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 find the couple (torque) acting on a magnet when it is deflected from a 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 of the magnet,
- \( B \) 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, \( m = 0.1 \, \text{A m}^2 \)
- Magnetic field intensity, \( B = 3 \times 10^{-4} \, \text{Wb/m}^2 \)
- Angle of deflection, \( \theta = 30^\circ \)

2. **Convert the angle to radians if necessary:**
- For this calculation, we can use degrees directly since the sine function can take degrees as input.

3. **Calculate the sine of the angle:**
- \( \sin(30^\circ) = \frac{1}{2} \)

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

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

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