A small circular loop of conducting wire has radius a and carries current I. It is placed in a uniform magnetic field B perpendicular to its plane such that when rotated slightly about its diameter and released, it starts performing simple harmonic motion of time period T. If the mass of the loop is m then :
Solution
$\tau$ = - MBsin $\theta$
<br><br>I$\alpha$ = - MBsin $\theta$
<br><br>for small $\theta$,
<br><br>$\alpha$ = $- {{MB} \over I}\theta$
<br><br>$\therefore$ ${\omega ^2}$ = ${{MB} \over I}$
<br><br>$\Rightarrow$ $\omega$ = $\sqrt {{{I\left( {\pi {R^2}} \right)B} \over {{{m{R^2}} \over 2}}}}$ = $\sqrt {{{2I\pi B} \over m}}$
<br><br>$\therefore$ T = ${{2\pi } \over \omega }$ = $\sqrt {{{2\pi m} \over {IB}}}$
About this question
Subject: Physics · Chapter: Magnetic Effects of Current · Topic: Biot-Savart Law
This question is part of PrepWiser's free JEE Main question bank. 96 more solved questions on Magnetic Effects of Current are available — start with the harder ones if your accuracy is >70%.