The magnetic moment of a magnet is `0.25 A-m^(2)` . It is suspended in a magnetic field of intesity `2 xx 10^(-5)` T . The couple acting on it when deflected by ` 30^(@)` from the magnetic fields is .

Options

A. `2.5xx 10^(-5) N- m`
B. `2.5 xx 10^(-6) N - m`
C. `1.25xx10^(-6) N - m`
D. `1.25 xx 10^(-5) N- m`

Explanation

To find the couple (torque) acting on a magnet when it is deflected from the magnetic field, we can use the formula for torque (τ) given by:

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

where:
- \( M \) is the magnetic moment,
- \( B \) is the magnetic field intensity,
- \( \theta \) is the angle between the magnetic moment and the magnetic field.

### Step-by-step Solution:

1. **Identify the given values**:
- Magnetic moment, \( M = 0.25 \, \text{A-m}^2 \)
- Magnetic field intensity, \( B = 2 \times 10^{-5} \, \text{T} \)
- Angle of deflection, \( \theta = 30^\circ \)

2. **Convert the angle to radians if necessary** (not required here since we can use degrees directly for sine):
- \( \sin(30^\circ) = \frac{1}{2} \)

3. **Substitute the values into the torque formula**:
\[ \tau = M \cdot B \cdot \sin(\theta) \]
\[ \tau = 0.25 \, \text{A-m}^2 \cdot 2 \times 10^{-5} \, \text{T} \cdot \sin(30^\circ) \]
\[ \tau = 0.25 \cdot 2 \times 10^{-5} \cdot \frac{1}{2} \]

4. **Simplify the expression**:
\[ \tau = 0.25 \cdot 2 \times 10^{-5} \cdot 0.5 \]
\[ \tau = 0.25 \cdot 10^{-5} \]
\[ \tau = 0.125 \times 10^{-5} \, \text{N-m} \]

5. **Convert to standard form**:
\[ \tau = 1.25 \times 10^{-6} \, \text{N-m} \]

### Final Answer:
The couple acting on the magnet when deflected by \( 30^\circ \) from the magnetic field is \( 1.25 \times 10^{-6} \, \text{N-m} \).