A bar magnet with magnetic moment `2.5xx10^(3)JT^(-2)` is rotating in horizontal plane in the space containing magnetic induction `B=4xx10^(5)T`. The work done in rotating the magnet slowly from a direction parallel to the field to a direction `45^(@)` from the field, is (in joule).

Options

A. 0
B. 0.2
C. 0.03
D. 0.02

Explanation

To solve the problem of calculating the work done in rotating a bar magnet from a direction parallel to the magnetic field to a direction 45 degrees from the field, we can use the formula for work done in rotating a magnetic moment in a magnetic field.

### Step-by-Step Solution:

1. **Understand the Magnetic Moment and Magnetic Field**:
- The magnetic moment \( M \) of the bar magnet is given as \( 2.5 \times 10^3 \, \text{JT}^{-2} \).
- The magnetic induction \( B \) is given as \( 4 \times 10^{-5} \, \text{T} \).

2. **Work Done Formula**:
- The work done \( W \) in rotating a magnetic moment in a magnetic field is given by the formula:
\[ W = -M B \cos \theta_f + M B \cos \theta_i \]
- Here, \( \theta_f \) is the final angle (45 degrees) and \( \theta_i \) is the initial angle (0 degrees, parallel to the field).

3. **Convert Angles to Radians**:
- Convert the angles from degrees to radians for calculations:
- \( \theta_i = 0^\circ = 0 \, \text{radians} \)
- \( \theta_f = 45^\circ = \frac{\pi}{4} \, \text{radians} \)

4. **Calculate Cosines**:
- Calculate \( \cos \theta_i \) and \( \cos \theta_f \):
- \( \cos \theta_i = \cos(0) = 1 \)
- \( \cos \theta_f = \cos\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}} \)

5. **Substitute Values into the Work Done Formula**:
- Substitute the values into the work done formula:
\[ W = -M B \left(\frac{1}{\sqrt{2}}\right) + M B (1) \]
- This simplifies to:
\[ W = MB \left(1 - \frac{1}{\sqrt{2}}\right) \]

6. **Calculate the Work Done**:
- Now substitute \( M = 2.5 \times 10^3 \, \text{JT}^{-2} \) and \( B = 4 \times 10^{-5} \, \text{T} \):
\[ W = (2.5 \times 10^3) \times (4 \times 10^{-5}) \left(1 - \frac{1}{\sqrt{2}}\right) \]
- Calculate \( 2.5 \times 4 = 10 \):
\[ W = 10 \times 10^{-2} \left(1 - \frac{1}{\sqrt{2}}\right) = 0.1 \left(1 - \frac{1}{\sqrt{2}}\right) \]

7. **Final Calculation**:
- Now calculate \( 1 - \frac{1}{\sqrt{2}} \):
\[ \frac{1}{\sqrt{2}} \approx 0.707 \Rightarrow 1 - 0.707 \approx 0.293 \]
- Therefore:
\[ W \approx 0.1 \times 0.293 \approx 0.0293 \, \text{J} \]

### Final Answer:
The work done in rotating the magnet slowly from a direction parallel to the field to a direction 45 degrees from the field is approximately \( 0.0293 \, \text{J} \).