Question

A short magnet produces a neutral point in end on position at $25cm$. The deflection produced by the magnet in $\tan B$ position of deflection magnetometer when used at same distance is: ($\tan {27}^{\circ }=1/2$)

• A. ${27}^o$
• B. ${54}^o$
• C. ${30}^o$
• D. ${45}^o$

Answer 1

The correct option is A ${27}^o$

Deflection in $\tan A$ position is,
$\frac{{\mu }_0\left(2M\right)}{4\pi d^3}=B_H\tan {\theta }_A$
Deflection in $\tan B$ position is,
$\frac{{\mu }_0\left(M\right)}{4\pi d^3}=B_H\tan {\theta }_B$
From above equations we get,
$\frac{\tan {\theta }_A}{\tan {\theta }_B}=2$
$\tan {\theta }_B=\frac{1}{2}\tan {\theta }_A$
since distance is same
$\tan {\theta }_B=\frac{1}{2}$
${\theta }_B={\tan }^{-1}\left(\frac{1}{2}\right)={27}^{\circ }$