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Step-by-Step Detailed Solution
Step 1: Understanding the problem
We are given two identical wires of equal length $l$, each carrying the same current $I$. One wire is bent into a circle (wire A) of radius $R$, and the other (wire B) is bent to form a square of side $a$. We need to find the ratio of the magnetic field at the center of the circle ($B_A$) to that at the center of the square ($B_B$), i.e., $B_A / B_B$.
Step 2: Relate the circle’s radius to the wire length
For the circular loop (wire A), the circumference is equal to its length:
$$
2\pi R = l \quad \Longrightarrow \quad R = \frac{l}{2\pi}.
$$
Step 3: Magnetic field at the center of the circular loop
The magnetic field at the center of a circular loop of radius $R$ carrying current $I$ is:
$$
B_A = \frac{\mu_0}{4\pi} \,\frac{2\pi I}{R}.
$$
Substituting $R = \frac{l}{2\pi}$:
$$
B_A = \frac{\mu_0}{4\pi} \times \frac{2\pi I}{\tfrac{l}{2\pi}}
= \frac{\mu_0}{4\pi} \times I \times \frac{2\pi \times 2\pi}{l}
= \frac{\mu_0 I}{4\pi} \times \frac{4\pi^2}{l}
= \frac{\mu_0 I \,\pi}{l}.
$$
Step 4: Relate the square’s side to the wire length
For the square loop (wire B), all four sides add up to the wire’s length:
$$
4a = l \quad \Longrightarrow \quad a = \frac{l}{4}.
$$
Step 5: Magnetic field at the center of the square loop
The distance from the center of the square to each side is $a/2$. Using the standard result (via the Biot–Savart law or Ampère’s law) for the magnetic field at the center due to each side and summing over four sides, one gets:
$$
B_B = 4 \times \left(\frac{\mu_0}{4\pi} \,\frac{I}{\tfrac{a}{2}}\,[\sin 45^\circ + \sin 45^\circ]\right).
$$
Because $\sin 45^\circ = \frac{1}{\sqrt{2}}$, we have $(\sin 45^\circ + \sin 45^\circ) = \sqrt{2}$. Therefore,
$$
B_B = 4 \times \frac{\mu_0 I}{4\pi \,\frac{a}{2}} \times \sqrt{2}
= \frac{4 \mu_0 I \sqrt{2}}{4\pi \,\frac{a}{2}}
= \frac{4 \mu_0 I \sqrt{2}}{2\pi a}
= \frac{2 \mu_0 I \sqrt{2}}{\pi a}.
$$
Substituting $a = \frac{l}{4}$:
$$
B_B = \frac{2 \mu_0 I \sqrt{2}}{\pi \,\frac{l}{4}}
= \frac{8 \mu_0 I \sqrt{2}}{\pi l}.
$$
Step 6: Compute the ratio $B_A / B_B$
$$
\frac{B_A}{B_B}
= \frac{\frac{\mu_0 I \,\pi}{l}}{\frac{8 \mu_0 I \sqrt{2}}{\pi l}}
= \frac{\mu_0 I \,\pi}{l} \times \frac{\pi l}{8 \mu_0 I \sqrt{2}}
= \frac{\pi^2}{8\sqrt{2}}.
$$
Step 7: Final Answer
Hence, the desired ratio is
$$
\frac{B_A}{B_B} = \frac{\pi^2}{8\sqrt{2}}.
$$
