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 solve the problem, we need to calculate the torque (or couple) acting on a magnet when it is deflected by an angle of \(30^\circ\) in a magnetic field. The formula for torque (\(\tau\)) in a magnetic field is 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 of deflection.

### 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:**
- In this case, we can directly use degrees since the sine function can take degrees as input.

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

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

5. **Simplify the equation:**
- The \(2\) in the numerator and the \(2\) in the denominator cancel out:
\[ \tau = 0.25 \cdot 10^{-5} \]

6. **Calculate the final value:**
\[ \tau = 0.25 \times 10^{-5} \, \text{N-m} = 2.5 \times 10^{-6} \, \text{N-m} \]

### Final Answer:
The couple acting on the magnet when deflected by \(30^\circ\) is \(2.5 \times 10^{-6} \, \text{N-m}\).

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