Question

A straight wire of length $\left({\pi }^2\right)$ metre is carrying a current of 2A and the magnetic field due to it is measured at a point distant 1 cm 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

The correct option is B 1 : 50

If a wire of length l is bant in the form of a circle of radius r then
$2\pi r=l\Rightarrow r=\frac{l}{2\pi }$
$\overrightarrow{F}=i(\overrightarrow{L}\times \overrightarrow{B})$
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 $B_2=\frac{{\mu }_0}{4\pi }.\frac{2\pi i}{r}=\frac{{\mu }_0}{4\pi }.\frac{2\pi \times 2}{\pi /2}\Rightarrow \frac{B_2}{B_1}=\frac{1}{50}$