Electromagnetic Induction — Page 5 | JEE Physics

Q41

The time taken for the magnetic energy to reach 25% of its maximum value, when a solenoid of resistance R, inductance L is connected to a battery, is :

A infinite

B

{L \over R}

ln10

C

{L \over R}

ln2

D

{L \over R}

ln5

Correct Answer

Option C

Solution

Magnetic energy, U =

{1 \over 2}

LI

_0^2

Given : U = 25% of U0. $\Rightarrow$

{1 \over 2}L{I^2} = {1 \over 4} \times {1 \over 2}LI_0^2

\Rightarrow {I^2} = {{I_0^2} \over 4} \Rightarrow I = {{{I_0}} \over 2}

$\therefore$

I = {I_0}(1 - {e^{ - t/\tau }})

\Rightarrow {{{I_0}} \over 2} = {I_0}(1 - {e^{ - t/\tau }})

\Rightarrow {1 \over 2} = {e^{ - t/\tau }}

\Rightarrow {e^{t/\tau }} = 2

\Rightarrow t = \tau \ln 2

\Rightarrow t = {L \over R}\ln 2

Q42

There are two long co-axial solenoids of same length

l

. The inner and outer coils have radii r1 and r2 and number of turns per unit length n1 and n2, respectively. The ratio of mutual inductance to the self - inductance of the inner-coil is :

A

{{{n_2}} \over {{n_1}}}.{{{r_2}^2} \over {{r_1}^2}}

B

{{{n_2}} \over {{n_1}}}

C

{{{n_1}} \over {{n_2}}}

D

{{{n_2}} \over {{n_1}}}.{{{r_1}} \over {{r_2}}}

Correct Answer

Option B

Solution

M = {\mu _0}\,{n_1}\,{n_2}\,\pi r_1^2

L = {\mu _0}\,n_1^2\,\pi r_1^2

\Rightarrow \,\,{M \over L} = {{{n_2}} \over {{n_1}}}

Q43

A coil is placed in a time varying magnetic field. If the number of turns in the coil were to be halved and the radius of wire doubled, the electrical power dissipated due to the current induced in the coil would be : (Assume the coil to be short circuited.)

A Halved

B Quadrupled

C The same

D Doubled

Correct Answer

Option D

Solution

The electrical power dissipated due to the current induced in a coil placed in a time-varying magnetic field can be determined by considering Faraday's Law of Induction and the resistance of the coil.

Faraday's Law of Induction The induced EMF ( $\mathcal{E}$ ) in a coil with $N$ turns experiencing a time-varying magnetic flux ( $\Phi_B$ ) is given by:

\mathcal{E} = -N \frac{d\Phi_B}{dt}

Resistance of the Coil The resistance ( $R$ ) of a wire depends on its length ( $L$ ) and cross-sectional area ( $A$ ) as well as the resistivity ( $\rho$ ) of the material:

R = \frac{\rho L}{A}

For a coil of radius $r$ and with wire of radius $a$ , assuming the wire is wound tightly with a length approximated by the circumference of the coil multiplied by the number of turns, we have:

L \approx 2\pi r N

And the cross-sectional area of the wire is given by:

A = \pi a^2

Thus, the resistance becomes:

R \approx \frac{\rho \cdot 2\pi r N}{\pi a^2} = \frac{2\rho r N}{a^2}

Power Dissipated The electrical power ( $P$ ) dissipated in the coil is related to the induced current ( $I$ ) and the resistance ( $R$ ):

P = I^2 R

Using Ohm's Law, the induced current $I$ can be expressed as:

I = \frac{\mathcal{E}}{R}

Substituting the expressions for $\mathcal{E}$ and $R$ :

I = \frac{N \frac{d\Phi_B}{dt}}{\frac{2\rho r N}{a^2}} = \frac{a^2}{2\rho r} \cdot \frac{d\Phi_B}{dt}

Hence, the power dissipated:

P = \left(\frac{a^2}{2\rho r} \cdot \frac{d\Phi_B}{dt}\right)^2 \cdot \frac{2\rho r N}{a^2}

P = \frac{a^4}{4\rho^2 r^2} \left( \frac{d\Phi_B}{dt} \right)^2 \cdot \frac{2\rho r N}{a^2}

P = \frac{a^2}{2\rho r} \cdot N \left( \frac{d\Phi_B}{dt} \right)^2

Case when the number of turns is halved and the radius of the wire is doubled Halving the number of turns:

N' = \frac{N}{2}

Doubling the radius of the wire:

a' = 2a

Substituting these into the power formula:

P' = \frac{(2a)^2}{2\rho r} \cdot \frac{N}{2} \left( \frac{d\Phi_B}{dt} \right)^2

P' = \frac{4a^2}{2\rho r} \cdot \frac{N}{2} \left( \frac{d\Phi_B}{dt} \right)^2

P' = \frac{4a^2}{2\rho r} \cdot \frac{N}{2} \left( \frac{d\Phi_B}{dt} \right)^2

P' = 2 \left( \frac{a^2}{2\rho r} \cdot N \left( \frac{d\Phi_B}{dt} \right)^2 \right)

P' = 2P

Thus, the electrical power dissipated in the coil would be: Option D Doubled

Q44

The flux linked with a coil at any instant

't'

is given by

\phi = 10{t^2} - 50t + 250

The induced

emf

at

t=3s

is

A

-190

V

B

-10

V

C

10

V

D

190

V

Correct Answer

Option B

Solution

\phi = 10{t^2} - 50t + 250

e = - {{d\phi } \over {dt}} = - \left( {20t - 50} \right)

{e_{t = 3}} = - 10\,V

Q45

A copper wire is wound on a wooden frame, whose shape is that of an equilateral triangle. If the linear dimension of each side of the frame is increased by a factor of 3, keeping the number of turns of the coil per unit length of the frame the same, then the self inductance of the coil:

A decreases by a factor of

9\sqrt 3

B increases by a factor of 27

C decreases by a factor of 9

D increases by a factor of 3

Correct Answer

Option D

Solution

Total length L will remain constant L = (3a) N (N = total turns) and length of winding = (d) N (d = diameter of wire) self inductance = $\mu$ 0n2A

\ell

= $\mu$ 0n2

\left( {{{\sqrt 3 {a^2}} \over 4}} \right)

dN $\propto$ a2 N $\propto$ a So self inductance will become 3 times

Q46

In a coil, the current changes from

-2 \mathrm{~A}

to

+2 \mathrm{~A}

in

0.2 \mathrm{~s}

and induces an emf of

0.1 \mathrm{~V}

. The self inductance of the coil is :

A 4 mH

B 2.5 mH

C 1 mH

D 5 mH

Correct Answer

Option D

Solution

To find the self-inductance of the coil, we can use the formula for induced electromotive force (emf), which is given by Faraday's law of electromagnetic induction as it applies to self-induction:

\text{emf} = - L \frac{\Delta I}{\Delta t}

Where: $\text{emf}$ = induced voltage in volts (V) $L$ = self-inductance of the coil in henries (H) $\Delta I$ = change in current in amperes (A) $\Delta t$ = time interval in seconds (s) over which the current change occurs Here, the problem gives us the following data: $\text{emf} = 0.1 \, \text{V}$ $\Delta I = 2 \, \text{A} - (-2 \, \text{A}) = 4 \, \text{A}$ $\Delta t = 0.2 \, \text{s}$ Substituting the given values into the formula, we get:

0.1 = -L \frac{4}{0.2}

Solving for $L$ , the equation becomes:

0.1 = -L \cdot 20

Therefore:

L = - \frac{0.1}{20}

Calculating the value of $L$ :

L = -0.005 \, \text{H}

Or, expressing $L$ in millihenries (mH):

L = -5 \, \text{mH}

However, considering the absolute value (since inductance is a magnitude and cannot be negative in this context):

L = 5 \, \text{mH}

Thus, the self-inductance of the coil is 5 mH, which corresponds to Option D.

Q47

A conducting circular loop made of a thin wire, has area 3.5

\times

10

-

3 m2 and resistance 10

\Omega

. It is placed perpendicular to a time dependent magnetic field B(t) = (0.4T)sin(50

\pi

t). The field is uniform in space. Then the net charge flowing through the loop during t = 0 s and t = 10 ms is close to :

A 0.14 mC

B 0.7 mC

C 0.21 mC

D 0.6 mC

Correct Answer

Option A

Solution

At t = 0 s B(0) = 0.4 sin (0) = 0 and at t = 10 ms B(10) = 0.4 sin (50 $\pi$ $\times$ 10 $\times$ 10-3) = 0.4 sin

\left( {{\pi \over 2}} \right)

= 0.4 As q =

{{\Delta \phi } \over R}

=

{{A\left[ {B\left( {10} \right) - B\left( 0 \right)} \right]} \over {10}}

=

{{3.5 \times {{10}^{ - 3}}\left[ {0.4 - 0} \right]} \over {10}}

= 0.14 mC

Q48

A solid metal cube of edge length 2 cm is moving in a positive y-direction at a constant speed of 6 m/s. There is a uniform magnetic field of 0.1 T in the positive z-direction. The potential difference between the two faces of the cube perpendicular to the x-axis, is -

A 2mV

B 12 mV

C 6 mV

D 1 mV

Correct Answer

Option B

Solution

We can apply Faraday's law of electromagnetic induction to solve this problem.

Faraday's law states that the induced electromotive force (emf) in any closed circuit is equal to the rate of change of the magnetic flux through the circuit.

The cube is moving through a magnetic field, so it's behaving like a conductor moving through a magnetic field.

The induced emf or voltage can be calculated by using the formula: emf = B $\times$ v $\times$ d Where: B is the magnetic field strength, v is the velocity of the conductor, and d is the length of the conductor perpendicular to the direction of motion and magnetic field.

In this case, the cube is moving in the y-direction and the magnetic field is in the z-direction.

So, the faces perpendicular to the x-axis are involved.

The length of the conductor (d) perpendicular to the motion and magnetic field is the edge length of the cube, which is 2 cm or 0.02 m.

So, plugging the given values into the formula: emf = 0.1 T $\times$ 6 m/s $\times$ 0.02 m = 0.012 V = 12 mV So, the potential difference between the two faces of the cube perpendicular to the x-axis is 12 mV.

Therefore, Option B is correct.

Q49

A rectangular loop of length

2.5 \mathrm{~m}

and width

2 \mathrm{~m}

is placed at

60^{\circ}

to a magnetic field of

4 \mathrm{~T}

. The loop is removed from the field in

10 \mathrm{~sec}

. The average emf induced in the loop during this time is

A

-2 \mathrm{~V}

B

+2 \mathrm{~V}

C

+1 \mathrm{~V}

D

-1 \mathrm{~V}

Correct Answer

Option C

Solution

According to Faraday's Law of Electromagnetic Induction, the induced emf in a closed circuit is equal to the negative rate of change of magnetic flux through the circuit.

Mathematically, it is expressed as:

\varepsilon = -\frac{d\Phi}{dt}

Where $\varepsilon$ is the induced emf, and $\Phi$ is the magnetic flux.

To find the magnetic flux $\Phi$ through the rectangular loop, we use the formula:

\Phi = B \cdot A \cdot \cos(\theta)

Where: $B$ is the magnetic field strength, $A$ is the area of the loop, and $\theta$ is the angle between the magnetic field lines and the normal (perpendicular) to the plane of the loop.

The area $A$ of the rectangular loop is:

A = \text{length} \times \text{width} = 2.5 \mathrm{~m} \times 2 \mathrm{~m} = 5 \mathrm{~m}^2

Given the angle $\theta = 60^\circ$ , we can calculate the initial magnetic flux $\Phi_{initial}$ :

\Phi_{initial} = B \cdot A \cdot \cos(60^\circ)

\Phi_{initial} = 4 \mathrm{~T} \cdot 5 \mathrm{~m}^2 \cdot \cos(60^\circ)

\Phi_{initial} = 4 \cdot 5 \cdot \frac{1}{2} = 10 \mathrm{~Wb}

(Since $\cos(60^\circ) = \dfrac{1}{2}$ ) When the loop is removed from the magnetic field, the final magnetic flux $\Phi_{final}$ is zero, because the loop is no longer within the magnetic field.

Thus, the change in magnetic flux $\Delta\Phi$ is:

\Delta\Phi = \Phi_{final} - \Phi_{initial} = 0 - 10 \mathrm{~Wb} = -10 \mathrm{~Wb}

The loop is removed from the field in $t = 10$ seconds, so the rate of change of magnetic flux is:

\frac{d\Phi}{dt} = \frac{\Delta\Phi}{\Delta t} = \frac{-10 \mathrm{~Wb}}{10 \mathrm{~s}} = -1 \mathrm{~Wb/s}

Now we can find the average induced emf $\varepsilon$ :

\varepsilon = -\frac{d\Phi}{dt}

\varepsilon = -(-1 \mathrm{~Wb/s})

\varepsilon = +1 \mathrm{~V}

Therefore, the average induced emf in the loop during this time is $+1 \mathrm{~V}$ . The correct answer is: Option C

+1 \mathrm{~V}

Q50

An ideal coil of

10H

is connected in series with a resistance of

5\Omega

and a battery of

5V

.

2

second after the connection is made, the current flowing in ampere in the circuit is

A

\left( {1 - {e^{ - 1}}} \right)

B

\left( {1 - e} \right)

C

e

D

{{e^{ - 1}}}

Correct Answer

Option A

Solution

KEY CONCEPT :

I = {I_0}\left( {1 - {e^{ - {R \over L}t}}} \right)

(When current is in growth in

LR

circuit)

= {E \over R}\left( {1 - {e^{ - {R \over L}t}}} \right)

= {5 \over 5}\left( {1 - {e^{ - {5 \over {10}} \times 2}}} \right)

= \left( {1 - {e^{ - 1}}} \right)