Question

The time of vibration of a dip needle vibration in the vertical plane in the magnetic meridian is $3 s$ . When the same magnetic needle is made to vibrate in the horizontal plane, the time of vibration is 3 $ \sqrt{2} $ s. Then angle of dip will be

• A. $ 90{}^\circ $
• B. $ 60{}^\circ $
• C. $ 45{}^\circ $
• D. $ 30{}^\circ $

Answer 1

The Correct Option is B

In vertical plane in magnetic meridian both horizontal and vertical components of magnetic field exist. When $M$ is magnetic moment of the magnet, $H$ and $V$ are the horizontal and vertical components of earth's magnetic field and $I$ is moment of inertia of magnet about its axis of vibration, then the time-period of magnet is $T=2 \pi \sqrt{\frac{I}{M H}}$ when horizontal component is taken $T^{'}=2 \pi \sqrt{\frac{I}{M B}}$ [as in vertical plane in magnetic meridian both $V$ and $H$ act on the needle]

Given, $T=3 \sqrt{2} s , T^{'}=3 s$ $\therefore \frac{T^{'}}{T}=\sqrt{\frac{H}{V}}=\frac{3}{3 \sqrt{2}}=\frac{1}{\sqrt{2}}$ Also the angle of dip at a place is the angle between the direction of earth's magnetic field and the horizontal in the magnetic meridian at that place $\therefore \sqrt{\frac{H}{V}}=\sqrt{\cos \phi}=\frac{1}{\sqrt{2}}$ $\therefore \cos \phi=\frac{1}{2}$ $\Rightarrow \phi=60^{\circ}$

Answer 2

To determine the angle of dip given the times of vibration of a dip needle in different planes, we use the relationship between the time periods of vibration and the components of the Earth's magnetic field.
Given Data:
- Time of vibration in the vertical plane (\( T_v \)): 3 s
- Time of vibration in the horizontal plane (\( T_h \)): \( 3\sqrt{2} \) s
Concept and Formula:
The time periods of vibration (\( T_v \) and \( T_h \)) are related to the magnetic field components in the following way:
\[ T_v \propto \frac{1}{\sqrt{B_v}} \]
\[ T_h \propto \frac{1}{\sqrt{B_h}} \]
Here, \( B_v \) and \( B_h \) are the vertical and horizontal components of the Earth's magnetic field respectively. They are related to the total magnetic field \( B \) and the angle of dip \( \delta \) by:
\[ B_v = B \cos \delta \]
\[ B_h = B \sin \delta \]
Relationship Between Time Periods:
From the proportionality,
\[ T_v \sqrt{B_v} = T_h \sqrt{B_h} \]
Substitute \( B_v = B \cos \delta \) and \( B_h = B \sin \delta \):
\[ T_v \sqrt{B \cos \delta} = T_h \sqrt{B \sin \delta} \]
Since \( B \) is common and can be cancelled out, we get:
\[ T_v \sqrt{\cos \delta} = T_h \sqrt{\sin \delta} \]
Substitute the Given Times:
\[ 3 \sqrt{\cos \delta} = 3\sqrt{2} \sqrt{\sin \delta} \]
Divide both sides by 3:
\[ \sqrt{\cos \delta} = \sqrt{2} \sqrt{\sin \delta} \]
Square both sides to remove the square roots:
\[ \cos \delta = 2 \sin \delta \]
Using the identity \( \sin^2 \delta + \cos^2 \delta = 1 \), substitute \( \cos \delta \):
\[ \cos^2 \delta = 4 \sin^2 \delta \]
Replace \( \cos^2 \delta \) with \( 1 - \sin^2 \delta \):
\[ 1 - \sin^2 \delta = 4 \sin^2 \delta \]
Combine like terms:
\[ 1 = 5 \sin^2 \delta \]
Solve for \( \sin^2 \delta \):
\[ \sin^2 \delta = \frac{1}{5} \]
Therefore,
\[ \sin \delta = \frac{1}{\sqrt{5}} \]
Use the identity \( \tan \delta = \frac{\sin \delta}{\cos \delta} \) to find \( \tan \delta \):
\[ \cos \delta = \sqrt{1 - \sin^2 \delta} = \sqrt{1 - \frac{1}{5}} = \sqrt{\frac{4}{5}} = \frac{2}{\sqrt{5}} \]
Thus,
\[ \tan \delta = \frac{\sin \delta}{\cos \delta} = \frac{\frac{1}{\sqrt{5}}}{\frac{2}{\sqrt{5}}} = \frac{1}{2} \]
The angle whose tangent is \(\frac{1}{2}\) is \( 60^\circ \).
Hence, the angle of dip is:
\[ \delta = 60^\circ \]