Question

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 $\hat{n}$ 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 $\alpha=\frac{A^2 \mu_0^2 m^2 I^2 \omega}{2 R}$.

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

Answer 1

Answer: (C)

(I) $\overrightarrow{ B }=\frac{\mu_0 mI }{\sqrt{2}}(\sin \omega t \hat{ j }+\cos \omega t \hat{ k }) $
$ \phi=\overrightarrow{ B } \cdot \overrightarrow{ A }=\frac{\mu_0 mI }{\sqrt{2}} \cos (\omega t ) \cdot A $
$ \varepsilon=\frac{ d \phi}{ dt }=\frac{\mu_0 mI \omega A }{\sqrt{2}} \sin (\omega t ) $
$ i =\frac{\varepsilon}{ 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 }) $
$ \phi=0, \varepsilon=0, i =0, t =0$
(III) $ \overrightarrow{ B }=\frac{\mu_0 mI }{\sqrt{2}}(\sin \omega t \hat{ i }+\cos \omega t \hat{ k }) $
$ \phi=\overrightarrow{ B } \cdot \overrightarrow{ A }=\frac{\mu_0 mI }{\sqrt{2}} \cdot \cos (\omega t ) \cdot A $
$ \varepsilon=-\frac{ d \phi}{ dt }=\frac{\mu_0 mI \omega A }{\sqrt{2}} \sin (\omega t )$
$ i =\frac{\varepsilon}{ 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 })$
$\vec{\tau}=\overrightarrow{ M } \times \overrightarrow{ B }=\frac{\mu_0 m ^2 I ^2 \omega A ^2}{2 R } \sin ^2(\omega t )(+\hat{ j })$
$ =\frac{\alpha}{4} \hat{ j }$
(IV) $ \overrightarrow{ B }=\frac{\mu_0 mI }{\sqrt{2}}(\cos \omega t \hat{ j }+\sin \omega t \hat{ k }) $
$ \phi=\overrightarrow{ B } \cdot \overrightarrow{ A }=\frac{\mu_0 mI }{\sqrt{2}} \cdot \sin (\omega t ) \cdot A $
$\varepsilon=-\frac{ d \phi}{ dt }=\frac{\mu_0 mI \omega A }{\sqrt{2}} \cos (\omega t ) $
$ i =\frac{\varepsilon}{ R }=-\frac{\mu_0 mI \omega A }{\sqrt{2} R } \cos (\omega t ) $
$ \overrightarrow{ M }= i \overrightarrow{ A }= iA (\hat{ k })=-\frac{\mu_0 mI ^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}{2 R } \cos ^2(\omega t )(-\hat{ i }) $
$ =\alpha \cdot \cos ^2\left(\frac{\pi}{6}\right) \hat{ i } $
$ =\frac{3 \alpha}{4} \hat{ i }$