Question

Derive an expression of the magnetic field at the centre of a circular current carrying coil.

Answer 1

Consider a circular coil of radius a and carrying current I in the direction shown in Figure. Suppose the loop lies in the plane of paper. It is desired to find the magnetic field at the centre O of the coil. Suppose the entire circular coil is divided into a large number of current elements, each of length dl. According to Biot-Savart law, the magnetic field $\overrightarrow{dB}$ at the centre O of the coil due to current element $I\overrightarrow{dl}$ is given by,

$\overrightarrow{dB}=\frac{{\mu }_{o}I(\overrightarrow{dl}\times \overrightarrow{r})}{4\pi r^3}$

where $\overrightarrow{r}$ is the position vector of point O from the current element. The magnitude of $\overrightarrow{dB}$ at the centre O is

$dB=\frac{{\mu }_{o}Idlasin\theta }{4\pi a^3}$

$\therefore dB=\frac{{\mu }_{o}Idlsin\theta }{4\pi a^2}$

The direction of $\overrightarrow{dB}$ is perpendicular to the plane of the coil and is directed inwards. Since each current element contributes to the magnetic field in the same direction, the total magnetic field B at the center O can be found by integrating the above equation around the loop i.e.

$\therefore B=⎰dB=⎰\frac{{\mu }_{o}Idlsin\theta }{4\pi a^2}$

For each current element, angle between $\overrightarrow{dl}$ and $\overrightarrow{r}$ is 90°. Also distance of each current element from the center O is a.

$\therefore B=\frac{{\mu }_{o}Isin{90}^o}{4\pi a^2}⎰dl$

But $⎰dl=2\pi a=$total length of the coil

$\therefore B=\frac{{\mu }_{o}I}{4\pi a^2}2\pi a$

$\therefore B=\frac{{\mu }_{o}I}{2a}$