Question

Magnetic field at a point on the axis of a circular current carrying coil —

Answer 1

Magnetic field at a point on the axis of a circular current carrying coil —

The magnetic field at a point on the axis of the circular current carrying coil is

$B=\frac{\mu_0}{4 \pi} \frac{2 \pi N I a^2}{\left(a^2+x^2\right)^{3 / 2}}$

where $a$ is the radius of coil, $x$ is the distance of the point on the axis from the centre of the coil, $N$ is the number of turns in the coil.

Special cases :

• If the point lies at the centre of the coil, i.e. $x=0$, then

$B=\frac{\mu_0}{4 \pi} \frac{2 \pi N I a^2}{\left(a^2\right)^{3 / 2}}=\frac{\mu_0}{4 \pi} \frac{2 \pi N I}{a}=\frac{\mu_0 N I}{2 a}$

• If $x>>a$, then

$B=\frac{\mu_0}{4 \pi} \frac{2 \pi N I a^2}{x^3}=\frac{\mu_0}{4 \pi} \frac{2 N I A}{x^3}=\frac{\mu_0}{4 \pi} \frac{2 M}{x^3}$

where NIA = M = magnetic dipole moment of current loop, $A=$ cross sectional area of loop.

Magnetic field at the centre due to current carrying circular arc —

The magnetic field at the centre $O$ of the circular arc of radius $a$ carrying current $I$ is

$B=\frac{\mu_0 I \phi}{4 \pi a}$