Question

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 the magnetic field at the centres of the circle and square respectively, then the ratio $\frac{B_A}{B_B}$ is :

• A. $\frac{{\pi }^2}{8\sqrt{2}}$
• B. $\frac{{\pi }^2}{8}$
• C. $\frac{{\pi }^2}{16\sqrt{2}}$
• D. $\frac{{\pi }^2}{16}$

Answer 1

The correct option is A $\frac{{\pi }^2}{8\sqrt{2}}$

Given

two identical length wire A,B

Wire A is bent into circle of radius R

Wire B is bent into square of side 'a'

$B_a$ is magnetic field of wire A at centre.

$B_b$ is magnetic field of wire B at centre

Solution

Case 1 wire A

L=length of wire=circumference of circle made of radius R

$L=2\pi R$

$R=\frac{L}{2\pi }$

Magnetic field at centre of circular wire is

$\frac{{\mu }_0\ast I}{2R}$

Putting $R=\frac{L}{2\pi }$

$\frac{{\mu }_0\ast I\ast \pi }{L}$

Case 2 wire B

L=length of wire=perimeter of square of side 'a'

L=4*a

$a=\frac{L}{4}$

Magnetic field of square at centre

$B_B=4\ast \frac{2^{\frac{1}{2}}\ast {\mu }_0\ast I}{4\pi \frac{a}{2}}$

Putting a=L/4

$B_B=\frac{8\ast 2^{\frac{1}{2}}\ast {\mu }_0\ast I}{\pi \ast L}$

Ratio of two wires magnetic fields is

$\frac{B_A}{B_B}=\frac{{\pi }^2}{8\ast 2^{\frac{1}{2}}}$

The correct option is A