Question

A current carrying thin uniform wire of length '$4l$' is bent like a square so that it produces a magnetic induction $B_1$ at the centre of the square. When the same wire is bent like a circle, it produces a magnetic induction $B_2$ at the centre of the circle. The ratio between $B_1$ and $B_2$ is:

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

Answer 1

Answer: (D)

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

When a current (let i) carrying wire of length $4l$ is bent like a square , side of the square , $a=l$ .

The magnetic induction $B_1$ at the center of a current carrying square of side $l$ is given by ,

$B_1=4\times \frac{{\mu }_0}{4\pi }\frac{2\sqrt{2}i}{l}$ .......................eq1

When a current (let i) carrying wire of length $4l$ is bent like a circle of radius $r$, circumference of circle ,

$2\pi r=4l$

or $r=\frac{2l}{\pi }$

The magnetic induction $B_1$ at the center of a current carrying circle of $2l/\pi$ is given by ,

$B_1=\frac{{\mu }_{0}i}{2\left(2l/\pi \right)}$ .......................eq2

dividing eq1 by eq2 , we get

$\frac{B_1}{B_2}=4\times \frac{{\mu }_{0}2\sqrt{2}i}{4\pi l}\times \frac{2\left(2l/\pi \right)}{{\mu }_{0}i}=\frac{8\sqrt{2}}{{\pi }^2}$

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