MCQ Questions for Class 12 Medical Physics Electromagnetic Induction Quiz 11

For previous objective, which of the following graphs is correct

Explanation

If at any instant, current through the circuit is i then applying Kirchoffs voltage law,

$iR + e = E \Rightarrow e = E - iR.$ Therefore,graph between e and i will be a straight line having negative slope and having a positive intercept.

When a bar magnet is entered into a coil, thee induced emf in the coil does not depend upon:

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Speed of magnet
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number of turns in coil
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Magnetic moment of magnet
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Resistance of wire in coil

If a conducting rod moves with a constant velocity

$v$ in a magnetic field, emf is induced between both its ends if:

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$v$ and $B$ are parallel
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$v$ is perpendicular to $\overrightarrow{B}$
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$v$ and $B$ are in opposite direction
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All of the above

Explanation

The emf will only be induced if

$v$ is perpendicular to

$\overrightarrow{B}$
Velocity

$(\overrightarrow{v})$ is perpendicular to magnetic field

$(\overrightarrow{B})$

Induced

$emf$ of electromagnetic induction depends upon:

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Resistance on conductor
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The value of magnetic field
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The direction of conductor $w.r.t.$ the magnetic field
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Rate of change of flux linked

What would be the coefficient of self-inductance of a coil

$100 turns,$ if

$5 A$ current flows through it? The magnetic flux is of

$5 \times 10^{3}$ Maxwell.

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$0.5 \times 10^{-3} H$
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$2 \times 10^{-3} H$
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$Zero$
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$10^{-3}H$

Explanation

Given,

$N=100, I=5 amp$

$\phi_B=5 \times 10^{3} Mx$

$5 \times 10^3 \times 10^{-8} wb$

coefficient of self induction

$L=-\dfrac{N \phi_B}{I}$

$=-\dfrac{100\times 5\times 10^{-5}}{(0-5)}$

$=10^{-3} H$

The phenomenon of electro-magnetic induction is

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the process of charging a body.
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the process of generating magnetic field due to a current passing through a coil.
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producing induced current in a coil due to relative motion between a magnet and the coil.
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the process of rotating a coil of an electric motor.

Electromagnetic induction is

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Charging a substance.
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Process of developing a magnetic field around a coil by passing electricity through a coil
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Process of rotating the armature of a generator
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Process of making electricity by the relative motion of a magnet or a coiled conductor.

Answer the question:
An e.m.f. is induced across a wire when it moves through the magnetic field between the poles of a magnet.
Which electrical device operates because of this effect ?

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a battery
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a cathode-ray tube
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a generator
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a motor

A conducting rod of length l moves with velocity v in x-direction axis parallel to a long wire carrying a steady current I. The axis of the rod is maintained perpendicular to the wire with near end a distance r away as shown in the fig. Find the emf induced in the rod.

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$\dfrac{\mu _{0}I\upsilon }{\pi }\, \, ln \left ( \frac{r+1}{r} \right )$
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$\dfrac{2\mu _{0}I\upsilon }{\pi }\, \, ln \left ( \frac{r+1}{r} \right )$
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$\dfrac{\mu _{0}I\upsilon }{\pi }\, \, ln \left ( \dfrac{l}{r+1} \right )$
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$\dfrac{\mu _{0}I\upsilon }{ 2 \pi }\, \, ln \left ( \dfrac{r+1}{r} \right )$

In a uniform magnetic field of induction

$B$ , a wire in the form of semicircle of radius

$r$ rotates about the diameter of the circle with angular velocity

$\omega$ . If the total resistance of the circuit is

$R$ , the mean power generated per period of rotation is :

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$\dfrac{B\pi r^{2}\omega }{2R}$
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$\dfrac{(B\pi r^{2}\omega)^{2} }{8R}$
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$\dfrac{(B\pi r\omega)^{2} }{2R}$
0%
$\dfrac{(B\pi r\omega^{2})^{2} }{8R}$

Explanation

Flux

$\phi = BA\cos \theta=B \times ( \dfrac{1}{2}\pi r^2) \cos \theta$

$| \varepsilon| =| \dfrac{ d \phi}{dt} |= \dfrac{1}{2}\pi Br^2 \dfrac{d(\cos \theta)}{dt}$

$| \varepsilon|= \dfrac{1}{2}\pi Br^2( \sin \theta) \dfrac{d\theta}{dt}$

$| \varepsilon| = \dfrac{1}{2}\pi Br^2( \sin \omega t)\omega$
Power:

$P = \dfrac{| \varepsilon| ^2}{R}= \dfrac{(\dfrac{1}{2}\pi Br^2\omega \sin \omega t )^2}{R}$
Mean Power

$P_{mean} = \dfrac{ \int_{0}^{T} \dfrac{| \varepsilon| ^2}{R}dt} {\int_{0}^{T}dt}= \dfrac{\pi ^2 B^2r^4\omega ^2 \int_{0}^{T} \sin ^2\omega t\: dt }{4TR}$

$= \dfrac{\pi ^2 B^2r^4\omega ^2}{4TR}\frac{T}{2}= \dfrac{\pi ^2 B^2r^4\omega ^2}{8R}$

A square metal wire loop of side 10 cm and of resistance 2

$\Omega$ moves with constant velocity in the presence of a uniform magnetic field of induction 4 T, perpendicular and into the plane of the loop. The loop is connected to a network of resistance as shown in the Figure. If the loop should have a steady current of 2 mA, the speed of the loop must be (in cm s

$^{-1}$ ) :

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$4$
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$2$
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$3$
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$6$

Explanation

As this is wheatstore bridge.

$R_{eq}=\dfrac{6\times3}{6+3}=2\Omega$

$Total resistance=2+2=4\Omega$
Now steady current is 2mA

$\therefore$ Voltage required

$=(.002)4$

$=0.008V=8mV$

$B=4T$

$E=8mV=\dfrac{-d\phi}{dt}$

$=\dfrac{-d4(Area)}{dt}$

$8\times10^{-3}=4\times(0.10)V$

$V=2\times 10^{-2}m/s$

$=2cm/s$

A wire shaped as a semicircle of radius a is rotating about an axis PQ with a constant angular velocity

$\omega =\frac{1}{\sqrt{LC}}$ , with the help of an external agent. A uniform magnetic field B exists in space and is directed into the plane of the figure. (circuit part remains at rest (left part is at rest) ) . Then,

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the rms value of current in the circuit is $\dfrac{\pi Ba^{2}}{R\sqrt{2LC}}$
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the rms value of current in the circuit is $\dfrac{\pi Ba^{2}}{2R\sqrt{LC}}$
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the maximum energy stored in the capacitor is $\dfrac{\pi^{2} B^{2}a^{4}}{8R^{2}C}$
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the maximum power delivered by the external agent is $\dfrac{\pi^{2} B^{2}a^{4}}{4LCR}$

Explanation

Let at time t the angle between magnetic field and area vector(semicircle) be

$\theta$ ,

Then

$\theta =\omega t$

Magnetic flux at time t is:

$\phi =\vec {B}.\vec {S}=\dfrac{\pi a^{2}B}{2}\cos \omega t$

Induced emf at time t is:

$\varepsilon =-\dfrac{d\phi }{dt}=\dfrac{\pi Ba^{2}\omega }{2}\sin \omega t$

$\varepsilon_o=\dfrac{\pi Ba^{2} }{2\sqrt{LC}}$

$\omega = \dfrac{1}{\sqrt {LC}}$ , the circuit is in resonance.

$\left | Z \right |=R$

$\Rightarrow i _{0}=\dfrac{\pi Ba^{2} }{2R\sqrt{LC}}$ which is the peak current.

$i_{rms}=\dfrac{i_{0}}{\sqrt{2}}$

$\Rightarrow i _{rms}=\dfrac{\pi Ba^{2} }{2R\sqrt{ 2 LC}}$

$V_{C}=\frac{1}{2}CV_{0}^{2}\rightarrow max. energy , V_{0}\rightarrow peak voltage$

$V_{0}= i_{0}X_{c}=\dfrac{i_{0}}{C\omega }=\dfrac{i_{0}\sqrt{LC}}{C}$

$V_{C}=\dfrac{1}{2}C\times\dfrac{\pi ^{2}B^{2}a^{4}}{4R^{2}C^{2}}= \dfrac{\pi ^{2}B^{2}a^{4}}{8R^{2}C}$

$P_{ext}=P_{dissipated}=\varepsilon _{0}i_{0}=\dfrac{\pi Ba^{2} }{2\sqrt{LC}}\times \dfrac{\pi Ba^{2} }{2R\sqrt{LC}}$

$P_{ext}=\dfrac{\pi ^{2}B^{2}a^{4}}{4LCR}$

A semicircle conducting ring of radius R is placed in the xy plane, as shown in the figure. A uniform magnetic field is set up along the x-axis. No emf, will be induced in the ring if

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it moves along the xaxis
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it moves along the yaxis
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it moves along the z-axis
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All the above

Explanation

emf induced (by lenz's law)

$\begin{array}{l}E = \frac{{ - d\phi }}{{dt}}\\E = \frac{{ - d}}{{dt}}\left( {BA} \right) = - B\frac{d}{{dt}}\left( A \right) - A\frac{d}{{dt}}\left( B \right)\end{array}$
either magnetic field or area must be changed w.r.t. time to generate emf, and here the magnetic field and plane of ring are parallel thus no emf induced in either situation.

Find the inductance of a unit length of two parallel wires, each of radius a, whose centers are at a distance d apart and carry equal currents in opposite directions. Neglect the flux within the wire:

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$\dfrac {\mu_0}{2\pi} ln \left (\dfrac {d-a}{a}\right )$
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$\dfrac {\mu_0}{\pi} ln \left (\dfrac {d-a}{a}\right )$
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$\dfrac {3\mu_0}{\pi} ln \left (\dfrac {d-a}{a}\right )$
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$\dfrac {\mu_0}{3\pi} ln \left (\dfrac {d-a}{a}\right )$

Explanation

Using Ampere's law for one of the wires to obtain the magnetic field in terms of current

$I$ and the variable radius

$r$ due to one of the wire

$\mu_0\ I = \oint \vec{B}.\vec{dl}$

$= \int_0^{2\pi} Br\ d \theta$

$\mu_0\ I = B(2\pi r)$

$\implies B=\dfrac{\mu_0 I}{2\pi r}$

Note that because of the symmetry of the setup, the total magnetic flux

$(\phi)$ will be two times generated by one of the wires. Then, we continue to use the relation between self inductance and flux

$\phi_{total}= 2 \int_A \vec{B}.\vec{dA}$

$=2 \int_a ^{d-a} \bigg(\dfrac{\mu_0 I}{2\pi r}\bigg ) l\ dr$

$\phi_{total}= \dfrac{\mu_0 Il}{\pi} ln\bigg(\dfrac{d-a}{a}\bigg)$

Inductance,

$L= \dfrac{\phi_{total}}I= \dfrac{\mu_0 l}{\pi} ln\bigg(\dfrac{d-a}{a}\bigg)$

Inductance per unit length ,

$\phi_{total}= \dfrac{\mu_0}{\pi} ln\bigg(\dfrac{d-a}{a}\bigg)$

Hence the correct option is

$(B)$

1503215_120781_ans_2eb52d481e654c1bb273df35de1b02f5.PNG

A rod AB moves with a uniform velocity

$v$ in a uniform magnetic field as shown in figure.

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The rod becomes electrically charged
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The end A becomes positively charged
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The end B become positively charged
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The rod becomes hot because of Joule heating

Explanation

Magnetic force

$F= q(\vec{v}\times \vec{B})= (-e)(v\hat{i} \times (-B)\hat{k})=-ebB\hat{j}$ where we have taken the charge of electron to be (-e) where e is positive. The direction of the magnetic field is taken to be

$-B\hat{k}$ i.e into the plane of the paper and velocity is taken along the +ve x-axis.
Thus electrons will move towards the end B and accumulate at end B.

Thus, the end A will become deficient in electrons and will be positively charged.

The current i in a coil varies with time as shown in the figure. The variation of induced emf with time would be :

Explanation

We know,

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

$0$ to

$T/4:\ \dfrac {di}{dt} =$ constant and

$e$ is negative

$T/4$ to

$T/2$ :

$\dfrac {di}{dt} = 0$ and

$e$ is zero

$T/2$ to

$3T/4$ :

$\dfrac {di}{dt} =$ constant and

$e$ is positive

A long solenoid having

$200$ turns per cm carries a current of

$1.5 amp.$ At the centre of it is placed a coil of

$100$ turns of cross-sectional are

$3.14\times 10^{-4}m^2$ having its axis parallel to the field produced by the solenoid. When the direction of current in the solenoid is reversed within

$0.05 sec,$ the induced e.m.f. in the coil is

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$0.48 V$
0%
$0.048 V$
0%
$0.0048 V$
0%
$48 V$

Explanation

$B=\mu_0ni=(4\pi \times 10^{-7})(200\times 10^{-2})\times 1.5$

$=3.8\times 10^{-2}Wb/m^2$

Magnetic flux through each turn of the coil

$\phi=BA=(3.8\times 10^{-2})(3.14\times 10^{-4})=1.2\times 10^{-5}weber$

When the current in the solenoid is reversed, the change in magnetic flux

$=2\times (1.2\times 10^{-5})=2.4\times 10^{-5}weber$

Induced e.m.f.

$=N\dfrac {d\phi}{dt}=100\times \dfrac {2.4\times 10^{-5}}{0.05}=0.048V$ .

A wire shaped as a circle of radius R rotates about the axis OO' with an angular velocity

$\omega$ as shown in figure. Resistance of the circuit is

$R$ . Find the mean thermal power generated in the loop during a period of a rotation.

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$\dfrac{(B \pi a^2 \omega)^2}{4R}$
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$\dfrac{(B \pi a^2 \omega)^2}{2R}$
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$\dfrac{(3B \pi a^2 \omega)^2}{2R}$
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None of these

Explanation

$A = \pi a^2 \cos \omega t; \phi = BA = B \pi a^2 \cos \omega t$

$\displaystyle i = \dfrac{\varepsilon}{R} = - \dfrac{d\phi}{dt} /R = \dfrac{B \pi a^2 \omega \sin \omega t}{R}$ and

$<P>= \displaystyle \dfrac{1}{T} \int_0^T i^2 Rdt = \dfrac{(B \pi a^2 \omega)^2}{2R}$

A rectangular loop with a slide wire of length

$l$ is kept in a uniform magnetic field as shown in figure (a). The resistance of slider is

$R$ . Neglecting self inductance of the loop find the current in the connector during its motion with a velocity

$v$ .

0%
$\displaystyle \dfrac{Blv}{R_1 + R_2 + R}$
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$\displaystyle \dfrac{Blv(R_1+ R_2)}{R(R_1 + R_2)}$
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$\displaystyle \dfrac{Blv(R_1 + R_2)}{RR_1 + RR_2 + R_1R_2}$
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$\displaystyle Blv \left ( \dfrac{1}{R_1} + \dfrac{1}{R_2} + \dfrac{1}{R_3}\right )$

Explanation

The equivalent circuit is shown in figure (c).
Obviously,

$I = \displaystyle \dfrac{Blv}{R + \dfrac{R_1R_2}{R_1 +R_2}} = \dfrac{Blv(R_1 + R_2)}{RR_1 + RR_2 + R_1 R_2}$

The potential difference across a

$150mH$ 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

$t=4.0ms$ ?

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$2.67 \times 10^{-4}$ A
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$3.67 \times 10^{-2}$ A
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$6.67 \times 10^{-2}$ A
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$9.67 \times 10^{-4}$ A

Explanation

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

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

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

$t=4ms$

$i={ \left( 150\times { 10 }^{ -3 } \right) }^{ -1 }\left( \cfrac { 1 }{ 2 } \times 4\times { 10 }^{ -3 }\times 5 \right) =6.67\times { 10 }^{ -2 }A\quad$

A solenoid has

$2000$ turns wound over a length of

$0.3 m$ . Its cross-sectional area is equal to

$1.2\times 10^{-3} m^2$ . Around its central cross-section, a coil of 300 turns is wound. If an initial current of

$2 A$ flowing in the solenoid is reversed in

$0.25 s$ , the emf induced in the coil is

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$0.6 mV$
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$60 mV$
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$40.2 mV$
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$0.48 mV$

Explanation

$emf=-\dfrac { \mu { N }_{ 1 }{ N }_{ 2 } }{ l } \dfrac { \Delta I }{ \Delta t }$

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

${ N }_{ 1 } =2000$

${ N }_{ 2 }=300$

$l=0.3m$

$\dfrac { \Delta I }{ \Delta t } =\dfrac { 4 }{ 0.25 } A/s$

$\Rightarrow \left| emf \right| = \dfrac {4\pi \times 10^{-7} \times 2000 \times300 \times 4}{ 0.3 \times0.25} =$

$40.212mV$

Find the inductance of a solenoid of length

$l_0$ , made of Cu windings of mass

$m$ . The winding resistance is equal to

$R$ . The diameter of solenoid <<

$l$ .

$\rho_0$ is resistivity of Cu and

$\rho$ is density of the Cu.

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$\displaystyle \dfrac{\mu_0 Rm}{2 \pi l_0 \rho \rho_0}$
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$\displaystyle \dfrac{\mu_0 Rm}{4 \pi l_0 \rho \rho_0}$
0%
$\displaystyle \dfrac{\mu_0 Rm}{3 \pi l_0 \rho \rho_0}$
0%
$\displaystyle \dfrac{2 \mu_0 Rm}{3 \pi l_0 \rho \rho_0}$

Explanation

$R=\cfrac{\rho_0 l}{A}$
Length of the wire

$l = \displaystyle \dfrac{RA}{\rho_0} = \dfrac{Rm}{\rho_0 \rho l}$
or

$\displaystyle l = \sqrt{\dfrac{Rm}{\rho \rho_0}}$ where

$\rho_0$ is

resistivity of the Cu and

$\rho$ is density of the

$Cu l = nl_0 2 \pi r$ and

$L = \mu_0 n^2 l_0 \pi r^2$ where

$l_0$ is length of

the solenoid and r is the radius of the solenoid. Then

$l = l_0 2 \pi r \displaystyle \sqrt{\dfrac{L}{\mu_0 l_0 \pi r^2}}$

$\displaystyle l = \sqrt{\dfrac{4 \pi L l_0}{\mu_0}}$

$\displaystyle l = \sqrt{\dfrac{Rm}{\rho \rho_0}}$

$\displaystyle L = \dfrac{\pi_0 Rm}{4 \pi l_0 \rho \rho_0}$

A closed loop of cross-sectional area

$10^{-2}m^2$ which has inductance

$L=10 mH$ and negligible resistance is placed in a time-varying magnetic field. Figure shows the variation of B with time for the interval of 4 s. The field is perpendicular to the plane of the loop (given at

$t=0, B=0, I=0)$ . The value of the maximum current induced in the loop is :

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0.1 mA
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10 mA
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100 mA
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Data insufficient

Explanation

Induced e.m.f.

$|e|=L\dfrac {dI}{dt}=A\dfrac {dB}{dt}$

$\Rightarrow \int_0^I dI=\int_0^B\dfrac {A}{L}dB\Rightarrow I=\dfrac {A}{L}B$

$\Rightarrow I_{max}=\dfrac {A}{L}B_{max}=\dfrac {10^{-2}}{10\times 10^{-3}}\times 0.1=0.1A=100 mA$

A metallic rod of length

$l$ is hinged at the point

$M$ and is rotating about an axis perpendicular to the plane of paper with a constant angular velocity

$\omega$ . A uniform magnetic field of intensity

$B$ is acting in the region (as shown in the figure) parallel to the plane of paper. The potential difference between the points

$M$ and

$N$ .

0%
is always zero
0%
varies between $\cfrac { 1 }{ 2 } B\omega { l }^{ 2 }$ to $0$
0%
is always $\cfrac { 1 }{ 2 } B\omega { l }^{ 2 }$
0%
is always $B\omega { l }^{ 2 }$

Explanation

Let an elementary part of width

$dx$ .

Total induced Emf across the rod is given by

$E = \int_0^l (\vec{v} \times \vec{B}) . \vec{dx}$

As according to the figure,

$\vec{B} = B \hat{x}$ and

$\vec{v} = -v \hat{y} = -wx \hat{y}$

$\therefore$

$\vec{v} \times \vec{B} = wxB \hat{k}$

$\implies (\vec{v} \times \vec{B}) . \vec{dx} = (wxB \hat{k}) . {dx}$

$\hat{x} = 0$

Note: At every instant,

$(\vec{v} \times \vec{B}) \perp$

$\vec{dx}$

Thus total Emf induced is always

$Zero.$

In the figure shown, a uniform magnetic field

$\left| \vec { B } \right| =0.5T$ is perpendicular to the plane of circuit. The sliding rod of length

$l=0.25m$ moves uniformly with constant speed

$v=4 m{s}^{-1}$ . If the resistance of the slides is

$2\Omega$ , then the current flowing through the sliding rod is :

0%
$0.1A$
0%
$0.17A$
0%
$0.08A$
0%
$0.03A$

Explanation

$e=Bvl$

$=0.5\times 4\times 0.25=0.5V$

$12\Omega$ and

$4\Omega$ are parallel. hence their net resistance

$R=3\Omega$

$i=\cfrac{e}{R+r}=\cfrac{0.5}{3+2}=0.1A$

An alternating current

$I$ in an inductance coil varies with time

$t$ according to the graph as shown:
Which one of the following graphs gives the variation of voltage with time?

Explanation

emf =

$L \dfrac{di}{dt}$
Rate of change of current is constant for one period at a positive value and is constant at negative value for the second time period.
Therefore emf is a constant positive value for first half and constant negative value for second half. Option C is correct.

A conducting straight wire

$PQ$ of length

$l$ is fixed along a diameter of a non-conducting ring as shown in the figure. The ring is given a pure rolling motion on a horizontal surface such that its centre of mass has a veleocity

$v$ . There exists a uniform horizontal magnetic field

$B$ in horizontal direction perpendicular to the plane of ring. The magnitude of induced emf in the wire

$PQ$ at the position shown in the figure will be :

0%
$Bvl$
0%
$2Bvl$
0%
$3Bvl/2$
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zero

Explanation

$\omega=\cfrac{v}{R}$

$=\cfrac{v}{l/2}=\cfrac{2v}{l}$

$e=\cfrac{B\omega {l}^{2}}{2}$

$=\cfrac { B\left( \cfrac { 2v }{ l } \right) { l }^{ 2 } }{ 2 } =Bvl$

Two metallic rings of radius

$R$ are rolling on a metallic rod. A magnetic field of magnitude

$B$ is applied in the region. The magnitude of potential difference between points

$A$ and

$C$ on the two rings (as shown), will be :

0%
$0$
0%
$4B\omega{R}^{2}$
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$8B\omega{R}^{2}$
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$2B\omega{R}^{2}$

Explanation

${V}_{A}-{V}_{0}=\cfrac { B\omega { \left( 2R \right) }^{ 2 } }{ 2 } =2B\omega { R }^{ 2 }......(i)$

${V}_{0}-{V}_{C}=\cfrac { B\omega { \left( 2R \right) }^{ 2 } }{ 2 } =2B\omega { R }^{ 2 }......(ii)$
Adding these tow equations we get,

${V}_{A}-{V}_{C}=4B\omega {R}^{2}$

Choose the correct options

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SI unit of magnetic flux is henry-ampere
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SI unit of coefficient of self-inductance is $J/A$
0%
SI unit of coefficient of self inductance is $\cfrac{volt-second}{ampere}$
0%
SI unit of magnetic induction is weber

Explanation

(a)

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

$\therefore$

$\phi=\cfrac{Li}{N}$
S, SI unit of fulx is henry-ampere
(b)

$L=\cfrac { -e }{ \Delta i/\Delta t } =\cfrac { -e\Delta t }{ \Delta i }$
Hence, SI unit of

$L$ is

$\cfrac{V-s}{ampere}$

A straight conducting rod

$PQ$ is executing SHM in

$xy$ plane from

$x=-d$ to

$x=+d$ . Its mean position is

$x=0$ and its length is along y-axis. There exists a uniform magnetic field

$B$ from

$x=-d$ to

$x=0$ pointing inward normal to the paper and from

$x=0$ to

$=+d$ there exists another uniform magnetic field of same magnitude

$B$ but pointing outward normal to the plane of the paper. At the instant

$t=0$ , the rod is at

$x=0$ and moving to the right. The induced emf

$(\varepsilon )$ across the rod

$PQ$ vs time

$(t)$ graph will be

Explanation

The displacement equation of the rod executing

$SHM$

$x = d$

$sinwt$ .......(1)

where

$d$ is the amplitude of oscillation.

Thus velocity equation is given by

$v = \dfrac{dx}{dt} = wd$

$coswt$

Let the length of the conduction rod

$PQ$ be

$L$

Induced EMF in the rod

$\mathcal{E} = BvL = BLwd$

$coswt$

$\implies$ At

$t = 0$ , Induced EMF is maximum i.e

$\mathcal{E_{max}} = BLwd$ and after that , it decreases.

$t = \dfrac{T}{4}$ , Induced EMF

$\mathcal{E} = 0$

Now the rod starts to move in

$-ve$ x direction and reach the mean position i.e

$x= 0$

$t \leq \dfrac{T}{2}$ , Induced EMF

$\mathcal{E} = -BLwd$

Now for

$t> \dfrac{T}{2}$ OR

$x<0$ ,

$B$ reverses its direction and hence

$\mathcal{E}$ also reverses its polarity.

$\implies$ At

$t > \dfrac{T}{2}$ , Induced EMF

$\mathcal{E} = BLwd$ and then it decreases with time again.

A conducting ring of radius

$r$ is rolling without slipping with a constant angular velocity

$\omega$ (figure). If the magnetic field strength is

$B$ and is directed into the page then the emf induced across

$PQ$ is

0%
$B\omega {r}^{2}$
0%
$\cfrac{B\omega{r}^{2}}{2}$
0%
$4B\omega{r}^{2}$
0%
$\cfrac{B\omega{r}^{2}}{4}$

Explanation

Let the velocity of each point due to rotation form an angle

$\theta$ with the vertical.

EMF induced across PQ

$=\int_{0}^{\pi/2} B(r\omega\cos\theta)(rd\theta)$

$=\int_0^{\pi/2}Br^2\omega\cos\theta d\theta\\ =B\omega r^2$

A rectangular loop of sides

$a$ and

$b$ is placed in

$x-y$ plane. A uniform but time varying magnetic field of strength

$\vec { B } =20t\hat { i } +10{ t }^{ 2 }\hat { j } +50\hat { k }$ is present in the region. The magnitude of induced emf in the loop at time

$t$ is :

0%
$20+20t$
0%
$20$
0%
$20t$
0%
zero

Explanation

$\vec { S } =(ab)\hat { k }$

$\rightarrow$ perpendicular to x-y plane

$\phi =\vec { B } \cdot \vec { S } =(50)(ab)=$ constant

$\cfrac { d\phi }{ dt } =0$

$\therefore \quad e=0$

A wooden stick of length

$3l$ is rotated about an end with constant angular velocity

$\omega$ in a uniform magnetic field

$B$ perpendicular to the plane of motion. If the upper one-third of its length is coated with copper, the potential difference across the whole length of the stick is

0%
$\dfrac{9B\omega {l}^{2}}{2}$
0%
$\dfrac{4B\omega {l}^{2}}{2}$
0%
$\dfrac{5B\omega {l}^{2}}{2}$
0%
$\dfrac{B\omega {l}^{2}}{2}$

Explanation

Force on one electron in the rod, at a point r form center, due to the motion =

$F= q \vec v \times \vec B = q_e r \omega B$

total emf =

$\int_{2l}^{3l}{\dfrac{F}{q_e} dr}= \int_{2l}{3l}{ r \omega B dr}$

$= B \omega (\dfrac{9l^2}{2} - \dfrac{4l^2}{2} ) = B \omega \dfrac{ 5 l^2}{2}$

A rod of length

$L$ rotates in the form of a conical pendulum with an angular velocity

$\omega$ about its axis as shown in figure. The rod makes an angle

$\theta$ with the axis. The magnitude of the motional emf developed across the two ends of the rod is

0%
$\cfrac{1}{2}B\omega {L}^{2}$
0%
$\cfrac{1}{2}B\omega {L}^{2}\tan ^{ 2 }{ \theta }$
0%
$\cfrac{1}{2}B\omega {L}^{2}\cos ^{ 2 }{ \theta }$
0%
$\cfrac{1}{2}B\omega {L}^{2}\sin ^{ 2 }{ \theta }$
0%
answer required

Explanation

Figure shows the Top view, in which the rod of length

$r$ is revolving in a circle about an axis passing through its one end.

From top view, effective length

$=r=l\sin { \theta }$

Induced emf

$e=\dfrac { 1 }{ 2 } B\omega { r }^{ 2 }=\dfrac { 1 }{ 2 } B\omega { l }^{ 2 }\sin ^{ 2 }{ \theta }$

A conducting rod of length

$0.5 m$ rotates about a horizontal axis passing through one of its ends and perpendicular to its length with angular velocity of

$4 rad/s$ . A uniform magnetic field of

$2.0T$ exists parallel to the axis of rotation. The emf developed across the ends of the conductor will be

0%
$0V$
0%
$1V$
0%
$2V$
0%
$0.5V$

Two identical conducting rings

$A$ and

$B$ of radius

$R$ are rolling over a horizontal conducting plane with same speed

$v$ but in opposite direction. A constant magnetic field

$B$ is present pointing into the plane of the paper. Then the potential difference between the highest points of the two rings is

0%
$0$
0%
$2BvR$
0%
$4BvR$
0%
none of these
0%
answer required

Explanation

Let us consider one ring, travelling in one direction. Each ring has two components of velocity, the linear component and the angular component.

The potential difference induced between the top and bottom points of each ring due to the angular component is zero.

The potential difference induced between the same points due to linear component

$=\int_{0\\-\pi/2}^{\pi/2} B(vcos\theta)Rd\theta$

$=2BvR$

This is the potential difference between top and bottom points of one ring.

Similarly the potential difference between top and bottom of other ring is

$2BvR$ , however the polarity is opposite since the rings travel in opposite directions. The potential of the bottom of both ring is equal(connected).

Hence potential difference between top of both the rings

$=2BvR-(-2BvR)=4BvR$

$AB$ is a resistanceless conducting rod which forms a diameter of a conducting ring of radius

$r$ rotating in a uniform magnetic field

$B$ as shown in figure. The resistors

${R}_{1}$ and

${R}_{2}$ do not rotate. Then the current through the resistor

${R}_{1}$

0%
$\cfrac { B\omega { r }^{ 2 } }{ 2{ R }_{ 1 } }$
0%
$\cfrac { B\omega { r }^{ 2 } }{ 2{ R }_{ 2 } }$
0%
$\cfrac { B\omega { r }^{ 2 } }{ 2{ R }_{ 1 }{ R }_{ 2 } } \left( { R }_{ 1 }{ +R }_{ 2 } \right) \quad$
0%
$\cfrac { B\omega { r }^{ 2 } }{ 2\left( { R }_{ 1 }{ +R }_{ 2 } \right) }$
0%
answer required

Explanation

Emf induced between A and center of circle or B and center of circle

$=\int_0^rB(r\omega)dr=\dfrac{1}{2}B\omega r^2$

This is the also the potential difference across resistor

$R_1$ .

Hence,

$\dfrac{1}{2}B\omega r^2=i(R_1)$

$\implies i=\dfrac{B\omega r^2}{2R_1}$

A circular loop of radius

$r$ , having

$N$ turns of a wire, is placed in a uniform and constant magnetic field

$B$ . The normal loop makes an angle

$\theta$ with the magnetic field. Its normal rotates with an angular velocity

$\omega$ such that the angle

$\theta$ is constant. Choose the correct statement from the following.

0%
emf in the loop is $NB\omega {r}^{2}/2 \cos{\theta}$
0%
emf induced in the loop is zero
0%
emf must be induced as the loop crosses magnetic lines.
0%
emf must not be induced as flux does not change with time.
0%
answer required

Magnetic flux linked with a stationary loop resistance

$R$ varies with respect to time during the time period

$T$ as follows:

$\phi=at(T-t)$
The amount of heat generated in the loop during that time (inductance of the coil is negligible) is

0%
$\cfrac{\alpha T}{3R}$
0%
$\cfrac{{a}^{2}{T}^{2}}{3R}$
0%
$\cfrac{{a}^{2}{T}^{2}}{R}$
0%
$\cfrac{{a}^{2}{T}^{3}}{3R}$

Explanation

Given ,

$\phi=at(T-t)$

The emf is,

$\varepsilon=-\dfrac{d\phi}{dt}=-at+2at=at$

Power,

$P=\varepsilon I=\varepsilon . \dfrac{\varepsilon}{R}=\dfrac{a^2t^2}{R}$

Heat generated,

$H=\int_0^T Pdt=\int^T_0\dfrac{a^2t^2}{R} dt=\dfrac{a^2T^3}{3R}$

In the figure, there exists a uniform magnetic field

$B$ into the plane of paper. Wire

$CD$ is in the shape of an arc and is fixed.

$OA$ and

$OB$ are the wires rotating with angular velocity

$\omega$ as shown in the figure in the same plane as that of the arc about point

$O$ . If at some instant,

$OA=OB=1$ and each wire makes angle

$\theta={30}^{o}$ with y-axis, then the current through resistance

$R$ is (wire

$OA$ and

$OB$ have no resistance)

0%
$0$
0%
$\cfrac{B\omega {l}^{2}}{R}$
0%
$\cfrac{B\omega {l}^{2}}{2R}$
0%
$\cfrac{B\omega {l}^{2}}{4R}$
0%
answer required

Explanation

EMF induced in wire OA =

$\dfrac{1}{2} B l^2 \omega$ with A positive wrt O

emf induced in wire OB =

$\dfrac{1}{2} B l^2 \omega$ with O positive wrt B

Therefore total emf in the loop consisting of wire OA , OB and the arc is

$B l^2 \omega$

current in loop =

$\dfrac{ B l^2 \omega}{R}$

The current through the coil in figure (i) varies as shown in figure (ii). Which graph best shows ammeter

$A$ reading as a function of time?

0%
0%
0%
0%
0%
answer required

Explanation

When current flows through first circuit, flux links the second circuit due to mutual inductance. when there is a change in current, there is a corresponding changing in linking flux between the two circuits and hence a emf is induced in the second circuit given by,
emf = rate of change of flux = M

$\times$ rate of change of current.
current in the first circuit changes twice with a constant rate, one time positive rate of change and second time negative rate of change.
Therefore emf will be constant positive for first period when current changes. And emf will be constant negative in the second period the current changes.

A semicircular wire of radius

$R$ is rotated with constant angular velocity about an axis passing through one end and perpendicular to the plane of the wire. There is a uniform magnetic field of strength

$B$ . The induced emf between the ends is

0%
$B\omega {R}^{2}/2$
0%
$2B\omega {R}^{2}$
0%
is variable
0%
none of these

Explanation

Induced motional emf in

$MNQ$ is equivalent to the motional emf in an imaginary wire

$MQ$ .

Therefore,

${ e }_{ MNQ }={ e }_{ MQ }$

$=\dfrac { 1 }{ 2 } { B\omega l }^{ 2 }$

$={ 2B\omega R }^{ 2 }\quad \left(\because l=2R \right)$

The magnetic flux

$\phi$ linked with a conducting coil depends on time as

$\phi=4{t}^{n}+6$ , where

$n$ is a positive constant. The induced emf in the coil is

$e$ .

0%
If $0< n< 1, e\ne 0$ and $\left| e \right|$ decreases with time
0%
If $n=1,e$ is constant
0%
If $n>1,\left| e \right|$ increases with time
0%
If $n>1, \left| e \right|$ decreases with time
0%
answer required

Explanation

Faraday's Law states:

$\epsilon = -\cfrac{d \phi}{dt}$

$\epsilon = -4nt^{n-1}$

$n=1$ ,

$|e| = 4$

$n>1$ ,

$|e| = 4nt^{n-1}$

$n-1 > 1$ , then

$e$ increases

else if

$n-1<1$ , e decreases

A disc of radius

$R$ is rolling without sliding on a horizontal surface with a velocity of center of mass

$v$ and angular velocity

$\omega$ in a uniform magnetic field

$B$ which is perpendicular to the plane of the disc as shown in figure.

$O$ is the center of the disc and

$P,Q,R$ and

$S$ are the four points on the disc. Which of the following statements is true?

0%
Due to translation, induced emf across $PS=Bvr$
0%
Due to rotation, induced emf across $QS=0$
0%
Due to translation, induced emf across $RO=0$
0%
Due to rotation, induced emf across $OQ=Bvr$

Explanation

Consider an infinitesimally small element on the surface of disc between point P and S of length

$rd\theta$ , where

$\theta$ is the angle line connecting that point with center forms with horizontal.

Thus emf induced across the element=

$Bvrd\theta\cos\theta$

Hence net emf between points P and S is

$\int_{0}^{\pi/2}BvR\cos\theta d\theta=Bvr$

Since points R and O lie along same line of along velocity of translation, there is no emf induced across RO. And, for the same reason, there is no emf induced across QS and OQ.

The self inductance of a coil having 500 turns is 5 mH. The magnetic flux through the cross - sectional area of the coil while current through it is 8 mA is found to be :

0%
$8\times 10^{-9}Wb$
0%
$0.8\times 10^{-9}Wb$
0%
$8\times 10^{-6}Wb$
0%
$80\times 10^{-3}Wb$

A long wire carries a steady current of 1 Ampere. A ''S'' shaped conducting rod AB consisting of two semicircles each of radius r is placed in such a way that the centre C of the conducting wire is at a distance 2r from the end of the wire.The rod AB moves with velocity of 5

$ms^{-1}$ along the direction of the current flow as shown in the figure. If the line joining the ends of the rod makes an angle

$60^{\circ}$ with the wire then,

0%
the emf induced between the ends of the rod is (ln 4) $\mu V$
0%
the end A is at higher potential than end B.
0%
the end A is at lower potential than end B.
0%
the emf induced between the ends of the rod is (ln 3) $\mu V.$

Explanation

$AB =4r$
Induced emf in the conductor will be same as in AB.
Thus, we have a rod of effective length

$l_{eff} =4r\cos \theta$ translating in a non-uniform magnetic field.
Consider a small element dx located at a distance x from the wire.

The magnetic field at this location is

$B=\dfrac{\mu_0I}{2 pi x}$

The potential difference across the element is

$dV=\dfrac{\mu_0I}{2 \pi x} v dx$

$\displaystyle V =\int_{d-2r\cos \theta}^{d + 2r\cos \theta} dV= \int_{d-2r\cos \theta}^{d + 2r\cos \theta} \frac{\mu_0I}{2 \pi x} v dx =\frac{\mu_0Iv}{2\pi}\ln(\frac{{d + 2r\cos \theta}}{{d - 2r\cos \theta}})=2 \times \frac{\mu_0}{4\pi} \times 1 \times 5 (\ln \frac { 2r + 2r \cos 30^0}{2r - 2r \cos 30^0})$

$\displaystyle \therefore V= 10^{-7} \times 10 \times \frac{ 1 + cos 30^0}{1- \cos 30^0}= 10^{-6} \times\ln( \frac{2 +\sqrt{3}}{ 2 - \sqrt{3}})=ln(\frac{2 +\sqrt{3}}{ 2 - \sqrt{3}}) \mu V$

A thin semicircular conducting ring of radius

$R$ is falling with its plane vertical in horizontal magnetic induction

$\vec { B }$ . At the position

$MNQ$ , the speed of the ring is

$V$ , and the potential difference developed across the ring is

0%
zero
0%
$BV\pi {R}^{2}/2$ and $M$ is at higher potential
0%
$\pi RBV$ and $Q$ is at higher potential
0%
$2RBV$ and $Q$ is at higher potential

Explanation

$E = \cfrac{d \phi}{dt}= B \cfrac{dA}{dt}$

When it is just about to move out:

$dA = 2R (vdt)$

$\cfrac{dA}{dt}= 2 Rv$

Hence,

$E = 2BRV$

The magnetic flux

$\phi$ linked with a coil depends on time t as

$\phi = at^n$ , where a and n are constants. The emfinduced in the coil is e

0%
If 0 < n < 1, e = 0
0%
If 0 < n < 1, e $\neq$ 0
0%
If n = 1, e is constant
0%
If n > 1, |e| increases with time

Explanation

$\displaystyle |E_{in}| = |e| = \frac {d \phi} {dt} \Rightarrow |e| = nat^{n\,-\,1}$

If 0 < n < 1 $\Rightarrow$ (n - 1) negative $\Rightarrow \displaystyle |e| = \frac {na} {t^{1\,-\,1}}$ hence |e| decay with time.

If n = 1 $\Rightarrow$ = na = constant; If n > 1 $\Rightarrow$ |e| increases with time.

In figure,

$R$ is a fixed conducting ring of negligible resistance and radius

$a$ . It is hinged at the center of the ring and rotated about this point in clockwise direction with a uniform angular velocity

$\omega$ . There is a uniform magnetic field of strength

$B$ pointing inward and

$r$ is a stationary resistance. Then

0%
current through $r$ is zero
0%
current through $r$ is $(2B\omega {a}^{2})/5r$
0%
direction of current in external resistance $r$ is from center to circumference
0%
direction of current in external resistance $r$ is from circumference to center
0%
answer required

Explanation

Current through

$r= I =\dfrac { e }{ r }$

$=\dfrac { induced\quad emf }{ Resistance }$

$e=-\dfrac { d\phi }{ dt } =-\dfrac { d }{ dt } BAcos\theta$

$e=\dfrac { 2Bw{ a }^{ 2 } }{ 5 }$

A

$=$ area swept in unit time.

$A=\dfrac { 2w{ a }^{ 2 } }{ 5 }$ (area of sector )

$\therefore I=\dfrac { 2Bw{ a }^{ 2 } }{ 5r }$

Since the

$\overrightarrow { B }$ is into the plane of the plane, the direction of current is external resistance'r' is from centre to circumference.

A uniform thin rod of length L is moving in a uniform magnetic field

$\displaystyle B_{0}$ such that velocity of its centre of mass is v and angular velocity is

$\displaystyle \omega=\frac{4v}{L}$ Then

0%
e.m.f. between end P and Q of the rod is $\displaystyle B_{0}lv$
0%
end P of the rod is at higher potential than end Q of the rod
0%
end Q of the rod is at higher potential than end P of the rod
0%
the electrostatic field induced in the rod has same direction at all points along the length of rod

Explanation

The charges inside the rod move under the magnetic force given by

$\vec{F}=q(\vec{v}\times \vec{B})$

Thus the charges experience force towards end P. Hence end P is at higher potential.

EMF induced along a infinitesimal element along rod of length

$dx$ is

$dV=Bv(dx)$

$B(x\omega)dx$

Hence emf between points P and Q=

$\int_0^L B(x\omega )dx$

$=\dfrac{1}{2}B\omega L^2$

$=2BvL$

CBSE Questions for Class 12 Medical Physics Electromagnetic Induction Quiz 11 - MCQExams.com

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Practice Class 12 Medical Physics Quiz Questions and Answers

CBSE Questions for Class 12 Medical Physics Electromagnetic Induction Quiz 11 - MCQExams.com

0:0:1

Practice Class 12 Medical Physics Quiz Questions and Answers

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