A bar magnet is held perpendicular to a uniform magnetic field. If the couple acting on the magnet is to be halved by rotating it, then the angle by which it is to be rotated is

Options

A. `30^(@)`
B. `45^(@)`
C. `60^(@)`
D. `90^(@)`

Explanation

To solve the problem, we need to determine the angle by which a bar magnet, initially held perpendicular to a uniform magnetic field, should be rotated to halve the torque acting on it.

### Step-by-Step Solution:

1. **Understanding Torque on a Bar Magnet**:
The torque (\( \tau \)) acting on a bar magnet in a magnetic field is given by the formula:
\[ \tau = mB \sin(\theta) \]
where:
- \( m \) is the magnetic moment of the bar magnet,
- \( B \) is the magnetic field strength,
- \( \theta \) is the angle between the magnetic moment and the magnetic field.

2. **Initial Conditions**:
Initially, the bar magnet is held perpendicular to the magnetic field, which means:
\[ \theta_1 = 90^\circ \]
Therefore, the torque at this position is:
\[ \tau_1 = mB \sin(90^\circ) = mB \]

3. **Halving the Torque**:
We want to find the angle \( \theta_2 \) such that the torque is halved:
\[ \tau_2 = \frac{1}{2} \tau_1 = \frac{1}{2} mB \]

4. **Setting Up the Equation**:
Using the torque formula for the new angle \( \theta_2 \):
\[ \tau_2 = mB \sin(\theta_2) \]
Setting this equal to the halved torque:
\[ mB \sin(\theta_2) = \frac{1}{2} mB \]
We can cancel \( mB \) from both sides (assuming \( mB \neq 0 \)):
\[ \sin(\theta_2) = \frac{1}{2} \]

5. **Finding the Angle**:
The angle \( \theta_2 \) that satisfies \( \sin(\theta_2) = \frac{1}{2} \) is:
\[ \theta_2 = 30^\circ \]

6. **Conclusion**:
Thus, the angle by which the bar magnet should be rotated to halve the torque is:
\[ \theta_2 = 30^\circ \]