The work done in turning a magnet of magnetic moment M by an angle of `90^(@)` from the magnetic meridian is n times the corresponding work done to turn through an angle of `60^(@)`. What is the value of n ?

Answer

To solve the problem, we need to find the value of \( n \) such that the work done in turning a magnet of magnetic moment \( M \) by an angle of \( 90^\circ \) from the magnetic meridian is \( n \) times the work done to turn it through an angle of \( 60^\circ \).

### Step 1: Understand the formula for work done
The work done \( W \) in turning a magnetic moment \( M \) through an angle \( \theta \) in a uniform magnetic field \( B \) is given by the formula:
\[ W = -M B \cos(\theta) \]
where \( \theta \) is the angle between the magnetic moment and the magnetic field.

### Step 2: Calculate the work done for \( 90^\circ \)
For \( \theta = 90^\circ \):
\[ W_{90} = -M B \cos(90^\circ) = -M B \cdot 0 = 0 \]
Thus, the work done in turning the magnet by \( 90^\circ \) is \( 0 \).

### Step 3: Calculate the work done for \( 60^\circ \)
For \( \theta = 60^\circ \):
\[ W_{60} = -M B \cos(60^\circ) = -M B \cdot \frac{1}{2} = -\frac{M B}{2} \]

### Step 4: Relate the two work done values
According to the problem, the work done in turning the magnet by \( 90^\circ \) is \( n \) times the work done in turning it by \( 60^\circ \):
\[ W_{90} = n \cdot W_{60} \]
Substituting the values we found:
\[ 0 = n \cdot \left(-\frac{M B}{2}\right) \]

### Step 5: Solve for \( n \)
Since the left side is \( 0 \), the only way for the equation to hold true is if \( n = 0 \):
\[ n = 0 \]

### Conclusion
The value of \( n \) is \( 0 \).

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