If a square frame of side L is made from a uniform wire and current I is flowing through it and its magnetic field intensity at the centre is `B_(1)`. When the same current is passed through the circular loop whose circumference is same as that of the square, the magnetic field at the centre is `B_(2)`. What will be `B_(1)//B_(2)` ?

Options

A. `(4 sqrt2)/pi^(3)`
B. `(8 sqrt2)/pi^(3)`
C. `(4 sqrt2)/pi^(2)`
D. `(8 sqrt2)/pi^(2)`

Explanation

To solve the problem, we need to find the ratio \( \frac{B_1}{B_2} \), where \( B_1 \) is the magnetic field intensity at the center of a square frame of side \( L \) carrying current \( I \), and \( B_2 \) is the magnetic field intensity at the center of a circular loop with the same current \( I \) and the same circumference as the square frame.

### Step-by-Step Solution:

1. **Calculate the Perimeter of the Square Frame:**
The perimeter \( P \) of a square frame with side length \( L \) is given by:
\[ P = 4L \]

2. **Determine the Radius of the Circular Loop:**
The circumference \( C \) of the circular loop is equal to the perimeter of the square frame:
\[ C = 2\pi R = 4L \]
Solving for \( R \):
\[ R = \frac{4L}{2\pi} = \frac{2L}{\pi} \]

3. **Calculate the Magnetic Field \( B_1 \) at the Center of the Square Frame:**
The magnetic field at the center of a square loop carrying current \( I \) is given by:
\[ B_1 = \frac{\mu_0 I}{4\pi} \cdot \frac{4}{L} \cdot \sqrt{2} \]
Simplifying this:
\[ B_1 = \frac{\mu_0 I \sqrt{2}}{\pi L} \]

4. **Calculate the Magnetic Field \( B_2 \) at the Center of the Circular Loop:**
The magnetic field at the center of a circular loop carrying current \( I \) is given by:
\[ B_2 = \frac{\mu_0 I}{2R} \]
Substituting \( R = \frac{2L}{\pi} \):
\[ B_2 = \frac{\mu_0 I}{2 \cdot \frac{2L}{\pi}} = \frac{\mu_0 I \pi}{4L} \]

5. **Find the Ratio \( \frac{B_1}{B_2} \):**
Now, we can find the ratio:
\[ \frac{B_1}{B_2} = \frac{\frac{\mu_0 I \sqrt{2}}{\pi L}}{\frac{\mu_0 I \pi}{4L}} \]
Simplifying this:
\[ \frac{B_1}{B_2} = \frac{\sqrt{2}}{\frac{\pi}{4}} = \frac{4\sqrt{2}}{\pi} \]

6. **Final Result:**
Thus, the ratio \( \frac{B_1}{B_2} \) is:
\[ \frac{B_1}{B_2} = \frac{8\sqrt{2}}{\pi^2} \]