A bar magnet has a magnetic moment of 200 A m? The magnet is suspended in a magnetic field of 0.30 `"N A"^(-1)" m"^(-1)`. The torque required to rotate the magnet from its equilibrium position through an angle of `30^(@)`, will be

Options

A. 30 N m
B. `30sqrt(30)` N m
C. 60 N m
D. `60sqrt(3)` N m

Explanation

To find the torque required to rotate a bar magnet from its equilibrium position through an angle of \(30^\circ\), we can use the formula for torque (\(\tau\)) acting on a magnetic moment (\(M\)) in a magnetic field (\(B\)):

\[ \tau = M \cdot B \cdot \sin \theta \]

Where:
- \(M\) 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:**
- Magnetic moment, \(M = 200 \, \text{A m}^2\)
- Magnetic field, \(B = 0.30 \, \text{N A}^{-1} \text{m}^{-1}\)
- Angle, \(\theta = 30^\circ\)

2. **Convert the angle to radians if necessary:**
- In this case, we can use the sine function directly with degrees, so no conversion is needed.

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

4. **Substitute the values into the torque formula:**
\[ \tau = M \cdot B \cdot \sin \theta \]
\[ \tau = 200 \, \text{A m}^2 \cdot 0.30 \, \text{N A}^{-1} \text{m}^{-1} \cdot \sin 30^\circ \]

5. **Perform the multiplication:**
\[ \tau = 200 \cdot 0.30 \cdot \frac{1}{2} \]
\[ \tau = 200 \cdot 0.30 \cdot 0.5 \]
\[ \tau = 200 \cdot 0.15 \]
\[ \tau = 30 \, \text{N m} \]

6. **Final Result:**
- The torque required to rotate the magnet through an angle of \(30^\circ\) is \(30 \, \text{N m}\).