The magnetic moment of a magnet is `0.1 ampxxm^(2)`. It is suspended in a magnetic field of intensity `3xx10^(-4)"weber"//m^(2)`. The couple acting upon it when deflected by `30^(@)` from the magnetic field is

Options

A. `1xx10^(-5)Nm`
B. `1.5xx10^(-5)Nm`
C. `2xx10^(-5)Nm`
D. `2.5xx10^(-5)Nm`

Explanation

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

### Step-by-step Solution:

1. **Identify the given values:**
- Magnetic moment, \(M = 0.1 \, \text{A m}^2\)
- Magnetic field intensity, \(B = 3 \times 10^{-4} \, \text{Wb/m}^2\)
- Angle of deflection, \(\theta = 30^\circ\)

2. **Convert the angle to radians (optional):**
- However, since we will use \(\sin(30^\circ)\), we can directly use the value.
- \(\sin(30^\circ) = \frac{1}{2}\)

3. **Substitute the values into the torque formula:**
\[ \tau = M \cdot B \cdot \sin(\theta) \]
\[ \tau = 0.1 \, \text{A m}^2 \cdot 3 \times 10^{-4} \, \text{Wb/m}^2 \cdot \sin(30^\circ) \]
\[ \tau = 0.1 \cdot 3 \times 10^{-4} \cdot \frac{1}{2} \]

4. **Calculate the torque:**
\[ \tau = 0.1 \cdot 3 \times 10^{-4} \cdot 0.5 \]
\[ \tau = 0.1 \cdot 1.5 \times 10^{-4} \]
\[ \tau = 1.5 \times 10^{-5} \, \text{N m} \]

5. **Final result:**
The couple acting upon the magnet when deflected by \(30^\circ\) from the magnetic field is:
\[ \tau = 1.5 \times 10^{-5} \, \text{N m} \]