Question

A straight wire of length $(\pi {)}^2$ metre is carrying a current of $2$ Amp and the magnetic field due to it is measured at a point distant $1cm$ from it. If the wire is to be bent into a circle and is to carry the same current as before, the ratio of the magnetic field at its centre to that obtained in the first case would be:

• A. $50:1$
• B. $1:50$
• C. $100:1$
• D. $1:100$

Answer 1

Answer: (B)

$1:50$

If a wire of length $l$ is bent in the form of a circle of radius $r$ then $2\pi r=l\Rightarrow$

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

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

Magnetic field due to straight wire $B_1=\frac{{\mu }_0}{4\pi }.\frac{2i}{r}=\frac{{\mu }_0}{4\pi }\times \frac{2\times 2}{1\times {10}^{-2}}$ also magnetic field due to circular loop $B_2=\frac{{\mu }_0}{4\pi }.\frac{2\pi i}{r}=\frac{{\mu }_0}{4\pi }.\frac{2\pi \times 2}{\pi /2}\Rightarrow$

$\frac{B_2}{B_1}=\frac{1}{50}$