A small circular loop of area $A$ and resistance $R$ is fixed on a horizontal $xy$-plane with the center of the loop always on the axis $\hat{n}$ of a long solenoid. The solenoid has $m$ turns per unit length and carries current $I$ counterclockwise as shown in the figure. The magnetic field due to the solenoid is in $\hat{n}$ direction. List-I gives time dependences of $\hat{n}$ in terms of a constant angular frequency $\omega$. List-II gives the torques experienced by the circular loop at time $t=\frac{\pi}{6 \omega}$. Let $\alpha=\frac{A^{2} \mu_{0}^{2} m^{2} I^{2} \omega}{2 R}$. | List-I | | List-II | | | --- | --- | --- | --- | | (I) | $\frac{1}{\sqrt{2}}(\sin \omega t \hat{j}+\cos \omega t \hat{k})$ | (P) | 0 | | (II) | $\frac{1}{\sqrt{2}}(\sin \omega t \hat{i}+\cos \omega t \hat{j})$ | (Q) | $-\frac{\alpha}{4} \hat{i}$ | | (III) | $\frac{1}{\sqrt{2}}(\sin \omega t \hat{i}+\cos \omega t \hat{k})$ | (R) | $\frac{3 \alpha}{4} \hat{i}$ | | (IV) | $\frac{1}{\sqrt{2}}(\cos \omega t \hat{i}+\sin \omega t \hat{k})$ | (S) | $\frac{\alpha}{4} \hat{j}$ | | | | (T) | $-\frac{3 \alpha}{4} \hat{i}$ | Which one of the following options is correct?

(A) I→Q, II→P, III→S, IV→T

(B) I→S, II→T, III→Q, IV→P

(D) I→T, II→Q, III→P, IV→R

Electromagnetic Induction — JEE Advanced · 2022 · 28 Aug 2022 · Paper 1 Question with Solution

A small circular loop of area

A

and resistance

R

is fixed on a horizontal

x y

-plane with the center of the loop always on the axis

\overset{n}{^}

of a long solenoid. The solenoid has

m

turns per unit length and carries current

I

counterclockwise as shown in the figure. The magnetic field due to the solenoid is in

\overset{n}{^}

direction. List-I gives time dependences of

\overset{n}{^}

in terms of a constant angular frequency

ω

. List-II gives the torques experienced by the circular loop at time

t = \frac{π}{6 ω}

. Let

α = \frac{A ^{2} μ _{0}^{2} m ^{2} I ^{2} ω}{2 R}

Which one of the following options is correct?

Answer: (C)

For each case, calculate flux

ϕ = B \cdot A

, induced emf

ε = - d ϕ / d t

, current

i = ε / R

, magnetic moment

M = i A

, and torque

τ = M \times B

.

(I)

\overset{n}{^} = \frac{1}{2} (sin ω t \hat{j} + cos ω t \hat{k})

, loop in

x y

-plane (normal

\hat{k}

ϕ = \frac{μ _{0} m I A}{2} cos ω t

ε = \frac{μ _{0} m I A ω}{2} sin ω t

τ = - \frac{μ _{0}^{2} m ^{2} I ^{2} ω A ^{2}}{2 R} sin^{2} ω t \hat{i} = - \frac{α}{4} \hat{i}

t = π / (6 ω)

I→Q

(II)

\overset{n}{^} = \frac{1}{2} (sin ω t \hat{i} + cos ω t \hat{j})

: No flux through loop in

x y

-plane.

τ = 0

II→P

(III)

\overset{n}{^} = \frac{1}{2} (sin ω t \hat{i} + cos ω t \hat{k})

τ = \frac{α}{4} \hat{j}

t = π / (6 ω)

III→S

(IV)

\overset{n}{^} = \frac{1}{2} (cos ω t \hat{i} + sin ω t \hat{k})

τ = α cos^{2} (π /6) \hat{i} = \frac{3 α}{4} \hat{i}

IV→R

Answer: (C) I→Q, II→P, III→S, IV→R

JEE Advanced · 2022 · 28 Aug 2022 · Paper 1 · Single correct option