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

Explanation

magnetic field produced a long the axis of the currnet carrying circular coil: Consider a current carrying ciruclar loop of radius R and let I be the current flowing through the wire in the direction. The magnetic field at a point P on the axis 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 d`vec(l)` at C and D. Let `vec(r)` be the vector joining the current element `(I d vec(l) )` at C to the point P.
PC = PD r = `sqrt(R^(2) + Z^(2))` and
angle `anlgeCPO = angleDPO = theta`
According to Biot - Savart.s law, the magnetic field at P due to the current element I `d vec(l)` is `d vec(B) = (mu_(0) I d vec(l) xx hat(r))/(4pi r^(2))`
The magnitude of magnetic field due to current element I d `vec(l )` at C and D are equal because
Of equal distance from the coil. the magnetic field d `vec(B)` due to each current element I `d vec(l)` is resolved into two components , dB sin `theta` along y-direction and dB cos `theta` along z-direction. Horizontal components of 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 d`vec(l)` around the loop , d`vec(B)` sweeps out a cone, then the net magnetic field `vec(B)` at point P is
`vec(B) = int d vec(B) = int " dB cos " theta hat(k)`
`vec(B) = (mu_(0)I)/(4 pi) int (dl)/(r^(2)) cos theta hat(k)`
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 `2piR`, we get `vec(B) = (mu_(0)I)/(2pi ) (R^(2))/((R^(2) + Z^(2))^((3)/(2))`
Not that the magnetic field `vec(B)` 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.