{2}-2\left ( r+R \right )R \right ]}{\left ( r+R \right )^{4}}=\frac{\varepsilon _{m}^{2}\left ( r-R \right )}{\left ( r+R \right )^{3}}=0\Rightarrow R=r$ .

An ac generator with emf amplitude $\xi _{m}=220V$ and operating at frequency $400Hz$ causes oscillations in a series RLC circuit having $R=200\Omega$ , $L=150 mH$ and $C=24.0\mu F$ . Find (a) the capacitive reactance $X_{C}$ , (b) the impedance $Z$ and (c) the current amplitude $I$ . A second capacitor of the same capacitance is then connected in series with the other components. Determine whether the values of (d) $X_{C}$ , (e) $Z$ and (f) $I$ increase, decrease or remain the same.

(a) The capacitive reactance is,

$X_{C}=\frac{1}{2\pi fC}=\frac{1}{2\pi \left ( 400Hz \right )\left ( 24.0\times10^{-6}F \right )}=16.6\Omega$ .

(b) The impedance is,

$Z=\sqrt{R^{2}+\left ( X_{L}-X_{C} \right )^{2}}=\sqrt{R^{2}+\left ( 2\pi fL-X_{C} \right )^{2}}$ ,

$=\sqrt{\left ( 220\Omega\right )^{2}+\left [ 2\pi \left ( 400Hz \right )\left ( 150\times10^{-3}H \right )-16.6\Omega\right ]^{2}}=422\Omega$ .

$I=\frac{\varepsilon _{m}}{Z}=\frac{220V}{422\Omega}=0.521A$ .
(d) Now

$X_{C}\propto C_{eq}^{-1}$ . Thus,

$X_{C}$ increases as

$C_{eq}$ decreases.

(e) Now

$C_{eq}=C/2$ , and the new impedance is,

$Z=\sqrt{\left ( 220\Omega\right )^{2}+\left [ 2\pi \left ( 400Hz \right )\left ( 150*10^{-3}H \right ) -2\left ( 16.6\Omega \right )\right ]}^{2}=408\Omega <422\Omega$
Therefore, the impedance decreases.

(f) Since

$I\propto Z^{-1}$ , it increases.

A long vertical wire carries an unknown current. Coaxial with the wire is a long, thin, cylindrical conducting surface that carries a current of $30 mA$ upward. The cylindrical surface has a radius of $3.0 mm$ . If the magnitude of the magnetic field at a point $5.0 mm$ from the wire is $1.0\mu T$ , what are the (a) size and (b) direction of the current in the wire?

$Given$ :- Radius of cylindrical surface

$= 3× 10^{-3}\space m$

Current in cylindrical conducting surface

$= 0.03 \space A$

Magnetic field

$(B=1.0 × 10^{-6} \space T)$ at distance

$5× 10^{-3}\space m$ from wire .

$To \space Find$ :- Magnitude and direction of current

$(i)$ in the wire

$Solution$ :- Using Ampere's law ,

Integral of

$B.ds = \mu_0I_{enclosed}$

$B(2πr) = \mu_0 I_{enclosed}$

$\implies$

$I_{enclosed} = \dfrac{B (2πr)}{\mu_0}$

$( r = 5× 10^{-3} \space m)$

$= 0.025$

Now ,

$I_{enclosed} = 0.03 + i$

$\implies$

$\underline{i= -5.0 \space mA}$

Hence ,

$\underline{i = -5.0\space mA \space in \space downward \space direction .}$

Figure shows a circuit having a coil of resistance $R=2.5 \Omega$ and inductance 'L' connected to a conducting rod PQ which can slide on a perfectly conducting circular ring of radius 10 cm with its centre at 'P'.
Assume that friction and gravity are absent and a constant uniform magnetic field of 5 T exists as shown in figure. At t =0,the circuit is switched on and simultaneously a time varying external torque is applied on the rod so that it rotates about 'P' with a constant angular velocity 40 rad/s.Find the magnitude of this torque ( in 'P') when current reaches half of its maximum value.Neglect the self inductance of the loop formed by the circuit.

Induced EMF =

$\dfrac{1}{2}B \omega l^2$

Maximum current :

$i_0=\dfrac{B \omega l^2}{2R}$

Torque about the hinge P is

$\tau =\int^l_0 i(dx)Bx$

$\Rightarrow \tau =\dfrac{1}{2}iBl^2$

Putting

$i=i_0/2,$

we get ;

$\tau =\dfrac{B^2 \omega l^4}{8R}=5 mNm$

(a) In an $RLC$ circuit, can the amplitude of the voltage across an inductor be greater than the amplitude of the generator emf? (b) Consider an $RLC$ circuit with emf amplitude $\xi _{m}=10V$ , resistance $R=10\Omega$ , inductance $L=1.0 H$ and capacitance $C=1.0\mu F$ . Find the amplitude of the voltage across the inductor at resonance.

(a) Yes, the voltage amplitude across the inductor can be much larger than the amplitude of the generator emf.

(b) The amplitude of the voltage across the inductor in an

$RLC$ series circuit is given by

$V_{L}=IX_{L}=I\omega L$ . At resonance, the driving angular frequency equals the natural angular frequency:

$\omega _{d}=\omega =1/\sqrt{LC}$ . For the given circuit,

$X_{L}=\frac{L}{\sqrt{LC}}=\frac{1.0H}{\sqrt{\left ( 1.0H \right )\left ( 1.0\times10^{-6}F \right )}}=1000\Omega$ .

At resonance the capacitive reactance has this same value, and the impedance reduces simply:

$Z = R$ . Consequently,

$I=\frac{\varepsilon _{m}}{Z}|_{resonance}=\frac{\varepsilon _{m}}{R}=\frac{10V}{10\Omega}=1.0A$ .

The voltage amplitude across the inductor is therefore

$V_{L}=IX_{L}=\left ( 1.0A \right )\left ( 1000\Omega \right )=1.0\times10^{3}V$ .
which is much larger than the amplitude of the generator emf.

$\overrightarrow {v}$ along two parallel metal rails, connected with a strip of metal at one end. A magnetic field of magnitude B = 0.350 T points out of the page.
(a) If the rails are separated by L = 25.0 cm and the speed of the rod is 55.0 cm/s, what emf is generated?
(b) If the rod has a resistance of $18.0\Omega$ and the rails and connector have negligible resistance, what is the current in the rod?
(c) At what rate is energy being transferred to thermal energy?

(a) Equation $\mathscr{E}=\dfrac{d\Phi _B}{dt}=\dfrac{d}{dt}BLx=BL\dfrac{dx}{dt}=BLv$ leads to

$\varepsilon=B L v=(0.350 \mathrm{T})(0.250 \mathrm{m})(0.55 \mathrm{m} / \mathrm{s})=0.0481 \mathrm{V}$

(b) By Ohm's law, the induced current is

$i=0.0481 \mathrm{V} / 18.0 \Omega=0.00267 \mathrm{A}$

By Lenz's law, the current is clockwise in Figure.

Faraday's Law of Induction (10)
Scientific work is currently under way to determine whether weak oscillating magnetic fields can affect human health. For example, one study found that drivers of trains had a higher incidence of blood cancer than other railway workers, possibly due to long exposure to mechanical devices in the train engine cab. Consider a magnetic field of magnitude $1.00 \times 10^{-3} \,T$ , oscillating sinusoidally at $60.0 \,Hz$ . If the diameter of a red blood cell is $8.00 \,mm$ , determine the maximum emf that can be generated around the perimeter of a cell in this field.

1979420_1872269_ans_051dbcba3ed546f38297b5df685ebb9d.jpeg

An $ac$ generator produces emf $\xi =\xi _{m}\sin \left ( \omega _{d}t-\pi /4 \right )$ , where $\xi _{m}=30.0V$ and $\omega _{d}=350rad/s$ . The current in the circuit attached to the generator is $i\left ( t \right )=I\sin \left ( \omega _{d}t+\pi /4 \right )$ , where $I=620 mA$ . (a) At what time after $t=0$ does the generator emf first reach a maximum? (b) At what time after $t=0$ does the current first reach a maximum? (c) The circuit contains a single element other than the generator. Is it a capacitor, an inductor, or a resistor? Justify your answer. (d) What is the value of the capacitance, inductance or resistance as the case may be?

(a) Let

$\omega t-\pi /4=\pi /2$ to obtain

$t=3\pi /4\omega =3\pi /\left [ 4\left ( 350rad/s \right ) \right ]=6.73\times10^{-3}s$ .

(b) Let

$\omega t+\pi /4=\pi /2$ to obtain

$t=\pi /4\omega =\pi /\left [ 4\left ( 350rad/s \right ) \right ]=2.24\times10^{-3}s$ .

(c) Since

$i$ leads

$ε$ in phase by

$π/2$ , the element must be a capacitor.

(d) We solve

$C$ from

$X_{C}=\left ( \omega C \right )^{-1}=\varepsilon _{m}/I$ :

$C=\frac{I}{\varepsilon _{m}\omega}=\frac{6.20\times10^{-3}A}{\left ( 30.0V \right )\left ( 350rad/s \right )}=5.90\times10^{-5}F$ .

The figure shows a loop
model (loop L) for a diamagnetic
material. (a) Sketch the magnetic
field lines within and about the material due to the bar magnet. What is the direction of (b) the loops net magnetic dipole moment $\vec \mu$ ,
(c) the conventional current i in the loop (clockwise or counterclockwise in the figure), and (d) the magnetic force on the loop?

(a) A sketch of the field lines (due to the presence of the bar magnet) in the vicinity of
the loop is shown below:

(b) The primary conclusion of Section 32-9 is two-fold:

$\vec u$ is opposite to

$\vec B$ , and the
effect of

$\vec F$ is to move the material toward regions of smaller |

$\vec B$ | values. The direction
of the magnetic moment vector (of our loop) is toward the right in our sketch, or in the +x
direction

(c) The direction of the current is clockwise (from the perspective of the bar magnet).

(d) Since the size of |

$\vec B$ | relates to the “crowdedness” of the field lines, we see that

$\vec F$ is
toward the right in our sketch, or in the +x direction.
1661199_1778630_ans_51fb921d4c6c4f00a5c7208d704c761e.jpg

A rectangular loop and a circular loop are moving out of a uniform magnetic field region to a field free region with a constant velocity. In which loop do you the induced emf to be a constant during the passage out of the field region? The field to the loop.

In rectangular coil the induced emf will remain constant because in this the case rate of change of area in the magnetic field region remains constant, while in circular coil the rate of change of area in the magnetic field region is not constant.

Answer the following question
A conduction rod of length $'I'$ with one end pivoted, is rotated with a uniform angular speed $'\omega '$ in a vertical plane, normal to a uniform magnetic field $'B'$ . Deduce an expression for the emf induced in this rod.
If resistance of rod is $R$ , what is the current induced in it?

Expression for Induced emf in a Rotating Rod
Consider a metallic rod

$OA$ of length

$I$ which is rotating with angular velocity

$'\omega '$ in a uniform magnetic field

$B$ , the plane of rotation being perpendicular to the magnetic field. A rod may be supposed to be formed of a large number of small elements. Consider a small element of length

$dx$ at a distance

$X$ from centre. If

$V$ is the linear velocity of this elements, then area swept by the element per second

$=v\ dx$
The emf induced across the ends of element

$d \varepsilon =B \dfrac{dA}{dt}=Bv\ dx$
But

$v=x \omega$

$\therefore d \varepsilon = B \times \omega \ dx$

$\therefore$ The emf induced across the rod

$\varepsilon = \displaystyle \int _0 ^1 B\ x \omega \ dx = B \omega \int _0 ^1\ x\ dx$

$B \omega \left[\dfrac{x^2}{2}\right] =B \omega \left[ \dfrac{l^2}{2} - 0 \right]= \dfrac{1}{2} \ B \omega I^2$
Current induced in rod

$I=\dfrac { \varepsilon }{R} = \dfrac{1}{2} \dfrac{B \omega l^2}{R}$ .
If circuit is closed, power dissipated

$=\dfrac { \varepsilon ^2 }{R} = \dfrac{B^2 \omega^2 l^2}{4R}$ .

A wheel with $8$ metallic spokes each $50\ cm$ long is rotated with a speed of $12\ rev/min$ in a plane normal to the horizontal component of the Earth's magnetic field. The Earth's magnetic field at the plane is $0.4\ G$ and the angle of dip is $60^{o}$ . Calculate the emf induced between the axle and the rim of the wheel.
How will the value of emf be affected if the number of spokes were increasde?

If a rod of length

$I$ rotates with angular speed

$\omega$ in uniform magnetic field

$B$

$\varepsilon=\dfrac{1}{2}BI^{2}\omega$
In case of earth's magnetic field

$B_{H}=|B_{e}|\cos \delta$
and

$B_{V}=|B_{e}|\sin \delta$

$\therefore \varepsilon=\dfrac{1}{2}|B_{e}|\cos \delta.l^{2}\omega$

$=\dfrac{1}{2}\times 0.4\times 10^{-4}\cos 60^{o}\times (0.5)^{2}\times 2\pi v$

$=\dfrac{1}{2}\times 0.4\times 10^{-4}\times \dfrac{1}{2}\times (0.5)^{2}\times 2\pi \times \left(\dfrac{120\ rev}{60s}\right)$

$=10^{-5}\times 0.25\times 2\times 3.14\times 2$ '

$=3.14\times 10^{-5}$ volt
Induced emf is independent of the number of spokes i.e. it remain same.

is coaxial with a solenoid of 220 turns/cm and diameter 3.2 cm. The
solenoid current drops from 1.5 A to zero in time interval $\Delta t=25ms$ . What current is induced in the coil during $\Delta$ ?

The total induced emf is given by

$\varepsilon =-N \dfrac{d \Phi_{B}}{d t}=-N A\left(\dfrac{d B}{d t}\right)=-N A \dfrac{d}{d t}\left(\mu_{0} n i\right)=-N \mu_{0} n A \dfrac{d i}{d t}=-N \mu_{0} n\left(\pi r^{2}\right) \dfrac{d i}{d t}$

$=-(120)\left(4 \pi \times 10^{-7} \mathrm{T} \cdot \mathrm{m} / \mathrm{A}\right)(22000 / \mathrm{m}) \pi(0.016 \mathrm{m})^{2}\left(\dfrac{1.5 \mathrm{A}}{0.025 \mathrm{s}}\right)$

$=0.16 \mathrm{V}$

Ohm's law then yields $i=|\varepsilon| / R=0.016 \mathrm{V} / 5.3 \Omega=0.030 \mathrm{A}$

In above figure, a three-phase generator G produces electrical power that is transmitted by means of three wires. The electric potentials (each relative to a common reference level) are $V_{1}=A\sin \omega _{d}t$ for wire $1$ , $V_{2}=A\sin \left ( \omega _{d}t-120^{0}\right )$ for wire $2$ , and $V_{3}=A\sin \left ( \omega _{d}t-240^{0}\right )$ for wire $3$ . Some types of industrial equipment (for example, motors) have three terminals and are designed to be connected directly to these three wires. To use a more conventional two-terminal device (for example, a lightbulb), one connects it to any two of the three wires. Show that the potential difference between any two of the wires (a) oscillates sinusoidally with angular frequency $\omega _{d}$ and (b) has an amplitude of $A\sqrt{3}$ .

(a) We consider the following combinations:

$\Delta V_{12}=V_{1}-V_{2}$ ,

$\Delta V_{13}=V_{1}-V_{3}$ and

$\Delta V_{23}=V_{2}-V_{3}$

For,

$\Delta V_{12}$ .

$\Delta V_{12}=A\sin \left ( \omega _{d}t \right )-A\sin\left ( \omega _{d}t-120^{0} \right )=2A\sin \left ( \frac{120^{0}}{2} \right )\cos \left ( \frac{2\omega _{d}t-120^{0}}{2} \right )=\sqrt{3}A\cos \left ( \omega _{d}t-60^{0} \right )$ .

where we use

$\sin \alpha -\sin \beta=2\sin \left [ \left ( \alpha -\beta \right )/2 \right ]\cos \left [ \left ( \alpha +\beta \right )/2 \right ]$ .
and,

$\sin 60^{0}=\sqrt{3}/2$ . Similarly,

$\Delta V_{13}=A\sin \left ( \omega _{d}t \right )-A\sin \left ( \omega _{d}t-240^{0} \right )=2A\sin \left ( \frac{240^{0}}{2} \right )\cos \left ( \frac{2\omega _{d}t-240^{0}}{2} \right )=\sqrt{3}A\cos \left ( \omega _{d}t-120^{0} \right )$ .
and,

$\Delta V_{23}=A\sin \left ( \omega _{d}t-120^{0} \right )-A\sin \left ( \omega _{d}t-240^{0} \right )=2A\sin \left ( \frac{120^{0}}{2} \right )\cos \left ( \frac{2\omega _{d}t-360^{0}}{2} \right )=\sqrt{3}A\cos \left ( \omega _{d}t-180^{0} \right )$ .
All three expressions are sinusoidal functions of

$t$ with angular frequency

$\omega _{d}$ .

(b) We note that each of the above expressions has an amplitude of

$\sqrt{3}A$ .

Answer the following questions
A rod of length $I$ is moved horizontally with a uniform velocity $'v'$ in a direction perpendicular to its length through a region in which a uniform magnetic field is acting vertically downward. Derive the expression for the emf induced across the ends of the rod.

Suppose a rod of length

$I$ moves with velocity

$v$ inward in the region having uniform magnetic field

$B$
Initial magnetic flux enclosed in the rectangular space is

$\varphi=|B|Ix$
As the rod moves with velocity

$-v=\dfrac{dx}{dt}$
Using Lenz's law

$\varepsilon=-\dfrac{d\varphi}{dt}=-\dfrac{d}{dt}(Blx)=Bl\left(-\dfrac{dx}{dt}\right)$

$\therefore \varepsilon =Blv$

The switch in Figure is closed on a at time t = 0.
What is the ratio $\mathscr{E}_L/\mathscr{E}$ of the inductor’s self-induced emf to the battery’s emf (a) just after t = 0 and
(b) at $t = 2.00\tau _L$ ? (c) At what multiple of $\tau _L$ will $\mathscr{E}_L/\mathscr{E}=0.500$ ?

(a) Immediately after the switch is closed, $\varepsilon-\varepsilon_{L}=i R .$
But $i=0$ at this instant, so $\varepsilon_{L}=\varepsilon,$ or $\varepsilon_{L} / \varepsilon=1.00$

(b) $\varepsilon_{L}(t)=\varepsilon e^{-t / \tau_{L}}=\varepsilon e^{-2.0 \tau_{L} / \tau_{L}}=\varepsilon e^{-2.0}=0.135 \varepsilon,$ or $\varepsilon_{L} / \varepsilon=0.135$

$\dfrac{t}{\tau_{L}}=\ln \left(\dfrac{\varepsilon}{\varepsilon_{L}}\right)=\ln 2 \Rightarrow t=\tau_{L} \ln 2=0.693 \tau_{L} \Rightarrow t / \tau_{L}=0.693$

Five long wires A,B,C,D and E each carrying current I are arranged to form edges of a pentagonal prism as shown in Fig.4.6 Each carries out of the plane of paper.
(a) what will be magnetic induction at a point on the axis O Axis is at a distance R from each wire.
(b) What will be the field if current in one of the wires (say A) is switched off
(c) What if current in one of the wire (say) A is reversed

(a) Figure shows that five conductors AA' , BB' ,CC' , DD' and EE' are along height of regular pentagonal prism ABCDE.

It is given that five identical conducting wires are along the heights of regular pentagon, represented in figure above by AA',BB' , CC' , DD' and EE' .

Axis of regular pentagon is OO' will be equidistant (R) from all five conductors, the current is passing through all five conductors are equal let (I).

(b) when current in AA' is switched off, then

$B_{1}$

$= 0$ and resultant becomes

$R= B_{2}+ B_{3} + B_{4} + B_{5} = 0$

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Figure shows a wire that has been bent into a circular arc of radius r = 24.0 cm, centered at O. A straight wire OP can be rotated about O and makes sliding contact with the arc at P. Another straight wire OQ completes the conducting loop. The three wires have cross-sectional area $1.20mm^2$ and resistivity $1.70 \times 10^{-8} \Omega \cdot m$ , and the apparatus lies in a uniform magnetic field of magnitude B = 0.150 T directed out of the figure.Wire OP begins from rest at angle $\theta =0$ and has constant angular acceleration of $12rad/s^2$ . As functions of $\theta$ (in rad), find (a) the loops resistance and (b) the magnetic flux through the loop. (c) For what $\theta$ is the induced current maximum and (d) what is that maximum?

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A uniform magnetic field $\overrightarrow {B}$ is perpendicular to the plane of a circular wire loop of radius r. The magnitude of the field varies with time according to $B=B_0e^{-t/\tau }$ , where $B_0$ and $\tau$ are constants. Find an expression for the emf in the loop as a function of time.

Faraday's law (for a single turn, with $B$ changing in time) gives

$\varepsilon=-\dfrac{d \Phi_{B}}{d t}=-\dfrac{d(B A)}{d t}=-A \dfrac{d B}{d t}=-\pi r^{2} \dfrac{d B}{d t}$

In this problem, we find $\dfrac{d B}{d t}=-\dfrac{B_{0}}{\tau} e^{-t l \tau} .$
Thus, $\varepsilon=\pi r^{2} \dfrac{B_{0}}{\tau} e^{-t / \tau}$

$ODBAC$ is a fixed rectangular conductor of negligible resistance $(CO$ is not connected) and $OP$ is a conductor which rotates clockwise with an angular velocity $\omega$ (Fig.). The entire system is in a uniform magnetic field $B$ whose direction is along the normal to the surface of the rectangular conductor $ABDC$ . The conductor $OP$ is in electric contact with $ABDC$ . The rotating conductor has a resistance of $\lambda$ per unit length. Find the current in the rotating conductor, as it rotates by $180^{\circ}$ .

(i) Let rotating conductor is in contact with

$BD$ at

$Q$ making angle

$0^{\circ} < \theta < 45^{\circ}$
Magnetic flux in

$\triangle ODQ$ is

$\phi$

$\phi = B.A$
The direction of

$B$ and

$A$ are

$0^{\circ}$ or

$180^{\circ}$

$Q = B \times \dfrac {1}{2} l\times QD$

$\therefore \phi = B.A\cos 0 = BA = B\dfrac {1}{2} \times l.l \tan \theta \left [\because \dfrac {QD}{l} = \tan \theta \right ] Qd = l\tan \theta$

$\phi = \dfrac {1}{2} Bl^{2}\tan \theta = \dfrac {1}{2}Bl^{2} \tan \omega t (\because \theta = \omega t)$
Induced e.m.f.

$\epsilon = \dfrac {-d\phi}{dt} = \dfrac {d}{dt} \dfrac {1}{2} Bl^{2} \tan \omega t$

$\epsilon = \dfrac {1}{2} Bl^{2}\omega \sec^{2} \omega t$

$I = \dfrac {\epsilon}{R} = \dfrac {Bl^{2}}{2R} \omega \sec^{2} \omega t\left [\because \cos \theta = \dfrac {1}{x}, x = \dfrac {1}{\cos \theta}\right ]$

$R$ of

$OQ = \lambda x = \dfrac {\lambda l}{\cos \theta}$

$R = \dfrac {\lambda l}{\cos \omega t}$

$I = \dfrac {Bl^{2}}{2.\lambda l}\omega \cos \omega t\sec^{2} \omega t = \dfrac {Bl\omega}{2\lambda \cos \omega t}$
(ii) Now rotating conductor rotates from

$B$ to

$A$ i.e.,

$45^{\circ}$ to

$135^{\circ}$ or

$\dfrac {\pi}{4} < \theta < \dfrac {3\pi}{4}$ .

$\phi = B . A = B$ or

$ODBQ = B$ .
Area of

$\triangle ORQ = \dfrac {1}{2} y\times l$

$\tan \theta = \dfrac {l}{y}$

$y = \dfrac {l}{\tan \theta}$

$\therefore$ Area of

$\triangle ORQ = \dfrac {1}{2} \dfrac {l^{2}}{\tan \theta} = \dfrac {l^{2}}{2\tan^{2}\omega t}$
Flux through

$OQBD$

$\phi = B.\left (lz + \dfrac {l^{2}}{2\tan \theta}\right )$

$= Blz + \dfrac {1}{2} Bl^{2} \cos \theta = Blz + \dfrac {1}{2} Bl^{2} \cot \omega t$

$\therefore \dfrac {d\phi}{dt} = \dfrac {d}{dt} Blz + \dfrac {1}{2} Bl^{2} (-cosec^{2} \omega t)\omega$
As

$l\ B$ and

$z$ are constant

$\therefore \dfrac {d(lBz)}{dt} = 0$

$\because \epsilon = \dfrac {-d\theta}{dt}$

$\therefore -\epsilon = 0 - \dfrac {1}{2} \dfrac {Bl^{2}\omega}{\sin^{2}\omega t}$

$\epsilon = \dfrac {1}{2} \dfrac {Bl^{2}}{\sin^{2}} \dfrac {\omega}{\omega t}$

$I = \dfrac {\epsilon}{R} = \dfrac {\epsilon}{\lambda x} = \dfrac {\epsilon \sin \omega t}{\lambda l}\left [\because \sin \theta = \dfrac {l}{x}, x = \dfrac {1}{\sin \omega t}\right ]$

$= \dfrac {\sin \omega t}{\lambda l} . \dfrac {1}{2} \dfrac {Bl^{2}\omega}{\sin^{2}\omega t}$

$I = \dfrac {Bl\omega}{2\lambda \sin \omega t}$
(iii) When

$\dfrac {3\pi}{4} < \theta < \dfrac {2\pi}{2}$ , the flux through

$OQABD = \phi = B.A$

$\phi = B.\left (2l^{2} + \dfrac {1}{2} ly\right )$
$$\phi = B . \left (2l^{2} + \dfrac {l^{2}\tan \omega t}{2}\right ) .... \begin{pmatrix}\tan (180 -\theta) = \dfrac {y}{l}\\y = l(-\tan \theta) \\y = -l \tan \theta
\end{pmatrix}$$

$\dfrac {d\phi}{dt} = \dfrac {d}{dt} \left [2l^{2} + \dfrac {l^{2}}{2} (\tan \omega t)\right ]B$

$-\epsilon = 0 + \dfrac {Bl^{2}}{2} \dfrac {d}{dt}\tan \omega t$

$-\epsilon = + \dfrac {Bl^{1}\omega}{2} \sec^{2} \omega t = -\dfrac {Bl^{2}\omega}{2\cos^{2} \omega t}$

$I = \dfrac {\epsilon}{R} = \dfrac {\epsilon}{\lambda x}$

$I = \dfrac {-Bl^{2}\omega}{2\cos^{2}\omega t}\dfrac {\cos \omega t}{\lambda $$(-1)} \left [\because \dfrac {l}{x}$

$= \cos (180 -\theta) \dfrac {1}{x}$

$$= -\cos \theta \Rightarrow x = \dfrac {-1}{\cos \omega t\right ]

I = \dfrac {Bl\omega}{2\lambda \cos \omega t}$$.

1687307_1796554_ans_3e898492bf2644febb281924a1ce547f.png

A long solenoid $'S'$ has $n$ turns per meter, with diameter $'a'$ . At the centre of this coil we place a smaller coil of $'N'$ turns and diameter $'b'$ (where $b < a$ ). If the current in the solenoid increases linearly, with time, what is the induced emf appearing in the smaller coil. Plot graph showing nature of variation in emf, if current varies as a function of $mt^{2} + c$ .

Varying magnetic field

$B(t)$ in solenoid is

$B_{1}(t) = \mu_{0} nI(t)$

This varying magnetic field changes flux in the smaller coil.

Magnetic flux in IInd coil

$= \mu_{0}nI (t) . \pi b^{2}$

Induced e.m.f. in second coil due to solenoid's varying magnetic field in

$1$ turn

$\epsilon' = \dfrac {-d\phi_{2}}{dt} = \dfrac {-d}{dt} \mu_{0} n \pi b^{2} I(t)$

$= -\mu_{0} n\pi b^{2} \dfrac {d}{dt} (mt^{2} + C)$

$= -\mu_{0} n \pi b^{2} . 2mt$

So net e.m.f. produced in

$N$ turns of smaller coil

$\epsilon = -\mu_{0} Nn\pi b^{2} 2mt$ .

(i)Two circular coils P and Q are kept close to each other, of which coil P carries a current. If coil P is moved towards Q, will some current be induced in coil Q? Give reason for your answer and name the phenomenon involved.
(ii) What happens if coil P is moved away from Q?
(iii) State any two methods of inducing current in a coil.

(i) When coil P is moved towards Q, current will be induced in coil Q. This is because on moving P the magnetic field associated with Q increases and so a current is induced. The Phenomenon is electromagnetic induction.
(ii) If P is moved away from Q, the field associated with Q will decrease and a current will be induced but in the opposite direction
(iii) Current can be induced in a coil by (a) moving a magnet towards or away from the coil (b) moving a coil towards or away from a magnet (c) rotating a coil within a magnetic field.

Two solenoids are part of the spark coil of an automobile. When the current in one solenoid falls from 6.0 A to zero in 2.5 ms, an emf of 30 kV is induced in the other solenoid. What is the mutual inductance M of the solenoids?

Given: current in solenoids falls from 6 A to 0 in just 2.5 ms, emf=30 kV

We use $\varepsilon_{2}=-M d i_{1} / d t \approx M \Delta i / \Delta t |$ to find $M$

$M=\left|\dfrac{\varepsilon}{\Delta i_{1} / \Delta t}\right|=\dfrac{30 \times 10^{3} \mathrm{V}}{6.0 \mathrm{A} /\left(2.5 \times 10^{-3} \mathrm{s}\right)}=13 \mathrm{H}$

Figure shows a uniform magnetic field $\overrightarrow {B}$ confined to a cylindrical volume of radius R. The magnitude of $\overrightarrow {B}$ is decreasing at a constant rate of 10 mT/s. In
unit-vector notation, what is the initial acceleration of an electron released at (a) point a (radial distance r = 5.0 cm), (b) point b (r = 0), and (c) point c (r = 5.0 cm)?

The induced electric field is given by Eq. $\oint \vec{E} \cdot d\vec{s}=-\dfrac{d\Phi _B}{dt}\text{ (Faraday’s Law) }$ :

$\oint \vec{E} \cdot d \vec{s}=-\dfrac{d \Phi_{B}}{d t}$

The electric field lines are circles that are concentric with the cylindrical
region. Thus,

$E(2 \pi r)=-\left(\pi r^{2}\right) \dfrac{d B}{d t} \Rightarrow E=-\dfrac{1}{2} \dfrac{d B}{d t} r$

The force
on the electron is $\vec{F}=-e \vec{E}$ , so by Newton's second law, the acceleration is

$\vec{a}=-e \vec{E} / m$

(a) At point $a$

$E=-\dfrac{r}{2}\left(\dfrac{d B}{d t}\right)=-\dfrac{1}{2}\left(5.0 \times 10^{-2} \mathrm{m}\right)\left(-10 \times 10^{-3} \mathrm{T} / \mathrm{s}\right)=2.5 \times 10^{-4} \mathrm{V} / \mathrm{m}$

With the normal taken to be into the page, in the direction of the magnetic field, the positive direction for $\vec{E}$ is clockwise. Thus, the direction of the electric field at point a is to the left, that is $\vec{E}=-\left(2.5 \times 10^{-4} \mathrm{V} / \mathrm{m}\right)$ i. The resulting acceleration is

$\vec{a}_{a}=\dfrac{-e \vec{E}}{m}=\dfrac{\left(-1.60 \times 10^{-19} \mathrm{C}\right)\left(-2.5 \times 10^{-4} \mathrm{V} / \mathrm{m}\right)}{\left(9.11 \times 10^{-31} \mathrm{kg}\right)} \hat{\mathrm{i}}=\left(4.4 \times 10^{7} \mathrm{m} / \mathrm{s}^{2}\right) \hat{\mathrm{i}}$

The acceleration is to the right.

(b) At point $b$ we have $r_{b}=0,$ so the acceleration is zero.

(c) The electric field at point $c$ has the same magnitude as the field in $a$ , but
with its direction reversed. Thus, the acceleration of the electron released at
point $c$ is

$\vec{a}_{c}=-\vec{a}_{a}=-\left(4.4 \times 10^{7} \mathrm{m} / \mathrm{s}^{2}\right) \hat{\mathrm{i}}$

In Figure a, a circular loop of wire is concentric with a solenoid and lies in a plane perpendicular to the solenoid’s central axis. The loop has radius 6.00 cm. The solenoid has radius 2.00 cm, consists of 8000 turns/m, and has a current $i_{sol}$ varying with time t as given in Figure b, where the vertical axis scale is set by $i_s=1.00A$ and the horizontal axis scale is set by $t_s=2.0s$ . Figure c shows, as a function of time, the energy $E_{th}$ that is transferred to thermal energy of the loop; the vertical axis scale is set by $E_s=100.0nJ$ .What is the loop’s resistance?

Equation $P=\dfrac{V^2}{R}$
gives $\varepsilon^{2} / R$ as the rate of energy transfer into thermal forms $\left(d E_{\mathrm{th}} / d t\right.$ which, from Figure c , is roughly $40 \mathrm{nJ} / \mathrm{s}$ ). Interpreting $\varepsilon$ as the induced emf (in absolute value) in the single-turn loop $(N=1)$ from Faraday's law, we have

$\varepsilon=\dfrac{d \Phi_{B}}{d t}=\dfrac{d(B A)}{d t}=A \dfrac{d B}{d t}$

Equation $B=\mu _0in$ gives $B=\mu_{0} n i$ for the solenoid (and note that the field is zero outside of the solenoid, which implies that $A=A_{\text {coil }}$ , so our expression for the magnitude of the induced emf becomes

$\varepsilon=A \dfrac{d B}{d t}=A_{\text {coil }} \dfrac{d}{d t}\left(\mu_{0} n_{\text { coil }}\right)=\mu_{0} n A_{\text { coil }} \dfrac{d i_{\text {coil }}}{d t}$

where Figure b, suggests that $d_{\mathrm{ coil }} / d t=0.5 \mathrm{A} / \mathrm{s} .$ With $n=8000$ (in SI units) and $A_{\mathrm{coil}}$ $=\pi(0.02)^{2}$ (note that the loop radius does not come into the computations of this problem, just the coil's), we find $\mathrm{V}=6.3 \mu \mathrm{V}$ . Returning to our earlier observations, we can now solve for the resistance:

$R=\varepsilon^{2} /\left(d E_{\mathrm{th}} / d t\right)=1.0 \mathrm{m} \Omega$

Describe the principle , construction and working of a single phase A.C. generator.

Working Principle: Single phase AC generator operates on Faraday's theory of electromagnetic induction which states that change in a magnetic field produces an electric current.

Construction:

It consists of 2-poles of a magnet in order to have a uniform magnetic field.

There is a rectangular shaped coil called an armature.

The armature is connected to 2 sets of slip rings, which helps in maintaining contact with the brushes.

These slip rings are attached to the carbon brushes.

The rotational motion of armature is free as its axis is perpendicular to the magnetic field.

Working:

Consider rectangular coil has 'N' no. of turns in a magnetic field rotating with

$\omega$ angular velocity.

Maximum flux linked with the coil when the coil's plane normal is aligned with magnetic field lines.

In time 't' seconds, coil rotates an angle

$\theta=\omega t$ .

In this deflected position, the component of flux which is perpendicular to the plane of the coil is :

$\phi=\phi_mcos(\omega t)$

Hence, the flux linkage at any time 't' is:

$\phi'=N\phi=N\phi_m\ cos(\omega t)$

Now, as we know that induced potential is:

$e=-\dfrac{d\phi'}{dt}$

$=-\dfrac{d(N\phi_m\ cos(\omega t))}{dt}$

$=-N\phi_m\ [-sin(\omega t)\cdot\omega]$

$\boxed{e=\omega N\phi_m\ sin(\omega t)}$

And we know that

$i=\dfrac eR$

So,

$i=\dfrac{\omega\ N\phi_m}{R}\ sin(\omega t)$

$\boxed{i=i_0\ sin(\omega t)}$ , where

$i_0=\dfrac{\omega N\phi_m}{R}$

1721431_1811699_ans_a301fb91233d46d1b743f0b906bdb965.png

A $\Pi-shaped$ conductor is located in a uniform magnetic field perpendicular to the plane of the conductor and varying with time at the rate $\overset {\cdot}{B} = 0.10\ T/s$ . A conducting connector starts moving with an acceleration $w = 10\ cm/s^{2}$ along the parallel bars of the conductor. The length of the connector is equal to $l = 20\ cm$ . Find the emf induced in the loop $t = 2.0\ s$ after the beginning of the motion, if at the moment $t = 0$ the loop area and the magnetic induction are equal to zero. The inductance of the loop is to be neglected.

The flux through the loop changes due to the variation in

$\vec {B}$ with time and also due to the movement of the conductor.
So,

$\xi_{in} = \left |\dfrac {d(\vec {B} \cdot \vec {S})}{dt}\right | = \left |\dfrac {d(BS)}{dt}\right |$ as

$\vec {S}$ and

$\vec {B}$ are colliniear
But,

$B$ , after

$t$ sec. of beginning of motion

$= Bt$ , and

$S$ becomes

$= l\dfrac {1}{2} wt^{2}$ , as connector starts moving from rest with a constant acceleration

$w$ .
So,

$\xi_{ind} = \dfrac {3}{2} B\ l\ w\ t^{2}$ .

In a Faraday disc dynamo, a metal disc of radius $R$ rotates with an angular velocity $\omega$ about an axis perpendicular to the plane of the disc and passing through its center. The disc is placed in a magnetic field $B$ acting perpendicular to the plane of the disc. Determine the induced emf between the rim and the axis of the disc.

$A=\pi R^{2}$

$\frac{dA}{dT}=\frac{\pi R^{2}}{\frac{2\pi}{W}}=\frac{WR^{2}}{2}$

emf

$\epsilon =-B*\frac{dA}{dt}=\frac{1}{2}(B\omega R^{2})$ .

A long straight wire carrying a current $I$ and a $\Pi-shaped$ conductor with sliding connector are located in the same plane as shown in Fig. The connection of length $l$ and resistance $R$ slides to the right with a constant velocity $v$ . Find the current induced in the loop as a function of separation $r$ between the connector and the straight wire. The resistance of the $\Pi-shaped$ conductor and the self-inductance of the loop are assumed to be negligible.

Field, due to the current carrying wire in the region, right to it, is directed into the plane of the paper and its magnitude is given by,

$B = \dfrac {\mu_{0}}{2\pi}\dfrac {i}{r}$ where

$r$ is the perpendicular distance from the wire.
As

$B$ is same along the length of the rod thus motional e.m.f.

$\xi_{in} = \left |-\int_{1}^{2} (\vec {v}\times \vec {B})\cdot d\vec {l}\right | = vBl$
and it is directed in the sense of

$(\vec {v} \times \vec {B})$
So, current (induced) in the loop,

$i_{in} = \dfrac {\xi_{in}}{R} = \dfrac {1}{2} \dfrac {\mu_{0}Ivi}{\pi R r}$ .

Above figure shows a metal rod $PQ$ resting on the smooth rails $AB$ and positioned between the poles of a permenent magnet. The rails, the rod, and the magnetic filed are in three mutual perpendicular directions. A galvanometer $G$ connects the rail through a switch $K$ . Length of the rod is $15cm$ , $B=0.50T$ , resistanbce of the closed-loop containing the rod is $9.0$ . Assume the filed to be uniform.
(a) Suppose $K$ is open and the rod is moved with a speed of $12cm$ $s^{-1}$ in the direction shown. Give the polarity and magnitude of the induced emf.
(b) Is there an excess charge build up at the ends of the rods when $K$ is open? What is $K$ is closed?
(c) With $K$ open and the rod moving uniformly, there is no net force on the electrons in the rod $PQ$ even through they do experience magnetic force due to the motion of the rod. Explain.
(d) What is the retarding force on the rod when $K$ is closed?
(e) How much power is required (by an external agent) to keep the rod moving at the same speed ( $=12cm$ $s^{-1}$ ) when $K$ is closed? How much power is required when $K$ is open?
(f) How much power is dissipated as heat in the closed-circuit? What is the source of this power?
(g) What is the induced emf in the moving rod if the magnetic field is parallel to the rails instead of being perpendicular?

Here,

$B=0.50T;l=15cm =15*10^{-2}m$ ;

$R=9.0mQ=9.0*10^{3}fl$
(a) Now,

$e=BVl$
Here,

$V=12cm$

$s^{-1}=12*10^{-2}ms^{-1}$

$\therefore e=0.50*12*10^{-2}*15*10^{2}=9*10^{-3}V$
If

$q$ is change on as electron, then the elctrons in the rod will experience magnetic Lorentz force

$-q[\underset{v}{\rightarrow}+\underset{B}{\rightarrow}]P.Q$ . Hence, the end

$P$ of the rod will become positive and the end

$Q$ will become negative.
(b) When the switch

$K$ is open, the elctron collect at the end

$Q$ . Therefore, excess change is build up at the end

$Q$ . However, when the switch

$K$ is closed, the accumulated charge at the end

$Q$ flows through the circuit.
(c) The magnetic Lorentz force on electron is cancelled by the electronic force acting on it due to the electronic filed set up accross the two ends due to accumulation of positive and negative charges at the end

$P$ and

$Q$ respectively.
(d) Retarding force,

$F=BIl=0.50*\frac{e}{R}*15*10^{-2}$

$=0.50*\frac{9*10^{-3}}{9*10^{-3}}*15*10^{-2}=7.5*10^{-2}N$ .
(e) When the switch

$K$ is closed, power required by theb external agent against the retarding force,

$P=Fv=7.5*10^{-2}*12*10^{-2}=9*10^{-3}W$ .
(f) Power dissipated as heat,

$\frac{e^{2}}{R}=\frac{[9*10^{-3}]^{2}}{9*10^{-3}}=9*10^{-3}W$ /
The sorce of this power is the power of the external agent against the retarding force,

$P=Fv=7.5*10^{-2}*12*10^{-2}=9*10^{-3}W$ .
(g) The motion of rod does not cut field lines, hence no induced e.m.f. is produced.

A long straight solenoid of cross-sectional diameter $d = 5\ cm$ and with $n = 20$ turns per one cm of its length has a round turn of copper wire of cross-sectional area $S = 1.0\ mm^{2}$ tightly put on its winding. Find the current flowing in the turn if the current in the solenoid winding is increased with a constant velocity $I = 100\ A/s$ . The inductance of the turn is to be neglected.

The e.m.f. induced in the turn is

$\mu_{0} n\overset {\cdot}{I} \pi \dfrac {d^{2}}{4}$
The resistance is

$\dfrac {\pi d}{S} \rho$ .
So, the current is

$\dfrac {\mu_{0} n\overset {\cdot}{I} S\ d}{4\rho} = 2\ mA$ , where

$\rho$ is the resistivity of copper.

A current $I$ flows along a thin wire shaped as a regular polygon with $n$ sides which can be inscribed into a circle of radius $R$ . Find the magnetic induction at the centre of the polygon. Analyse the obtained expression at $n\rightarrow \infty$ .

$\angle AOB = \dfrac {2\pi}{n}, OC$ or perpendicular distance of any segment from centre equals

$R\cos \dfrac {\pi}{n}$ . Now magnetic induction at

$O$ , due to the right current carrying element

$AB$

$= \dfrac {\mu_{0}}{4\pi} \dfrac {i}{R\cos \dfrac {\pi}{n}} 2\sin \dfrac {\pi}{n}$
(From Biot-Savart's law, the magnetic field at

$O$ due to any section such as

$AB$ is perpendicular to the plane of the figure and has the magnitude.)

$B = \int \dfrac {\mu_{0}}{4\pi} i \dfrac {dx}{r^{2}} \cos \theta = \int_{-\dfrac {\pi}{n}}^{\dfrac {\pi}{n}} \dfrac {\mu_{0}i}{4\pi} \dfrac {R\cos \dfrac {\pi}{n}\sec^{2} \theta d\theta}{R^{2} \cos \dfrac {2\pi}{n}\sec^{2} \theta} \cos \theta = \dfrac {\mu_{0} i}{4\pi} \dfrac {1}{R\cos \dfrac {\pi}{n}} 2\sin \dfrac {\pi}{n}$
As there are

$n$ number of sides and magnetic induction vectors, due to each side at

$O$ , are equal in magnitude and direction. So,

$B_{0} = \dfrac {\mu_{0}}{4\pi} \dfrac {nr}{R\cos \dfrac {\pi}{n}} 2\sin \dfrac {\pi}{n}\cdot n$

$= \dfrac {\mu_{0}}{2\pi} \dfrac {ni}{R}\tan \dfrac {\pi}{n}$ and for

$n\rightarrow \infty$

$B_{0} = \dfrac {\mu_{0}}{2} \dfrac {i}{R} \displaystyle Lt_{n\rightarrow \infty} \left (\dfrac {\tan \dfrac {\pi}{n}}{\pi/ n}\right ) = \dfrac {\mu_{0}}{2} \dfrac {i}{R}$ .
1787926_1854742_ans_c5da7750395b42c0920b6c6188041a99.jpg

1787926_1854742_ans_c5da7750395b42c0920b6c6188041a99.jpg

In the middle of a long solenoid there is a coaxial ring of square cross-section, made of conducting material with resistivity $\rho$ . The thickness of the ring is equal to $h$ , its inside and outside radii are equal to $a$ and $b$ respectively. Find the current induced in the ring if the magnetic induction produced by the solenoid varies with time as $B = \beta t$ where $\beta$ is a constant. The inductance of the ring is to be neglected.

Take an elementary ring of radius

$r$ and width

$dr$ .
The e.m.f. induced in this elementary ring is

$\pi r^{2} \beta$ .
Now the conductance of this ring is.

$d\left (\dfrac {1}{R}\right ) = \dfrac {hdr}{\rho 2\pi s}$ so

$d\ I = \dfrac {h\ r\ dr}{2\rho} \beta$
Integrating we get the total current,

$I = \int_{a}^{b} \dfrac {hr\ dr}{2\rho} \beta = \dfrac {h\beta (b^{2} - a^{2})}{4\rho}$ .
1798259_1855091_ans_af931c293ba0403fabbc899f43abf00e.jpg

Faraday's Law of Induction (7)
To monitor the breathing of a hospital patient, a thin belt is girded around the patient's chest. The belt is a $200$ -turn coil. When the patient inhales, the area encircled by the coil increases by $39.0 \,cm^2$ . The magnitude of the Earth's magnetic field is $50.0 \mu T$ and makes an angle of $28.0^{o}$ with the plane of the coil. Assuming a patient takes $1.80 \,s$ to inhale, find the average inducede mf in the coil during this time interval.

1979416_1872260_ans_e31b67fbdf9f4ec098d78493d34cae8a.jpg

Conceptual Questions(5)
A circular loop of wire is located in a uniform and constant magnetic field. Describe how an emf can be induced in the loop in this situation.

1979393_1871892_ans_f92ba52f6d714cb78b58a521256c57ea.jpeg

Faraday's Law of Induction (8)
A strong electromagnet produces a uniform magnetic field of $1.60 \,T$ over a cross-sectional area of $0.200 m^2$ . A coil having $200$ turns and a total resistance of $20.0 \Omega$ is placed around the electromagnet. The current in the electromagnet is then smoothly reduced until it reaches zero in $20.0 \,ms$ . What is the current induced in the coil?

1979418_1872261_ans_b29cb257acd94a8093c0799fb8eed2b4.jpg

A quadratic undeformabe superconducting loop of mass $m$ and side $a$ lies in a horizontal plane in a nonuniform magnetic field whose induction varies in space according to the law ${B}_{x}=-\alpha x$ , ${B}_{y}=0$ , ${B}_{z}=\alpha z+{B}_{0}$ (Figure). The inductance of the loop is $L$ . At the initial moment, the centre of the loop coincides with the origin $O$ , and its sides are parallel to the $x-$ and $y-$ axes. The current in the loop is zero, and it is released.
How will it move and where will it be in time $t$ after the beginning of motion?

The magnetic flux across the surface bounded by the superconducting loop is constant. Indeed,

$\Delta \Phi / \Delta t$ , but

$\varepsilon =IR=0$ (Since

$R=0$ )
and hence

$\Phi =const.$
The magnetic flux through the surface bounded by the loop is the sum of the external magnetic flux and the flux of the magnetic field produced by the current

$I$ passing through the loop. Therefore, the magnetic flux across the loop at any instant is

$\Phi ={ a }^{ 2 }{ B }_{ 0 }+{ a }^{ 2 }\alpha z+LI$
since

$\Phi={B}_{0}{a}^{2}$ at the initial moment (

$z=0$ and

$I=0$ ) the current

$I$ at any other instant will be determined by the relation

$LI=-\alpha z{ a }^{ 2 },I=-\cfrac { \alpha z{ a }^{ 2 } }{ L }$
The resultant force exerted by the magnetic field on the current loop is the sum of the forces acting on the sides of the loop which are parallel to the y-axis is

$F=2a\left| \alpha x \right| I={ a }^{ 2 }\alpha I$
and is directed along the z-axis
Therefore, the equation of motion for the loop has the form

$\overset { \bullet \bullet }{ mz } =-mg+{ a }^{ 2 }\alpha I=-\cfrac { mg-{ a }^{ 4 }{ \alpha }^{ 2 }z }{ L }$
This equation is similar to the equation of vibrations of a body of mass

$m$ suspended on a spring of rigidity

$k={ a }^{ 4 }{ \alpha }^{ 2 }/L$

$\overset { \bullet \bullet }{ mz } =-mg-kz$
This analogy shows that the loop will perform harmonic oscillations along the z-axis near the equilibrium position determined by the condition

$\cfrac { { a }^{ 4 }{ \alpha }^{ 2 }z }{ L } { z }_{ 0 }=-mg;{ z }_{ 0 }=-\cfrac { mgL }{ { a }^{ 4 }{ \alpha }^{ 2 } }$
The frequency of these oscillations will be

$\omega =\cfrac { { a }^{ 2 }\alpha }{ \sqrt { Lm } }$
The coordinate of the loop in a certain time

$t$ after the beginning of motion will be

$\quad z=-\cfrac { mgL }{ { a }^{ 4 }{ \alpha }^{ 2 } } \left[ -1+\cos { \left( \cfrac { { a }^{ 2 }\alpha }{ \sqrt { Lm } } t \right) } \right]$

Faraday's Law of Induction (1)
A flat loop of wire consisting of a single turn of cross sectional area $8.00 \,cm^2$ is perpendicular to a magnetic field that increases uniformly in magnitude from $0.500 \,T$ to $2.50 \,T$ in $1.00 \,s$ . What is the resulting induced current if the loop has a resistance of $2.00 \,\Omega$ ?

1979411_1871899_ans_ced6bbf31bdb444dad6edc66056e5e55.jpg

A direct current flowing through the winding of a long cylindrical solenoid of radius $R$ produces in it a uniform magnetic field of induction $B$ . An electron flies into the solenoid along the radius between its turns (at right angles to the solenoid axis) at a velocity $v$ (figure). After a certain time, the electron deflected by the magnetic field leaves the solenoid.
Determine the time $t$ during which the electron moves in the solenoid.

The magnetic of the solenoid is directed along its axis. Therefore, the Lorentz force acting on the electron at any instant of time will lie in the plane perpendicular to the solenoid axis. Since the electron velocity at the initial moment is directed at right angles to the solenoid axis, the electron trajectory will lie in the plane perpendicular to the solenoid axis. The Lorentz force can be found from the formula

$F=evB$

The trajectory of the electron in the solenoid is an arc of the circle whose radius can be determined from the relation

$evB=m{v}^{2}/r$ , whence

$r=\cfrac { mv }{ eB }$

The trajectory of the electron is shown in figure (top view), wher e

${O}_{1}$ is the centre of the arc

$AC$ described by the electron,

$v'$ is the velocity at which the electron leaves the solenoid. the segments

$OA$ and

$OC$ are tangents to the electron trajectory at points

$A$ and

$C$ . The angle between

$v$ and

$v'$ is obviously

$\phi =\angle A{ O }_{ 1 }C$ since

$\angle OA{ O }_{ 1 }=\angle OC{ O }_{ 1 }$

In order to find

$\phi$ , let us consider the right triangle

$OA{O}_{1}$ : side

$OA=R$ and side

$A{O}_{1}=r$ . Therefore,

$\tan { \left( \phi /2 \right) } =r/r=eBR/(mv)$ . Therefore

$\phi =2arc\tan { \left( \cfrac { eBR }{ mv } \right) }$

Obviously, the magnitude of the velocity remains unchanged over the entire trajectory since the Lorentz force is perpendicular to the velocity at any instant. Therefore, the transit time of electron in the solenoid can be determined from the relation

$t=\cfrac { r\phi }{ v } =\cfrac { m\phi }{ eB } =\cfrac { 2m }{ eB } arc\tan { \left( \cfrac { eBR }{ mv } \right) }$

Magnetron is a device consisting of a filament of radius $a$ and a coaxial cylindrical anode of radius $b$ which are located in a uniform magnetic field parallel to the filament. An accelerating potential difference $V$ is applied between the filament and the anode. Find the value of magnetic induction at which the electrons leaving the filament with zero velocity reach the anode.

When a current

$I$ flows along the axis, a magnetic field

$B_{\varphi} = \dfrac {\mu_{0}I}{2\pi \rho}$ is set up where

$\rho^{2} = x^{2} + y^{2}$ . In terms of components,

$B_{x} = -\dfrac {\mu_{0}Iy}{2\pi \rho^{2}}, B_{y} = \dfrac {\mu_{0}Ix}{2\pi \rho^{2}}$ and

$B_{z} = 0$
Suppose a p.d.

$V$ is set up between the inner cathode and the outer anode. This means a potential function of the form

$\varphi = V \dfrac {ln \rho/b}{ln a/b}, a > \rho > b$ ,
as one can check by solving Laplace equation.
The electric field corresponding to this is,

$E_{x} = -\dfrac {Vx}{\rho^{2} ln a/b}, E_{y} = \dfrac {Vy}{\rho^{2} ln a/b}, E_{z} = 0$ .
The equations of motion are,

$\dfrac {d}{dt} mv_{x} = + \dfrac {|e|Vz}{\rho^{2} ln a/b} + \dfrac {|e|\mu_{0}I}{2\pi \rho^{2}} x\overset {\cdot}{z}$

$\dfrac {d}{dt}mv_{y} = + \dfrac {|e|Vy}{\rho^{2} ln a/b} + \dfrac {|e|\mu_{0}I}{2\pi \rho^{2}} y\overset {\cdot}{z}$
and

$\dfrac {d}{dt} mv_{z} = -|e| \dfrac {\mu_{0}I}{2\pi \rho^{2}} (x\overset {\cdot}{x} + y \overset {\cdot}{y}) = -|e| \dfrac {\mu_{0}I}{2\pi} \dfrac {d}{dt} ln \rho$

$(-|e|)$ is the charge on the electron.
Integrating the last equation,

$mv_{z} = -|e| \dfrac {\mu_{0}I}{2\pi} ln \rho / a = m\overset {\cdot}{z}$ .
since

$v_{z} = 0$ where

$\rho = a$ . We now substitute this

$\overset {\cdot}{z}$ in the other equations to get

$\dfrac {d}{dt} \left (\dfrac {1}{2} mv_{x}^{2} + \dfrac {1}{2} mv_{y}^{2}\right )$

$= \left [\dfrac {|e|V}{ln a/b} - \dfrac {|e|^{2}}{m} \left (\dfrac {\mu_{0} I}{2\pi}\right )^{2} ln \rho/ b\right ] \cdot \dfrac {x\overset {\cdot}{x} + y \overset {\cdot}{y}}{\rho^{2}}$

$= \left [\dfrac {|e|V}{ln \dfrac {a}{b}} - \dfrac {|e|^{2}}{m} \left (\dfrac {\mu_{0}I}{2\pi}\right )^{2} ln \dfrac {\rho}{b}\right ] \cdot \dfrac {1}{2\rho^{2}} \dfrac {d}{dt} \rho^{2}$

$= \left [\dfrac {|e|V}{ln \dfrac {a}{b}} - \dfrac {|e|^{2}}{m} \left (\dfrac {\mu_{0}I}{2\pi}\right )^{2} ln \dfrac {\rho}{b}\right ] \dfrac {d}{dt} ln \dfrac {\rho}{b}$
Integrating and using

$v^{2} = 0$ , at

$\rho = b$ , we get,

$\dfrac {1}{2} mv^{2} = \left [\dfrac {|e|V}{ln \dfrac {a}{b}} ln \dfrac {\rho}{b} - \dfrac {1}{2m} |e|^{2} \left (\dfrac {\mu_{0}I}{2\pi}\right )^{2} \left (ln \dfrac {\rho}{b}\right )\right ]$
The RHS must be positive, for all

$a > \rho > b$ . The condition for this is,

$V \geq \dfrac {1}{2} \dfrac {|e|}{m}\left (\dfrac {\mu_{0}I}{2\pi}\right )^{2} ln \dfrac {a}{b}$
This differs from in

$(a \leftrightarrow b)$ and the magnetic field is along the z-direction. Thus

$B_{x} = B_{y} = 0, B_{z} = B$
Assuming as usual the charge of the electron to be

$-|e|$ , we write the equation of motion

$\dfrac {d}{dt}mv_{x} = \dfrac {|e|V_{x}}{\rho^{2} ln \dfrac {b}{a}} - |e| B\overset {\cdot}{y}, \dfrac {d}{dt} mv_{y} = \dfrac {|e|V_{y}}{\rho^{2} ln \dfrac {b}{a}} + |e| B\overset {\cdot}{x}$
and

$\dfrac {d}{dt} mv_{z} = 0\Rightarrow z = 0$
The motion is confined to the plane

$z = 0$ . Eliminating

$B$ from the first two equations,

$\dfrac {d}{dt}\left (\dfrac {1}{2} mv^{2}\right ) = \dfrac {|e|V}{ln b/a} \dfrac {x\overset {\cdot}{x} + y\overset {\cdot}{y}}{\rho^{2}}$
or,

$\dfrac {1}{2} mv^{2} = |e| V \dfrac {ln \rho/ a}{ln b/a}$
so, as expected, since magnetic forces do not work,

$v = \sqrt {\dfrac {2|e|V}{m}}$ , at

$\rho = b$ .
On the other hand, eliminating

$V$ , we also get,

$\dfrac {d}{dt} m(xv_{y} - yv_{x}) = |e| B(x\overset {\cdot}{x} + y\overset {\cdot}{y})$
i.e.

$(xv_{y} - yv_{x}) = \dfrac {|e|B}{2m} \rho^{2} + constant$
The constant is easily evaluated, since

$v$ is zero at

$\rho = a$ . Thus,

$(xv_{y} - yv_{x}) = \dfrac {|e|B}{2m} (\rho^{2} - a^{2}) > 0$
At

$\rho = b, (xy_{y} - yv_{x}) \leq vb$
Thus,

$vb \geq \dfrac {|e|B}{2m} (b^{2} - a^{2})$
or,

$B \leq \dfrac {2mb}{b^{2} - a^{2}} \sqrt {\dfrac {2|e|V}{m}} \times \dfrac {1}{|e|}$
or,

$B \leq \dfrac {2b}{b^{2} - a^{2}} \sqrt {\dfrac {2mB}{|e|}}$ .

Faraday's Law of Induction (18)
When a wire carries an AC current with a known frequency, you can use a Rogowski coil to determine the amplitude $I_{max}$ of the current without disconnecting the wire to shunt the current through a meter. The Rogowski coil, shown in Figure, simply clips around the wire. It consists of a toroidal conductor wrapped around a circular return cord. Let $n$ represent the number of turns in the toroid per unit distance along it. Let A represent the cross-sectional area of the toroid. Let $I(t) = I_{max} \sin \omega t$ represent the current to be measured.
(a) Show that the amplitude of the emf induced in the Rogowski coil is $\varepsilon_{max}= \mu_0 n A\omega I_{max}$ .
(b) Explain why the wire carrying the unknown current need not be at the center of the Rogowski coil and why the coil will not respond to nearby currents that it does not enclose.

1982064_1872313_ans_423c14e080ba47e29f4bc4bfd93993e8.jpg

Lenz's Law (31)
Review. Figure shows a bar of mass $m =0.200 \,kg$ that can slide without friction on a pair of rails separated by a distance $l= 1.20 \,m$ and located on an inclined plane that makes an angle $\theta = 25.0^{o}$ with respect to the ground. The resistance of the resistor is $R = 1.00 \Omega$ and a uniform magnetic field of magnitude $B = 0.500 \,T$ is directed downward, perpendicular to the ground, over the entire region through which the bar moves. With what constant speed $v$ does the bar slide along the rails?

1982069_1872402_ans_aec5999077bd43799b0c3c3dedf41422.jpg

Lenz's Law (21)
A helicopter (Figure) has blades of length $3.00 \,m$ , extending out from a central hub and rotating at $2.00 \,rev/s$ . If the vertical component of the Earths magnetic field is $50.0 \mu T$ , what is the emf induced between the blade tip and the center hub?

1982047_1872361_ans_bc742d38ec8747c1b7320b8a24940f8f.jpg

Electromagnetic Induction - Class 12 Medical Physics - Extra Questions

Define coefficient of mutual induction. If in the primary coil of a transformer, the current decreases from 0.8A0.8A to 0.2A0.2A in 44 milliseconds, calculate the induced e.m.f in the secondary coil. Mutual inductance is 1.76H1.76H.

A current of 1.0A1.0A is established in a tightly wound solenoid of radius 2cm2cm having 10001000 turns/metre. Find the magnetic energy stored in each metre of the solenoid.

An average emf of 20V20V is induced in an inductor when the current in it is changed from 2.5A2.5A in one direction to the same value in the opposite direction in 0.1s0.1s. Find the self-inductance of the inductor.

How many times, the current produced by ACAC Generator reverses its direction in one complete revolution of its coil?

How many times ACAC supply in India reversed its direction in one second.

A plot of magnetic flux (φ)(\varphi ) versus current (I)(I) is shown in the figure for two inductors AA and BB. Which of the two has larger value of self inductance?

Solve a problem differing from the foregoing one by a magnetic field with induction B=0.8 TB = 0.8\ T replacing the electric field.

An inductor with an inductance of 3.00H3.00H and a resistance of 7.00Ω7.00\Omega is connected to the terminals of a battery with an emf of 12.0V12.0V and negligible internal resistance. Find the initial rate of increase of current in the circuit.

Which end of the inductor, aa or bb, is at a higher potential?

The potential difference across a 150mH150mH inductor as a function of time is shown in figure. Assume that the initial value of the current in the inductor is zero. What is the current when t=2.0mst=2.0ms?

Interpret K' - K.

A frame ABCDABCD is rotating with an angular velocity ω\omega about an axis passing through point OO perpendicular to the plane of paper as shown in the figure. A uniform magnetic field →B\vec { B } is applied into the plane of the paper in the region as in the figure. Match the following.

A conducting rod ABAB moves parallel to the x-axis in a uniform magnetic field pointing in the positive z direction. The end AA of the rod gets positively charged. explain.

The magnetic flux linked with the armature coil changed in a generator =Φ(t)=xNBAcos(ωt)\Phi (t) = xNBA\cos (\omega t) then x=

The voltage applied to a purely inductive coil of self inductance 15.9mH15.9mH is given by the equation V=100sin 314tV=100sin 314t + 75sin942t75sin942t + 450sin1570t.450sin1570t. Find the equation of current wave.

A solenoid of length 1 m1 m and 0.05 m0.05 m diameter has 500500 turns. If a current of 2 A2 A passes through the coil, calculate the co-efficient of self induction of the coil.

On what factors does the induced electromotive force depend?

Give two definitions of mutual inductance and write its unit.

What is an AC generator? Obtain an expression for the sinusoidal emf induced in the coil of ac generator, rotating with a uniform angular speed in a uniform magnetic field.

When the number of turn in a coil is doubled without, any change in the length of the coil , its self inductance becomes

A conducting circular loop having a radius of 5.0 5.0 cm, is placed perpendicular to a magnetic field of 0.50 T0.50 T. It is removed from the field in 0.50 s0.50 s. Find the average emf produced in the loop during this time.

A circular coil of 200200 turns and of radius 20 cm20 cm carries a current of 5A5A. Calculate the magnetic induction at a point along its axis, at a distance three times the radius of the coil from its centre.

A stiff semi-circular wire of radius RR is rotated in a uniform magnetic field BB about an axis passing through its ends. If the frequency of rotation of wire is ff, calculate the amplitude of alternating emf induced in the wire.

A coil of insulated copper wire is connected to a galvanometer. What will happen if a bar magnet is pushed into the coil?

A pair of adjacent coils has a mutual inductance of 1.5 H1.5 H. If the current in one coil changes from 00 to 10 A10 A in 0.2 s0.2 s, what is the change of flux linkage with the other coil?

In a fluorescent lamp choke (a small transformer) 100100V of reverse voltage is produced when the choke current changes uniformly from 0.250.25A to 00 in a duration of 0.0250.025ms. The self-inductance of the choke (in mH) is estimated to be _______.

What is a solenoid? Draw a diagram with a solenoid connected in a circuit. How can you increase the strength of a solenoid?

The current in a long solenoid of radius RR and having nn turns per unit length is given by i={i}_{0}\sin \omega ti={i}_{0}\sin \omega t. A coil having NN turns is wound around it near the centre. Find (a) the induced emf in the coil and (b) the mutual inductance between the solenoid and the coil.

Name one device which converts electrical energy into mechanical energy.

Name the phenomenon which is made use of in an electric generator.

If you hold a coil of wire next to a magnet, no current will flow in the coil. What else in needed to induce a current?

How can the magnitude of the induced current be increased?

Suggest two ways in an a.c. generator to produce a higher e.m.f.

A magnetic substance of volume 30 cm^{3}30 cm^{3} is placed in a magnetising field of 55 oersted. It produces a magnetic moment of 6 A/m^{2}6 A/m^{2} . Calculate the magnetic induction.

Two identical loops, one of copper and another of aluminium are rotated with the same speed in the same magnetic field. In which case, the (a) The induced emf and(b) induced current will be more and why?

A horizontal wire 20 m20 m long extending from east to west is falling with a velocity of 10 m/s10 m/s normal to the Earths magnetic field of 0.5*10^{-4} T0.5*10^{-4} T. What is the value of induced emf in the wire?

Maximum voltage produced in an AC generator completing 6060 cycles in 3030 seconds is 250\,V.250\,V.What is the maximum emf produced when the armature completes 18001800 rotation ?

A pair of adjacent coil has a mutual inductance of 1.5 \,H1.5 \,H. If the current in one coil changes from 00 to 20 \,A20 \,A in 0.50.5 sec, what is the change of flux linkage with the other coil?

Derive the expression for the self inductance of a long solenoid of cross sectional area AA and length ll, having nn turns per unit length.

Define self-inductance of a coil. Show that magnetic energy required to build up the current I in coil of coil of self inductance L is given by \displaystyle \frac{1}{2}LI^{2}\displaystyle \frac{1}{2}LI^{2}

Two concentric circular coils of radii r and R are placed coaxially with centres coiciding. If R >> r then calculate the mutual inductance between the coils.

An aircraft having a wingspan of 20.48 m20.48 m flies due north at a speed of 40{ ms }^{ -1 }40{ ms }^{ -1 }. If the vertical component of earth's magnetic field at the place is 2\times { 10 }^{ -5 }T2\times { 10 }^{ -5 }T, calculate the emf induced between the ends of the wings.

The unit of self-inductance in SI system is ________ .

Name and state the principle of a simple a.c. generator. What is its use?

State one factor that determines the magnitude of induced e.m.f.

What is the source of energy associated with the current obtained in part when a magnet is moved towards a coil having a galvanometer at its ends?

Match \: the \: statements \: in \: Column \: A, \: with \: those \: in \: Column \: B.Match \: the \: statements \: in \: Column \: A, \: with \: those \: in \: Column \: B.

Define the term 'self-inductance' of a coil. Write its S.I. unit.

Fill in the blanks.The type of electric current that changes its direction twice during one cycle of the dynamo is called _____________ .

State Faraday's laws of electromagnetic induction.

What is motional emf? State any two factors on which it depends.

Two coils A and B have mutual inductance 2\times { 10 }^{ -2 }2\times { 10 }^{ -2 } henry. If the current in the primary is i=5sin(10\pi t)i=5sin(10\pi t), then the maximum value of e.m.f. induced in coil B is :

What is a generator state the principle on which generators work

Describe the activity that shows that a current-carrying conductor experiences a force perpendicular to its length and the external magnetic field. How does Flemings' left-hand rule help us to find the direction of the force acting on the current-carrying conductor?

The current through an indicator of 1H1H is given by i=3t\sin { t } i=3t\sin { t } . Find the voltage across the inductor.

the self inductance of the primary coil.

At the instant when the current in an inductor is increasing at a rate of 0.0640A/s0.0640A/s, the magnitude of the self-induced emf is 0.0160V0.0160V. If the inductor is a solenoid with 400400 turns, what is the average magnetic flux through each turn when the current is 0.720A0.720A?

When a current of 4A4A between two coils changes to 12A12A in 0.5s0.5s in primary and induces an emf of 50mV50mV in the secondary. Calculate the mutual inductance between the two coils.

At the instant when the current in an inductor is increasing at a rate of 0.0640A/s0.0640A/s, the magnitude of the self-induced emf is 0.0160V0.0160V. What is the inductance of the inductor?

The current (in Ampere) in an inductor is given by I=5+16tI=5+16t, where tt is in seconds. The self-induced emf in it is 10mV10mV. Find the self-inductance.

Two bar magnets are placed side by side by side that the north pole of one magnet is next to the south pole of the other magnet. If these magnets are then pushed toward a coil of wire, would you expect an emf to be induced in the coil. Explain your answer.

emf induced in the arm PQ

Current in a circuit falls from 5.0 A5.0 A to 0.0 A0.0 A in 0.1\ s0.1\ s. If an average emf of 200\ V200\ V induced, find an estimate of the self-inductance of the circuit.

The property of a conductor which enables to induce an EMF due to change of current in the same coil is ____________.

Derive Faradays law of induction from law of conservation of energy.

An emf of 2V is induced in a coil when current in it is changed from 0A to 10A in 0.40 sec. Find the coefficient of self-inductance of the coil.

A coil of self inductance 2\cdot 5 H2\cdot 5 H and resistance 20\Omega20\Omega is connected to a battery of emf 120V having internal resistance of 5 \Omega5 \Omega. Find the current in the circuit in steady state.

Write Faraday's law of electromagnetic induction.

Write the Faraday's law of electromagnetic induction.

Current in a circuit fall from 5.05.0A to zero in 0.10.1S. If an average emf of 100100 Volt is induced then calculate self-inductance of a inductor in the circuit.

State Faraday's laws of electromagnetic induction and Lenz's law.

Explain self-induction of a coil. Arrive at an expression for the induced emf in a coil and the rate of change of current in it.

Current in a circuit falls from 5.0 A5.0 A to 0.0 A0.0 A in 0.1 sec0.1 sec. If an average emf of 200 V200 V is induced, give an estimate of the self-inductance of the circuit.

What are the methods of producing induced emf?

What is the self inductance of a straight conductor?

Explain seld induction and mutual induction.

Draw a labelled diagram of an ac generator. Obtain the expression for the emf induced in the rotating coil of N turns each of cross-sectional area A, in the presence of a magnetic field \overrightarrow{B}\overrightarrow{B}.

Current in a circuit falls from 5.0 A to 0.0 A in 0.1 s. If an average emf of 200V induced, calculate the self-inductance of the circuit.

What is the effect on self inductance of a solenoid, if a core of soft iron is placed in it?

Define coefficient of mutual induction. If in the primary coil of a transformer, the current decreases from $0.8A$ to $0.2A$ in $4$ milliseconds, calculate the induced e.m.f in the secondary coil. Mutual inductance is $1.76H$ .

A current of $1.0A$ is established in a tightly wound solenoid of radius $2cm$ having $1000$ turns/metre. Find the magnetic energy stored in each metre of the solenoid.

An average emf of $20V$ is induced in an inductor when the current in it is changed from $2.5A$ in one direction to the same value in the opposite direction in $0.1s$ . Find the self-inductance of the inductor.

How many times, the current produced by $AC$ Generator reverses its direction in one complete revolution of its coil?

How many times $AC$ supply in India reversed its direction in one second.

A plot of magnetic flux $(\varphi )$ versus current $(I)$ is shown in the figure for two inductors $A$ and $B$ . Which of the two has larger value of self inductance?

Solve a problem differing from the foregoing one by a magnetic field with induction $B = 0.8\ T$ replacing the electric field.

An inductor with an inductance of $3.00H$ and a resistance of $7.00\Omega$ is connected to the terminals of a battery with an emf of $12.0V$ and negligible internal resistance. Find the initial rate of increase of current in the circuit.

Which end of the inductor, $a$ or $b$ , is at a higher potential?

What is the current when $t=2.0ms$ ?

A frame $ABCD$ is rotating with an angular velocity $\omega$ about an axis passing through point $O$ perpendicular to the plane of paper as shown in the figure. A uniform magnetic field $\vec { B }$ is applied into the plane of the paper in the region as in the figure. Match the following.

A conducting rod $AB$ moves parallel to the x-axis in a uniform magnetic field pointing in the positive z direction. The end $A$ of the rod gets positively charged. explain.

The magnetic flux linked with the armature coil changed in a generator = $\Phi (t) = xNBA\cos (\omega t)$ then x=

The voltage applied to a purely inductive coil of self inductance $15.9mH$ is given by the equation $V=100sin 314t$ + $75sin942t$ + $450sin1570t.$
Find the equation of current wave.

A solenoid of length $1 m$ and $0.05 m$ diameter has $500$ turns. If a current of $2 A$ passes through the coil, calculate the co-efficient of self induction of the coil.

A conducting circular loop having a radius of $5.0$ cm, is placed perpendicular to a magnetic field of $0.50 T$ . It is removed from the field in $0.50 s$ . Find the average emf produced in the loop during this time.

A circular coil of $200$ turns and of radius $20 cm$ carries a current of $5A$ . Calculate the magnetic induction at a point along its axis, at a distance three times the radius of the coil from its centre.

A stiff semi-circular wire of radius $R$ is rotated in a uniform magnetic field $B$ about an axis passing through its ends. If the frequency of rotation of wire is $f$ , calculate the amplitude of alternating emf induced in the wire.

A pair of adjacent coils has a mutual inductance of $1.5 H$ . If the current in one coil changes from $0$ to $10 A$ in $0.2 s$ , what is the change of flux linkage with the other coil?

In a fluorescent lamp choke (a small transformer) $100$ V of reverse voltage is produced when the choke current changes uniformly from $0.25$ A to $0$ in a duration of $0.025$ ms. The self-inductance of the choke (in mH) is estimated to be _______.

The current in a long solenoid of radius $R$ and having $n$ turns per unit length is given by $i={i}_{0}\sin \omega t$ . A coil having $N$ turns is wound around it near the centre. Find (a) the induced emf in the coil and (b) the mutual inductance between the solenoid and the coil.

A magnetic substance of volume $30 cm^{3}$ is placed in a magnetising field of $5$ oersted. It produces a magnetic moment of $6 A/m^{2}$ . Calculate the magnetic induction.

Two identical loops, one of copper and another of aluminium are rotated with the same speed in the same magnetic field. In which case, the
(a) The induced emf and
(b) induced current will be more and why?

A horizontal wire $20 m$ long extending from east to west is falling with a velocity of $10 m/s$ normal to the Earths magnetic field of $0.5*10^{-4} T$ . What is the value of induced emf in the wire?

Maximum voltage produced in an AC generator completing $60$ cycles in $30$ seconds is $250\,V.$
What is the maximum emf produced when the armature completes $1800$ rotation ?

A pair of adjacent coil has a mutual inductance of $1.5 \,H$ . If the current in one coil changes from $0$ to $20 \,A$ in $0.5$ sec, what is the change of flux linkage with the other coil?

Derive the expression for the self inductance of a long solenoid of cross sectional area $A$ and length $l$ , having $n$ turns per unit length.

Define self-inductance of a coil. Show that magnetic energy required to build up the current I in coil of coil of self inductance L is given by $\displaystyle \frac{1}{2}LI^{2}$

An aircraft having a wingspan of $20.48 m$ flies due north at a speed of $40{ ms }^{ -1 }$ . If the vertical component of earth's magnetic field at the place is $2\times { 10 }^{ -5 }T$ , calculate the emf induced between the ends of the wings.

$Match \: the \: statements \: in \: Column \: A, \: with \: those \: in \: Column \: B.$

Fill in the blanks.
The type of electric current that changes its direction twice during one cycle of the dynamo is called _____________ .

Two coils A and B have mutual inductance $2\times { 10 }^{ -2 }$ henry. If the current in the primary is $i=5sin(10\pi t)$ , then the maximum value of e.m.f. induced in coil B is :

The current through an indicator of $1H$ is given by $i=3t\sin { t }$ . Find the voltage across the inductor.

At the instant when the current in an inductor is increasing at a rate of $0.0640A/s$ , the magnitude of the self-induced emf is $0.0160V$ . If the inductor is a solenoid with $400$ turns, what is the average magnetic flux through each turn when the current is $0.720A$ ?

When a current of $4A$ between two coils changes to $12A$ in $0.5s$ in primary and induces an emf of $50mV$ in the secondary. Calculate the mutual inductance between the two coils.

At the instant when the current in an inductor is increasing at a rate of $0.0640A/s$ , the magnitude of the self-induced emf is $0.0160V$ . What is the inductance of the inductor?

The current (in Ampere) in an inductor is given by $I=5+16t$ , where $t$ is in seconds. The self-induced emf in it is $10mV$ . Find the self-inductance.

Current in a circuit falls from $5.0 A$ to $0.0 A$ in $0.1\ s$ . If an average emf of $200\ V$ induced, find an estimate of the self-inductance of the circuit.

A coil of self inductance $2\cdot 5 H$ and resistance $20\Omega$ is connected to a battery of emf 120V having internal resistance of $5 \Omega$ . Find the current in the circuit in steady state.

Current in a circuit fall from $5.0$ A to zero in $0.1$ S. If an average emf of $100$ Volt is induced then calculate self-inductance of a inductor in the circuit.

Current in a circuit falls from $5.0 A$ to $0.0 A$ in $0.1 sec$ . If an average emf of $200 V$ is induced, give an estimate of the self-inductance of the circuit.

Draw a labelled diagram of an ac generator. Obtain the expression for the emf induced in the rotating coil of N turns each of cross-sectional area A, in the presence of a magnetic field $\overrightarrow{B}$ .

A rectangular coil has $60$ turns and its length and width is $20\ cm$ . and $10\ cm$ respectively. The coil rotates at a speed of $1800$ rotation per minute in a uniform magnetic field of $0.5\ T$ about its one of the diameter. Calculate maximum induced emf will be

A charged particle oscillates about its equilibrium position with an frequency of $100\ MHz$ . What is the frequency of electromagnetic waves produced by the oscillator?

A circular coil of cross-sectional area $200cm^2$ and 20 turns is rotated about the vertical diameter with angular speed of $50 rad {s}^{-1}$ in a uniform magnetic field of magnetic $3.0 \times10^{-2}T$ . Calculate the maximum value of the current in the coil.

A circular coil of cross-sectional area $20 cm^2$ and $20$ turns is rotated about the diameter with angular speed of 50 rad $s^{-1}$ in a uniform magnetic field of magnetic $3.0 \times 10^{-2}T$ . Calculate the maximum value of the current in the coil.

A solenoid of $500$ turns, diameter 2 $0 cm$ and resistance $2$ $\Omega$ is rotated about its vertical diameter through $\pi$ radian in 1/4 s, when a horizontal field of $3 \times 10^{-5} T$ acts normal to its plane. Find the emf induced and current thereof.

What will be the magnitude and polarity of induced emf across the rod, if it is rotating about an axis at a distance $\cfrac{L}{4}$ from one of its ends?

A semicircle loop $PQ$ of radius $'R'$ is moved with velocity $'v'$ in transverse magnetic field as shown in figure. The value of induced emf. at the end of loop is :-

A $200km$ long telegraph wire has a capacitance of $0.014\mu F/km$ . If it carries an ac of $5kHz$ , what should be the inductance required to be connected in series, so that the impedance is minimum?
Take $\pi=\sqrt {10}$

A direct current $I$ flows along a lengthy straight wire. From the point $O$ (Fig.) the current spreads radially all over an infinite conducting plane perpendicular to the wire. Find the magnetic induction at all points of space.

A rectangular coil of area $A$ , having the number of turns $N$ is rotated at $f$ revolutions per second in a uniform magnetic field $B$ the field is perpendicular to the coil. Prove that the maximum emf induced in the coil is $2 \pi fNBA$ .

A coil having resistance $20\Omega$ and inductance $2H$ is connected to a battery of emf $4.0V$ . Find (a) the current at $0.20s$ after the connection is made and (b) the magnetic field energy in the coil at the instant

A conducting circular loop of area $1$ nm is placed compulsary with a long, straight wire at a distance of current which changes from $10$ A to zero in $0.1$ s. Find the average emf induced in the loop in $0.1$ s.

A coil draws a current of 1ampere and a power of 100 watt from an AC.Source of 100 volt and $\frac { 5\sqrt { 22 } }{ \pi }$ Hertz.Find the inductance and resistance of the coil

A copper disc of radius $0.1$ m is rotated about its center with $20$ revolution per second in a uniform magnetic field of $0.1$ T with its plane perpendicular to the field. The emf induced across the radius of the disc is?

On what principle $AC$ generator works?

A conducting disc of radius r rotates with a small constant angular velocity $\omega$ about its axis. A uniform magnetic field B exists parallel to the axis of rotations. Find the motional emf between the center and the periphery of the disc.

An LR circuit has $L=1.0H$ and $R=20\Omega$ . It is connected across an emf of $2.0V$ at $t=0$ What are the values of the self-induced emf in the circuit at the times indicated there in?

Consider a small cube of volume $1{mm}^{2}$ at the centre of a circular loop of radius $10cm$ carrying a current of $4A$ . Find the magnetic energy stored inside the cube.

The magnetic field at a point inside a $2.0mH$ inductor-coil becomes $0.80$ of its maximum value in $20\mu s$ when the inductor is joined to a battery. Find the resistance of the circuit.

A long wire carries a current of $4.00A$ . Find the energy stored in the magnetic field inside a volume of $1.00{mm}^{2}$ at a distance of $10.0cm$ from the wire.

Explain the various part of $AC$ generator.

A bar magnet $M$ is dropped so that it falls vertically through the coil $C$ . The graph obtained for voltage produced across the coil versus time is showing in figure(b)
Explain the shape of the graph.

Answer the following questions
Define mutual inductance.

In the given diagram, a coil $B$ is connected to low voltage bulb $L$ and placed parallel to another coil $A$ as shown. Explain bulb gets dimmer if the coil $B$ moves upwards.

Define $1$ henry.