Q. The needle of a deflection galvanometer shows a deflection of $60{^\circ}$ due to a short bar magnet at a certain distance in tan A position. If the distance is double the deflection is
A) $\sin^{- 1}\left\lbrack \frac{\sqrt{3}}{8} \right\rbrack$
B) $\cos^{- 1}\left\lbrack \frac{\sqrt{3}}{8} \right\rbrack$
C) $\tan^{- 1}\left\lbrack \frac{\sqrt{3}}{8} \right\rbrack$
D) $\cot^{- 1}\left\lbrack \frac{\sqrt{3}}{8} \right\rbrack$
Explanation
(c) For short bar magnet in tan A position
$\frac{\mu_{0}}{4\pi}\frac{2M}{d^{3}} = H\ tan\ \theta\ \ \ \ \ \ \ \ \ \ \ \ \ $…(i)
When distance is doubled, then new deflection $\theta'$ is given by
$\ \ \ \ \frac{\mu_{0}}{4\pi}\frac{2M}{(2d)^{3}} = H\tan{\theta'}$ ….(ii)
$\therefore\ \ \ \ \frac{\tan{\theta'}}{\tan\theta} = \frac{1}{8}$
$\Rightarrow \ \ \ \ \ \ \ \ \theta^{'} = \frac{tan\theta}{8} = \frac{\tan{60{^\circ}}}{8} = \frac{\sqrt{3}}{8}$
$\Rightarrow \ \ \ \ \ \ \ \ \theta^{'} = \tan^{- 1}\left\lbrack \frac{\sqrt{3}}{8} \right\rbrack$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.