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Electromagnetic Induction

NEET Physics · 87 questions · Page 5 of 9 · Click an option or "Show Solution" to reveal answer

Q41
At the center of a fixed large circular coil of radius R, a much smaller circular coil of radius r is placed. The two coils are concentric and are in the same plane. The larger coil carries a current I. The smaller coil is set to rotate with a constant angular velocity ω\omega about an axis along their common diameter. Calculate the emf induced in their smaller coil after a time t of its start of rotation.
A μoI2 R{{{\mu _o}{\rm I}} \over {2\,R}} ω\omega π\pi r2 sinω\omega t
B μoI4 R{{{\mu _o}{\rm I}} \over {4\,R}} ω\omega π\pi r2 sinω\omega t
C μoI4 R{{{\mu _o}{\rm I}} \over {4\,R}} ω\omega r2 sinω\omega t
D μoI2 R{{{\mu _o}{\rm I}} \over {2\,R}} ω\omega r2 sinω\omega t
Correct Answer
Option A
Solution

We know that electric flux

ϕ=B→ . A→\phi = \overrightarrow B \,.\,\overrightarrow A
⇒ϕ=BAcos⁡ωt\Rightarrow \phi = BA\cos \omega t

Now,

B=μ02IRB = {{{\mu _0}} \over 2}{I \over R}

is magnetic field due to circular coil of radius R and

A=πr2A = \pi {r^2}

is area of circular coil of radius r. Therefore,

ϕ=μ02IRπr2cos⁡ωt\phi = {{{\mu _0}} \over 2}{I \over R}\pi {r^2}\cos \omega t

Now induced emf

ε=−dϕdt=−ddt(μ02IRπr2cos⁡ωt)\varepsilon = {{ - d\phi } \over {dt}} = {{ - d} \over {dt}}\left( {{{{\mu _0}} \over 2}{I \over R}\pi {r^2}\cos \omega t} \right)
⇒ε=μ02IRπr2sin⁡ωt\Rightarrow \varepsilon = {{{\mu _0}} \over 2}{I \over R}\pi {r^2}\sin \omega t
Q42
Two coils 'P' and 'Q' are separated by some distance. When a current of 3 A flows through coil 'P', a magnetic flux of 10–3 Wb passes through 'Q'. No current is passed through 'Q'. When no current passes through 'P' and a current of 2 A passes through 'Q', the flux through 'P' is :-
A 3.67 × 10–4 Wb
B 3.67 × 10–3 Wb
C 6.67 × 10–4 Wb
D 6.67 × 10–3 Wb
Correct Answer
Option C
Solution

Mutual induction

ϕq=MIp{\phi _q} = M{I_p}

10–3 = M(3)

⇒M=13×10−3\Rightarrow M = {1 \over 3} \times {10^{ - 3}}
ϕp=MIq{\phi _p} = M{I_q}
⇒ϕp=6.67×10−4Wb\Rightarrow {\phi _p} = 6.67{\rm{ }} \times {\rm{ }}{10^{-4}}{\rm{ }}Wb
Q43
The total number of turns and cross-section area in a solenoid is fixed. However, its length L is varied by adjusting the separation between windings. The inductance of solenoid will be proportional to :
A 1/L2
B 1/L
C L
D L2
Correct Answer
Option B
Solution

ϕ\phi = NBA = LI N μ\mu0 nIπ\piR2 = LI

Nμ0NllπR2=LIN{\mu _0}{N \over l}l\pi {R^2} = LI

N and R constant Self inductance (L)

∝1l∝1length\propto {1 \over l} \propto {1 \over {length}}
Q44
An inductor coil stores 64 J of magnetic field energy and dissipates energy at the rate of 640 W when a current of 8A is passed through it. If this coil is joined across an ideal battery, find the time constant of the circuit in seconds :
A 0.4
B 0.8
C 0.125
D 0.2
Correct Answer
Option D
Solution
U=12Li2=64⇒L=2U = {1 \over 2}L{i^2} = 64 \Rightarrow L = 2
i2R=640{i^2}R = 640
R=640(8)2=10R = {{640} \over {{{(8)}^2}}} = 10
τ=LR=15=0.2\tau = {L \over R} = {1 \over 5} = 0.2
Q45
The electric current in a circular coil of 2 turns produces a magnetic induction B1 at its centre. The coil is unwound and in rewound into a circular coil of 5 tuns and the same current produces a magnetic induction B2 at its centre. The ratio of B2B1{{{B_2}} \over {{B_1}}} is
A 52{5 \over 2}
B 254{25 \over 4}
C 54{5 \over 4}
D 252{25 \over 2}
Correct Answer
Option B
Solution
B=nμ0I2RB = {{n{\mu _0}I} \over {2R}}
B1=2μ0I2R1{B_1} = {{2{\mu _0}I} \over {2{R_1}}}
B2=5μ0I2R2{B_2} = {{5{\mu _0}I} \over {2{R_2}}}
R2=2R15{R_2} = {{2{R_1}} \over 5}
⇒B2B1=52×R1R2=254\Rightarrow {{{B_2}} \over {{B_1}}} = {5 \over 2} \times {{{R_1}} \over {{R_2}}} = {{25} \over 4}
Q46
A coil is placed in magnetic field such that plane of coil is perpendicular to the direction of magnetic field. The magnetic flux through a coil can be changed : A. By changing the magnitude of the magnetic field within the coil. B. By changing the area of coil within the magnetic field. C. By changing the angle between the direction of magnetic field and the plane of the coil. D. By reversing the magnetic field direction abruptly without changing its magnitude. Choose the most appropriate answer from the options given below :
A A, B and D only
B A, B and C only
C A and B only
D A and C only
Correct Answer
Option B
Solution

The magnitude of magnetic flux is given by :

Φ=BAcos⁡θ\Phi = BA \cos \theta

where

BB

is the magnitude of the magnetic field,

AA

is the area of the coil, and θ\theta is the angle between the normal to the area and the direction of the magnetic field.

The correct option is A, B, and C only.

A.

The magnetic flux through a coil can be changed by changing the magnitude of the magnetic field within the coil.

A stronger magnetic field will increase the magnetic flux, while a weaker magnetic field will decrease the magnetic flux.

B.

The magnetic flux through a coil can also be changed by changing the area of the coil within the magnetic field.

A larger area of the coil will result in a greater magnetic flux, while a smaller area will result in a smaller magnetic flux.

C.

The magnetic flux through a coil can also be changed by changing the angle between the direction of the magnetic field and the plane of the coil.

When the angle is perpendicular to the plane of the coil, the magnetic flux is at its maximum.

When the angle is parallel to the plane of the coil, the magnetic flux is zero.

D.

Reversing the magnetic field direction abruptly without changing its magnitude will not change the magnetic flux through the coil.

Magnetic flux is proportional to the dot product of the magnetic field and the area vector of the coil.

If the magnitude of the magnetic field remains the same and the direction is reversed, the dot product remains the same and the magnetic flux remains unchanged.

Q47
A square loop of area 25 cm2^2 has a resistance of 10 Ω\Omega. The loop is placed in uniform magnetic field of magnitude 40.0 T. The plane of loop is perpendicular to the magnetic field. The work done in pulling the loop out of the magnetic field slowly and uniformly in 1.0 sec, will be
A 1.0×10−3 J\mathrm{1.0\times10^{-3}~J}
B 5×10−3 J\mathrm{5\times10^{-3}~J}
C 2.5×10−3 J\mathrm{2.5\times10^{-3}~J}
D 1.0×10−4 J\mathrm{1.0\times10^{-4}~J}
Correct Answer
Option A
Solution

From energy conservation Work done to pull the loop out = Energy is lost in the resistance Emf in the loop

=dϕdt=B×At=40×25×10−41s=0.1 V= {{d\phi } \over {dt}} = {{B \times A} \over t} = {{40 \times 25 \times {{10}^{ - 4}}} \over {1s}} = 0.1\,V

Energy lost

=emf2R=(0.1)210=10−3 J= {{em{f^2}} \over R} = {{{{(0.1)}^2}} \over {10}} = {10^{ - 3}}\,J
Q48
Given below are two statements: one is labelled as Assertion A\mathbf{A} and the other is labelled as Reason R\mathbf{R} Assertion A: A bar magnet dropped through a metallic cylindrical pipe takes more time to come down compared to a non-magnetic bar with same geometry and mass. Reason R: For the magnetic bar, Eddy currents are produced in the metallic pipe which oppose the motion of the magnetic bar. In the light of the above statements, choose the correct answer from the options given below
A A is true but R is false
B A is false but R is true
C Both A and R are true but R is NOT the correct explanation of A
D Both A and R are true and R is the correct explanation of A
Correct Answer
Option D
Solution

Assertion A is true because a bar magnet dropped through a metallic cylindrical pipe takes more time to come down compared to a non-magnetic bar with the same geometry and mass.

This is due to the effect of the magnet's magnetic field on the metallic pipe.

Reason R is also true because when the magnetic bar moves through the metallic pipe, it induces a changing magnetic field in the pipe.

This changing magnetic field, in turn, induces Eddy currents in the pipe.

According to Lenz's law, these Eddy currents produce their own magnetic field, which opposes the motion of the magnetic bar, causing it to fall more slowly through the pipe.

Since both Assertion A and Reason R are true and R provides the correct explanation for A, the correct answer is Both A and R are true and R is the correct explanation of A.

Q49
An emf of 0.08 V0.08 \mathrm{~V} is induced in a metal rod of length 10 cm10 \mathrm{~cm} held normal to a uniform magnetic field of 0.4 T0.4 \mathrm{~T}, when moves with a velocity of:
A 20 ms−120 \mathrm{~ms}^{-1}
B 2 ms−12 \mathrm{~ms}^{-1}
C 3.2 ms−13.2 \mathrm{~ms}^{-1}
D 0.5 ms−10.5 \mathrm{~ms}^{-1}
Correct Answer
Option B
Solution

The emf induced in a rod moving through a magnetic field is given by Faraday's law of electromagnetic induction, specifically, in the form of motional emf, which states that: emf=B⋅L⋅v \text{emf} = B \cdot L \cdot v where: (B) is the magnetic field strength, (L) is the length of the rod, and (v) is the velocity of the rod.

In this case, we are given the emf, (B), and (L), and we need to solve for (v).

Rearranging the equation gives: v=emfB⋅L v = \dfrac{\text{emf}}{B \cdot L} Substituting the given values: v=0.08 V0.4 T×0.1 m=2 m/s v = \dfrac{0.08 \, \text{V}}{0.4 \, \text{T} \times0.1 \, \text{m}} = 2 \, \text{m/s}

Q50
Consider I1 and I2 are the currents flowing simultaneously in two nearby coils 1 & 2, respectively. If L1 = self inductance of coil 1, M12 = mutual inductance of coil 1 with respect to coil 2, then the value of induced emf in coil 1 will be :
A e1 = -L1dI2dt\dfrac{dI_2}{dt} - M12dI1dt\dfrac{dI_1}{dt}
B e1 = -L1dI1dt\dfrac{dI_1}{dt} + M12dI2dt\dfrac{dI_2}{dt}
C e1 = -L1dI1dt\dfrac{dI_1}{dt} - M12dI1dt\dfrac{dI_1}{dt}
D e1 = -L1dI1dt\dfrac{dI_1}{dt} - M12dI2dt\dfrac{dI_2}{dt}
Correct Answer
Option D
Solution
ϕ1=L1I1+M12I2ε1=−dϕ1dt=−L1dI1dt−M12dIdt\begin{aligned} & \phi_1=\mathrm{L}_1 \mathrm{I}_1+\mathrm{M}_{12} \mathrm{I}_2 \\ & \varepsilon_1=-\frac{\mathrm{d} \phi_1}{\mathrm{dt}}=-\mathrm{L}_1 \frac{\mathrm{dI}_1}{\mathrm{dt}}-\mathrm{M}_{12} \frac{\mathrm{dI}}{\mathrm{dt}} \end{aligned}
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