Two identical wires `A and B` , each of length 'l', carry the same current `I`. Wire A is bent into a circle of radius `R and wire B` is bent to form a square of side 'a' . If ` B_(A) and B_(B)` are the values of magnetic field at the centres of the circle and square respectively , then the ratio `(B_(A))/(B_(B))` is :

Options

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

Explanation

To find the ratio of the magnetic fields \( B_A \) and \( B_B \) at the centers of the circular and square wire configurations, we will follow these steps:

### Step 1: Determine the magnetic field \( B_A \) at the center of the circular wire
1. The length of wire A is given as \( L \).
2. When bent into a circle, the radius \( R \) can be expressed as:
\[ R = \frac{L}{2\pi} \]
3. The formula for the magnetic field at the center of a circular loop carrying current \( I \) is:
\[ B_A = \frac{\mu_0 I}{2R} \]
4. Substituting the expression for \( R \) into the formula:
\[ B_A = \frac{\mu_0 I}{2 \cdot \frac{L}{2\pi}} = \frac{\mu_0 I \cdot \pi}{L} \]

### Step 2: Determine the magnetic field \( B_B \) at the center of the square wire
1. The length of wire B is also \( L \).
2. When bent into a square, each side \( a \) can be expressed as:
\[ a = \frac{L}{4} \]
3. The magnetic field at the center of a square loop can be calculated using the formula:
\[ B_B = \frac{\mu_0 I}{4\pi} \cdot \left(4 \cdot \frac{1}{\sqrt{2}} \cdot \frac{1}{\frac{a}{2}}\right) \]
where \( \frac{1}{\frac{a}{2}} \) accounts for the distance from the center to the midpoint of a side.
4. Substituting \( a = \frac{L}{4} \):
\[ B_B = \frac{\mu_0 I}{4\pi} \cdot \left(4 \cdot \frac{1}{\sqrt{2}} \cdot \frac{2}{L/4}\right) \]
5. Simplifying this expression:
\[ B_B = \frac{\mu_0 I}{4\pi} \cdot \left(4 \cdot \frac{8}{L \sqrt{2}}\right) = \frac{8 \mu_0 I}{4\pi L \sqrt{2}} = \frac{2 \mu_0 I}{\pi L \sqrt{2}} \]

### Step 3: Calculate the ratio \( \frac{B_A}{B_B} \)
1. Now we can find the ratio of the magnetic fields:
\[ \frac{B_A}{B_B} = \frac{\frac{\mu_0 I \cdot \pi}{L}}{\frac{2 \mu_0 I}{\pi L \sqrt{2}}} \]
2. Simplifying this ratio:
\[ \frac{B_A}{B_B} = \frac{\mu_0 I \cdot \pi}{L} \cdot \frac{\pi L \sqrt{2}}{2 \mu_0 I} = \frac{\pi^2 \sqrt{2}}{2} \]

### Final Result
Thus, the ratio of the magnetic fields is:
\[ \frac{B_A}{B_B} = \frac{\pi^2 \sqrt{2}}{2} \]