When a bar magnet is placed at `90^(@)` to a uniform magnetic field, it is acted upon by a couple which is maximum. For the couple to be half of the maximum value, at what angle should the magnet be inclined to the magnetic field `(B)`?

Options

A. (a)`30^@`
B. (b)`45^@`
C. (c)`60^@`
D. (d)`90^@`

Answer

To solve the problem, we need to determine the angle at which the couple (torque) acting on a bar magnet in a magnetic field is half of its maximum value.

### Step-by-Step Solution:

1. **Understanding the Torque on a Bar Magnet**:
The torque (τ) acting on a bar magnet in a magnetic field is given by the formula:
\[ τ = mB \sin(\theta) \]
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. **Maximum Torque**:
The maximum torque occurs when the angle \( \theta \) is \( 90^\circ \) (i.e., the magnet is perpendicular to the magnetic field). At this angle:
\[ τ_{\text{max}} = mB \sin(90^\circ) = mB \cdot 1 = mB \]

3. **Finding Half of the Maximum Torque**:
We want to find the angle \( \theta_2 \) such that the torque is half of the maximum torque:
\[ τ = \frac{1}{2} τ_{\text{max}} = \frac{1}{2} mB \]

4. **Setting Up the Equation**:
From the torque equation, we can set up the following:
\[ mB \sin(\theta_2) = \frac{1}{2} mB \]

5. **Simplifying the Equation**:
We can cancel \( mB \) from both sides (assuming \( mB \neq 0 \)):
\[ \sin(\theta_2) = \frac{1}{2} \]

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

### Final Answer:
The angle at which the magnet should be inclined to the magnetic field for the couple to be half of the maximum value is \( 30^\circ \).

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