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

Options

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

Answer

To solve the problem, we need to determine the external torque required to maintain a magnetic needle in a position after it has been turned through an angle of \(60^\circ\) in a magnetic field. We will follow these steps:

### Step 1: Understand the Torque Formula
The torque (\(\tau\)) acting on a magnetic needle in a magnetic field can be expressed as:
\[ \tau = mB \sin \theta \]
where:
- \(m\) is the magnetic moment of the needle,
- \(B\) is the magnetic field strength,
- \(\theta\) is the angle between the magnetic moment and the magnetic field.

### Step 2: Identify the Initial and Final Angles
Initially, the magnetic needle is parallel to the magnetic field, which means:
- Initial angle \(\theta_1 = 0^\circ\)
- Final angle \(\theta_2 = 60^\circ\)

### Step 3: Calculate the Torque at \(60^\circ\)
Using the torque formula, we substitute \(\theta = 60^\circ\):
\[ \tau = mB \sin 60^\circ \]
Since \(\sin 60^\circ = \frac{\sqrt{3}}{2}\), we have:
\[ \tau = mB \cdot \frac{\sqrt{3}}{2} = \frac{\sqrt{3}}{2} mB \]

### Step 4: Relate Work Done to Torque
The work done (\(W\)) in turning the magnetic needle from \(0^\circ\) to \(60^\circ\) can be expressed as:
\[ W = mB \left( \cos \theta_1 - \cos \theta_2 \right) \]
Substituting \(\theta_1 = 0^\circ\) and \(\theta_2 = 60^\circ\):
\[ W = mB \left( \cos 0^\circ - \cos 60^\circ \right) \]
Knowing that \(\cos 0^\circ = 1\) and \(\cos 60^\circ = \frac{1}{2}\), we get:
\[ W = mB \left( 1 - \frac{1}{2} \right) = mB \cdot \frac{1}{2} = \frac{1}{2} mB \]

### Step 5: Solve for \(mB\) in Terms of \(W\)
From the equation \(W = \frac{1}{2} mB\), we can express \(mB\) as:
\[ mB = 2W \]

### Step 6: Substitute \(mB\) into the Torque Equation
Now we substitute \(mB\) back into the torque equation:
\[ \tau = \frac{\sqrt{3}}{2} mB = \frac{\sqrt{3}}{2} \cdot (2W) = \sqrt{3} W \]

### Conclusion
The external torque required to maintain the magnetic needle in the position after turning through \(60^\circ\) is:
\[ \tau = \sqrt{3} W \]