Electromagnetic Induction — Page 3 | JEE Physics

Q21

An emf of

0.08 \mathrm{~V}

is induced in a metal rod of length

10 \mathrm{~cm}

held normal to a uniform magnetic field of

0.4 \mathrm{~T}

, when moves with a velocity of:

A

20 \mathrm{~ms}^{-1}

B

2 \mathrm{~ms}^{-1}

C

3.2 \mathrm{~ms}^{-1}

D

0.5 \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: $\text{emf} = B \cdot L \cdot v$ 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: $v = \dfrac{\text{emf}}{B \cdot L}$ Substituting the given values: $v = \dfrac{0.08 \, \text{V}}{0.4 \, \text{T} \times0.1 \, \text{m}} = 2 \, \text{m/s}$

Q22

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 = -L1

\dfrac{dI_2}{dt}

- M12

\dfrac{dI_1}{dt}

B e1 = -L1

\dfrac{dI_1}{dt}

+ M12

\dfrac{dI_2}{dt}

C e1 = -L1

\dfrac{dI_1}{dt}

- M12

\dfrac{dI_1}{dt}

D e1 = -L1

\dfrac{dI_1}{dt}

- M12

\dfrac{dI_2}{dt}

Correct Answer

Option D

Solution

\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}

Q23

Match with \cdot \vec{d} a=\frac{1}{\varepsilon_0} \int \rho d V$$

List - I		List - II
(B)	Faraday's law of electro magnetic induction	(II)	$\oint \vec{B} \cdot \vec{d} a=0$
(C)	Ampere's law	(III)	$\int \vec{E} \cdot \vec{d} l=\dfrac{-d}{d t} \int \vec{B} \cdot \vec{d} a$
(D)	Gauss's law of electrostatics	(IV)	$\oint \vec{B} \cdot \vec{d} l=\mu_0 I$

A A-III, B-IV, C-I, D-II

B A-IV, B-II, C-III, D-I

C A-II, B-III, C-IV, D-I

D A-I, B-III, C-IV, D-II

Correct Answer

Option C

Solution

Let's identify each law listed in List I and match it with the corresponding mathematical expression listed in List II.

Gauss's law of magnetostatics states that the total magnetic flux through a closed surface is zero, as magnetic monopoles do not exist.

This is given by the formula:

\oint \vec{B} \cdot \vec{da} = 0

So, (A) matches with (II).

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

It is given by:

\oint \vec{E} \cdot \vec{dl} = -\frac{d}{dt} \int \vec{B} \cdot \vec{da}

Hence, (B) matches with (III).

Ampere's law relates the integrated magnetic field around a closed loop to the electric current passing through the loop.

The mathematical expression given in the context of magnetostatics (without the displacement current) is:

\oint \vec{B} \cdot \vec{dl} = \mu_0 I

Consequently, (C) matches with (IV).

Lastly, Gauss's law of electrostatics states that the total electric flux out of a closed surface is proportional to the charge enclosed within the surface:

\oint \vec{E} \cdot \vec{da} = \frac{1}{\varepsilon_0} \int \rho dV

Therefore, (D) matches with (I).

Based on these matches, the correct answer must link A-II, B-III, C-IV, and D-I: The correct option is: Option C A-II, B-III, C-IV, D-I

Q24

Given below are two statements: Statement I : If the number of turns in the coil of a moving coil galvanometer is doubled then the current sensitivity becomes double. Statement II : Increasing current sensitivity of a moving coil galvanometer by only increasing the number of turns in the coil will also increase its voltage sensitivity in the same ratio In the light of the above statements, choose the correct answer from the options given below :

A Statement I is true but Statement II is false

B Statement I is false but Statement II is true

C Both Statement I and Statement II are false

D Both Statement I and Statement II are true

Correct Answer

Option A

Solution

Statement I: If the number of turns in the coil of a moving coil galvanometer is doubled then the current sensitivity becomes double.

This statement is true.

The formula for current sensitivity ( $I_s$ ) of a moving coil galvanometer is given by: $I_s = \dfrac{NAB}{k}$ where: $N$ is the number of turns in the coil, $A$ is the area of the coil, $B$ is the magnetic field strength, and $k$ is the spring constant of the coil.

From this formula, you can see that the current sensitivity is directly proportional to the number of turns (N).

If $N$ is doubled, then the current sensitivity will also double.

Statement II: Increasing current sensitivity of a moving coil galvanometer by only increasing the number of turns in the coil will also increase its voltage sensitivity in the same ratio.

This statement is false.

The formula for voltage sensitivity ( $V_s$ ) of a moving coil galvanometer is given by: $V_s = I_s R = \dfrac{NAB}{k} R$ where: $R$ is the resistance of the coil.

From this formula, you can see that the voltage sensitivity is proportional to the number of turns ( $N$ ) but also inversely proportional to the coil resistance ( $R$ ).

If you double the number of turns ( $N$ ), you also double the length of the wire making up the coil, and thus, you double the resistance ( $R$ ) of the coil.

The doubling of $N$ is offset by the doubling of $R$ , so the overall voltage sensitivity remains the same.

Therefore, increasing the current sensitivity by only increasing the number of turns in the coil will not increase the voltage sensitivity in the same ratio.

Q25

In a uniform magnetic field of induction

B

a wire in the form of a semicircle of radius

r

rotates about the diameter of the circle with an angular frequency

\omega .

The axis of rotation is perpendicular to the field. If the total resistance of the circuit is

R,

the mean power generated per period of rotation is

A

{{{{\left( {B\pi r\omega } \right)}^2}} \over {2R}}

B

{{{{\left( {B\pi {r^2}\omega } \right)}^2}} \over {8R}}

C

{{B\pi {r^2}\omega } \over {2R}}

D

{{{{\left( {B\pi r{\omega ^2}} \right)}^2}} \over {8R}}

Correct Answer

Option B

Solution

\phi = \overrightarrow B .\overrightarrow A ;\phi = BA\cos \,\omega t

\varepsilon = - {{d\phi } \over {dt}} = \omega BA\sin \,\omega t;\,\,

i = {{\omega BA} \over R}\sin \,\omega t

{P_{inst}} = {i^2}R = {\left( {{{\omega BA} \over R}} \right)^2} \times R{\sin ^2}\omega t

{p_{avg}} = {{\int\limits_0^T {{P_{inst}} \times dt} } \over {\int\limits_0^T {dt} }}

= {{{{\left( {\omega BA} \right)}^2}} \over R}{{\int\limits_0^T {{{\sin }^2}\omega tdt} } \over {\int\limits_0^T {dt} }}

= {1 \over 2}{{{{\left( {\omega BA} \right)}^2}} \over R}

$\therefore$

{P_{abg}} = {{{{\left( {\omega B\pi {r^2}} \right)}^2}} \over {8R}}\,\,\,\,\,\left[ {A = {{\pi {r^2}} \over 2}} \right]

Q26

A small square loop of side 'a' and one turn is placed inside a larger square loop of side b and one turn (b >> a). The two loops are coplanar with their centres coinciding. If a current I is passed in the square loop of side 'b', then the coefficient of mutual inductance between the two loops is :

A

{{{\mu _0}} \over {4\pi }}8\sqrt 2 {{{a^2}} \over b}

B

{{{\mu _0}} \over {4\pi }}{{8\sqrt 2 } \over a}

C

{{{\mu _0}} \over {4\pi }}8\sqrt 2 {{{b^2}} \over a}

D

{{{\mu _0}} \over {4\pi }}{{8\sqrt 2 } \over b}

Correct Answer

Option A

Solution

B = \left[ {{{{\mu _0}} \over {4\pi }}{I \over {b/2}} \times 2\sin 45} \right] \times 4

\phi = 2\sqrt 2 {{{\mu _0}} \over \pi }{I \over b} \times {a^2}

$\therefore$

M = {\phi \over I} = {{2\sqrt 2 {\mu _0}{a^2}} \over {\pi b}} = {{{\mu _0}} \over {4\pi }}8\sqrt 2 {{{a^2}} \over b}

Option (a)

Q27

A uniform magnetic field of 0.4 T acts perpendicular to a circular copper disc 20 cm in radius. The disc is having a uniform angular velocity of 10

\pi

rad s-1 about an axis through its centre and perpendicular to the disc. What is the potential difference developed between the axis of the disc and the rim?

(\pi=3.14)

A 0.5024 V

B 0.0628 V

C 0.2512 V

D 0.1256 V

Correct Answer

Option C

Solution

To determine the potential difference between the center and the rim of the disc (often called a Faraday disc or unipolar generator), we use the formula for motional EMF in a rotating disc:

V = \frac{1}{2} B \omega R^2

where:

B

is the magnetic field strength, $\omega$ is the angular velocity,

R

is the radius of the disc. Given:

B = 0.4\ \text{T}

,

\omega = 10\pi\ \text{rad/s}

,

R = 20\ \text{cm} = 0.2\ \text{m}

, we substitute these values into the formula: First calculate

R^2

:

R^2 = (0.2\ \text{m})^2 = 0.04\ \text{m}^2

Substitute into the formula:

V = \frac{1}{2} \times 0.4 \times (10\pi) \times 0.04

Multiply step-by-step:

0.4 \times 10\pi = 4\pi

, Then,

4\pi \times 0.04 = 0.16\pi

, Finally, multiply by

\frac{1}{2}

:

V = \frac{1}{2} \times 0.16\pi = 0.08\pi

Substitute

\pi \approx 3.14

:

V \approx 0.08 \times 3.14 = 0.2512\ \text{V}

Thus, the potential difference developed between the axis and the rim is approximately

0.2512\ \text{V}

, which corresponds to Option C.

Q28

A coil of inductance

300

mH

and resistance

2\,\Omega

is connected to a source of voltage

2

V

. The current reaches half of its steady state value in

A

0.1

s

B

0.05

s

C

0.3

s

D

0.15

s

Correct Answer

Option A

Solution

KEY CONCEPT : The charging of inductance given by,

i = {i_0}\left( {1 - {e^{ - {{Rt} \over L}}}} \right)

{{{i_0}} \over 2} = {i_0}\left( {1 - {e^{ - {{Rt} \over L}}}} \right) \Rightarrow {e^{ - {{Rt} \over L}}} = {1 \over 2}

Taking log on both the sides,

- {{Rt} \over L} = \log 1 - \log 2

\Rightarrow t = {L \over R}\log 2 = {{300 \times {{10}^{ - 3}}} \over 2} \times 0.69

\Rightarrow t - 0.1\,\sec .

Q29

A long solenoid of radius R carries a time (t) - dependent current I(t)=I0t(1 - t). A ring of radius 2R is placed coaxially near its middle. During the time interval 0

\le

t

\le

1, the induced current (IR) and the induced EMF(VR) in the ring change as :

A Direction of IR remains unchanged and VR is zero at t = 0.25

B Direction of IR remains unchanged and VR is maximum at t = 0.5

C At t = 0.25 direction of IR reverses and VR is maximum

D At t = 0.5 direction of IR reverses and VR is zero

Correct Answer

Option D

Solution

I(t) = I0t(1 - t) We know, $\phi$ = BA $\Rightarrow$ $\phi$ = $\mu$ 0nIA $\Rightarrow$ $\phi$ = $\mu$ 0nAI0(t - t2) Also VR =

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

= - $\mu$ 0nAI0(1 - 2t) VR = 0 when 1 - 2t = 0 $\Rightarrow$ t = 0.5 Also we know, VR = IRr $\Rightarrow$ IR =

{{{\mu _0}nA{I_0}\left( {1 - 2t} \right)} \over r}

after t = 0.5, IR reverses its direction.

Q30

A coil of area A and N turns is rotating with angular velocity

\omega

in a uniform magnetic field

\vec{B}

about an axis perpendicular to

\vec{B}

. Magnetic flux

\varphi

and induced emf

\varepsilon

across it, at an instant when

\vec{B}

is parallel to the plane of coil, are :

A φ = AB, φ = NABω

B φ = AB, φ = 0

C φ = 0, ε = 0

D φ = 0, ε = NABω

Correct Answer

Option D

Solution

\begin{aligned} &\begin{aligned} \phi & =\mathrm{BAN} \cdot \cos (\omega \mathrm{t}) \\ \varepsilon & =\frac{-\mathrm{d} \phi}{\mathrm{dt}}=\mathrm{BA} \omega \mathrm{~N} \cdot \sin (\omega \mathrm{t}) \end{aligned}\\ &\text{ When } B \text{ is parallel to plane, } \underline{\underline{\omega}} \mathrm{t}=\frac{\pi}{2}\\ &\Rightarrow \phi=0, \varepsilon=\mathrm{BA} \omega \mathrm{~N} \end{aligned}