Question

A deflection magnetometer is arranged with its arms along east -west direction. A small bar magnet is placed lengthwise with its north pole pointing east along the eastern arm with its centre 25 cm away from the centre of the needle of the magnetometer. The needle is then found to be deflected through an angle of 30${}^0$. If the arms of the magnetometer are rotated through 45${}^0$, the deflection of the needle would be :

• A. 0${}^0$
• B. 30${}^0$
• C. 45${}^0$
• D. $ta{n}^{-1}\frac{1}{\sqrt{6}+1}$

Answer 1

The correct option is D $ta{n}^{-1}\frac{1}{\sqrt{6}+1}$

From Tangent law: $B=B_H\tan \theta =\frac{B_H}{\sqrt{3}}$

When the arms are rotated by ${45}^0$, the resultant magnetic field will be,

${\overrightarrow{B}}_{Resultant}=B\sin \theta \hat{i}+(B\cos \theta +B_H)\hat{j}$

The resulting angle will be:

$\tan \alpha =\frac{B\sin {45}^{\circ }}{B\cos {45}^{\circ }+B_H}$

$\Rightarrow$ $\tan \alpha =\frac{1}{\sqrt{6}+1}$