Question 137: Magnetism and Matter

Question

A compass needle oscillates 20 times per minute at a place where the dip is $$30^{\circ}$$ and 30 times per minute where the dip is $$60^{\circ}$$. The ratio of total magnetic field due to the earth at two places respectively is $$\frac{4}{\sqrt{x}}$$. The value of $$x$$ is

Accepted Answer

$$
\begin{aligned}
& 	ext {Period of oscillation } \alpha \frac{1}{\sqrt{B_{\mathrm{H}}}} \\
& \mathrm{T} \alpha \frac{1}{\sqrt{\mathrm{B} \cos 	heta}} \Rightarrow \frac{T_1}{\mathrm{~T}_2}=\sqrt{\frac{\mathrm{B}_2 \cos 	heta_2}{\mathrm{~B}_1 \cos 	heta_1}} \\
& \Rightarrow \frac{60 / 20}{60 / 30}=\sqrt{\frac{\mathrm{B}_2}{\mathrm{~B}_1} \frac{\cos 60^{\circ}}{\cos 30^{\circ}}} \Rightarrow \frac{3}{2}=\sqrt{\frac{\mathrm{B}_2}{\sqrt{3} \mathrm{~B}_1}} \\
& \Rightarrow \frac{9}{4}=\frac{\mathrm{B}_2}{\sqrt{3} \mathrm{~B}_1} \Rightarrow \frac{\mathrm{B}_1}{\mathrm{~B}_2}=\frac{4}{9 \sqrt{3}}=\frac{4}{\sqrt{243}}
\end{aligned}
$$

Magnetism and Matter

A compass needle oscillates 20 times per minute at a place where the dip is $30^{\circ} and 30 times per minute where the dip is 60^{\circ}. The ratio of total magnetic field due to the earth at two places respectively is \frac{4}{\sqrt{x}}. The value of x$ is

Choose the correct answer: