Question

When a plane dip circle is along a magnetic meridian, the period of oscillation of the dip needle is $ {T_1} $ . When the plane of the dip circle is perpendicular to the magnetic meridian, the period of oscillation is $ {T_2} $ . Then $ \dfrac{{{T_1}}}{{{T_2}}} $ is: (Angle of dip $ = {30^0} $ )
(A) $ \sqrt 2 $
(B) $ 1 $
(C) $ 2 $
(D) $ \dfrac{1}{{\sqrt 3 }} $

Answer 1

Hint: To find the time period for the needle when it is in the horizontal plane and vertical plane we can use the corresponding time period formula. Given that in the first case the plane of the dip circle is along the magnetic meridian, therefore, we substitute the formula for a horizontal plane perpendicular to the magnetic moment. In the second case, the plane of the dip circle is perpendicular to the magnetic meridian therefore we need to substitute the formula for a vertical plane parallel to the magnetic moment.

Complete Step By Step Answer:
A dip circle is a device that is used to measure the angle of the dip. This device consists of a vertical circular scale that is further divided into four quadrants.
Now the time period in the horizontal plane is,
 $ \Rightarrow {T_1} = 2\pi \sqrt {\dfrac{I}{{M{B_H}}}} $
The time period in the vertical plane is,
 $ \Rightarrow {T_2} = 2\pi \sqrt {\dfrac{I}{{M{B_V}}}} $
Here, $ I $ is the moment of inertia
 $ M $ is said to be the magnetic moment of the needle
 $ {B_H} $ and $ {B_V} $ are the horizontal components and the vertical components from the magnetic field respectively.
We need to find the ratio of $ {T_1} $ to $ {T_2} $ . The relation between this ratio and the angle of dip is given by,
 $ \Rightarrow \dfrac{{{T_1}}}{{{T_2}}} = \tan \theta $
Here $ \theta $ is the angle of the dip.
Angle of dip is given $ {30^0} $ .
Substituting the formula and the value of the angle of dip we get,
 $ \Rightarrow \dfrac{{2\pi \sqrt {\dfrac{I}{{M{B_H}}}} }}{{2\pi \sqrt {\dfrac{I}{{M{B_V}}}} }} = \tan \theta $
Cancelling out the common terms,
 $ \Rightarrow \sqrt {\dfrac{I}{{M{B_H}}}} \times \sqrt {\dfrac{{M{B_V}}}{I}} = \tan \theta $
 $ \Rightarrow \dfrac{{{B_V}}}{{{B_H}}} = \tan \theta = \tan {30^0} $
 $ \therefore \dfrac{{{B_V}}}{{{B_H}}} = \dfrac{1}{{\sqrt 3 }} $
Therefore, the correct option is D.

Note:
The angle of dip is also called the magnetic dip. It is defined as the angle between which is made by the earth’s magnetic field lines with the horizontal. When the magnetic field points downwards then the angle of dip is said to be positive and if the magnetic field points towards upwards then the angle of dip is said to be negative.