A metallic wire is folded to form a square loop a side `'a'`. It carries a current `'I'` and is kept perpendicular to a uniform magnetic field. If the shape of the loop is changed from square to a circle without changing the length of thw wire and current, the amount of work done in doing so is

Options

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

Answer

To solve the problem of finding the work done in changing the shape of a metallic wire from a square loop to a circular loop while keeping the length of the wire and the current constant, we can follow these steps:

### Step-by-Step Solution:

1. **Determine the Length of the Wire**:
The length of the wire when it is folded into a square loop with side length 'a' is given by:
\[ L = 4a \]

2. **Relate the Length of the Wire to the Circle**:
When the wire is reshaped into a circular loop, the circumference of the circle must equal the length of the wire:
\[ 2\pi r = 4a \]
From this, we can solve for the radius \( r \):
\[ r = \frac{4a}{2\pi} = \frac{2a}{\pi} \]

3. **Calculate the Area of the Square Loop**:
The area \( A_s \) of the square loop is:
\[ A_s = a^2 \]

4. **Calculate the Area of the Circular Loop**:
The area \( A_c \) of the circular loop is:
\[ A_c = \pi r^2 = \pi \left(\frac{2a}{\pi}\right)^2 = \pi \cdot \frac{4a^2}{\pi^2} = \frac{4a^2}{\pi} \]

5. **Calculate the Initial Potential Energy**:
The potential energy \( U_i \) of the square loop in a magnetic field \( B \) is given by:
\[ U_i = -\mu \cdot B = -I \cdot A_s \cdot B = -I \cdot a^2 \cdot B \]

6. **Calculate the Final Potential Energy**:
The potential energy \( U_f \) of the circular loop is:
\[ U_f = -\mu \cdot B = -I \cdot A_c \cdot B = -I \cdot \frac{4a^2}{\pi} \cdot B \]

7. **Calculate the Work Done**:
The work done \( W \) in changing the shape of the loop is the difference between the final and initial potential energies:
\[ W = U_f - U_i \]
Substituting the values we found:
\[ W = \left(-I \cdot \frac{4a^2}{\pi} \cdot B\right) - \left(-I \cdot a^2 \cdot B\right) \]
Simplifying this:
\[ W = -I \cdot \frac{4a^2}{\pi} \cdot B + I \cdot a^2 \cdot B = I \cdot a^2 \cdot B \left(1 - \frac{4}{\pi}\right) \]

### Final Answer:
Thus, the amount of work done in changing the shape of the loop from a square to a circle is:
\[ W = I \cdot a^2 \cdot B \left(1 - \frac{4}{\pi}\right) \]