Question

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 $t=\frac{\pi }{6\omega }\text{Let}\alpha =\frac{A^2{\mu }_0^2{m}^2{I}^2\omega }{2R}.$

	List-I		List-II
(I)	$\frac{1}{\sqrt{2}}(\sin \omega t\hat{j}+\cos \omega t\hat{k})$	(P)	$0$
(II)	$\frac{1}{\sqrt{2}}\left(\sin \omega t\widehat{i+\cos \omega t\hat{j}}\right)$	(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)	$\frac{\alpha }{4}\hat{j}$
		(T)	$-\frac{3\alpha }{4}\hat{i}$

Which one of the following options is correct

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

Answer 1

The correct option is A $I\to Q,II\to P,III\to S,IV\to R$

I.

$\varphi =BA\hat{k}\cdot \hat{n}$

$\varphi =\frac{BA}{\sqrt{2}}\cos \left(\omega t\right)$

$\epsilon =\frac{BA\omega }{\sqrt{2}}\sin \left(\omega t\right)$

$i=\frac{BA\omega }{\sqrt{2}R}\sin \left(\omega t\right)$

$\overrightarrow{m}=iA\hat{k}=\frac{B{A}^2\omega }{\sqrt{2}R}\sin \left(\omega t\right)\hat{k}$

$\overrightarrow{\tau }=\overrightarrow{m}\times \overrightarrow{B}=\frac{B^2{A}^2\omega }{\sqrt{2}R}\sin \left(\omega t\right)(\hat{k}\times \hat{n})$

$=-\frac{B^2{A}^2\omega }{2R}\left[\hat{i}\right]{\sin }^2\left(\omega t\right)$

$\tau =-\frac{B^2{A}^2\omega }{2R}\left[{\sin }^2\left(\frac{\pi }{6}\right)\right]=\frac{-\alpha }{4}\hat{i}$

(I) $\to Q$

II.

$\varphi =O$

(II) $\to P$

III.

$\varphi =\frac{BA}{\sqrt{2}}\cos \left(\omega t\right)$

$i=\frac{BA\omega }{\sqrt{2}R}\sin \left(\omega t\right)$

$\overrightarrow{m}=\frac{B{A}^2\omega }{\sqrt{2}R}\sin \left(\omega t\right)\hat{k}$

$\overrightarrow{\tau }=\overrightarrow{m}\times \overrightarrow{B}=\frac{B^2{A}^2\omega }{\sqrt{2}\times \sqrt{2}R}\sin \left(\omega t\right)(\hat{k}\times (\sin \omega t\hat{i}+\cos \omega t\hat{k}\left)\right)$

$\tau =\frac{B^2{A}^2\omega \sin \left(\omega t\right)}{2R}\sin \left(\omega t\right)\hat{j}$

$=\frac{B^2{A}^2\omega }{2R}{\sin }^2\left(\omega t\right)\hat{j}$

$=\frac{\alpha }{4}\hat{j}$

(III) $\to S$

IV.

$\varphi =\frac{BA}{\sqrt{2}}\sin \left(\omega t\right)$

$i=-\frac{BA\omega }{\sqrt{2}R}\cos \left(\omega t\right)$

$\overrightarrow{m}=-\frac{B{A}^2\omega }{\sqrt{2}R}\cos \omega t\left(\hat{k}\right)$

$\overrightarrow{\tau }=\overrightarrow{m}\times \overrightarrow{B}=-\frac{B^2{A}^2\omega }{2R}(\hat{k}\times \hat{j}){\cos }^2\left(\omega t\right)$

$\tau =+\frac{B^2{A}^2\omega }{2R}\left(\hat{i}\right).{\cos }^2\left(\frac{\pi }{6}\right)$

$=+\frac{3}{4}\alpha \hat{i}$

$\left(IV\right)\to R.$
Hence, option $\left(C\right)$ is correct.