Question

A magnet is suspended in such a way that it oscillates in the horizontal plane. It makes $50$ oscillations per minute at a place whose dip angle is ${45}^{\circ }$ and $150$ oscillations per minute at a place whose dip angle is ${30}^{\circ }$. The ratio of earth's total magnetic field at the two places is

• A. $1:\sqrt{6}$
• B. $2\sqrt{3}:9$
• C. $1:6\sqrt{6}$
• D. $1:3\sqrt{6}$

Answer 1

The correct option is D $1:3\sqrt{6}$

Given:-

$f_1=50{\text{min}}^{-1}$

$f_2=150{\text{min}}^{-1}$

Let $B_1$ and $B_2$ be the earth's total magnetic field at the two places.

Frequency of oscillations is given by

$f=\frac{1}{2\pi }\sqrt{\frac{M{B}_H}{I}}$

where $B_H$ is the horizontal component of earth's total magnetic field

$\therefore f_1=\frac{1}{2\pi }\sqrt{\frac{M{B}_1\cos {\varphi }_1}{I}}\text{and}{f}_2=\frac{1}{2\pi }\sqrt{\frac{M{B}_2\cos {\varphi }_2}{I}}$

Here ${\varphi }_1={45}^{\circ },{\varphi }_2={30}^{\circ }$

$\therefore \frac{f_1}{f_2}=\sqrt{\frac{B_1\cos {\varphi }_1}{B_2\cos {\varphi }_2}}$

$\Rightarrow \frac{B_1}{B_2}={\left(\frac{f_1}{f_2}\right)}^2\frac{\cos {\varphi }_2}{\cos {\varphi }_1}$

$={\left(\frac{50}{150}\right)}^2\frac{\cos {30}^{\circ }}{\cos {45}^{\circ }}=\frac{1}{9}\times \frac{\frac{\sqrt{3}}{2}}{\frac{1}{\sqrt{2}}}$

$\therefore B_1:B_2=1:3\sqrt{6}$

Hence, option $\left(D\right)$ is correct.