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

Options

A. `9.2xx10^(-3) J`
B. `7.2xx10^(-2) J`
C. `6.4xx10^(-2) J`
D. `11.7xx10^(-3) J`

Answer

To solve the problem step by step, we will follow the concepts of torque and work done on a magnetic dipole in an external magnetic field.

### Step 1: Understand the Equilibrium Positions
- The stable equilibrium position of the magnet is when it is aligned with the magnetic field, which corresponds to an angle \( \theta = 0^\circ \).
- The unstable equilibrium position is when the magnet is anti-aligned with the magnetic field, which corresponds to an angle \( \theta = 180^\circ \).

### Step 2: Torque on the Magnet
The torque \( \tau \) experienced by a magnetic dipole in a magnetic field is given by the formula:
\[ \tau = mB \sin \theta \]
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 3: Calculate the Magnetic Moment
Given that the torque \( \tau = 0.018 \, \text{Nm} \), the magnetic field \( B = 0.06 \, \text{T} \), and the angle \( \theta = 30^\circ \), we can rearrange the torque formula to find the magnetic moment \( m \):
\[ m = \frac{\tau}{B \sin \theta} \]
Substituting the known values:
\[ m = \frac{0.018}{0.06 \cdot \sin(30^\circ)} \]
Since \( \sin(30^\circ) = 0.5 \):
\[ m = \frac{0.018}{0.06 \cdot 0.5} = \frac{0.018}{0.03} = 0.6 \, \text{A m}^2 \]

### Step 4: Calculate the Work Done
The work done \( W \) to move the magnet from stable equilibrium to unstable equilibrium is given by:
\[ W = U_{\text{unstable}} - U_{\text{stable}} \]
Where the potential energy \( U \) of a magnetic dipole in a magnetic field is given by:
\[ U = -mB \cos \theta \]
- At stable equilibrium (\( \theta = 0^\circ \)):
\[ U_{\text{stable}} = -mB \cos(0) = -mB \]
- At unstable equilibrium (\( \theta = 180^\circ \)):
\[ U_{\text{unstable}} = -mB \cos(180) = mB \]
Thus, the work done is:
\[ W = (mB) - (-mB) = 2mB \]
Substituting the values:
\[ W = 2 \cdot 0.6 \cdot 0.06 = 0.072 \, \text{J} = 7.2 \times 10^{-2} \, \text{J} \]

### Final Answer
The minimum work done required to move the magnet from stable equilibrium to unstable equilibrium is:
\[ \boxed{7.2 \times 10^{-2} \, \text{J}} \]