Question

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

Answer 1

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

$T_1=2\pi \sqrt{\frac{I}{M{B}_V}}.........\left(1\right)$

In the horizontal plane,

$T_2=2\pi \sqrt{\frac{I}{M{B}_H}}........\left(2\right)$

Given: $T_1=T_2=x\text{s}$

Equating equation $\left(1\right)$ and $\left(2\right)$ we conclude that $B_V=B_H$ at the given place.

Now, the tangent of the angle of dip is,

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

Accepted answers$:1,1.0,1.00$