Q.
A current $I$ flowing through the loop as shown in figure. The magnetic field at centre $O$ is
A) $\frac{7\ \mu_{0}I}{16R}\bigotimes$
B) $\frac{7\ \mu_{0}I}{16R}\bigodot$
C) $\frac{7\ \mu_{0}I}{16R}\bigodot$
D) $\frac{5\ \mu_{0}I}{16R}\bigodot$
Explanation
(a) The effective magnetic field at $O$
$B = B_{PE} + B_{RS} = \frac{\mu_{0}}{4\pi}.\frac{3}{2}\frac{\pi I}{R} + \frac{\mu_{0}}{4\pi}.\frac{\pi}{2}.\frac{I}{2R}$
$\Rightarrow B = \frac{\mu_{0}I}{4R}\left\lbrack \frac{3}{2} + \frac{1}{4} \right\rbrack = \frac{7}{16}\frac{\mu_{0}I}{R}$
As per Fleming’s Right Hand rule, direction of magnetic field is perpendicular and in the plane of paper
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.