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

Answer

To solve the problem, we need to find the ratio of the magnetic fields at the centers of two different shapes formed by identical wires carrying the same current. Let's break down the solution step by step.

### Step 1: Determine the magnetic field at the center of the circular wire (Wire A)

The formula for the magnetic field at the center of a circular loop of radius \( R \) carrying a current \( I \) is given by:

\[ B_A = \frac{\mu_0 I}{2R} \]

### Step 2: Relate the radius \( R \) to the length of the wire

The length of wire A when bent into a circle is equal to the circumference of the circle:

\[ L = 2\pi R \]

From this, we can express \( R \):

\[ R = \frac{L}{2\pi} \]

### Step 3: Substitute \( R \) back into the expression for \( B_A \)

Now, substituting \( R \) into the expression for \( B_A \):

\[ B_A = \frac{\mu_0 I}{2 \left(\frac{L}{2\pi}\right)} = \frac{\mu_0 I \cdot 2\pi}{2L} = \frac{\mu_0 I \pi}{L} \]

### Step 4: Determine the magnetic field at the center of the square wire (Wire B)

For a square wire of side length \( a \), the magnetic field at the center can be calculated by considering the contribution from each side of the square. The magnetic field due to one straight wire segment at a distance \( r \) from the wire is given by:

\[ B = \frac{\mu_0 I}{4\pi r} \sin \theta \]

In this case, each side of the square contributes to the magnetic field at the center. The distance from the center to each side is \( \frac{a}{2} \), and the angle \( \theta \) for each wire segment is \( 45^\circ \).

The total magnetic field at the center due to one side of the square is:

\[ B_{single} = \frac{\mu_0 I}{4\pi \left(\frac{a}{2}\right)} \sin 45^\circ = \frac{\mu_0 I}{4\pi \left(\frac{a}{2}\right)} \cdot \frac{1}{\sqrt{2}} = \frac{\mu_0 I}{2\pi a \sqrt{2}} \]

Since there are 4 sides contributing to the magnetic field, we multiply this by 4:

\[ B_B = 4 \cdot \frac{\mu_0 I}{2\pi a \sqrt{2}} = \frac{2\mu_0 I}{\pi a \sqrt{2}} \]

### Step 5: Relate the side length \( a \) to the length of the wire

The total length of wire B when bent into a square is:

\[ L = 4a \]

From this, we can express \( a \):

\[ a = \frac{L}{4} \]

### Step 6: Substitute \( a \) back into the expression for \( B_B \)

Now substituting \( a \) into the expression for \( B_B \):

\[ B_B = \frac{2\mu_0 I}{\pi \left(\frac{L}{4}\right) \sqrt{2}} = \frac{8\mu_0 I}{\pi L \sqrt{2}} \]

### Step 7: Find the ratio \( \frac{B_A}{B_B} \)

Now we can find the ratio of the magnetic fields:

\[ \frac{B_A}{B_B} = \frac{\frac{\mu_0 I \pi}{L}}{\frac{8\mu_0 I}{\pi L \sqrt{2}}} \]

Simplifying this gives:

\[ \frac{B_A}{B_B} = \frac{\pi^2}{8\sqrt{2}} \]

### Final Answer

Thus, the ratio of the magnetic fields at the centers of the circular and square wires is:

\[ \frac{B_A}{B_B} = \frac{\pi^2}{8\sqrt{2}} \]

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