Question

A compass needle oscillates $20$ times per minute at a place where the dip is ${45}^o$ and $30$ times per minute where the dip is ${30}^o$. The ratio of the total magnetic field of earth at second place to the first place is:

• A. $1.51$
• B. $1.83$
• C. $1.63$
• D. $1.23$

Answer 1

The correct option is B $1.83$

We know that time period of oscillation $T\propto \sqrt{B_H}$ where $B_H$ is the magnetic field in the horizontal direction and $B_H=B\cos \theta$ where $B$ is the magnetic fiels of the earth and $\theta$ is the angle of dip
Let the magnetic fields at the two places be $B_1$ and $B_2$ respectively.
$\therefore (\frac{T_1}{T_2}{)}^2=\frac{B_1\cos {\theta }_1}{B_2\cos {\theta }_2}$
$\therefore \frac{{\left(20\right)}^2}{{\left(30\right)}^2}=\frac{B_1\cos {45}^{\circ }}{B_2\cos {30}^{\circ }}$
$\therefore \frac{B_2}{B_1}=\frac{{30}^2\times 2}{{20}^2\times \sqrt{3}\times \sqrt{2}}\approx 1.83$