Question

A magnet oscillating in a horizontal plane has a time period of $2\text{sec}$ at a place where the angle of dip is ${30}^{\circ }$ and $3\text{sec}$ at another place where the angle of dip is ${60}^{\circ }$. The ratio of resultant magnetic fields at the two places is:

• A. $\frac{4\sqrt{3}}{7}$
• B. $\frac{4}{9\sqrt{3}}$
• C. $\frac{9}{4\sqrt{3}}$
• D. $\frac{9}{\sqrt{3}}$

Answer 1

The correct option is C $\frac{9}{4\sqrt{3}}$

The time period of the magnet is,

$T=2\pi \sqrt{\frac{I}{M{B}_H}}=2\pi \sqrt{\frac{I}{MB\cos \delta }}$

Here, $T\propto \frac{1}{\sqrt{B_H}}$

$\Rightarrow T\propto \frac{1}{\sqrt{B\cos \delta }}$

$\therefore \frac{T_1}{T_2}=\sqrt{\frac{B_2\cos {\delta }_2}{B_1\cos {\delta }_1}}$

So that, $\frac{B_1}{B_2}=\frac{T_2^2\cos {\delta }_2}{T_1^2\cos {\delta }_1}$

$\frac{B_1}{B_2}=\frac{(3{)}^2\cos {60}^{\circ }}{(2{)}^2\cos {30}^{\circ }}$

$\therefore \frac{B_1}{B_2}=\frac{9}{4\sqrt{3}}$

Hence, option $\left(C\right)$ is the correct answer.