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 wire loop $PQRSP$ formed by joining two semicircular wires of radii $R_1$ and $R_2$ carries a current $I$ as shown. The magnitude of the magnetic field at the centre '$O$' is

Two identical circular current loops of radius $R$ carrying equal currents $I$ are placed such that their axes are inclined at $45^\circ$ to each other. The distance from the center of each loop to point $P$ is $\sqrt{3}R$. The resultant magnetic field at $P$ is:

$A$ long wire is bent into a circular coil of one turn and then into a circular coil of smaller radius having $n$ turns. If the same current passes in both the cases,the ratio of magnetic fields produced at the centre for one turn to that of $n$ turns is

The magnetic field on the axis of a circular loop of radius $R = 100\,cm$ carrying current $I = \sqrt{2}\,A$,at a point $x = 1\,m$ away from the centre of the loop is given by:

The magnetic field at the centre $O$ of a wire loop formed by two semicircular wires of radii $R_1=2 \pi \text{ m}$ and $R_2=4 \pi \text{ m}$ carrying current $I=4 \text{ A}$ as per the figure given below is $\alpha \times 10^{-7} \text{ T}$. The value of $\alpha$ is . . . . . . . (Centre $O$ is common for all segments)

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