A bar magnet having a magnetic moment of `2 xx 10^4 JT^(-1)` is free to rotate in a horizontal plane. A horizontal magnetic field `B = 6 xx 10^(-4)` T exists in the space. The work done in taking the magnet slowly from a direction parallel to the field to a direction `60^@` from the field is

Options

A. 12J
B. 6J
C. 2J
D. 0.6J

Explanation

To solve the problem step by step, we will follow these steps:

### Step 1: Understand the given values
- Magnetic moment of the bar magnet, \( M = 2 \times 10^4 \, \text{JT}^{-1} \)
- Magnetic field strength, \( B = 6 \times 10^{-4} \, \text{T} \)
- Initial angle, \( \theta_1 = 0^\circ \) (parallel to the field)
- Final angle, \( \theta_2 = 60^\circ \)

### Step 2: Write the expression for work done
The work done \( W \) in rotating the magnet in a magnetic field can be expressed as:
\[ W = \int_{\theta_1}^{\theta_2} \tau \, d\theta \]
where \( \tau \) is the torque acting on the magnet.

### Step 3: Calculate the torque
The torque \( \tau \) on the magnet in a magnetic field is given by:
\[ \tau = M \cdot B \cdot \sin(\theta) \]
Thus, we can substitute this into the work done expression:
\[ W = \int_{\theta_1}^{\theta_2} M B \sin(\theta) \, d\theta \]

### Step 4: Substitute the limits and constants
Since \( M \) and \( B \) are constants, we can take them out of the integral:
\[ W = M B \int_{\theta_1}^{\theta_2} \sin(\theta) \, d\theta \]
Now substituting the limits:
\[ W = M B \left[ -\cos(\theta) \right]_{\theta_1}^{\theta_2} \]

### Step 5: Evaluate the integral
Substituting the limits \( \theta_1 = 0^\circ \) and \( \theta_2 = 60^\circ \):
\[ W = M B \left[ -\cos(60^\circ) + \cos(0^\circ) \right] \]
Using the values \( \cos(60^\circ) = \frac{1}{2} \) and \( \cos(0^\circ) = 1 \):
\[ W = M B \left[ -\frac{1}{2} + 1 \right] = M B \left[ \frac{1}{2} \right] \]

### Step 6: Substitute the values of \( M \) and \( B \)
Now substituting the values of \( M \) and \( B \):
\[ W = (2 \times 10^4 \, \text{JT}^{-1}) \cdot (6 \times 10^{-4} \, \text{T}) \cdot \frac{1}{2} \]
Calculating this gives:
\[ W = (2 \times 6 \times 10^4 \times 10^{-4}) \cdot \frac{1}{2} = (12 \times 10^0) \cdot \frac{1}{2} = 6 \, \text{J} \]

### Final Answer
The work done in taking the magnet from parallel to the field to \( 60^\circ \) from the field is \( \boxed{6 \, \text{J}} \).

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