Q. Due to the flow of current in a circular loop of radius $R$, the magnetic induction produced at the centre of the loop is $B$. The magnetic moment of the loop is ($\mu_{0} =$ permeability constant)
A) $BR^{3}/2\pi\mu_{0}$
B) $2\pi BR^{3}/\mu_{0}$
C) $BR^{2}/2\pi\mu_{0}$
D) $2\pi BR^{2}/\mu_{0}$
Explanation
(b) $B = \frac{\mu_{0}i}{2R}\ \Rightarrow i = \frac{B \times 2R}{\mu_{0}}$
Now, $M = i \times A = i\pi R^{2} = \frac{B \times 2R}{\mu_{0}} \times \pi R^{2} = \frac{2\pi BR^{3}}{\mu_{0}}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.