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

Column I		Column 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. $I\to Q,II\to P,III\to S,IV\to T$
• 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. $I\to T,II\to Q,III\to P,IV\to R$

Answer 1

Answer: (C)

(I) $\overrightarrow{B}=\frac{{\mu }_{0}mI}{\sqrt{2}}(\sin \omega t\hat{j}+\cos \omega t\hat{k})$
$\varphi =\overrightarrow{B}\cdot \overrightarrow{A}=\frac{{\mu }_{0}mI}{\sqrt{2}}\cos (\omega t)\cdot A$
$\epsilon =\frac{d\varphi }{dt}=\frac{{\mu }_{0}mI\omega A}{\sqrt{2}}\sin (\omega t)$
$i=\frac{\epsilon }{R}=\frac{{\mu }_{0}mI\omega A}{\sqrt{2}R}\sin (\omega t)$
$\overrightarrow{M}=i\overrightarrow{A}=iA(\hat{k})=\frac{{\mu }_{0}mI\omega A^2}{\sqrt{2}R}\sin (\omega t)(\hat{k})$
$\vec{\tau }=\overrightarrow{M}\times \overrightarrow{B}=\frac{{\mu }_0{m}^2{I}^2\omega A^2}{\sqrt{2}R}{\sin }^2(\omega t)(-\hat{i})$
$=-\left(\frac{\alpha }{4}\right)\hat{i}$
(II) $\overrightarrow{B}=\frac{{\mu }_{0}mI}{\sqrt{2}}(\sin \omega t\hat{i}+\cos \omega t\hat{j})$
$\varphi =0,\epsilon =0,i=0,t=0$
(III) $\overrightarrow{B}=\frac{{\mu }_{0}mI}{\sqrt{2}}(\sin \omega t\hat{i}+\cos \omega t\hat{k})$
$\varphi =\overrightarrow{B}\cdot \overrightarrow{A}=\frac{{\mu }_{0}mI}{\sqrt{2}}\cdot \cos (\omega t)\cdot A$
$\epsilon =-\frac{d\varphi }{dt}=\frac{{\mu }_{0}mI\omega A}{\sqrt{2}}\sin (\omega t)$
$i=\frac{\epsilon }{R}=\frac{{\mu }_{0}mI\omega A}{\sqrt{2}R}\sin (\omega t)$
$\overrightarrow{M}=i\overrightarrow{A}=iA(\hat{k})=\frac{{\mu }_{0}mI\omega A^2}{\sqrt{2}R}\sin (\omega t)(\hat{k})$
$\vec{\tau }=\overrightarrow{M}\times \overrightarrow{B}=\frac{{\mu }_0{m}^2{I}^2\omega A^2}{2R}{\sin }^2(\omega t)(+\hat{j})$
$=\frac{\alpha }{4}\hat{j}$
(IV) $\overrightarrow{B}=\frac{{\mu }_{0}mI}{\sqrt{2}}(\cos \omega t\hat{j}+\sin \omega t\hat{k})$
$\varphi =\overrightarrow{B}\cdot \overrightarrow{A}=\frac{{\mu }_{0}mI}{\sqrt{2}}\cdot \sin (\omega t)\cdot A$
$\epsilon =-\frac{d\varphi }{dt}=\frac{{\mu }_{0}mI\omega A}{\sqrt{2}}\cos (\omega t)$
$i=\frac{\epsilon }{R}=-\frac{{\mu }_{0}mI\omega A}{\sqrt{2}R}\cos (\omega t)$
$\overrightarrow{M}=i\overrightarrow{A}=iA(\hat{k})=-\frac{{\mu }_{0}m{I}^2{A}^2}{\sqrt{2}R}\cos (\omega t)(\hat{k})$
$\vec{\tau }=\overrightarrow{M}\times \overrightarrow{B}=-\frac{{\mu }_0{m}^2{I}^2\omega A^2}{2R}{\cos }^2(\omega t)(-\hat{i})$
$=\alpha \cdot {\cos }^2\left(\frac{\pi }{6}\right)\hat{i}$
$=\frac{3\alpha }{4}\hat{i}$