Question

A dip needle vibrates in the vertical plane perpendicular to the magnetic meridian. The time period of vibration is found to be$$2s$$ . The same needle is then allowed to vibrate in the horizontal plane , and the time period is again found to be $$2s$$. Then the angle of dip is
A. $0^{\circ }$
B. ${30}^{\circ }$
C. ${45}^{\circ }$
D. ${90}^{\circ }$

Answer 1

Hint: Use the formula of time period for the needle when it is in the vertical plane perpendicular to magnetic moment. Also use the same formula for the needle when it is in the horizontal plane perpendicular to magnetic moment.
Note that the time periods are the same for both cases.
The tangent of the angle of the dip equals the ratio of the vertical component and horizontal component of the magnetic field. Use this formula to find the angle of the dip.

Formula used:
The time period of the needle in vertical plane, $T_1=2\pi \sqrt{{\displaystyle \frac{I}{M{B}_V}}}$
And, The time period of the needle in horizontal plane, $T_2=2\pi \sqrt{{\displaystyle \frac{I}{M{B}_H}}}$
$I=$The moment of inertia of the needle.
$M=$The magnetic moment of the needle.
$B_V$ And $B_H$ are the horizontal components and vertical components from the magnetic field respectively.

Complete step by step answer:
In the first case, A needle is vibrating in the vertical plane that is perpendicular to the magnetic meridian or magnetic moment. So due to this vibration frequency as well as time-period exist. So the relation between the time period of the needle in vertical plane and the magnetic moment is, $T_1=2\pi \sqrt{{\displaystyle \frac{I}{M{B}_V}}}$
$I=$The moment of inertia of the needle.
$M=$The magnetic moment of the needle.
$B_V=$ the vertical component of the magnetic field.
Given, The time period $T_1=2\sec$

In the second case, the same needle is vibrating in the horizontal plane. Similarly, The time period of the needle in horizontal plane, $T_2=2\pi \sqrt{{\displaystyle \frac{I}{M{B}_H}}}$
$B_H$ is the horizontal component of the magnetic field.
Given, The time period $T_2=2\sec$
So, $T_1=T_2$
$\Rightarrow 2\pi \sqrt{{\displaystyle \frac{I}{M{B}_V}}}=2\pi \sqrt{{\displaystyle \frac{I}{M{B}_H}}}$
$\Rightarrow B_V=B_H$

Since, $B_V$And $B_H$ are the vertical and horizontal components of the magnetic field respectively, for the angle of dip be $\theta$ ,
$\tan \theta ={\displaystyle \frac{B_V}{B_H}}$
From the condition, $B_V=B_H$
$\Rightarrow \tan \theta =1$
$\Rightarrow \theta ={45}^{\circ }$
So, the angle of dip is ${45}^{\circ }$.

Hence, the correct answer is option (C).

Note: The magnetic field intensity $B$ is a vector quantity. So, the geomagnetic field intensity can be divided into two components – vertical and horizontal. Clearly these two components are situated in the magnetic meridian. If the resultant vector of these two components be $R$ and the angle of dip is $\theta$,
Then, $B_V=R\sin \theta$ and $B_H=R\cos \theta$
$\Rightarrow {\displaystyle \frac{B_V}{B_H}}=\tan \theta$
And, ${B_V}^2+{B_H}^2=R^2{\sin }^2\theta +R^2{\cos }^2\theta =R^2$
$\Rightarrow R=\sqrt{{B_V}^2+{B_H}^2}$.