Magnetic field at the centre of a circular loop of area $A$ is $B$. Then the magnetic moment of the loop is ($\mu_0$ is the permeability of free space).

  • A
    $\frac{B A^2}{\mu_0 \pi}$
  • B
    $\frac{B A \sqrt{A}}{\mu_0}$
  • C
    $\frac{B A \sqrt{A}}{\mu_0 \pi}$
  • D
    $\frac{2 B A \sqrt{A}}{\mu_0 \sqrt{\pi}}$

Explore More

Similar Questions

An electron in a circular orbit of radius $0.05 \ nm$ performs $10^{16}$ revolutions per second. The magnetic moment due to this rotation of the electron is (in $A \ m^{2}$):

If a current of $15 \,A$ passes through a solenoid of length $25 \,cm$,radius $2 \,cm$ and number of turns $500$,then the magnetic moment of the solenoid is

The relation between magnetic moment $(M)$ of a current-carrying circular coil and the length $(L)$ of the wire used is:

Due to the flow of current in a circular loop of radius $R$,the magnetic field produced at the centre of the loop is $B$. The magnetic moment of the loop is:

$A$ loop carrying current $I$ lies in the $x$-$y$ plane as shown in the figure. The unit vector $\hat{k}$ is coming out of the plane of the paper. The magnetic moment of the current loop is:

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo