The magnetic moment of a short bar magnet placed with its magnetic axis at `30^(@)` to an external field of 900 G and experiences a torque of 0.02 N m is

Options

A. `0.35Am^(2)`
B. `0.44Am^(2)`
C. `2.45Am^(2)`
D. `1.5Am^(2)`

Explanation

To find the magnetic moment \( m \) of a short bar magnet placed at an angle of \( 30^\circ \) to an external magnetic field of \( 900 \, \text{G} \) experiencing a torque of \( 0.02 \, \text{N m} \), we can use the formula for torque:

\[ \tau = m B \sin \theta \]

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. **Convert the Magnetic Field from Gauss to Tesla:**
\[ B = 900 \, \text{G} = 900 \times 10^{-4} \, \text{T} = 0.09 \, \text{T} \]

2. **Identify the Given Values:**
- Torque \( \tau = 0.02 \, \text{N m} \)
- Magnetic field \( B = 0.09 \, \text{T} \)
- Angle \( \theta = 30^\circ \)

3. **Calculate \( \sin \theta \):**
\[ \sin 30^\circ = \frac{1}{2} \]

4. **Substitute the Known Values into the Torque Equation:**
\[ 0.02 = m \cdot 0.09 \cdot \frac{1}{2} \]

5. **Rearranging the Equation to Solve for \( m \):**
\[ 0.02 = m \cdot 0.045 \]
\[ m = \frac{0.02}{0.045} \]

6. **Calculating \( m \):**
\[ m = \frac{0.02}{0.045} \approx 0.4444 \, \text{A m}^2 \]

7. **Final Answer:**
\[ m \approx 0.44 \, \text{A m}^2 \]

### Summary:
The magnetic moment \( m \) of the short bar magnet is approximately \( 0.44 \, \text{A m}^2 \).