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(C) The magnetic field of the solenoid is $\vec{B} = \mu_0 m I \hat{n}$. The loop is in the $xy$-plane,so its area vector is $\vec{A} = A\hat{k}$.$(I)

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$I \rightarrow Q, II \rightarrow P, III \rightarrow S, IV \rightarrow R$

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(C) The magnetic field of the solenoid is $\vec{B} = \mu_0 m I \hat{n}$. The loop is in the $xy$-plane,so its area vector is $\vec{A} = A\hat{k}$.$(I)$ $\hat{n} = \frac{1}{\sqrt{2}}(\sin \omega t \hat{j} + \cos \omega t \hat{k})$. Flux $\phi = \vec{B} \cdot \vec{A} = \frac{\mu_0 m I A}{\sqrt{2}} \cos \omega t$. Induced $EMF$ $\varepsilon = -\frac{d\phi}{dt} = \frac{\mu_0 m I A \omega}{\sqrt{2}} \sin \omega t$. Current $i = \frac{\varepsilon}{R}$. Magnetic moment $\vec{M} = i A \hat{k} = \frac{\mu_0 m I A^2 \omega}{\sqrt{2} R} \sin \omega t \hat{k}$. Torque $\vec{	au} = \vec{M} 	imes \vec{B} = \frac{\mu_0^2 m^2 I^2 A^2 \omega}{2R} \sin^2 \omega t (\hat{k} 	imes \hat{j}) = \alpha \sin^2 \omega t (-\hat{i})$. At $t = \frac{\pi}{6\omega}$,$\sin^2(\pi/6) = 1/4$,so $\vec{	au} = -\frac{\alpha}{4} \hat{i}$ $(Q)$.$(II)$ $\hat{n} = \frac{1}{\sqrt{2}}(\sin \omega t \hat{i} + \cos \omega t \hat{j})$. Here $\vec{B} \cdot \hat{k} = 0$,so $\phi = 0$,$\varepsilon = 0$,$i = 0$,$\vec{	au} = 0$ $(P)$.$(III)$ $\hat{n} = \frac{1}{\sqrt{2}}(\sin \omega t \hat{i} + \cos \omega t \hat{k})$. Flux $\phi = \frac{\mu_0 m I A}{\sqrt{2}} \cos \omega t$. Similar to $(I)$,$i = \frac{\mu_0 m I A \omega}{\sqrt{2} R} \sin \omega t$. $\vec{M} = i A \hat{k}$. $\vec{	au} = \vec{M} 	imes \vec{B} = \alpha \sin^2 \omega t (\hat{k} 	imes \hat{i}) = \alpha \sin^2 \omega t \hat{j}$. At $t = \frac{\pi}{6\omega}$,$\vec{	au} = \frac{\alpha}{4} \hat{j}$ $(S)$.$(IV)$ $\hat{n} = \frac{1}{\sqrt{2}}(\cos \omega t \hat{j} + \sin \omega t \hat{k})$. Flux $\phi = \frac{\mu_0 m I A}{\sqrt{2}} \sin \omega t$. $\varepsilon = -\frac{d\phi}{dt} = -\frac{\mu_0 m I A \omega}{\sqrt{2}} \cos \omega t$. $i = -\frac{\mu_0 m I A \omega}{\sqrt{2} R} \cos \omega t$. $\vec{M} = i A \hat{k}$. $\vec{	au} = \vec{M} 	imes \vec{B} = \alpha \cos^2 \omega t (-\hat{k} 	imes \hat{j}) = \alpha \cos^2 \omega t \hat{i}$. At $t = \frac{\pi}{6\omega}$,$\cos^2(\pi/6) = 3/4$,so $\vec{	au} = \frac{3\alpha}{4} \hat{i}$ $(R)$.

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

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