Question

Two identical wires A and B have the same length L and 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 $B_1$ and $B_2$ are the values of magnetic induction at the centre of the square and the circle respectively, then the ratio $\frac{B_1}{B_2}$ is :

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

Answer 1

Answer: (B)

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

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

$B_2=\frac{{\mu }_{0}I}{2R}=\frac{\pi {\mu }_{0}I}{L}$

$B_1$=$4B$ where $B$=magnetic field due to any one side of square of length($\frac{L}{4}$).

$B=\frac{{\mu }_{0}I}{4\pi d}[\cos 45°-\cos (180°-45°)]$

$=\frac{{\mu }_{0}I}{4\pi \left(\frac{L}{8}\right)}(\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}})=\frac{2{\mu }_{0}I}{\pi L}\times \sqrt{2}$

$B_1=\frac{8\sqrt{2}{\mu }_{0}I}{\pi L}$

$\frac{B_1}{B_2}==\frac{{\pi }^2}{8\sqrt{2}}$