JEE Advanced 2022 — Paper 1 — PHY-III solution

JEE Advanced2022Paper 1PHY-III

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

List-I	List-II
(I) $\frac{1}{2} (sin ω t \hat{j} + cos ω t \hat{k})$	(P) $0$
(II) $\frac{1}{2} (sin ω t \hat{i} + cos ω t \hat{j})$	(Q) $- \frac{α}{4} \hat{i}$
(III) $\frac{1}{2} (sin ω t \hat{i} + cos ω t \hat{k})$	(R) $\frac{3 α}{4} \hat{i}$
(IV) $\frac{1}{2} (cos ω t \hat{j} + sin ω t \hat{k})$	(S) $\frac{α}{4} \hat{j}$
	(T) $- \frac{3 α}{4} \hat{i}$

Which one of the following options is correct?

A
I $\to$ Q, II $\to$ P, III $\to$ S, IV $\to$ T
B
I $\to$ S, II $\to$ T, III $\to$ Q, IV $\to$ P
C
I $\to$ Q, II $\to$ P, III $\to$ S, IV $\to$ R
D
I $\to$ T, II $\to$ Q, III $\to$ P, IV $\to$ R

Solution

Flux through loop: $ϕ = B A \hat{k} \cdot \overset{n}{^}$ where $B = μ_{0} m I$ . The induced current is $i = - \frac{1}{R} \frac{d ϕ}{d t}$ , and the magnetic moment is $m = i A \hat{k}$ . The torque is $τ = m \times B$ where $B = B \overset{n}{^}$ .

(I) $\overset{n}{^}$ lies in the $y z$ -plane. $ϕ = \frac{B A}{2} cos ω t$ , $i = \frac{B A ω}{2 R} sin ω t$ , $m = i A \hat{k}$ . Then $\hat{k} \times \overset{n}{^} \propto - \hat{i} sin ω t$ , so $τ = - \frac{B ^{2} A ^{2} ω}{2 R} sin^{2} ω t \hat{i}$ . At $ω t = π /6$ : $τ = - \frac{α}{4} \hat{i}$ . I $\to$ Q.

(II) $\overset{n}{^}$ lies in the $x y$ -plane, perpendicular to $\hat{k}$ , so $ϕ = 0$ and $τ = 0$ . II $\to$ P.

(III) $\overset{n}{^}$ in the $x z$ -plane. $ϕ = \frac{B A}{2} cos ω t$ . Compute $\hat{k} \times \overset{n}{^}$ for the $\hat{i}$ and $\hat{k}$ components: $\hat{k} \times \hat{i} = \hat{j}$ , $\hat{k} \times \hat{k} = 0$ . Torque $\propto sin^{2} ω t \hat{j}$ . At $ω t = π /6$ : $τ = \frac{α}{4} \hat{j}$ . III $\to$ S.

(IV) $\overset{n}{^}$ in the $y z$ -plane with sine and cosine swapped. $ϕ = \frac{B A}{2} sin ω t$ , $i \propto - cos ω t$ . $\hat{k} \times \hat{j} = - \hat{i}$ , giving $τ \propto + cos^{2} ω t \hat{i}$ . At $ω t = π /6$ : $cos^{2} (π /6) = 3/4$ , so $τ = \frac{3 α}{4} \hat{i}$ . IV $\to$ R.

The correct matching is option (C).

← Back to full paper

Practice more PHY-III

Concept-wise practice with instant solutions on Fundamenthol.

Start practicing →