The magnetic field at the centre of a circular coil of radius $r$,due to current $I$ flowing through it,is $B$. The magnetic field at a point along the axis at a distance $r/2$ from the centre is

  • A
    $B / 2$
  • B
    $2 B$
  • C
    $\left(\frac{2}{\sqrt{5}}\right)^{3} B$
  • D
    $\left(\frac{2}{\sqrt{3}}\right)^{3} B$

Explore More

Similar Questions

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:

If the radius of a coil is halved and the number of turns is doubled,then the magnetic field at the centre of the coil,for the same current,will

$A$ current of $1\,A$ is flowing through the sides of an equilateral triangle of side $4.5\times10^{-2}\,m$. The magnetic field at the centre of the triangle will be

$B_{x}$ and $B_{y}$ are the magnetic fields at the center of two coils $x$ and $y$ respectively,each carrying equal current. If coil $x$ has $200$ turns and $20 \ cm$ radius,and coil $y$ has $400$ turns and $20 \ cm$ radius,the ratio of $B_{x}$ and $B_{y}$ is $:-$

If the strength of the magnetic field produced $10\,cm$ away from an infinitely long straight conductor is ${10^{ - 5}}\,Wb/m^2$,the value of the current flowing in the conductor will be........$A$.

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