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

The magnetic field at the centre $O$ of a square loop of side $a$ carrying current $I$ as shown in the figure is:

Describe Oersted's observation.

In the hydrogen atom,the electron is making $6.6 \times 10^{15} \, r.p.s.$ If the radius of the orbit is $0.53 \times 10^{-10} \, m,$ then the magnetic field produced at the centre of the orbit is (in $Tesla$):

The magnetic field due to a current in a straight wire segment of length $L$ at a point on its perpendicular bisector at a distance $r$ $(r >> L)$ is:

An electron moves in a circular orbit with uniform speed $v$. It produces a magnetic field $B$ at the centre of the circle. The radius of the circle is (where $\mu_{0} =$ permeability of free space,$e =$ electronic charge):

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