An electron moving in a circular orbit of radius $r$ makes $n$ rotations per second. The magnetic field produced at the centre has magnitude

  • A
    $\frac{\mu_0 ne}{2\pi r}$
  • B
    zero
  • C
    $\frac{\mu_0 n^2 e}{r}$
  • D
    $\frac{\mu_0 ne}{2r}$

Explore More

Similar Questions

The magnitude of the magnetic field at $O$ due to a current-carrying loop as shown in the figure is, where $O$ is the center of two circular portions with radii $1 \, cm$ and $2 \, cm$ respectively. (Take the value of current $I = \frac{1.2}{\pi} \, A$)

$A$ charge $q$ coulomb moves in a circle at $n$ revolutions per second and the radius of the circle is $r$ metre; then the magnetic field at the centre of the circle is

$A$ current of $1 \,A$ is flowing along the sides of an equilateral triangle of side $a = 4.5 \times 10^{-2} \,m$. The magnetic field at the centroid of the triangle is $(\mu_0 = 4 \pi \times 10^{-7} \,T \cdot m/A)$.

Two very long,straight and insulated wires are kept at a $90^o$ angle from each other in the $xy$-plane as shown in the figure. These wires carry a current of equal magnitude $I$,whose directions are shown in the figure. The net magnetic field at point $P$ will be:

$A$ $10 \ A$ current is passing through a very long wire of radius $5 \ cm$. The magnetic field at a distance of $2 \ cm$ inside from its curved surface is . . . . . . $\times 10^{-5} \ T$.

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