A wire carrying a current i is first bent in the form of a square of side a and placed at right angle to a uniform magnetic field of induction B. The work done in changing its shape' into a circle is

Options

A. `ia^(2)B(pi+2)`
B. `ia^(2)B(pi-2)`
C. `ia^(2)B(4/pi-1)`
D. `ia^(2)B(1-4/pi)`

Answer

To solve the problem of calculating the work done in changing the shape of a wire carrying current from a square to a circle while placed in a magnetic field, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Initial Shape and Area**:
- The wire is initially bent into a square with each side of length \( a \).
- The area of the square, \( A_{\text{initial}} \), is given by:
\[ A_{\text{initial}} = a^2 \]

2. **Determine the Final Shape and Area**:
- The wire is reshaped into a circle.
- The circumference of the square is \( 4a \), which will be equal to the circumference of the circle.
- The circumference of the circle is given by \( 2\pi r \), where \( r \) is the radius of the circle.
- Setting these equal gives:
\[ 4a = 2\pi r \implies r = \frac{2a}{\pi} \]
- The area of the circle, \( A_{\text{final}} \), is given by:
\[ A_{\text{final}} = \pi r^2 = \pi \left(\frac{2a}{\pi}\right)^2 = \frac{4a^2}{\pi} \]

3. **Calculate the Work Done**:
- The work done in changing the shape of the wire can be calculated using the change in potential energy in the magnetic field.
- The potential energy \( U \) in a magnetic field is given by:
\[ U = -i A B \cos(\theta) \]
- Since the wire is placed at a right angle to the magnetic field, \( \cos(\theta) = 1 \).
- Therefore, the initial potential energy \( U_{\text{initial}} \) is:
\[ U_{\text{initial}} = -i A_{\text{initial}} B = -i a^2 B \]
- The final potential energy \( U_{\text{final}} \) is:
\[ U_{\text{final}} = -i A_{\text{final}} B = -i \left(\frac{4a^2}{\pi}\right) B \]
- The work done \( W \) is the change in potential energy:
\[ W = U_{\text{final}} - U_{\text{initial}} = \left(-i \frac{4a^2}{\pi} B\right) - \left(-i a^2 B\right) \]
- Simplifying this gives:
\[ W = -i B \left(\frac{4a^2}{\pi} - a^2\right) = -i B a^2 \left(\frac{4}{\pi} - 1\right) \]
- Factoring out \( -i B a^2 \):
\[ W = -i B a^2 \left(1 - \frac{4}{\pi}\right) \]

4. **Final Expression for Work Done**:
- The work done in changing the shape of the wire from a square to a circle is:
\[ W = i a^2 B \left(1 - \frac{4}{\pi}\right) \]

### Conclusion:
The work done in changing the shape of the wire from a square to a circle is:
\[ W = i a^2 B \left(1 - \frac{4}{\pi}\right) \]