JEE Advanced 2022 · Paper 1 · Physics Q15

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PhysicsList match+3 / -1Question 33 of 54

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 $t = \dfrac{\pi}{6\omega}$ . Let $\alpha = \dfrac{A^2\mu_0^2m^2I^2\omega}{2R}$ .

Which one of the following options is correct?

A long solenoid with its axis along a unit vector n̂ tilted from the vertical z-axis; the current I circulates counterclockwise about n̂. A small horizontal circular loop sits on the solenoid axis inside it. Axes x, y, z shown.

List-I

I
$\frac1{\sqrt2}(\sin\omega t\,\hat j + \cos\omega t\,\hat k)$
II
$\frac1{\sqrt2}(\sin\omega t\,\hat i + \cos\omega t\,\hat j)$
III
$\frac1{\sqrt2}(\sin\omega t\,\hat i + \cos\omega t\,\hat k)$
IV
$\frac1{\sqrt2}(\cos\omega t\,\hat j + \sin\omega t\,\hat k)$

List-II

P
$0$
Q
$-\frac\alpha4\,\hat i$
R
$\frac{3\alpha}4\,\hat i$
S
$\frac\alpha4\,\hat j$
T
$-\frac{3\alpha}4\,\hat i$

I → Q, II → P, III → S, IV → T

I → S, II → T, III → Q, IV → P

I → Q, II → P, III → S, IV → R

I → T, II → Q, III → P, IV → R

Answer and worked solution

Official answer: Dropped: full marks to everyone

This question was dropped (marks to all) in the official key. Below is the standard working. The organisers did not publish a reason.

Flux and current. The solenoid's field is $\vec B=\mu_0mI\,\hat n$ , and the loop's normal is $\hat k$ . Let $\beta=\mu_0mIA$ .

Flux: $\Phi=\beta\,n_z$ .
Current: $i=-\frac{\beta\dot n_z}{R}$ .
Moment $\vec\mu=iA\hat k$ , so the torque is
$\vec\tau=\vec\mu\times\vec B=-\frac{\beta^2}{R}\dot n_z\,(\hat k\times\hat n)=-\frac{2\alpha}{\omega}\dot n_z\,(\hat k\times\hat n).$

At $\omega t=\frac\pi6$ .

Case	$n_z$	Torque	Match
(I)	$\frac{\cos\omega t}{\sqrt2}$	$-\alpha\sin^2\omega t\,\hat i=-\frac\alpha4\hat i$	Q
(II)	0	0	P
(III)	$\frac{\cos\omega t}{\sqrt2}$	$\alpha\sin^2\omega t\,\hat j=\frac\alpha4\hat j$	S
(IV)	$\frac{\sin\omega t}{\sqrt2}$	$\alpha\cos^2\omega t\,\hat i=\frac{3\alpha}4\hat i$	R

This is option (C). Option (A) differs only in (IV), which suggests the expected sign was disputed.

Topics:Faraday's and Lenz's laws Torque on a current loop Ampere's circuital law

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JEE Advanced 2022 · Paper 1, Physics question 15