Discussion Forum



1)

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 BA and BB are the values of the magnetic field at the centers of the circle and square respectively. then the ratio $\frac{B_{A}}{B_{B}}$ is


A) $\frac{\pi^{2}}{8}$

B) $\frac{\pi^{2}}{16\sqrt{2}}$

C) $\frac{\pi^{2}}{16}$

D) $\frac{\pi^{2}}{8\sqrt{2}}$

Answer:

Option D

Explanation:

B at centre of a circle =  $\frac{\mu_{0}I}{2R}$

B at centre of a square 

     $=4\times \frac{\mu I}{4\pi.\frac{l}{2}}[\sin 45 ^{\circ} + \sin 45 ^{\circ}]$

     $=4\sqrt{2}\frac{\mu I}{2\pi l}$

Now, $R=\frac{L}{2\pi} $   and   $l= \frac{L}{4}$

                                                                 (As L=$2\pi R= 4l$ )

 Where , L= length of wire

   $B_{A}=\frac{\mu_{0}I}{2.\frac{L}{2\pi}}=\frac{\pi\mu_{0}I}{L}=\pi [\frac{\mu_{0}I}{L}]$

  $B_{B}=4\sqrt{2}\frac{\mu_{0}I}{2\pi(\frac{L}{4})}$

$=\frac{8\sqrt{2}\mu_{0}I}{\pi L}=\frac{8\sqrt{2}}{\pi}[\frac{\mu_{0}I}{L}]$

    $\frac{B_{A}}{B_{B}}=\pi^{2}:8\sqrt{2}$