A magnetic dipole is under the influence of two magnetic fields. The angle between the field direction is `60^(@)` and one of the fields has magnitude of `1.2 xx 10^(-2)T`. If the dipole comes to stable equilibrium at an angle of `30^(@)` with this Held, then the magnitude of the field is

Options

A. `1.2xx10^(-4)T`
B. `2.4xx10^(-4)T`
C. `1.2xx10^(-2)T`
D. `2.4xx10^(-2)T`

Explanation

To solve the problem of finding the magnitude of the second magnetic field when a magnetic dipole is under the influence of two magnetic fields, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Problem**:
We have a magnetic dipole in two magnetic fields, with one field having a magnitude of \( B_1 = 1.2 \times 10^{-2} \, T \) and the angle between the two fields being \( 60^\circ \). The dipole is in stable equilibrium at an angle of \( 30^\circ \) with respect to one of the fields.

2. **Using Torque Equation**:
The torque \( \tau \) experienced by a magnetic dipole in a magnetic field is given by:
\[ \tau = \vec{m} \times \vec{B} \]
where \( \vec{m} \) is the magnetic moment and \( \vec{B} \) is the magnetic field.

3. **Condition for Stable Equilibrium**:
For stable equilibrium, the net torque must be zero. This occurs when the potential energy is minimized. The potential energy \( U \) of a magnetic dipole in a magnetic field is given by:
\[ U = -\vec{m} \cdot \vec{B} \]
The condition for equilibrium can be expressed as:
\[ m B_1 \sin(30^\circ) = m B_2 \sin(60^\circ) \]
Here, \( B_2 \) is the magnitude of the second magnetic field.

4. **Substituting Values**:
We know:
- \( \sin(30^\circ) = \frac{1}{2} \)
- \( \sin(60^\circ) = \frac{\sqrt{3}}{2} \)

Substituting these values into the equation gives:
\[ m B_1 \cdot \frac{1}{2} = m B_2 \cdot \frac{\sqrt{3}}{2} \]

5. **Canceling \( m \)**:
Since \( m \) appears on both sides, we can cancel it out (assuming \( m \neq 0 \)):
\[ B_1 \cdot \frac{1}{2} = B_2 \cdot \frac{\sqrt{3}}{2} \]

6. **Solving for \( B_2 \)**:
Rearranging the equation to solve for \( B_2 \):
\[ B_2 = \frac{B_1}{\sqrt{3}} \]

7. **Substituting \( B_1 \)**:
Now substituting \( B_1 = 1.2 \times 10^{-2} \, T \):
\[ B_2 = \frac{1.2 \times 10^{-2}}{\sqrt{3}} \approx \frac{1.2 \times 10^{-2}}{1.732} \approx 6.93 \times 10^{-3} \, T \]

### Final Answer:
The magnitude of the second magnetic field \( B_2 \) is approximately \( 6.93 \times 10^{-3} \, T \).