The work done in rotating the magnet form the direction of uniform field to the opposite direction to the field is`W`. The work done in rotating the magnet form the field direction to half the maximum couple position is

Options

A. `2W`
B. `(sqrt3W)/(2)`
C. `(W)/(4)(2-sqrt3)`
D. `(W)/(2)(1-sqrt3)`

Explanation

To solve the problem of finding the work done in rotating a magnet from the direction of a uniform magnetic field to half the maximum couple position, we can follow these steps:

### Step 1: Understand the Initial Conditions
The work done in rotating the magnet from the direction of the uniform magnetic field to the opposite direction is given as \( W \). This means we are rotating the magnet from an angle of \( 0^\circ \) (aligned with the field) to \( 180^\circ \) (opposite to the field).

### Step 2: Calculate Work Done from 0° to 180°
The potential energy \( U \) of a magnetic dipole in a magnetic field is given by:
\[ U = -mB \cos(\theta) \]
Where:
- \( m \) is the magnetic moment of the magnet,
- \( B \) is the magnetic field strength,
- \( \theta \) is the angle between the magnetic moment and the magnetic field.

At \( \theta = 0^\circ \):
\[ U_0 = -mB \cos(0) = -mB \]

At \( \theta = 180^\circ \):
\[ U_{180} = -mB \cos(180) = mB \]

The work done \( W \) in rotating from \( 0^\circ \) to \( 180^\circ \) is the change in potential energy:
\[ W = U_{180} - U_0 = mB - (-mB) = 2mB \]

### Step 3: Determine Half the Maximum Couple Position
The maximum torque \( \tau_{\text{max}} \) occurs at \( \theta = 90^\circ \):
\[ \tau_{\text{max}} = mB \]
Half of the maximum torque is:
\[ \tau = \frac{mB}{2} \]

### Step 4: Find the Angle Corresponding to Half the Maximum Torque
The torque \( \tau \) is given by:
\[ \tau = mB \sin(\theta) \]
Setting this equal to half the maximum torque:
\[ mB \sin(\theta) = \frac{mB}{2} \]
Dividing both sides by \( mB \) (assuming \( mB \neq 0 \)):
\[ \sin(\theta) = \frac{1}{2} \]
Thus, \( \theta = 30^\circ \).

### Step 5: Calculate Work Done from 0° to 30°
Now we calculate the work done from \( 0^\circ \) to \( 30^\circ \):
\[ W_{0 \to 30} = U_{30} - U_0 \]
Calculating \( U_{30} \):
\[ U_{30} = -mB \cos(30) = -mB \cdot \frac{\sqrt{3}}{2} \]
Thus:
\[ W_{0 \to 30} = \left(-mB \cdot \frac{\sqrt{3}}{2}\right) - (-mB) = mB \left(1 - \frac{\sqrt{3}}{2}\right) \]

### Step 6: Relate \( mB \) to \( W \)
From our earlier calculation, we know:
\[ W = 2mB \implies mB = \frac{W}{2} \]
Substituting this into the work done from \( 0^\circ \) to \( 30^\circ \):
\[ W_{0 \to 30} = \frac{W}{2} \left(1 - \frac{\sqrt{3}}{2}\right) = \frac{W}{2} \cdot \frac{2 - \sqrt{3}}{2} = \frac{W(2 - \sqrt{3})}{4} \]

### Final Answer
The work done in rotating the magnet from the direction of the field to half the maximum couple position is:
\[ W_{0 \to 30} = \frac{W(2 - \sqrt{3})}{4} \]