The flux density obtained at the centre of a circular coil of radius $R$ which carries a current $i$ is $B_0$. At a distance $pR$ from the centre on the axis,the flux density will be

  • A
    $\frac{B_0}{(p^2+1)^{3/2}}$
  • B
    $\frac{B_0}{(p^2+1)^{1/2}}$
  • C
    $\frac{B_0}{(1 + p^2)^{3/2}}$
  • D
    $\frac{B_0}{(p^2+1)^{2}}$

Explore More

Similar Questions

$A$ current $I$ flows in a wire bent in the shape of a quarter ring of radius $R$ as shown in the figure. The magnetic field induction $B$ at point $O$ is

Write the formula for the magnetic field at a perpendicular distance $r$ from a finite current-carrying wire.

The magnetic induction at any point due to a long straight wire carrying a current is

$PQ$ and $RS$ are long parallel conductors separated by a certain distance. $M$ is the midpoint between them (see the figure). The net magnetic field at $M$ is $B$. Now,the current $2 \text{ A}$ is switched off. The field at $M$ now becomes

The ratio of the magnetic field at the centre of a current-carrying circular wire and the magnetic field at the centre of a square coil made from the same length of wire will be

Difficult
View Solution

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