Short bar magnet is place 30degrees with the external magnetic field 0.06T which experiences a torque of 0.018Nm then the minimum work done required to move from its stable equilibrium to unstable equilibrium is

Answer

To solve the problem of finding the minimum work done required to move a short bar magnet from stable equilibrium to unstable equilibrium in an external magnetic field, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Given Information**:
- The angle θ between the magnet and the magnetic field is 30 degrees.
- The external magnetic field (B) is 0.06 T.
- The torque (τ) experienced by the magnet is 0.018 Nm.

2. **Finding the Magnetic Moment (m)**:
The torque on a magnetic dipole in a magnetic field is given by the formula:
\[ \tau = mB \sin(\theta) \]
Rearranging this to find the magnetic moment (m):
\[ m = \frac{\tau}{B \sin(\theta)} \]
Substituting the known values:
\[ m = \frac{0.018}{0.06 \sin(30^\circ)} \]
Since \(\sin(30^\circ) = 0.5\):
\[ m = \frac{0.018}{0.06 \times 0.5} = \frac{0.018}{0.03} = 0.6 \, \text{A m}^2 \]

3. **Calculating the Work Done (W)**:
The work done in moving the magnet from stable equilibrium (θ = 0°) to unstable equilibrium (θ = 180°) can be calculated using the potential energy difference:
\[ W = U_2 - U_1 \]
Where:
- \(U_1 = -mB \cos(0^\circ) = -mB\)
- \(U_2 = -mB \cos(180^\circ) = mB\)

Therefore:
\[ W = mB - (-mB) = 2mB \]
Substituting the values of m and B:
\[ W = 2 \times 0.6 \times 0.06 \]
\[ W = 2 \times 0.036 = 0.072 \, \text{J} \]

### Final Answer:
The minimum work done required to move the magnet from stable equilibrium to unstable equilibrium is **0.072 Joules**.