Question

A dip needle vibrates in the vertical plane perpendicular to the magnetic meridian. The time period of vibration is found to be $2\text{sec}$. The same needle is then allowed to vibrate in the horizontal plane and the time period is again found to be $2\text{sec}$. Then the angle of dip is

• A. $0^{\circ }$
• B. ${30}^{\circ }$
• C. ${45}^{\circ }$
• D. ${90}^{\circ }$

Answer 1

The correct option is C ${45}^{\circ }$

In a vertical plane, perpendicular to magnetic meridian the time period is given by

$T=2\pi \sqrt{\frac{I}{M{B}_V}}.........\left(i\right)$

In the horizontal plane,

$T=2\pi \sqrt{\frac{I}{M{B}_H}}........\left(ii\right)$

Given, $T=2$ in both the cases.

Equating equation $\left(i\right)$ and $\left(ii\right)$ we conclude that $B_V=B_H$ at the given place.

Now $\tan \delta =\frac{B_V}{B_H}=1$

$\therefore \delta ={45}^{\circ }$

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