Using Biot Savart's law, derive the expression for the magnetic field at a point on the axis of a circular current loop.

Explanation

Consider a circular loop of radius R carrying a current I. Let P be a point at a distance x from the centre the , of the loop on axis. AB is current element of length dl. The distance of the point P from current element is r.
`therefore` From figure
`r=sqrt(R^2+x^2) sin theta=R/r`
According to Biot-Savart.s law, magnetic field at the pont P due to the current element AB is
`dB =mu_0/(4pi)(Idl)/r^2 "sin"phi`
Since `phi=90^@`
`dB=mu_0/(4pi)(Idl)/r^2` sin `90^@` ( `because ` sin `90^@=1`)
`dB=mu_0/(4pi)(I dl)/r^2 ` along PM
dB due to the current element A.B. at the point P is `dB=mu_0/(4pi)(I dl)/r^2` along PN.
The magnetic field dB along PM can be resolved into two rectangular components dB cos `phi` along Py and dB sin `phi` along Px. Similarly the magnetic field `d_B` along PN can be resolved into two components .
If we consider more small elements cos `theta` components cancel each other out .
`SigmadB cos phi=0 " " alphaSigmad_B cos theta # 0`
`B=intdB sin phi`
`B=intmu_0/(4pi)(Idl)/r^2 sin phi`
Substituting the values of `r=sqrt(R^2+x^2)` and sin `phi=R/r` in the equation , we get
`B=intmu_0/(4pi)(Idl)/(R^2+x^2) R/r = int mu_0/(4pi) (I dl)/(R^2+x^2) R/sqrt(R^2+x^2) =mu_0/(4pi) (IR)/((R^2+x^2)^(3//2)) int dl`
But `intdl=2piR to` circumference of the loop
`therefore B=(mu_0IR)/(4pi(x^2+R^2)^(3//2))xx2piR`
`therefore B=mu_0/(4pi)(2piIR^2)/((x^2+R^2)^(3//2))`