EMI_JEE-Advanced-2022_S1 - Competishun – Live JEE & NEET Courses and Exam Prep

JEE Advanced 2022

Paper-1 2022

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 filed 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 $\mathrm{t}=\frac{\pi}{6 \omega}$, Let $\theta=\frac{A^2 \mu_0^2 \mathrm{~m}^2 \mathrm{l}^2 \omega}{2 R}$.

Which one of the following options is correct?

Select the correct option:

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

✓ Correct! Well done.

✗ Incorrect. Try again or view the solution.

Solution

$ \begin{aligned} & \phi=B A \hat{k} \cdot \hat{n} \\ & \phi=\frac{B A}{\sqrt{2}} \cos (\omega t) \\ & \varepsilon=\frac{B A \omega}{\sqrt{2}} \sin (\omega t) \\ & i=\frac{B A \omega}{\sqrt{2} R} \sin (\omega t) \\ & \vec{m}=i A \hat{k}=\frac{B A^2 \omega}{\sqrt{2} R} \sin (\omega t) \hat{k} \\ & \vec{\tau}=\vec{m} \times \vec{B}=\frac{B^2 A^2 \omega}{\sqrt{2} R} \sin (\omega t)(\hat{k} \times \hat{n}) \\ & =-\frac{B^2 A^2 \omega}{2 R}[\hat{i}] \sin ^2(\omega t) \\ & \tau=-\frac{B^2 A^2 \omega}{2 R}\left[\sin ^2\left(\frac{\pi}{6}\right)\right]=\frac{-\alpha}{4} \hat{i} \end{aligned} $
II. $\phi=0$
(II) $\rightarrow$ Q
III. $\phi=\frac{\mathrm{BA}}{\sqrt{2}} \cos (\omega \mathrm{t})$
$ \begin{aligned} & i=\frac{B A \omega}{\sqrt{2} R} \sin (\omega t) \\ & \vec{m}=\frac{B A^2 \omega}{\sqrt{2} R} \sin (\omega t) \hat{k} \\ & \vec{\tau}=\vec{m} \times \vec{B}=\frac{B^2 A^2 \omega}{\sqrt{2} \times \sqrt{2} R} \sin \omega t(\hat{k} \times(\sin \omega t \hat{i}+\cos \omega t \hat{k})) \\ & \tau=\frac{B^2 A^2 \omega \sin (\omega t)}{2 R} \sin (\omega t) \hat{j} \\ & =\frac{B^2 A^2 \omega}{2 R} \sin ^2(\omega t) \hat{j} \\ & =\frac{\alpha}{4} \hat{j} \\ & \text { III } \rightarrow S \end{aligned} $
IV.
$ \begin{aligned} & \phi=\frac{B A}{\sqrt{2}} \sin (\omega t) \\ & i=-\frac{B A \omega}{\sqrt{2} R} \cos (\omega t) \\ & \vec{m}=-\frac{B A^2 \omega}{\sqrt{2} R} \cos \omega t(\hat{k}) \\ & \vec{\tau}=\vec{m} \times \vec{B}=-\frac{B^2 A^2 \omega}{2 R}(\hat{k} \times \hat{j}) \cos ^2(\omega t) \\ & \tau=+\frac{B^2 A^2 \omega}{2 R}(\hat{i}) \cdot \cos ^2\left(\frac{\pi}{6}\right)=+\frac{3}{4} \alpha \hat{j} \\ & \text { (IV) } \rightarrow R \end{aligned} $

Question Tags

JEE Advance

Physics

Easy

Start Preparing for JEE with Competishun

Report Issue