We can calculate magnetic field on the axis of the current carrying circular coil.
Suppose a circular coil of radius ‘a’ is carrying current ‘I’ as shown in figure. Take a point ‘P’ on the axis of coil at which the magnetic field is to be determined. Let XY is equal to X’ Y’= dl be the small element of the conductor at the opposite end of diameter as shown in figure.
Let dB and dB^’ be the magnetic field due the element XY and X’ Y’ at point P
Now, from Biot and Savart law,
Magnetic field at P due the element XY and X’ Y’ is:
Here, = 10-7wbm-1A-1 called the permeability of air.
Resolving the component of dB and dB’ we get, as shown in figure. Here, dBcosθ and DB’cosθ are equal, opposite and acting at a point. Thus they cancel each other. Thus the total magnetic field due to the coil is sum of dBsinθ. We can calculate the magnetic field by integrating from 0 to 2 πa. Now,
If there are ‘N’ turns of coil, then:
When point P is at the centre of the coil, then x =0, so:
When P is very far away from the centre of the coil, then x is very greater than a, so:
Here, the magnetic field is along the axis of the coil.