#### Question

Obtain an expression for magnetic flux density B at the centre of a circular coil of radius R, having N turns and carrying a current I

#### Solution

Magnetic flux density B at the centre of a circular coil of radius r having N turns and carrying a current I.

Consider a circular coil of radius r and carrying the current I in the direction as shown in the figure. 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 `vec(dB)` at the centre O of the coil due to current element `Ivec(dl)` is given by

`vec(dB) = mu_0/(4pi) (I(vec(dl) xx vecr))/r^3`

The magnitude of `vec(db)` at the centre O is

`dB = mu_0/(4pi) (Idl xx r sin theta)/(r^3)`

`db = mu_0/(4pi) (I dl sin theta)/r^2`

`.B = int dB`

`= int mu_0/(4pi) (i dl sin theta)/r^2`

`theta= 90^@` `∴ sin 90^@ = 1`

`:. B = mu_o/(4pi) I/r^2 int dldl`

`intdl`= total length of the coil = `2pir`

`B = mu_o/(4pi) I/r^2 (2pir)`

`B =(mu_oI)/(2r)`

If the coil N turns

`:. B = (mu_oNI)/(2r)`