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

Explanation

Let r be the radius of coil with centre O. Let I be current which flows through it in clockwise direction. Let dB is the magnetic field intensity produce at the centre by a small element XY of length dl, then according to Biot-Savart.s law,
`dB = (mu_0)/(4pi) (I dl sin theta)/(r^2)`
But in this case, `theta = 90^@` and `sin 90 = 1 `
Total magnetic field at the centre of the coil due to the whole coil.
`int dB = int (mu_0)/(4pi)(I dl)/(r^2)`
`implies B = (mu_0)/(4pi) I/(r^2) int dl`
`implies B = (mu_0 I)/(4pi)(I)/(r^2)(2pi r)`
where `l = 2 pi r` = length of the ring.
`B = (mu_0)/(4pi) (2pi l)/(r) " or " B = (mu_0 I)/(2r)`
If there are n turns in the coil then
`B = (mu_0)/(4pi) (2pi n I)/r`
or `B = (mu_0 n I)/(2r)`
According to Right Hand thumb Rule, magnetic field at the centre of currect carrying circular coil is perpendicular to the plane of loop and in vertically downward direction.