A short bar magnet placed with its axis at `30^(@)` to a uniform magnetic field of 0.2 T experience torque of 0.6 Nm. Then magnetic moment will be:

Options

A. `1.0 Am^(2)`
B. `1.5 Am^(2)`
C. `0.6 Am^(2)`
D. `0.2 Am^(2)`

Explanation

To find the magnetic moment of a short bar magnet placed at an angle to a uniform magnetic field, we can use the formula for torque. The torque (\( \tau \)) experienced by a magnetic moment (\( m \)) in a magnetic field (\( B \)) is given by the equation:

\[ \tau = m \cdot B \cdot \sin(\theta) \]

Where:
- \( \tau \) is the torque,
- \( m \) is the magnetic moment,
- \( B \) is the magnetic field strength,
- \( \theta \) is the angle between the magnetic moment and the magnetic field.

### Step-by-Step Solution:

1. **Identify the given values:**
- Torque (\( \tau \)) = 0.6 Nm
- Magnetic field strength (\( B \)) = 0.2 T
- Angle (\( \theta \)) = 30°

2. **Convert the angle to radians if necessary:**
- Since we are using sine, we can directly use degrees.
- \( \sin(30°) = \frac{1}{2} \)

3. **Substitute the values into the torque formula:**
\[ \tau = m \cdot B \cdot \sin(\theta) \]
\[ 0.6 = m \cdot 0.2 \cdot \sin(30°) \]
\[ 0.6 = m \cdot 0.2 \cdot \frac{1}{2} \]

4. **Simplify the equation:**
\[ 0.6 = m \cdot 0.1 \]

5. **Solve for the magnetic moment (\( m \)):**
\[ m = \frac{0.6}{0.1} = 6 \text{ Ampere meter}^2 \]

### Final Answer:
The magnetic moment of the bar magnet is \( 6 \, \text{A m}^2 \).