A bar magnet of length `10cm` and having the pole strength equal to `10^(-3)` weber is kept in a magnetic field having magnetic induction (B) equal to `4pixx10^(-3)` Tesla. It makes an angle of `30^(@)` with the direction of magnetic induction. The value of the torque acting on the magnet is
`(mu_(0)=4pixx10^(-7) weber//ampxxm)`

Options

A. `2pixx10^(-7) Nxxm`
B. `2pixx10^(-5) Nxxm`
C. `0.5 Nxxm`
D. `0.5xx10^(2) Nxxm`

Explanation

To find the torque acting on the bar magnet, we can use the formula for torque (\( \tau \)) in a magnetic field:

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

Where:
- \( m \) is the pole strength of the magnet,
- \( B \) is the magnetic induction,
- \( L \) is the length of the magnet,
- \( \theta \) is the angle between the magnet and the magnetic field.

### Step 1: Identify the given values
- Length of the magnet, \( L = 10 \, \text{cm} = 0.1 \, \text{m} \)
- Pole strength, \( m = 10^{-3} \, \text{Wb} \)
- Magnetic induction, \( B = 4\pi \times 10^{-3} \, \text{T} \)
- Angle, \( \theta = 30^\circ \)

### Step 2: Convert angle to radians if necessary
For this problem, we can use the sine function directly in degrees, so we do not need to convert it to radians.

### Step 3: Calculate \( \sin(\theta) \)
\[ \sin(30^\circ) = \frac{1}{2} \]

### Step 4: Substitute the values into the torque formula
Now we can substitute the values into the torque formula:

\[ \tau = (10^{-3} \, \text{Wb}) \cdot (4\pi \times 10^{-3} \, \text{T}) \cdot (0.1 \, \text{m}) \cdot \sin(30^\circ) \]

### Step 5: Calculate the torque
Substituting \( \sin(30^\circ) = \frac{1}{2} \):

\[ \tau = (10^{-3}) \cdot (4\pi \times 10^{-3}) \cdot (0.1) \cdot \left(\frac{1}{2}\right) \]

Calculating the constants:

\[ \tau = 10^{-3} \cdot 4\pi \cdot 10^{-3} \cdot 0.1 \cdot 0.5 \]

\[ \tau = 10^{-3} \cdot 4\pi \cdot 0.05 \cdot 10^{-3} \]

\[ \tau = 2\pi \cdot 10^{-6} \, \text{N m} \]

### Step 6: Final value of torque
Using the approximate value of \( \pi \approx 3.14 \):

\[ \tau \approx 2 \cdot 3.14 \cdot 10^{-6} \approx 6.28 \times 10^{-6} \, \text{N m} \]

Thus, the torque acting on the magnet is approximately:

\[ \tau \approx 6.28 \times 10^{-6} \, \text{N m} \]