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

Explanation

To solve the problem, we need to calculate the magnetic fields \( B_A \) and \( B_B \) at the centers of the circular and square wire configurations, respectively, and then find the ratio \( \frac{B_A}{B_B} \).

### Step 1: Calculate \( B_A \) for the circular wire
1. **Circumference of the circle**: The length of wire \( L \) is equal to the circumference of the circle.
\[ L = 2\pi R \]
From this, we can express the radius \( R \):
\[ R = \frac{L}{2\pi} \]

2. **Magnetic field at the center of the circle**: The formula for the magnetic field at the center of a circular loop carrying current \( I \) is given by:
\[ B_A = \frac{\mu_0 I}{2R} \]
Substituting \( R = \frac{L}{2\pi} \) into the equation:
\[ B_A = \frac{\mu_0 I}{2 \cdot \frac{L}{2\pi}} = \frac{\mu_0 I \cdot \pi}{L} \]

### Step 2: Calculate \( B_B \) for the square wire
1. **Length of one side of the square**: Since the wire is bent into a square, the total length \( L \) is equal to the perimeter of the square:
\[ L = 4a \implies a = \frac{L}{4} \]

2. **Magnetic field at the center of the square**: The magnetic field at the center due to one side of the square is given by:
\[ B_1 = \frac{\mu_0 I}{4\pi} \cdot \frac{a}{2} \cdot \sin(45^\circ) \]
Since \( \sin(45^\circ) = \frac{1}{\sqrt{2}} \):
\[ B_1 = \frac{\mu_0 I}{4\pi} \cdot \frac{L/4}{2} \cdot \frac{1}{\sqrt{2}} = \frac{\mu_0 I L}{32\pi \sqrt{2}} \]

3. **Total magnetic field \( B_B \)**: Since there are four sides contributing to the magnetic field at the center:
\[ B_B = 4B_1 = 4 \cdot \frac{\mu_0 I L}{32\pi \sqrt{2}} = \frac{\mu_0 I}{8\pi \sqrt{2}} \]

### Step 3: Calculate the ratio \( \frac{B_A}{B_B} \)
Now we can find the ratio of the magnetic fields:
\[ \frac{B_A}{B_B} = \frac{\frac{\mu_0 I \pi}{L}}{\frac{\mu_0 I}{8\pi \sqrt{2}}} \]
Cancelling \( \mu_0 I \) from the numerator and denominator:
\[ \frac{B_A}{B_B} = \frac{\pi}{\frac{1}{8\pi \sqrt{2}}} = \pi \cdot 8\pi \sqrt{2} = \frac{\pi^2}{8\sqrt{2}} \]

### Final Answer
Thus, the ratio \( \frac{B_A}{B_B} \) is:
\[ \frac{B_A}{B_B} = \frac{\pi^2}{8\sqrt{2}} \]