The work done in turning a magnet of magnetic moment 'M' by an angle of `90^(@)` from the meridian is 'n' times the corresponding work done to turn it through an angle of `60^(@)`, where 'n' is given by

Options

A. 1
B. 2
C. `1//2`
D. `1//4`

Answer

To solve the problem, we need to calculate the work done in turning a magnet of magnetic moment 'M' through angles of 90 degrees and 60 degrees from the meridian. The relationship between these two works will help us find the value of 'n'.

### Step-by-Step Solution:

1. **Understanding Work Done in Turning the Magnet**:
The work done \( W \) in turning a magnet is related to the change in potential energy. 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, \( B \) is the magnetic field strength, and \( \theta \) is the angle between the magnetic moment and the magnetic field.

2. **Calculating Work Done for 90 Degrees**:
For the first case, when the magnet is turned from the meridian (0 degrees) to 90 degrees:
- Initial angle \( \theta_i = 0^\circ \)
- Final angle \( \theta_f = 90^\circ \)

The work done \( W_1 \) is given by:
\[ W_1 = U_i - U_f = (-MB \cos 0^\circ) - (-MB \cos 90^\circ) \]
\[ W_1 = -MB(1) - 0 = -MB \]

3. **Calculating Work Done for 60 Degrees**:
For the second case, when the magnet is turned from 0 degrees to 60 degrees:
- Initial angle \( \theta_i = 0^\circ \)
- Final angle \( \theta_f = 60^\circ \)

The work done \( W_2 \) is given by:
\[ W_2 = U_i - U_f = (-MB \cos 0^\circ) - (-MB \cos 60^\circ) \]
\[ W_2 = -MB(1) - (-MB \cdot \frac{1}{2}) = -MB + \frac{MB}{2} = -\frac{MB}{2} \]

4. **Finding the Ratio of Work Done**:
We know from the problem statement that:
\[ W_1 = n \cdot W_2 \]
Substituting the values we found:
\[ -MB = n \left(-\frac{MB}{2}\right) \]
Dividing both sides by \(-MB\) (assuming \(MB \neq 0\)):
\[ 1 = n \cdot \frac{1}{2} \]
Therefore:
\[ n = 2 \]

### Final Answer:
The value of \( n \) is \( 2 \).

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