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

Options

A. `sqrt(3) W`
B. `W`
C. `(sqrt(3))/(2) W`
D. `2W`

Explanation

To solve the problem, we need to find the torque required to maintain a magnetic needle in a position after it has been turned through 60 degrees in a magnetic field. We start by using the relationship between work done and torque.

### Step-by-Step Solution:

1. **Understanding Work Done**:
The work done \( W \) in rotating a magnetic needle in a magnetic field is given by the formula:
\[ W = MB \left( \cos \theta_1 - \cos \theta_2 \right) \]
where:
- \( M \) is the magnetic moment of the needle,
- \( B \) is the magnetic field strength,
- \( \theta_1 \) is the initial angle (0 degrees, since it is parallel),
- \( \theta_2 \) is the final angle (60 degrees).

2. **Calculating Work Done**:
Substituting the values into the work done formula:
\[ W = MB \left( \cos 0^\circ - \cos 60^\circ \right) \]
We know that:
- \( \cos 0^\circ = 1 \)
- \( \cos 60^\circ = \frac{1}{2} \)

Thus:
\[ W = MB \left( 1 - \frac{1}{2} \right) = MB \left( \frac{1}{2} \right) = \frac{MB}{2} \]

3. **Finding Torque**:
The torque \( \tau \) required to maintain the needle in the new position is given by:
\[ \tau = MB \sin \theta \]
where \( \theta \) is the angle at which the needle is held, which is \( 60^\circ \) in this case.

4. **Calculating Torque**:
Substituting \( \theta = 60^\circ \):
\[ \tau = MB \sin 60^\circ \]
We know that:
- \( \sin 60^\circ = \frac{\sqrt{3}}{2} \)

Therefore:
\[ \tau = MB \cdot \frac{\sqrt{3}}{2} = \frac{\sqrt{3} MB}{2} \]

5. **Relating Torque to Work Done**:
From the previous step, we found that \( W = \frac{MB}{2} \). Thus, we can express \( MB \) in terms of \( W \):
\[ MB = 2W \]

Now substituting this back into the torque equation:
\[ \tau = \frac{\sqrt{3}}{2} (2W) = \sqrt{3} W \]

### Final Answer:
The torque needed to maintain the needle in this position will be:
\[ \tau = \sqrt{3} W \]