Obtain a relation for the magnetic induction at a point along the axis of a circular coil carrying current.

Explanation

Consider a current carrying circular loop of radius R and let I be the current flowing through the wire in the direction as shown in Figure.
The magnetic field at a point P on the exis of the circular coil at a distance z from its center of the coil O . It is computed by taking two diametrically opposite line elements of the coil each of length `vec(dl)` at C and D. Let `hat r` be the vector joining the current element `(I vec(dl))` at C to the point P .
We know that
` PC = PD = r = sqrt(R^(2) + Z^(2)) " and angle " angleCPO = angleDPO = theta `

According to Biot - Savart's law, the magnetic field at P due to the current element I `vec(dl) ` is
`vec(dB) = (mu_(0))/(4 pi)(Ivec(dl) xx hat r)/r^(2) `
The magnitude of magnetic field due to current element I `vec(dl) ` at C and D are equal because of equal distance from the coil. The magnetic field `vec(dB) ` due to each current element I `vec(dl)` is resolved into two components, dB sin `theta` along y - direction and dB `cos theta` along z - direction.
Horizontal components each current element cancels out while the vertical components `(dB cos theta hat k)` alone contribute to total magnetic field at the point P .

If we integrate `vec(dl) ` around the loop, `vec(dB) ` sweeps out a cone as shown in Figure, then the net magnetic field `vecB` at point P is
`vecB = int vec(dB) = int dB cos theta hatk`
` vecB = (mu_(0)I)/(4 pi)int (dl)/(r^(2)) cos theta hatk`
But `cos theta = R/((R^(2) + z^(2))^(1/2)) `, using Pythagorous theorem `r^(2) = R^(2) + z^(2)` and integrating line element from 0 to `2 pi R`, we get
`vecB = (mu_(0)I)/(2 pi) R^(2)/((R^(2) + z^(2))^(3/2))hatk`
Note that the magnetic field `vecB` points along the direction from the point O to P . Suppose if the current flows in clockwise direction , then magnetic field points in the direction from the point P to O .