Q. 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 distance $1\ cm$ 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) $1\ :100$
Explanation
(b) 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
$B_{2} = \frac{\mu_{0}}{4\pi}.\frac{2\pi i}{r} = \frac{\mu_{0}}{4\pi}.\frac{2\pi \times 2}{\pi/2} \Rightarrow \frac{B_{2}}{B_{1}} = \frac{1}{50}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.