A bar magnet is initially parallel to the magnetic field in which it is placed. Work done u rotate it by `60^(@)` is 2.6 J. Work done in rotating it further by `30^(@)` will be

Options

A. 1.3 J
B. zero
C. 2.6 J
D. 5.43 J

Answer

To solve the problem, we need to calculate the work done in rotating a bar magnet in a magnetic field. Let's break down the steps:

### Step 1: Understand the Work Done Formula
The work done \( W \) in rotating a magnetic moment \( M \) in a magnetic field \( B \) is given by the formula:
\[ W = MB \left( \cos \theta_1 - \cos \theta_2 \right) \]
where:
- \( M \) is the magnetic moment of the magnet,
- \( B \) is the magnetic field strength,
- \( \theta_1 \) is the initial angle,
- \( \theta_2 \) is the final angle.

### Step 2: Initial Conditions
Initially, the bar magnet is parallel to the magnetic field, which means:
- \( \theta_1 = 0^\circ \)
- \( \cos \theta_1 = \cos 0^\circ = 1 \)

### Step 3: First Rotation
The magnet is rotated by \( 60^\circ \). Therefore:
- \( \theta_2 = 60^\circ \)
- \( \cos \theta_2 = \cos 60^\circ = \frac{1}{2} \)

Using the work done for the first rotation:
\[ W_1 = MB \left( \cos 0^\circ - \cos 60^\circ \right) = MB \left( 1 - \frac{1}{2} \right) = MB \left( \frac{1}{2} \right) \]
Given that \( W_1 = 2.6 \, J \):
\[ MB \left( \frac{1}{2} \right) = 2.6 \implies MB = 5.2 \]

### Step 4: Second Rotation
Now, we need to find the work done in rotating the magnet further by \( 30^\circ \). The final angle after this rotation will be:
- \( \theta_3 = 60^\circ + 30^\circ = 90^\circ \)
- \( \cos \theta_3 = \cos 90^\circ = 0 \)

Using the work done for the second rotation:
\[ W_2 = MB \left( \cos 60^\circ - \cos 90^\circ \right) = MB \left( \frac{1}{2} - 0 \right) = MB \left( \frac{1}{2} \right) \]

### Step 5: Calculate Work Done
Substituting \( MB = 5.2 \):
\[ W_2 = 5.2 \left( \frac{1}{2} \right) = 2.6 \, J \]

### Conclusion
The work done in rotating the magnet further by \( 30^\circ \) is \( 2.6 \, J \).

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