In the following figure,a wire bent in the form of a regular polygon of $n$ sides is inscribed in a circle of radius $a$. The net magnetic field at the centre will be

  • A
    $\frac{{{\mu _0}i}}{{2\pi a}}\tan \frac{\pi }{n}$
  • B
    $\frac{{{\mu _0}ni}}{{2\pi a}}\tan \frac{\pi }{n}$
  • C
    $\frac{2}{\pi }\frac{{ni}}{a}{\mu _0}\tan \frac{\pi }{n}$
  • D
    $\frac{{ni}}{{2a}}{\mu _0}\tan \frac{\pi }{n}$

Explore More

Similar Questions

In the following figure, the magnitude of the magnetic field at point '$O$' will be

$A$ long curved conductor carries a current $I$. $A$ small current element of length $dl$ on the wire induces a magnetic field at a point away from the current element. If the position vector between the current element and the point is $\vec{r}$,making an angle $\theta$ with the current element,then the induced magnetic field density $d\vec{B}$ at the point is $(\mu_0 = \text{permeability of free space})$:

The magnetic field due to a current-carrying square loop of side $a$ at a point located symmetrically at a distance of $a/2$ from its centre (as shown in the figure) is:

$A$ circular coil $A$ has a radius $R$ and the current flowing through it is $I$. Another circular coil $B$ has a radius $2R$ and the current flowing through it is $2I$. The ratio of the magnetic fields at the centre of the circular coils $(B_A : B_B)$ is: (in $:1$)

$1 \ T = \dots \text{Gauss}$.

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