Q.
The figure shows the cross-section of a long cylindrical conductor of radius $a$ carrying a uniformly distributed current $i.$ The magnetic field due to current at $P$ is
A) $\mu_{0}ir/(2\pi a^{2})\ $
B) $\mu_{0}ir^{2}/(2\pi a)$
C) $\mu_{0}ia/(2\pi r^{2})$
D) $\mu_{0}ia^{2}/(\pi r^{2})$
Explanation
(a) The current enclosed with in the circle
$\frac{i}{\pi a^{2}}\ ,\ \pi r^{2} = \frac{i}{a^{2}}\ r^{2}$
Ampere’s law $\oint_{}^{}{\mathbf{B.dl} = \mu_{0}i'}$ gives
$B\ .2\pi r = \frac{\mu_{0}ir^{2}}{a^{2}}$
or $B = \ \frac{\mu_{0}ir}{2\pi a^{2}}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.