Q. A straight conductor of length $I$ carrying a current $I,$ is bent in the form of a semicircle. The magnetic field (in tesla) at the centre of the semicircle is
A) $\frac{\pi^{2}I}{l} \times 10^{- 7}$
B) $\frac{\pi I}{l} \times 10^{- 7}$
C) $\frac{\pi I}{l^{2}} \times 10^{- 7}$
D) $\frac{\pi I^{2}}{l} \times 10^{- 7}$
Explanation
(a) Radius, $r = \frac{1}{\pi}\ $
Magnetic field at the centre, $B = \frac{\mu_{0}I}{4r}$
$= \frac{4\pi \times 10^{- 7} \times I}{4 \times \frac{l}{\pi}}$
$= \frac{\pi^{2}I \times 10^{- 7}}{l}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.