A uniform horizontal magnetic field of `7.5xx10^(-2)T` is set up at an angle of `30^(@)` with the axis of an solenoid and the magnetic moment associated with it is `1.28JT^(-1)`. Then the torque on it is

Options

A. `4.8xx10^(-2)Nm`
B. `1.6xx10^(-2)Nm`
C. `1.2xx10^(-2)Nm`
D. `4.8xx10^(-4)Nm`

Explanation

To find the torque on the solenoid in a magnetic field, we can use the formula for torque (\( \tau \)) given by:

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

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, \( m = 1.28 \, \text{J T}^{-1} \)
- Magnetic field, \( B = 7.5 \times 10^{-2} \, \text{T} \)
- Angle, \( \theta = 30^\circ \)

2. **Convert the angle to radians if necessary:**
- In this case, we can use the sine value directly for \( 30^\circ \) without converting to radians.
- \( \sin(30^\circ) = \frac{1}{2} \)

3. **Substitute the values into the torque formula:**
\[ \tau = m \cdot B \cdot \sin(\theta) \]
\[ \tau = 1.28 \, \text{J T}^{-1} \cdot 7.5 \times 10^{-2} \, \text{T} \cdot \sin(30^\circ) \]

4. **Calculate \( \tau \):**
\[ \tau = 1.28 \cdot 7.5 \times 10^{-2} \cdot \frac{1}{2} \]
\[ \tau = 1.28 \cdot 7.5 \times 10^{-2} \cdot 0.5 \]
\[ \tau = 1.28 \cdot 0.0375 \]
\[ \tau = 0.048 \, \text{N m} \]

5. **Express the torque in scientific notation:**
\[ \tau = 4.8 \times 10^{-2} \, \text{N m} \]

### Final Answer:
The torque on the solenoid is \( 4.8 \times 10^{-2} \, \text{N m} \).