$ ext{A square frame of side 'L' carries a current } I. ext{ The magnetic field at its centre is } B. ext{ The same current is passed through a circular coil having the same perimeter as the square. The magnetic field at the centre of the circular coil is } B'. ext{ The ratio of } \frac{B}{B'} ext{ is :}$

Q: $\text{A square frame of side 'L' carries a current } I. \text{ The magnetic field at its centre is } B. \text{ The same current is passed through a circular coil having the same perimeter as the square. The magnetic field at the centre of the circular coil is } B'. \text{ The ratio of } \frac{B}{B'} \text{ is :}$

$\text{A square frame of side 'L' carries a current } I. \text{ The magnetic field at its centre is } B. \text{ The same current is passed through a circular coil having the same perimeter as the square. The magnetic field at the centre of the circular coil is } B'. \text{ The ratio of } \frac{B}{B'} \text{ is :}$

Question

$	ext{A square frame of side 'L' carries a current } I. 	ext{ The magnetic field at its centre is } B. 	ext{ The same current is passed through a circular coil having the same perimeter as the square. The magnetic field at the centre of the circular coil is } B'. 	ext{ The ratio of } \frac{B}{B'} 	ext{ is :}$

Accepted Answer

To determine the ratio $\frac{B}{B'}$, where $B$ is the magnetic field at the center of a square frame and $B'$ is the magnetic field at the center of a circular coil with the same perimeter, we need to analyze the magnetic fields produced by both geometries.

### Explanation:

1. **Square Frame:**
   - The side of the square is $L$.
   - The perimeter of the square is $4L$.
   - The magnetic field at the center of a square loop carrying current $I$ is given by:
     $$
     B = \frac{2\sqrt{2} \mu_0 I}{\pi L}
     $$
   - This formula is derived from the Biot-Savart law, considering the contribution of each side of the square to the magnetic field at the center.

2. **Circular Coil:**
   - The perimeter of the circular coil is the same as that of the square, i.e., $2\pi R = 4L$.
   - Solving for $R$, the radius of the circle, we get $R = \frac{2L}{\pi}$.
   - The magnetic field at the center of a circular loop carrying current $I$ is given by:
     $$
     B' = \frac{\mu_0 I}{2R}
     $$
   - Substituting $R = \frac{2L}{\pi}$ into the equation for $B'$, we get:
     $$
     B' = \frac{\mu_0 I \pi}{4L}
     $$

3. **Ratio $\frac{B}{B'}$:**
   - Now, we calculate the ratio $\frac{B}{B'}$:
     $$
     \frac{B}{B'} = \frac{\frac{2\sqrt{2} \mu_0 I}{\pi L}}{\frac{\mu_0 I \pi}{4L}}
     $$
   - Simplifying the expression:
     $$
     \frac{B}{B'} = \frac{2\sqrt{2}}{\pi} \times \frac{4L}{\pi L} = \frac{8\sqrt{2}}{\pi^2}
     $$

### Conclusion:

- **Option 1: $\frac{8\sqrt{2}}{\pi^2}$** is indeed the correct answer as derived from the calculations above.

### Explanation of Incorrect Options:

- **Option 2: $\frac{8\sqrt{2}}{\pi^3}$**
  - This option incorrectly includes an extra factor of $\pi$ in the denominator, which does not match the derived expression.

- **Option 3: $\frac{8\sqrt{2}}{\pi}$**
  - This option lacks the necessary $\pi^2$ in the denominator, leading to an incorrect ratio.

- **Option 4: $\frac{4\sqrt{2}}{\pi^2}$**
  - This option incorrectly halves the numerator, which does not align with the derived expression.

By following the logical derivation and calculations, Option 1 is confirmed as the correct answer.