A magnetic needle lying parallel to a magnetic field requires X units of work to turn it through `60^(@)` . The torque necessary to maintain the needle in this position is

Options

A. `2 X `
B. X
C. `(sqrt3)/(2)`X
D. `sqrt3` X

Answer

To solve the problem step by step, we will analyze the work done in rotating a magnetic needle and calculate the torque required to maintain it in the new position.

### Step-by-Step Solution:

1. **Understanding the Work Done**:
The work done \( W \) to rotate a magnetic dipole (or needle) from a position parallel to the magnetic field to an angle \( \theta \) is given by the formula:
\[ W = mB(1 - \cos \theta) \]
where \( m \) is the magnetic moment, \( B \) is the magnetic field strength, and \( \theta \) is the angle through which the needle is turned.

2. **Substituting the Given Values**:
In this case, the needle is turned through \( 60^\circ \). Therefore, substituting \( \theta = 60^\circ \) into the work done formula gives:
\[ W = mB(1 - \cos 60^\circ) \]
Since \( \cos 60^\circ = \frac{1}{2} \), we can substitute this value:
\[ W = mB\left(1 - \frac{1}{2}\right) = mB\left(\frac{1}{2}\right) = \frac{mB}{2} \]

3. **Relating Work Done to Given Value**:
According to the problem, the work done is \( X \) units. Therefore, we can equate:
\[ \frac{mB}{2} = X \]
From this, we can express \( mB \) in terms of \( X \):
\[ mB = 2X \]

4. **Calculating the Torque**:
The torque \( \tau \) required to maintain the needle at an angle \( \theta \) is given by:
\[ \tau = mB \sin \theta \]
Substituting \( mB = 2X \) and \( \theta = 60^\circ \) (where \( \sin 60^\circ = \frac{\sqrt{3}}{2} \)):
\[ \tau = 2X \sin 60^\circ = 2X \cdot \frac{\sqrt{3}}{2} = \sqrt{3}X \]

5. **Final Answer**:
Therefore, the torque necessary to maintain the needle in this position is:
\[ \tau = \sqrt{3}X \]

### Summary:
The torque necessary to maintain the needle in the position after turning it through \( 60^\circ \) is \( \sqrt{3}X \).