A magnet of dipole moment `2Am^(2)` is deflected through `30^(@)` from magnetic meridian. The required deflecting torque is `(B_(H)=0.4xx10^(-4)T)`

Options

A. `0.4xx10^(-4)Nm`
B. `0.4 Nm`
C. `0.2 xx10^(-4) Nm`
D. `0.8 Nm`

Answer

To solve the problem, we need to calculate the deflecting torque (τ) on a magnetic dipole moment (μ) when it is deflected at an angle (θ) in a magnetic field (B_H).

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Magnetic dipole moment, \( \mu = 2 \, \text{Am}^2 \)
- Magnetic field strength, \( B_H = 0.4 \times 10^{-4} \, \text{T} \)
- Angle of deflection, \( \theta = 30^\circ \)

2. **Use the Torque Formula:**
The torque (τ) experienced by a magnetic dipole in a magnetic field is given by the formula:
\[ \tau = \mu \cdot B_H \cdot \sin(\theta) \]

3. **Calculate the Sine of the Angle:**
For \( \theta = 30^\circ \):
\[ \sin(30^\circ) = \frac{1}{2} \]

4. **Substitute the Values into the Torque Formula:**
Now, substituting the values into the torque formula:
\[ \tau = 2 \, \text{Am}^2 \cdot (0.4 \times 10^{-4} \, \text{T}) \cdot \sin(30^\circ) \]
\[ \tau = 2 \cdot (0.4 \times 10^{-4}) \cdot \frac{1}{2} \]

5. **Simplify the Expression:**
\[ \tau = 2 \cdot 0.4 \cdot 10^{-4} \cdot \frac{1}{2} \]
The \( 2 \) in the numerator and denominator cancels out:
\[ \tau = 0.4 \times 10^{-4} \, \text{N m} \]

6. **Final Result:**
Thus, the required deflecting torque is:
\[ \tau = 0.4 \times 10^{-4} \, \text{N m} \]