The magnetic dipole moment of a coil is `5.4xx10^(-6)` joule/tesla and it is lined up with an external magnetic field whose strength is `0.80 T`. Then the work done in rotating the coil (for `theta=180^(@))` is

Options

A. `4.32 muJ`
B. `2.16 muJ`
C. `8.6 muJ`
D. None of these

Explanation

To find the work done in rotating the coil through an angle of \(180^\circ\) in a magnetic field, we can use the relationship between potential energy and work done. Here’s the step-by-step solution:

### Step 1: Understand the formula for potential energy
The potential energy \(U\) of a magnetic dipole in a magnetic field is given by:
\[ U = -\mathbf{M} \cdot \mathbf{B} = -M B \cos \theta \]
where:
- \(M\) is the magnetic dipole moment,
- \(B\) is the magnetic field strength,
- \(\theta\) is the angle between the magnetic moment and the magnetic field.

### Step 2: Calculate the initial potential energy
When the coil is aligned with the magnetic field (\(\theta = 0^\circ\)):
\[ U_{\text{initial}} = -M B \cos(0^\circ) = -M B \]
Substituting the values:
\[ U_{\text{initial}} = - (5.4 \times 10^{-6} \, \text{J/T}) \times (0.8 \, \text{T}) = -4.32 \times 10^{-6} \, \text{J} \]

### Step 3: Calculate the final potential energy
When the coil is rotated \(180^\circ\) (\(\theta = 180^\circ\)):
\[ U_{\text{final}} = -M B \cos(180^\circ) = -M B (-1) = M B \]
Substituting the values:
\[ U_{\text{final}} = (5.4 \times 10^{-6} \, \text{J/T}) \times (0.8 \, \text{T}) = 4.32 \times 10^{-6} \, \text{J} \]

### Step 4: Calculate the change in potential energy
The work done \(W\) in rotating the coil is equal to the change in potential energy:
\[ W = U_{\text{final}} - U_{\text{initial}} \]
Substituting the values:
\[ W = 4.32 \times 10^{-6} \, \text{J} - (-4.32 \times 10^{-6} \, \text{J}) = 4.32 \times 10^{-6} \, \text{J} + 4.32 \times 10^{-6} \, \text{J} = 8.64 \times 10^{-6} \, \text{J} \]

### Step 5: Convert to microjoules
To express the work done in microjoules:
\[ W = 8.64 \times 10^{-6} \, \text{J} = 8.64 \, \mu\text{J} \]

### Final Answer
The work done in rotating the coil through \(180^\circ\) is:
\[ \boxed{8.64 \, \mu\text{J}} \]