The effective radius of a circular coil is R and number of turns is N. The current through it is i ampere. The work done is rotating the coil from angle `theta = 0^(@)` to `theta = 180^(@)` in an external magnetic field B will be -

Options

A. `piNIR^(2)B`
B. `2piNiR^(2)B`
C. `((2NiB))/(piR ^(2))`
D. Zero

Answer

To find the work done in rotating a circular coil from an angle of \( \theta = 0^\circ \) to \( \theta = 180^\circ \) in an external magnetic field \( B \), we can follow these steps:

### Step 1: Understand the Magnetic Moment
The magnetic moment \( m \) of a circular coil is given by the formula:
\[ m = N \cdot I \cdot A \]
where \( N \) is the number of turns, \( I \) is the current, and \( A \) is the area of the coil. For a circular coil, the area \( A \) is:
\[ A = \pi R^2 \]
Thus, the magnetic moment becomes:
\[ m = N \cdot I \cdot \pi R^2 \]

### Step 2: Calculate Initial Potential Energy
The potential energy \( U \) in a magnetic field is given by:
\[ U = -m \cdot B \cdot \cos(\theta) \]
For \( \theta = 0^\circ \):
\[ U_{\text{initial}} = -m \cdot B \cdot \cos(0) = -m \cdot B \cdot 1 = -N \cdot I \cdot \pi R^2 \cdot B \]

### Step 3: Calculate Final Potential Energy
For \( \theta = 180^\circ \):
\[ U_{\text{final}} = -m \cdot B \cdot \cos(180) = -m \cdot B \cdot (-1) = m \cdot B \]
Thus:
\[ U_{\text{final}} = N \cdot I \cdot \pi R^2 \cdot B \]

### Step 4: Calculate 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 we calculated:
\[ W = \left( N \cdot I \cdot \pi R^2 \cdot B \right) - \left( -N \cdot I \cdot \pi R^2 \cdot B \right) \]
\[ W = N \cdot I \cdot \pi R^2 \cdot B + N \cdot I \cdot \pi R^2 \cdot B \]
\[ W = 2 \cdot N \cdot I \cdot \pi R^2 \cdot B \]

### Final Answer
Thus, the work done in rotating the coil from \( \theta = 0^\circ \) to \( \theta = 180^\circ \) is:
\[ W = 2 \cdot N \cdot I \cdot \pi R^2 \cdot B \]