Electromagnetic Induction — Page 4 | JEE Physics

Q31

Spherical insulating ball and a spherical metallic ball of same size and mass are dropped from the same height. Choose the correct statement out of the following {Assume negligible air friction}

A Metal ball will reach the earth's surface earlier than the insulating ball

B Both will reach the earth's surface simultaneously.

C Insulating ball will reach the earth's surface earlier than the metal ball

D Time taken by them to reach the earth's surface will be independent of the properties of their materials

Correct Answer

Option C

Solution

The correct answer is option C: the insulating ball will reach the earth's surface earlier than the metal ball.

When the two balls are dropped from the same height, they will experience the same gravitational force, which will cause them to accelerate downwards.

However, According to Faraday’s law of electromagnetic induction motion of metal is opposed by earth magnetic field.

Which will slightly reduce its acceleration.

On the other hand, the insulating ball will not experience any electromagnetic induction.

Therefore, it will accelerate downwards at a slightly faster rate than the metal ball.

As a result, the insulating ball will reach the earth's surface earlier than the metal ball.

So, option C is the correct statement.

Q32

A metallic conductor of length 1 m rotates in a vertical plane parallel to east-west direction about one of its end with angular velocity 5 rad s

-

1. If the horizontal component of earth's magnetic field is 0.2

\times

10

-

4 T, then emf induced between the two ends of the conductor is :

A 5

\mu

V

B 50

\mu

V

C 5 mV

D 50 mv

Correct Answer

Option B

Solution

Emf = {1 \over 2}B\omega {l^2}

= {1 \over 2} \times 0.2 \times {10^{ - 4}} \times 5 \times {1^2}

V = 0.5 $\times$ 10 $-$ 4 V = 50 $\mu$ V

Q33

Certain galvanometers have a fixed core made of non magnetic metallic material. The function of this metallic material is

A to bring the coil to rest quickly

B to produce large deflecting torque on the coil

C to oscillate the coil in magnetic field for longer period of time

D to make the magnetic field radial

Correct Answer

Option A

Solution

The function of the non-magnetic metallic core in a galvanometer is to provide a path for the induced current generated by the moving coil in the magnetic field.

This induced current opposes the motion of the coil and brings it to rest quickly due to the effect known as eddy current damping.

When the coil swings past its equilibrium position, a change in magnetic flux occurs.

According to Faraday's law of electromagnetic induction, this change in magnetic flux generates an induced current, known as an eddy current.

The eddy current creates its own magnetic field which opposes the original change in flux, causing the coil to quickly come to rest.

Q34

The induced emf can be produced in a coil by A. moving the coil with uniform speed inside uniform magnetic field B. moving the coil with non uniform speed inside uniform magnetic field C. rotating the coil inside the uniform magnetic field D. changing the area of the coil inside the uniform magnetic field Choose the correct answer from the options given below:

A A and C only

B C and D only

C B and D only

D B and C only

Correct Answer

Option B

Solution

If the coil is simply moved at uniform or non-uniform speed in a uniform magnetic field without changing the orientation of the coil or the area of the coil enclosed by the magnetic field, the magnetic flux through the coil does not change, and no emf is induced according to Faraday's Law of electromagnetic induction.

Q35

A coil is suspended in a uniform magnetic field, with the plane of the coil parallel to the magnetic lines of force. When a current is passed through the coil it starts oscillating; It is very difficult to stop. But if an aluminium plate is placed near to the coil, it stops. This is due to :

A development of air current when the plate is placed

B induction of electrical charge on the plate

C shielding of magnetic lines of force as aluminium is a para-magnetic material.

D electromagnetic induction in the aluminium plate giving rise to electromagnetic damping.

Correct Answer

Option D

Solution

Because of the Lenz's law of conservation of energy.

Q36

Regarding self-inductance: A. The self-inductance of the coil depends on its geometry. B. Self-inductance does not depend on the permeability of the medium. C. Self-induced e.m.f. opposes any change in the current in a circuit. D. Self-inductance is electromagnetic analogue of mass in mechanics. E. Work needs to be done against self-induced e.m.f. in establishing the current. Choose the correct answer from the options given below:

A A, B, C, E only

B A, B, C, D only

C A, C, D, E only

D B, C, D, E only

Correct Answer

Option C

Solution

Let's analyze each statement:

\textbf{A. The self-inductance of the coil depends on its geometry.}

This is true because the self-inductance of a coil is given by formulas that include its geometrical parameters (number of turns, cross-sectional area, length, etc.).

For example, for a solenoid,

L = \mu_0 \mu_r \frac{N^2 A}{l},

where

N

is the number of turns,

A

is the cross-sectional area, and

l

is the length.

\textbf{B. Self-inductance does not depend on the permeability of the medium.}

This is false. The permeability of the medium (represented by

\mu = \mu_0 \mu_r

) directly influences the inductance, as seen in the formula above.

So, any change in the medium’s permeability will affect the inductance.

\textbf{C. Self-induced e.m.f. opposes any change in the current in a circuit.}

This is true, and it is a direct consequence of Lenz's law.

The induced electromotive force (e.m.f.) always acts in a direction such that it opposes the change in current that produced it.

\textbf{D. Self-inductance is the electromagnetic analogue of mass in mechanics.}

This is true.

Just as mass resists changes in velocity (inertia), inductance resists changes in current, which is why it is often compared to the inertial mass in mechanical systems.

\textbf{E. Work needs to be done against self-induced e.m.f. in establishing the current.}

This is true because, when you try to change the current, you must do work to overcome the opposing self-induced e.m.f., storing energy in the magnetic field of the inductor.

Based on the above reasoning, the true statements are A, C, D, and E.

Thus, the correct answer is: Option C A, C, D, E only

Q37

The magnetic flux through a coil perpendicular to its plane is varying according to the relation

\phi = (5{t^3} + 4{t^2} + 2t - 5)

Weber. If the resistance of the coil is 5 ohm, then the induced current through the coil at t = 2 s will be,

A 15.6 A

B 16.6 A

C 17.6 A

D 18.6 A

Correct Answer

Option A

Solution

$\phi=5 \mathrm{t}^3+4 \mathrm{t}^2+2 \mathrm{t}-5$ $|\mathrm{e}|=\dfrac{\mathrm{d} \phi}{\mathrm{dt}}=15 \mathrm{t}^2+8 \mathrm{t}+2$ At $\mathrm{t}=2,|\mathrm{e}|=15 \times 2^2+8 \times 2+2$ $\Rightarrow \mathrm{e}=78 \mathrm{~V}$ $\Rightarrow \mathrm{I}=\dfrac{\mathrm{e}}{\mathrm{R}}=\dfrac{78}{5}=15.60$

Q38

A uniform magnetic field B exists in a direction perpendicular to the plane of a square loop made of a metal wire. The wire has a diameter of 4 mm and a total length of 30 cm. The magnetic field changes with time at a steady rate

{{dB} \over {dt}}

= 0.032 Ts–1. The induced current in the loop is close to (Resistivity of the metal wire is 1.23

\times

10–8

\Omega

m)

A 0.53 A

B 0.43 A

C 0.34 A

D 0.61 A

Correct Answer

Option D

Solution

We know,

\phi = BA

Also,

E = {{d\phi } \over {dt}} = {{AdB} \over {dt}}

E = {l^2}{{dB} \over {dt}}

i = {E \over R}

=

{{{l^2}{{dB} \over {dt}}} \over {{{\rho l} \over A}}}

= {{{l^2}} \over {pl}}{{dB} \over {dt}}A

=

{{{{l^2}\pi {R^2}} \over {\rho l}}{{dB} \over {dt}}}

$\therefore$

i = {{30} \over 4} \times {{30} \over 4} \times {{{{10}^{ - 4}} \times 0.032 \times 4 \times {{10}^{ - 6}} \times \pi } \over {1.23 \times {{10}^{ - 8}} \times 30 \times {{10}^{ - 2}} \times {{10}^3}}}

$\Rightarrow$

i = {{240 \times \pi \times {{10}^{ - 10}}} \over {1.23 \times {{10}^{ - 7}}}}

$\Rightarrow$

i = {{240 \times 3.14 \times {{10}^{ - 3}}} \over {1.23}}

= {{753.6} \over {1.23}} \times {10^{ - 3}}

$\Rightarrow$

i = 612.68 \times {10^{ - 3}} = 0.61A

Q39

A planar loop of wire rotates in a uniform magnetic field. Initially at t = 0, the plane of the loop is perpendicular to the magnetic field. If it rotates with a period of 10 s about an axis in its plane then the magnitude of induced emf will be maximum and minimum, respectively at :

A 2.5 s and 7.5 s

B 5.0 s and 10.0 s

C 5.0 s and 7.5 s

D 2.5 s and 5.0 s

Correct Answer

Option D

Solution

Flux $\phi$ =

\overrightarrow B .\overrightarrow A

= BAcos $\omega$ t Induced emf = e =

- {{d\phi } \over {dt}}

= -BA $\omega$ (-)sin $\omega$ t = BA $\omega$ sin $\omega$ t e will be maximum at $\omega$ t =

{\pi \over 2}

,

{{3\pi } \over 2}

$\Rightarrow$

{{2\pi } \over T}t

=

{\pi \over 2}

,

{{3\pi } \over 2}

$\Rightarrow$ t =

{T \over 4}

or

{3T \over 4}

i.e. 2.5 s or 7.5 s. For induced emf to be minimum i.e zero.

{{2\pi t} \over T}

= n $\pi$ $\Rightarrow$ t =

n{\pi \over 2}

$\Rightarrow$ Induced emf is zero at t = 5 s, 10 s

Q40

A conducting metal circular-wire-loop of radius r is placed perpendicular to a magnetic field which varies with time as B = B0e

{^{{{ - t} \over r}}}

, where B0 and

\tau

are constants, at time t = 0. If the resistance of the loop is R then the heat generated in the loop after a long time (t

\to

\infty

) is :

A

{{{\pi ^2}{r^4}B_0^4} \over {2\tau R}}

B

{{{\pi ^2}{r^4}B_0^2} \over {2\tau R}}

C

{{{\pi ^2}{r^4}B_0^2R} \over \tau }

D

{{{\pi ^2}{r^4}B_0^2} \over {\tau R}}

Correct Answer

Option B

Solution

Given, B = B0e

^{ - {t \over \tau }}

Area of the circular loop, A = $\pi$ r2 $\therefore$ Flux $\phi$ = BA = $\pi$ r2 B0 e

^{ - {t \over \tau }}

Induced emf in the loop,

\varepsilon

= $-$

{{d\phi } \over {dt}}

= $\pi$ r2B0

{1 \over \tau }

e

^{ - {t \over \tau }}

Heat generated =

\int\limits_0^ \propto {{i^2}R\,dt}

=

\int\limits_0^ \propto {{{{\varepsilon ^2}} \over R}} \,dt

=

{1 \over R}{{{\pi ^2}{r^4}B_0^2} \over {{\tau ^2}}}\int\limits_0^ \propto {{e^{ - {{2t} \over \tau }}}} \,dt

=

{{{\pi ^2}{r^4}B_0^2} \over {{\tau ^2}R}} \times {1 \over {\left( { - {2 \over \tau }} \right)}}\left[ {{e^{ - {{2t} \over \tau }}}} \right]_0^ \propto

=

{{ - {\pi ^2}{r^4}B_0^2} \over {2{\tau ^2}R}} \times \tau \left( {0 - 1} \right)

=

{{{\pi ^2}{r^4}B_0^2} \over {2\tau R}}