A square frame of side `l` carries a current produces a field `B` at its centre. The same current is passed through a circular loop having same perimeter as the square. The field at its centre is `B'`, the ratio of `B//B'` is

Options

A. `(8)/(pi^(2))`
B. `(8sqrt(2))/(pi^(2))`
C. `(16)/(pi^(2))`
D. `(16 sqrt(2))/(pi^(2))`

Answer

To solve the problem, we need to find the ratio of the magnetic field \( B \) produced by a square frame carrying a current at its center to the magnetic field \( B' \) produced by a circular loop with the same perimeter carrying the same current.

### Step-by-Step Solution:

1. **Determine the Magnetic Field \( B \) at the Center of the Square Frame:**
- The magnetic field at the center of a square loop of side length \( l \) carrying a current \( I \) is given by the formula:
\[ B = \frac{\mu_0 I}{4 \pi l} \cdot \sqrt{2} \]
- Since there are 4 sides, the total magnetic field becomes:
\[ B = \frac{4 \mu_0 I}{4 \pi l} \cdot \sqrt{2} = \frac{\mu_0 I \sqrt{2}}{\pi l} \]

2. **Determine the Radius \( r \) of the Circular Loop:**
- The perimeter of the square is \( 4l \).
- The perimeter of the circular loop is \( 2\pi r \).
- Setting these equal gives:
\[ 4l = 2\pi r \implies r = \frac{4l}{2\pi} = \frac{2l}{\pi} \]

3. **Determine the Magnetic Field \( B' \) at the Center of the Circular Loop:**
- The magnetic field at the center of a circular loop of radius \( r \) carrying a current \( I \) is given by:
\[ B' = \frac{\mu_0 I}{2r} \]
- Substituting the value of \( r \):
\[ B' = \frac{\mu_0 I}{2 \cdot \frac{2l}{\pi}} = \frac{\mu_0 I \pi}{4l} \]

4. **Calculate the Ratio \( \frac{B}{B'} \):**
- Now, we can find the ratio of the magnetic fields:
\[ \frac{B}{B'} = \frac{\frac{\mu_0 I \sqrt{2}}{\pi l}}{\frac{\mu_0 I \pi}{4l}} \]
- Simplifying this expression gives:
\[ \frac{B}{B'} = \frac{\sqrt{2}}{\frac{\pi}{4}} = \frac{4\sqrt{2}}{\pi} \]

5. **Final Result:**
- Thus, the ratio \( \frac{B}{B'} \) is:
\[ \frac{B}{B'} = \frac{4\sqrt{2}}{\pi} \]