Question

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 doubled, the deflection is

• A. ${\tan }^{-1}\left(\frac{\sqrt{3}}{8}\right)$
• B. ${\sin }^{-1}\left(\frac{\sqrt{3}}{8}\right)$
• C. ${\cos }^{-1}\left(\frac{\sqrt{3}}{8}\right)$
• D. ${\cot }^{-1}\left(\frac{\sqrt{3}}{8}\right)$

Answer 1

The correct option is A ${\tan }^{-1}\left(\frac{\sqrt{3}}{8}\right)$

For a shorty bar magnet is $\tan A$ position
$\frac{{\mu }_0}{4\pi }\frac{2M}{d^3}=H\tan \theta ....\left(i\right)$
When distance is doubled, the new deflection ${\theta }^{\prime }$ is
$\frac{{\mu }_0}{4\pi }\frac{2M}{(2d{)}^3}=H\tan {\theta }^{\prime }....\left(ii\right)$
$\therefore \frac{\tan {\theta }^{\prime }}{\tan \theta }=\frac{1}{8}$
$\tan {\theta }^{\prime }=\frac{\tan \theta }{8}=\frac{\tan {60}^{\circ }}{8}=\frac{\sqrt{3}}{8}$
${\theta }^{\prime }={\tan }^{-1}\left(\frac{\sqrt{3}}{8}\right)$.