MCQExams
0:0:1
CBSE
JEE
NTSE
NEET
Practice
Homework
×
CBSE Questions for Class 12 Medical Physics Electromagnetic Induction Quiz 7 - MCQExams.com
CBSE
Class 12 Medical Physics
Electromagnetic Induction
Quiz 7
In figure when key is pressed the ammeter A reads i ampere. The charge passing in the galvanometer circuit of total resistance R is Q. The mutual inductance of the two coils is :
Report Question
0%
$$\cfrac{Q}{R}$$
0%
$$QR$$
0%
$$\cfrac{QR}{i}$$
0%
$$\cfrac{i}{QR}$$
What is the magnetomotive force (mmf) of a wire with 8 turns carrying three amperes of current?
Report Question
0%
2,400 At
0%
240 At
0%
24 At
0%
2.4 At
Explanation
Current flowing through the wire $$i = 3 \ A$$
Number of turns $$N = 8$$
Thus magneto motive force $$ = Ni = 8\times 3 = 24 \ At$$
A metal disc of radius R rotates with an angular velocity, $$\omega = 1 rad/s$$ about an axis perpendicular to its plane passing through its centre in a magnetic field of induction B acting perpendicular to the plane of the disc. The induced e.m.f. between the rim and axis of the disc is:
Report Question
0%
$$BR^2$$
0%
$$ 2B^2R^2$$
0%
$$BR^3$$
0%
$$BR^2/2$$
Explanation
As shown in figure 1.AB is the axis of rotation, $$\vec { B } $$ is directed into the plane of disc consider a ring between the disc of thickness $$dx$$ at a distance $$x$$ from centre 0. The velocity of a ring $$V=wx$$ where w = angular velocity of a disc.
Now emf induced due to this ring is
$$de=VBdx\\ =Bwxd\\ \int _{ 0 }^{ e }{ de } =\int _{ 0 }^{ R }{ Bxdx } \\ e=B\left[ \dfrac { { x }^{ 2 } }{ 2 } \right] ^{ R }_{ 0 }\\ e=\dfrac { B{ R }^{ 2 } }{ 2 } $$as w= 1
Two parallel, long wires carry currents t, and $$i_1$$ with $$i_1 > i_2$$. When the currents are in the same direction, the magnetic field at a point midway between the wires is $$10 pT$$. If the direction of $$i_2$$ is reversed, the field becomes $$30 pT$$. The ratio $$\frac{i_1}{i_2}$$ is
Report Question
0%
4
0%
3
0%
2
0%
1
Two coils A and B have 200 and 400 turns respectively. A current of 1 A in coil A causes a flux per turn of $$10^{-3}$$
Wb to link with A and a flux per turn of $$0.8 \times 10^{-3}$$
Wb through B. The ratio of self-inductance of A and the mutual inductance of A and B is :
Report Question
0%
5/4
0%
1/1.6
0%
1.6
0%
1
Explanation
Two coils A and B have 200 and 400 turns respectively. A current of 1 A in coil A causes a flux per turn of 10−3 Wb to link with A and a flux per turn of 0.8×10−3 Wb through B. The ratio of self-inductance of A and the mutual inductance of A and B is $$\dfrac{L_1}{L_2}=\dfrac{200*10^{-3}}{400*0.8*10^{-3}}=1/1.6$$
Electromagnetic induction is not used in :
Report Question
0%
Transformer
0%
Room heater
0%
AC generator
0%
Choke coil
Explanation
Room heater works on the principle of Joule's heating effect of current. It states that heat is produced in the conductor (or resistor) by passing an electric current through it for some time.
While transformer, AC generator, and choke coil, all work on the principle of Electromagnetic Induction.
Hence, the correct option is B
If an emf of $$25 V$$ is induced when the current in the coil changes at the rate of $$100\ As^{-1}$$, then the coefficient of self induction of the coil is
Report Question
0%
$$0.3\ H$$
0%
$$0.25\ H$$
0%
$$2.5\ H$$
0%
$$0.25\ mH$$
Explanation
Rate of change of current, $$\dfrac{di}{dt} = 100$$ $$As^{-1}$$
Magnitude of emf induced in the coil, $$|\mathcal{E}| = 25$$ V
Using: $$|\mathcal{E}| = L\dfrac{di}{dt}$$ where $$L$$ is the coefficient of self inductance
$$\implies$$
$$25 = L\times 100$$
$$\implies$$ $$L = 0.25$$ $$H$$
Reactance of a coil is $$157\Omega$$. On connecting the coil across a source of frequency $$ 100Hz$$, the current lags behind e.m.f. by $${ 45 }^{ o }$$. The inductance of the coil is _________.
Report Question
0%
$$0.25 H$$
0%
$$0.5 H$$
0%
$$4H$$
0%
$$314 H$$
Explanation
Since the phase angle is $$45^{\circ}$$,
$$\dfrac{X_L}{R}=tan\phi=tan45^{\circ}=1$$
$$\implies X_L=R$$
$$\implies \omega L=R$$
$$\implies 2\pi f L=R$$
$$\implies L=\dfrac{R}{2\pi f}$$
$$=\dfrac{157}{2\pi\times 100}H$$
$$=0.25H$$
The phenomenon of producing an emf in a circuit whenever the magnetic flux linked with a coil changes is ................................
Report Question
0%
Electro-magnetic induction
0%
Inducing current
0%
Inducing voltage
0%
Change in current
Explanation
The phenomenon of producing an emf in a circuit whenever the magnetic flux linked with a coil changes is electromagnetic induction. Electromagnetic induction is the production of an electromotive force (i.e., voltage) across an electrical conductor due to its dynamic interaction with a magnetic field. It has found many applications in technology, including electrical components such as inductors and transformers, and devices such as electric motors and generators.
In a coil of self inductance $$0.5$$ henry, the current varies at a constant rate from zero to 10 amperes in 2 seconds. The e.m.f.generated in the coil is
Report Question
0%
$$10$$ volts
0%
$$5$$ volts
0%
$$2.5$$ volts
0%
$$1.25$$ volts
Explanation
$$\dfrac{\Delta i}{\Delta t} = \dfrac{10}{2} = 5 \,A / sec $$
$$\Rightarrow e = L \dfrac{\Delta i}{\Delta t} = 0.5 \times 5 = 2.5 \,volts$$
For a current carrying inductor, emf associated in $$20mV$$. Now, current through it changes from $$6A$$ to $$2A$$ in $$2s$$. The coefficient of mutual inductance is
Report Question
0%
$$20mH$$
0%
$$10mH$$
0%
$$1mH$$
0%
$$2mH$$
Explanation
$$\displaystyle \left | e \right |=L\frac{dI}{dt}$$
Here, $$\displaystyle e=20mV=20\times 10^{-3}V$$
Coefficient of mutual inductance,
$$ 20\times 10^{-3}=L\times 2$$
$$\displaystyle \therefore L=10\times 10^{-3}=10mH$$
A solenoid $$30 cm$$ long is made by winding $$2000$$ loops of wire on an iron rod whose cross-section is $$1.5{ cm }^{ 2 }$$. If the relative permeability of the iron is $$6000$$. What is the self-inductance of the solenoid?
Report Question
0%
$$15 H$$
0%
$$25 H$$
0%
$$35 H$$
0%
$$5 H$$
Explanation
Given : $$\mu_r =6000$$, $$l = 30 cm = 0.3 m$$, $$N = 2000$$
Cross section of solenoid $$A = 1.5 cm^2 = 1.5\times 10^{-4} m^2$$
As we know, Self-inductance of the solenoid
$$L=\dfrac { { \mu }_{ r }\cdot { \mu }_{ 0 }{ N }^{ 2 }A }{ l } $$
$$=\dfrac { 600 0\times 4\pi \times { 10 }^{ -7 }\times { \left( 2000 \right) }^{ 2 }\times \left( 1.5 \right) \times { 10 }^{ -4 } }{ 0.3 } $$
$$=15 H$$
Two coils have a mutual inductance $$0.55H$$. The current changes in the first coil according to equation $$I={ I }_{ 0 }\sin { \omega t }$$.
where, $$ { I }_{ 0 }=10A$$ and $$\omega =100\pi { rad }/{ s }$$.
The maximum value of emf in the second coil is
Report Question
0%
$$2\pi $$
0%
$$5\pi $$
0%
$$\pi $$
0%
$$4\pi $$
Explanation
E.M.F. $$e=M\dfrac { di }{ dt } =0.005\times \dfrac { d }{ dt } \left( { i }_{ 0 }\sin { \omega t } \right) =0.005\times { i }_{ 0 }\cos { \omega t } $$
$${ e }_{ max }=0.005\times 10\times 100\pi =5\pi $$
In the figure shown, the magnetic field induction as the point $$O$$ will be
Report Question
0%
$$\dfrac { { \mu }_{ 0 }i }{ 2\pi r } $$
0%
$$\left( \dfrac { { \mu }_{ 0 } }{ 4\pi } \right) \left( \dfrac { i }{ r } \right) \left( \pi +2 \right) $$
0%
$$\left( \dfrac { { \mu }_{ 0 } }{ 4\pi } \right) \left( \dfrac { i }{ r } \right) \left( \pi +1 \right) $$
0%
$$\dfrac { { \mu }_{ 0 }i }{ 4\pi r } \left( \pi -2 \right) $$
Explanation
Field due to a straight wire of infinite length is $$\dfrac { { \mu }_{ 0 }i }{ 4\pi r }$$ if the point is on a line perpendicular to its length while at the centre of a semicircular coil is $$ \dfrac { { \mu }_{ 0 }\pi i }{ 4\pi r } $$
$$\therefore B={ B }_{ a }+{ B }_{ b }+{ B }_{ c }$$
$$=\dfrac { { \mu }_{ 0 } }{ 4\pi } \dfrac { i }{ r } +\dfrac { { \mu }_{ 0 } }{ 4\pi } \dfrac { \pi i }{ r } +\dfrac { { \mu }_{ 0 } }{ 4\pi } \dfrac { i }{ r }$$
$$ =\dfrac { { \mu }_{ 0 }i }{ 4\pi r } \left( \pi +2 \right) $$ out of the phase
The phenomenon of producing an emf in a circuit whenever the magnetic flux linked with a coil changes is ________.
Report Question
0%
Electromagnetic induction
0%
Inducing current
0%
Inducing voltage
0%
Change in current
Explanation
Electromagnetic induction is the phenomenon of producing an emf in a circuit whenever there is a change in magnetic flux linked with the coil. Lenz's law states that if the magnetic flux linked with the coil changes, an emf is induced in the coil in such a way so as to oppose that change in magnetic flux.
An induced emf has
Report Question
0%
A direction same as field direction
0%
A direction opposite to the field direction
0%
No direction of its own
0%
Non of these
Explanation
From Lentz's law, the direction of induced emf in a circuit is such that it opposes the magnetic flux that produces it.
So, if the magnetic flux linked with a closed circuit increases the induced current flows in a direction so as to develop a magnetic flux in the opposite direction of original flux.
If the magnetic flux linked with a closed circuit decreases then the induced current flows in the same direction of the original flux. So, the induced emf has not direction of its own.
A straight line conductor of length $$0.4\ m$$ is moved with a speed of $$ 7 ms^{-1} $$ perpendicular to magnetic field of intensity of $$0.9\ Wbm^{-2} . $$ The induced emf across the conductor is :
Report Question
0%
$$25.2 \,V$$
0%
$$5.04 \,V$$
0%
$$2.52 \,V$$
0%
$$1.26 \,V$$
Explanation
The induced emf across the conductor $$E = Blv$$
$$ = 0.9 \times 0.4 \times 7 = 2.52 V $$
Find the inductance L of a solenoid of length l whose windings are made of material of density D and resistivity $$ \rho . $$ The winding resistance is R :
Report Question
0%
$$ \dfrac {\mu_0}{4 \pi l} . \dfrac {R_m}{\rho D } $$
0%
$$ \dfrac {\mu_0}{4 \pi R } . \dfrac {l_m}{\rho D } $$
0%
$$ \dfrac {\mu_0}{4 \pi l } . \dfrac {R^2 m}{\rho D } $$
0%
$$ \dfrac {\mu_0}{2 \pi R } . \dfrac {l_m}{\rho D } $$
Explanation
For a solenoid $$ L = \mu_0 N^2 \dfrac {A}{l} 2 yx $$ is the length of the wire and a is the of cross-section, then
$$ R = \dfrac {px}{a} $$
and $$ m = axD $$
$$ R_m = \rho \dfrac {x}{a} = axD $$
$$ x = \sqrt{\dfrac{R_m}{\rho D }} $$
$$ x = 2 \pi r N , N = \dfrac {x}{2 \pi r} $$
$$ L = \mu_0 \left( \dfrac {x^2}{2 \pi r} \right)^2 \dfrac { \pi r^2}{l} $$
$$ = \dfrac {\mu_0}{4 \pi l} . \dfrac {R_m}{\rho D } $$
When the self-inductance of the primary and secondary coil is doubled, then the mutual inductance of the two coils is :
Report Question
0%
Halved
0%
Doubled
0%
Quadrupled
0%
Reduced to one-fourth
Explanation
$$M=K\sqrt{L_1L_2}$$
If self-inductance is doubled, then mutual inductance is also doubled.
Two coils $$A$$ and $$B$$ have mutual inductance $$2 \times {10}^{-2}$$ henry. If the current in the primary is $$i = 5 \sin{\left(10 \pi t\right)}$$ then the maximum value of e.m.f. induced in coil $$B$$ is
Report Question
0%
$$\pi\ volt$$
0%
$$\dfrac{\pi}{2}\ volt$$
0%
$$\dfrac{\pi}{3}\ volt$$
0%
$$\dfrac{\pi}{4}\ volt$$
Explanation
Current in the primary coil $$i = 5\sin (10 \pi t)$$
Rate of change of current $$\dfrac{di}{dt} = 50 \pi \cos(10\pi t)$$
Maximum rate of change of current
$$\dfrac{di}{dt} \bigg|_{max}= 50 \pi$$
Mutual inductance between the coils $$L = 2\times 10^{-2}$$ H
Thus maximum emf induced $$\mathcal{E}_{max} = L\dfrac{di}{dt}\bigg|_{max}$$
$$\therefore$$
$$\mathcal{E}_{max} = 2\times 10^{-2 }\times 50\pi = \pi $$ volts
Alternating current of peak value $$\left( \dfrac{2}{\pi} \right )$$ ampere flows through the primary coil of the transformer. The coefficient of mutual inductance between primary and secondary coil is 1 henry. The peak e.m.f. induced in secondary coil is
(Frequency of a.c. = 50 Hz)
Report Question
0%
100 V
0%
200 V
0%
300 V
0%
400 V
Explanation
Given : $$I_o = \dfrac{2}{\pi}$$ ampere $$\nu = 50 $$ Hz $$L = 1 H$$
Thus $$w = 2\pi \nu = 2\pi (50) = 100 \pi$$
Alternating current flowing through the coil is given by $$I = I_o \sin wt$$
Differentiating it wr.t. time we get $$\dfrac{dI}{dt} = I_o w \cos wt$$
$$\therefore$$
$$\dfrac{dI}{dt}\bigg|_{max} = I_o w= \dfrac{2}{\pi} \times 100 \pi = 200$$ ampere per second
Peak e.m.f induced $$\mathcal{E} = L\dfrac{dI}{dt}$$
$$\implies$$ $$\mathcal{E} = 1\times 200 = 200$$ V
A conducting rod of mass $$m$$ and length $$l$$ is free to move without friction on two parallel long conducting rails, as shown below. There is a resistance $$R$$ across the rails. In the entire space around, there is a uniform magnetic field $$B$$ normal to the plane of the rod and rails. The rod is given an impulsive velocity $${ v }_{ 0 }$$. Finally, the initial energy $$\dfrac { 1 }{ 2 } m{ v }_{ 0 }^{ 2 }$$
Report Question
0%
Will be converted fully into heat energy in the resistor
0%
Will enable rod to continue to move with velocity $${v}_{0}$$ since the rails are frictionless
0%
Will be converted fully into magnetic energy due to induced current
0%
Will be converted into the work done against the magnetic field
Explanation
Magnetic force equals to $$F=ilB$$ ,will act opposite to the direction of $$v_o$$.The reduce in kinetic energy will be converted into heat energy by resistor.
The figure shows a bar magnet coil. Consider four situations.
(I) Moving the magnet away from the coil.
(II) Moving the coil towards the magnet.
(III) Rotating the coil about the vertical diameter.
(IV) Rotating the coil about its axis.
An emf in the coil will be generated for the following situations.
Report Question
0%
(I) and (II) only
0%
(I), (II) and (IV) only
0%
(I), (II) and (III) only
0%
(I), (II), (III) and (IV) only
Explanation
To generate EMF we should have magnetic flux which changes with time. So when ever coil is moves away or close the flux will vary with time. Also rotating coil about vertical diameter will result in projected area and flux will change but when it is rotated about the axis then area and flux remain same .So the option 1,2 and 3 is correct.
If '$$N$$' is the number of turns in a circular coil then the value of self inductance varies as.
Report Question
0%
$$N^0$$
0%
$$N$$
0%
$$N^2$$
0%
$$N^{-2}$$
Explanation
Self-inductance of the coil is given by the relation
$$L = \dfrac{\mu_o N^2 A}{l}$$
here $$N$$ is the number of turns in the coil, $$A$$ and $$l$$ are the area and length of the coil, respectively.
$$\implies$$ $$L\propto N^2$$
Alternating current is flowing in inductance L and resistance R. The frequency of source is $$\displaystyle\frac{\omega}{2\pi}$$. Which of the following statement is correct.
Report Question
0%
For low frequency the limiting value of impedance is L
0%
For high frequency the limiting value of impedance is $$L\omega$$
0%
For high frequency the limiting value of impedance is R
0%
For low frequency the limiting value of impedance is $$L\omega$$
Explanation
$$\begin{array}{l} \, \, As\, frequency\, approaches\, zero\, or\, DC,\, the\, inducators\, reac\tan ce\, would\, decrease\, tozero\, , \\ acting\, like\, a\, short\, circuit.\, this\, means\, inductive\, reac\tan ce\, is\, proportional\, to\, fequency \\ \, \, \, \, \, \, \, \, \, \, \, \, \, so\, ,\, for\, low\, frequency\, the\, { { limimiting } }\, \, value\, of\, impedance\, is\, L,\, and\, \, alternating\, \\ current\, is\, flowing\, in\, inductance\, L\, and\, resistance\, R.\, \, The\, frequency\, of\, source\, is\, \frac { \omega }{ { 2\pi } } . \\ so\, the\, correct\, option\, is\, A. \end{array}$$
A square loop of side length $$a$$ having $$n$$ turns is kept in a horizontal plane. A uniform magnetic field $$B$$ exists in vertical direction as shown in figure. Now, the loop is rotated with constant angular speed $$\omega$$ as shown below.
Which of the following statement is correct?
Report Question
0%
Same emf is induced in both cases (i) and (ii)
0%
Maximum emf is induced in case (i)
0%
Emf induced in case (ii) is more than (i)
0%
No emf induced in case (ii)
Explanation
In the configuration (i), the coil rotates in horizontal plane about an axis passing through the centre and perpendicular to plane of the coil. During this movement of the coil, magnetic flux associated with the coil remains constant. Hence, emf induced is zero.
In the configuration (ii), magnetic flux associated with the coil changes sinusoidally and hence the emf is induced.
In a transformer, coefficient of mutual inductance between primary and secondary coil is 0.2 H. When current changes by 5 Na in the primary, then: the induced era in the secondary will be
Report Question
0%
0.5 V
0%
1 V
0%
1.5 V
0%
2.0 V
Explanation
$$e = M\dfrac{di}{dt} = 0.2\times5 = 1 V$$
In electromagnetic induction, the induced charge in a coil is independent of
Report Question
0%
Time
0%
Change in flux
0%
Resistance in the circuit
0%
None of the above
Explanation
We know that, $$e = \dfrac {d\phi}{dt}$$
But $$e = iR$$
And $$i = \dfrac {dq}{dt}$$
$$\therefore \dfrac {dq}{dt}\cdot R = \dfrac {d\phi}{dt}$$
$$dq = \dfrac {d\phi}{R}$$.
A rod of $$10\ cm$$ length is moving perpendicular to uniform magnetic field of intensity $$5\times 10^{-4}Wb/m^{2}$$. If the acceleration of the rod is $$5m/s^{2}$$, then the rate of increase of induced emf is _____.
Report Question
0%
$$25\times 10^{-4}Vs^{-1}$$
0%
$$2.5\times 10^{-4}Vs^{-1}$$
0%
$$20\times 10^{-4}Vs^{-1}$$
0%
$$2.0\times 10^{-4}Vs^{-1}$$
Explanation
Consider a rod of $$10\ cm$$ length is moving perpendicular to uniform magnetic field as shown below.
The emf induced in the rod
$$\epsilon = vBl$$
Differentiating with respect to time $$t$$
Rate of increase of induced $$emf = \dfrac {d\epsilon}{dt}$$
$$= \dfrac {dv}{dt} Bl = aBl$$
$$= 5\times 5\times 10^{-4}\times 10\times 10^{-2}$$
$$=250\times 10^{-6}$$
$$=2.5\times 10^{-4}Vs^{-1}$$.
Two Circular cells can be arranged in any of the three following situations as shown in figure. Their mutual inductance will be Maximum in which arrangement ?
Report Question
0%
(A)
0%
(B)
0%
(C)
0%
Same in all conditions
Explanation
The mutual inductance between two coils depends on their degree of flux linkage i.e. the fraction of flux linked with one coil which is also linked to the other coil.
Here, the two coils in arrangement (A) are placed with their planes parallel which allows maximum flux.
The flux linked with a circuit is given by $$\phi = t^3 + 3t - 7$$. The graph between time (x-axis) and induced emf (y-axis) will be
Report Question
0%
A straight line through the origin
0%
Straight line with positive intercept
0%
Straight line with negative intercept
0%
Parabola not through origin
Explanation
We given the flux linked with circuit $$\displaystyle \phi = t^3 + 3t-7$$
as, $$\displaystyle e=-\frac{d \phi}{dt}$$
$$\displaystyle =-\frac{d }{dt} (t^3 + 3t -7) = -3(t^2 + 1)$$
Clearly the graph between e and t (along x axis) will be a parabola.
At $$\displaystyle t=e, e=-3 (\neq 0)$$
$$\therefore$$ The parabola would not pass through the origin.
State whether true or false :
The three-phase alternator has three single-phase windings spaced such that the voltage induced in any one phase is displaced by 120 from the other two.
Report Question
0%
True
0%
False
What should be the value of self inductance of an indicator that should be connected to $$220$$V, $$50$$Hz supply so that a maximum current of $$0.9$$A flows through it?
Report Question
0%
$$11$$H
0%
$$2$$H
0%
$$1.1$$H
0%
$$5$$H
Explanation
The emf across the inductor is given by:
$$|e|=\displaystyle\frac{Ldi}{dt}$$
$$220=L\times \frac{0.9}{1/50}$$
$$L=4.88=5$$H.
EMF developed by generator depends upon :
Report Question
0%
size of magnet
0%
length of rotating wire
0%
radius of wire
0%
none of these
Explanation
EMF developed by generator depends upon the length of rotating wire.An electric generator generates emf by changing the number of magnetic flux lines $$\phi$$ passing through a wire coil.When the coil is rotated between the poles of the magnet by cranking the handle,an AC voltage waveform is produced.
$$\varepsilon =-N\dfrac { d\phi }{ dt } $$
$$\varepsilon \alpha N$$ and $$N \alpha$$length of the coil
$$\varepsilon \alpha$$length of the coil.
In A.C generator increasing no. of turns in coil :
Report Question
0%
decreases the EMF
0%
EMF remains same
0%
increases the EMF
0%
EMF becomes zero
Explanation
We know that emf generated in an AC generator is given by
$$E=NBA\omega \cos\omega t$$
Hence,on increasing number of turns of coil, EMF increases.
Answer-(C)
In alternative current generator, AC current reverses its direction :
Report Question
0%
20 times per second
0%
50 times per second
0%
once per second
0%
twice per second
Explanation
The frequency of an AC generator is 50 times per second.
Hence, the current reverses its direction 50 times.
Answer-(B)
Which of the following is wrong statement
Report Question
0%
An emf can be induced between the ends of a straight conductor by moving it through a uniform magnetic field
0%
The self induced emf produced by changing current in a coil always tends to decrease the current
0%
Inserting an iron core in a coil increases its coefficient of self induction
0%
According to Lenz's law, the direction of the induced current is such that it opposes the flux change that causes it
The part of the AC generator that passes the current from the coil to the external circuit is?
Report Question
0%
Field magnet
0%
Split rings
0%
Slip rings
0%
Brushes
Explanation
The three phase AC generator has three equally spaced windings and three output voltages that are $$120^{0}$$ out of phase of one another.
It is based on the principle of faraday law of electromagnetic induction
Field magnet passes the current from the coil to the external circuit in AC generator.
If coil is placed perpendicular to field lines then number of lines passing through coil are :
Report Question
0%
minimum
0%
maximum
0%
zero
0%
may be max. or min.
Explanation
The magnitude of the induced emf in a circuit is proportional to the time rate of magnetic flux linked with the circuit.
Consider a rectangular loop which rotates in a uniform magnetic field, around an axis in the plane of the loop and passing through the centres of opposite sides.
If the area of the loop is A and the field strength is B and the normal to the loop makes an angle phi with the field direction at any instant, the flux through the loop at any time is B A cos phi. The flux is a maximum when the loop is normal to the field $$(\phi =0)$$ and zero when the loop is aligned parallel to the field $$(\phi =90).$$
If the loop rotates with angular velocity w, then we can substitute phi = wt, and we can then write flux at any instant $$= A B cos wt.$$
The induced emf is $$E =\dfrac{ - d (A B cos wt)}{dt}$$
$$E= A B w sin wt.$$
The answer to your question follows from that last expression. The induced emf is a sinusoidal function. It will be a minimum (that is zero), when phi = wt = 0, when the plane of the coil is normal to the field, and it will be maximum when $$\phi = wt = 90$$, or when the plane of the loop is parallel to the field
If given arrangement is moving towards left with speed v, then potential difference between B and D and current in the loop are respectively.
Report Question
0%
BvR and non-zero
0%
2BvR and zero
0%
4BvR and non-zero
0%
4BvR and zero
Explanation
Curve wire moving transnational can be equalize by straight wire jointed to its free end.
Then potential difference between B & d will be =
$$ { V }_{ d }-{ V }_{ b }=\quad BV\times 4R\\ { V }_{ d }{ -V }_{ b }=4BVR$$
The current in the loop will be zero because we know that net emf in any closed loop is zero.
So option (D) is the correct answer.
If circular coil with $$N_{1}$$ turns is changed in to a coil of $$N_{2}$$ turns. What will be the ratio of self inductances in both cases.
Report Question
0%
$$\dfrac {N_{1}}{N_{2}}$$
0%
$$\dfrac {N_{2}}{N_{1}}$$
0%
$$\dfrac {N_{1}^{2}}{N_{2}^{2}}$$
0%
$$\sqrt {\dfrac {N_{1}}{N_{2}}}$$
Explanation
If circular coil with N1 turns is changed in to a coil of N2 turns.
$$L=\dfrac{Nd \phi}{dt} $$
$$L_1=\dfrac{N_1d \phi}{dt} $$
$$L_2=\dfrac{N_2d \phi}{dt} $$
L is in Henries
N is the Number of Turns
Φ is the Magnetic Flux
Ι is in Amperes
$$\dfrac{L_1}{L_2}=\dfrac{N_1}{N_2} $$
A wheel with 10 metallic spokes each 0.5 m long rotated with a speed of 120 rpm in a plane normal to the horizontal component of earth's magnetic field $$B_h$$ at a place. If $$B_h$$ = 0.4 G at the place. What is the induced emf between the axle and the rim of the wheel? (1 G = $$10^4$$ T)
Report Question
0%
0 V
0%
0.628 mV
0%
0.628 $$\mu$$V
0%
62.8 $$\mu$$V
Explanation
Given,
A wheel with $$410$$ metallic spokes.
Length of metallic spoke $$=0.5\,m$$
Angular speed of the wheel $$=120\,rev/min$$
Earth's magnetic field, $$B=0.4\,G$$
The area covered by an angle $$4\theta$$,
$$A=\pi r^2\dfrac{\theta}{2\pi}=\dfrac{r^2}{2}\theta$$
The induces emf,
$$E=\dfrac{d\phi}{dt}=\dfrac{d(BA)}{dt}$$
$$=B\dfrac{r^2}{2}\dfrac{d\phi}{dt}=B\dfrac{r^2}{2}\omega$$
$$v=120\,rev/min=2\,rev/s$$
$$\omega=2\pi r=4\pi rad/s$$
$$r=0.5\,m$$
$$B=0.4\,G=0.4\times10^{-4}\,T$$
Therefore,
Induced emf is given by,
$$E=\dfrac12 Br^2\omega$$
$$E=\dfrac{2\pi\times 2\times 0.4\times 10^{-4}\times (0.5^2)}{2}$$
$$=6.28\times 10^{-5}\,V$$
Each spoke will act as a parallel source of emf.
Hence, the emf will be $$0.628\,mV$$
A conducting rod of Length $$L=0.1$$m is moving with a uniform speed $$v=0.2$$ m/s on conducting rails in a magnetic field $$B=0.5$$T as shown. On one side, the end of the rails is connected to a capacitor of capacitance $$C=20\mu F$$. Then the charges on the capacitor plates are.
Report Question
0%
$$q_A=0=q_B$$
0%
$$q_A=+20\mu C$$ and $$q_B=-20\mu C$$
0%
$$q_A=+0.2\mu C$$ and $$q_B=-0.2\mu C$$
0%
$$q_A=-0.2\mu C$$ and $$q_B=+0.2\mu C$$
Explanation
$$L=0.1m$$ , $$v=0.2 ms^{-1}$$ ,$$B=0.5T$$ ,$$C=20 \mu F$$
$$V=BvL$$
=$$0.5 \times 0.2 \times 0.1$$=$$0.01$$
$$q=cv=20\mu F\times 0.01=0.2\mu C$$
$$q_A=+0.2 \mu C$$
$$q_B=-0.2 \mu C$$
A proton of mass 'm' moving with a speed v ($$< <$$c, velocity of light in vacuum) completes a circular orbit in time 'T' in a uniform magnetic field. If the speed of the proton is increased to $$\sqrt{2}$$v, what will be time needed to complete the circular orbit?
Report Question
0%
$$\sqrt{2}T$$
0%
T
0%
$$\displaystyle\frac{T}{\sqrt{2}}$$
0%
$$\displaystyle\frac{T}{2}$$
Explanation
$$T=\displaystyle\frac{2\pi m}{qB}$$
T is independent of v.
A solenoid is connected to a battery so that a steady current flows through it. If an iron core is inserted into the solenoid, the current will
Report Question
0%
increase
0%
decreases
0%
remain same
0%
first increases then decreases
Explanation
The current will decrease. This is because on inserting an iron core in the
solenoid, the magnetic field increases and hence magnetic flux linked with the solenoid increases. As per Lenzs law, the emf induced in the solenoid will oppose this increase, which can be achieved by a decrease in Current.
A copper rod of length $$l$$ rotates about its end with angular velocity $$\omega$$ in a uniform magnetic field B. The emf developed between the ends of the rod if the field is normal to the plane of rotation is:
Report Question
0%
$$B\omega l^2$$
0%
$$\dfrac{1}{2}B\omega l^2$$
0%
2 $$B\omega l^2$$
0%
$$\dfrac{1}{4}B\omega l^2$$
Explanation
$$dE=B\omega xdx\\E=\int _{ 0 }^{ L }{ B\omega xdx } \\ \quad =B\omega\cfrac{x^2}{2}\\ \quad=\cfrac{1}{2}B \omega l^2$$
The magnetic induction at any point due to a long straight wire carrying a current is
Report Question
0%
Proportional to the distance from the wire
0%
Inversely proportional to the distance from wire
0%
Inversely proportional proportional to the square of the distance from the wire
0%
Does not depend on distance
Explanation
Magnetic induction is given by
$$B=\dfrac{\mu_0 i}{2\pi r}$$
So, $$B \propto \dfrac 1r$$
The mutual inductance $$M_{12}$$ of a coil 1 with respect to coil 2
Report Question
0%
increases when they are brought nearer
0%
depends on the current passing through the coils.
0%
increases when one of them is rotated about an axis.
0%
both (a) and (b) are correct
Explanation
Mutual Induction: Whenever the current passing through a coil or circuit changes, the magnetic flux linked with a neighbouring coil or circuit will also change. Hence an emf will be induced in the neighbouring coil or circuit. This phenomenon is called ‘mutual induction’.
If the two coils $$1$$ and $$2$$ are present with mutual inductance $$M_1$$ and $$M_2$$. Then the mutual inductance of the coil 1 due to 2 increases when they are bought near since, mutual inductance is proportional to the flux passed through the coil.
The mutual induction of $$M_{12}$$ is same as $$M_{21}$$
Two circular coils can be arranged in any of three situations as shown in the figure. Their mutual inductance will be:
Report Question
0%
maximum in situation (i)
0%
maximum in situation (ii)
0%
maximum in situation (iii)
0%
same in all situation
Explanation
Mutual induction $$∝$$ flux linkage between coils
Flux linkage in coil $$(i)$$ is maximum because both coils have same axis.
Flux linkage between coils is 0 because thus axis are perpendicular to each other.
Flux linkage between coils is 0 because their axis are perpendicular to each other .
So, mutual inductance between coils is maximum in case $$(i)$$.
The total charge induced in a conduction loop when it is moved in magnetic field depends on
Report Question
0%
The rate of change of magnetic flux
0%
Initial magnetic flux only
0%
The total change in magnetic flux
0%
Final magnetic flux only
0:0:1
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
0
Answered
0
Not Answered
0
Not Visited
Correct : 0
Incorrect : 0
Report Question
×
What's an issue?
Question is wrong
Answer is wrong
Other Reason
Want to elaborate a bit more? (optional)
Practice Class 12 Medical Physics Quiz Questions and Answers
<
>
Support mcqexams.com by disabling your adblocker.
×
Please disable the adBlock and continue.
Thank you.
Reload page
