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.

Obtain an expression for magnetic field (B) at the centre of a circular coil having ‘n’ turns of radius ‘R’ when it is carrying a current ‘I’.

Answer 1

Let there be a loop of radius r, carrying a current I. Let O be the centre of the loop, at which the field is required. By Biot-Savart’s law, the magnitude of the magnetic field at 0 due to a small element dl of the loop is:

dB = `(mu_0)/(4pi) (I dl sin θ)/r^2`

where θ is the angle between the length of the element (dl) and the line joining the element to the point 0. Here, θ = 90° (every element of a circle is perpendicular to the radius). Therefore,

dB = `(mu_0)/(4pi) (I dl)/r^2`

The direction of the field will be the same for all elements of the coil. Therefore, the magnitude of the  field `vecB` at O due to the entire coil is

B = `(mu_0)/(4pi) I/r^2 ∑dl`

But ∑dl = 2πr (the length of the circumference of the coil).

B = `(mu_0)/(4pi) I/r^2 (2pir)`

= `(mu_0)/(4pi) I/r (2pi)`

= `(mu_0 I)/(2r)`

If the circular coil has N turns, then the field at the centre would be

B = `(mu_0 NI)/(2r)`