Question

A magnetic compass needle oscillates $30$ times per minute at a place where the dip is ${45}^{\circ }$, and $40$ times per minute where the dip is ${30}^{\circ }$. If $B_1\text{and}{B}_2$ are respectively the total magnetic field due to the earth at the two places, then the ratio $\frac{B_1}{B_2}$ is best given by :

• A. $3.6$
• B. $2.2$
• C. $0.7$
• D. $1.8$

Answer 1

The correct option is C $0.7$

We know that, the time period of oscillation is given by,

$T=2\pi \sqrt{\frac{I}{M{B}_x}}$

$\Rightarrow \text{frequency,}f=\frac{1}{2\pi }\sqrt{\frac{M{B}_x}{I}}$

$\Rightarrow \frac{f_1^2}{f_2^2}=\frac{B_{x_1}}{B_{x_2}}$

$\Rightarrow {\left(\frac{30}{40}\right)}^2=\frac{B_1\cos {45}^{\circ }}{B_2\cos {30}^{\circ }}$

$\Rightarrow \frac{9}{16}=\frac{B_1\times 2}{\sqrt{2}\times B_2⋊\sqrt{3}}$

$\Rightarrow \frac{B_1}{B_2}=\frac{9\sqrt{6}}{32}\approx 0.7$

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