Electromagnetic Induction — Page 6 | JEE Physics

Q51

Match with \cdot \overrightarrow{d l}=\mu_o i_c+\mu_o \varepsilon_o \frac{d \phi_E}{d t}$$

List - I		List - II
(B)	$\oint \vec{E} \cdot \overrightarrow{d l}=\dfrac{d \phi_B}{d t}$	(I)	Gauss' law for electricity
(C)	$\oint \vec{E} \cdot \overrightarrow{d A}=\dfrac{Q}{\varepsilon_o}$	(II)	Gauss' law for magnetism
(D)	$\oint \vec{B} \cdot \overrightarrow{d A}=0$	(III)	Faraday law
()		(IV)	Ampere - Maxwell law

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

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

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

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

Correct Answer

Option A

Solution

Ampere-Maxwell law

\rightarrow \oint \overrightarrow{\mathrm{B}} \cdot \overrightarrow{\mathrm{dl}}=\mu_0 \mathrm{i}_{\mathrm{c}}+\mu_0 \varepsilon_0 \frac{\mathrm{d} \phi_{\mathrm{E}}}{\mathrm{dt}}

Faraday law

\rightarrow \oint \vec{E} \cdot \overrightarrow{d l}=\frac{d \phi_{\mathrm{B}}}{d t}

Gauss' law for electricity

\rightarrow \oint \overrightarrow{\mathrm{E}} \cdot \overrightarrow{\mathrm{dA}}=\frac{\mathrm{Q}}{\varepsilon_0}

Gauss ' law for magnetism

\rightarrow \oint \vec{B} \cdot \overrightarrow{d A}=0

Q52

A coil having

n

turns and resistance

R\Omega

is connected with a galvanometer of resistance

4R\Omega .

This combination is moved in time

t

seconds from a magnetic field

{W_1}

weber to

{W_2}

weber. The induced current in the circuit is

A

{{\left( {{W_2} - {W_1}} \right)} \over {Rnt}}

B

- {{n\left( {{W_2} - {W_1}} \right)} \over {5\,\,Rt}}

C

- {{\left( {{W_2} - {W_1}} \right)} \over {5\,\,Rnt}}

D

- {{n\left( {{W_2} - {W_1}} \right)} \over {Rt}}

Correct Answer

Option B

Solution

{{\Delta \phi } \over {\Delta t}} = {{\left( {{W_2} - {W_1}} \right)} \over t}

{R_{tot}} = \left( {R + 4R} \right)\Omega = 5R\Omega

i = {{nd\phi } \over {{R_{tot}}dt}} = {{ - n\left( {{W_2} - {W_1}} \right)} \over {5Rt}}

( as

{W_2}\,\,\& \,\,{W_1}

are magnetic flux )

Q53

12 \mathrm{~V}

battery connected to a coil of resistance

6 \Omega

through a switch, drives a constant current in the circuit. The switch is opened in

1 \mathrm{~ms}

. The emf induced across the coil is

20 \mathrm{~V}

. The inductance of the coil is :

A

5 ~ \mathrm{mH}

B

8 ~\mathrm{mH}

C

10~ \mathrm{mH}

D

12 ~\mathrm{mH}

Correct Answer

Option C

Solution

When the switch is closed, the circuit is a simple DC circuit and the current in the circuit is given by Ohm's law: $I = \dfrac{V}{R} = \dfrac{12\text{V}}{6\Omega} = 2\text{A}$ .

When the switch is opened, the current in the circuit drops to zero instantaneously.

However, the magnetic field generated by the current in the coil does not disappear immediately, and it continues to produce a back EMF that opposes the change in current.

This back EMF induces a voltage across the coil that can be calculated using Faraday's law of induction: $\mathcal{E} = -L\dfrac{\Delta I}{\Delta t}$ , where $\mathcal{E}$ is the induced voltage, $L$ is the inductance of the coil, and $\Delta I/\Delta t$ is the rate of change of current in the coil.

In this case, we know that the induced voltage is $20\text{V}$ and the rate of change of current is $\Delta I/\Delta t = -2\text{A}/(1\text{ms}) = -2\times 10^3\text{A/s}$ .

Substituting these values into the equation above, we get: $20\text{V} = -L\times(-2\times 10^3\text{A/s})$ .

Solving for $L$ , we get: $L = \dfrac{20\text{V}}{2\times 10^3\text{A/s}} = 0.01\text{H}$ .

Therefore, the inductance of the coil is $0.01\text{H}$ , or 10 mH.

Q54

Consider a circular coil of wire carrying constant current I, forming a magnetic dipole. The magnetic flux through an infinite plane that contains the circular coil and excluding the circular coil area is given by

\phi

i. The magnetic flux through the area of the circular coil area is given by

\phi

0. Which of the following option is correct?

A

\phi

i =

\phi

0

B

\phi

i <

\phi

0

C

\phi

i

>

\phi

0

D

\phi

i = -

\phi

0

Correct Answer

Option D

Solution

As, magnetic field lines forms a closed loop, hence each line from circular area will pass through outer area in opposite direction hence $\phi$ i = - $\phi$ 0.

Q55

When the current changes from

+ 2A

to

-2A

in

0.05

second, an

e.m.f.

of

8

V

is inducted in a coil. The coefficient of self- induction of the coil is

A

0.2H

B

0.4H

C

0.8

H

D

0.1

H

Correct Answer

Option D

Solution

e = - {{\Delta \phi } \over {\Delta t}} = {{ - \Delta \left( {LI} \right)} \over {\Delta t}} = - L{{\Delta I} \over {\Delta t}}

$\therefore$

\left| e \right| = L{{\Delta I} \over {\Delta t}} \Rightarrow 8 = L \times {4 \over {0.05}}

\Rightarrow L = {{8 \times 0.05} \over 4} = 0.1H

Q56

A coil is places perpendicular to a magnetic field of

5000 \mathrm{~T}

. When the field is changed to

3000 \mathrm{~T}

in

2 \mathrm{~s}

, an induced emf of

22 \mathrm{~V}

is produced in the coil. If the diameter of the coil is

0.02 \mathrm{~m}

, then the number of turns in the coil is:

A 35

B 70

C 7

D 140

Correct Answer

Option B

Solution

\begin{aligned} \varepsilon & =\mathrm{N}\left(\frac{\Delta \phi}{\mathrm{t}}\right) \\ \Delta \phi & =(\Delta \mathrm{B}) \mathrm{A} \\ \mathrm{B}_{\mathrm{i}} & =5000 \mathrm{~T}, \\ \mathrm{~B}_{\mathrm{f}} & =3000 \mathrm{~T} \\ \mathrm{~d} & =0.02 \mathrm{~m} \\ \mathrm{r} & =0.01 \mathrm{~m} \\ \Delta \phi & =(\Delta \mathrm{B}) \mathrm{A} \\ & =(2000) \pi(0.01)^2=0.2 \pi \\ \varepsilon & =\mathrm{N}\left(\frac{\Delta \phi}{\mathrm{t}}\right) \Rightarrow 22=\mathrm{N}\left(\frac{0.2 \pi}{2}\right) \\ \mathrm{N} & =70 \end{aligned}

Q57

A transformer has an efficiency of

80 \%

and works at

10 \mathrm{~V}

and

4 \mathrm{~kW}

. If the secondary voltage is

240 \mathrm{~V}

, then the current in the secondary coil is :

A

1.33 \mathrm{~A}

B

13.33 \mathrm{~A}

C

1.59 \mathrm{~A}

D

15.1 \mathrm{~A}

Correct Answer

Option B

Solution

To find the current in the secondary coil of the transformer, we first need to calculate the output power, taking into account the efficiency.

The efficiency ( $\eta$ ) of the transformer is given by the ratio of the output power (

P_{\text{out}}

) to the input power (

P_{\text{in}}

) times 100%. The given efficiency is

\eta = 80\%

or

\eta = 0.8

in decimal form. The input power is also given as

P_{\text{in}} = 4 \text{kW}

or

P_{\text{in}} = 4000 \text{W}

. Let's calculate

P_{\text{out}}

using the efficiency formula:

P_{\text{out}} = \eta \times P_{\text{in}} = 0.8 \times 4000 \text{W} = 3200 \text{W}

Now that we have

P_{\text{out}}

, we can calculate the secondary current (

I_{\text{secondary}}

) using the formula:

P_{\text{out}} = V_{\text{secondary}} \times I_{\text{secondary}}

We are given

V_{\text{secondary}} = 240 \text{V}

. Isolating

I_{\text{secondary}}

gives us:

I_{\text{secondary}} = \frac{P_{\text{out}}}{V_{\text{secondary}}}

Substituting the known values:

I_{\text{secondary}} = \frac{3200 \text{W}}{240 \text{V}}

I_{\text{secondary}} = \frac{3200}{240}

I_{\text{secondary}} = 13.33 \text{A}

Therefore, the current in the secondary coil is

13.33 \text{A}

, which corresponds to Option B.

Q58

A 10 m long horizontal wire extends from North East to South West. It is falling with a speed of 5.0 ms–1, at right angles to the horizontal component of the earth's magnetic field of 0.3

\times

10–4 Wb/m2. The value of the induced emf in wire is :

A 0.3

\times

10–3 V

B 2.5

\times

10–3 V

C 1.5

\times

10–3 V

D 1.1

\times

10–3 V

Correct Answer

Option D

Solution

Induied emf = Bv

\ell

sin 45o = 0.3 $\times$ 10 $-$ 4 $\times$ 5 $\times$ 10 $\times$ sin 45o = 1.1 $\times$ 10 $-$ 3 V

Q59

Two coaxial solenoids are made by winding thin insulated wire over a pipe of cross-sectional area

A=

10\,\,c{m^2}

and length

=20

cm

. If one of the solenoid has

300

turns and the other

400

turns, their mutual inductance is

\left( {{\mu _0} = 4\pi \times {{10}^{ - 7}}\,Tm\,{A^{ - 1}}} \right)

A

2.4\pi \times {10^{ - 5}}H

B

4.8\pi \times {10^{ - 4}}H

C

4.8\pi \times {10^{ - 5}}H

D

2.4\pi \times {10^{ - 4}}H

Correct Answer

Option D

Solution

M = {{{\mu _0}{N_1}{N_2}A} \over \ell }

= {{4\pi \times {{10}^{ - 7}} \times 300 \times 400 \times 100 \times {{10}^{ - 4}}} \over {0.2}}

= 2.4\pi \times {10^{ - 4}}H