Q.
Current $I$ is flowing in conductor shaped as shown in the figure. The radius of the curved part is $r$ and the length of straight portion is very large. The value of the magnetic field at the centre $O$ will be
A) $\frac{\mu_{0}I}{4\pi r}\left( \frac{3\pi}{2} + 1 \right)$
B) $\frac{\mu_{0}I}{4\pi r}\left( \frac{3\pi}{2} - 1 \right)$
C) $\frac{\mu_{0}I}{4\pi r}\left( \frac{\pi}{2} + 1 \right)$
D) $\frac{\mu_{0}I}{4\pi r}\left( \frac{\pi}{2} - 1 \right)$
Explanation
(a) $B_{A} = 0$
[image:4_image137.emf]
$B_{B} = \ \frac{\mu_{0}}{4\pi}\frac{(2\pi - \pi/2)I\ }{r}\ \otimes = \frac{\mu_{0}}{4\pi}\ \frac{3\pi I}{2r}$
$B_{C} = \frac{\mu_{0}I}{4\pi r} \otimes \ $
So, net magnetic field at the centre
$= B_{A} + B_{B} + \ B_{C}$
$= 0 + \frac{\mu_{0}}{4\pi}\frac{3\pi I}{2r} + \frac{\mu_{0}I}{4\pi r} = \frac{\mu_{0}}{4\pi}\frac{I}{r}\left( \frac{3\pi}{2} + 1 \right)$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.