Apply Biot-Savart law to find the magnetic field due to a circular current carrying loop at a point on the axis of the loop.

Explanation

1. Consider current carrying loop with radius R as shown in figure.

2. Centre point of loop is O and loop lies on X-axis.
3. We wish to calculate the magnetic field at the point P on this axis.
4. Loop is kept perpendicular to the plane of loop.
5. Let x be the distance of P from the centre o of the loop.
6. The magnitude `dvecB` of the magnetic field due to `dvecl` is given by the Biot-Savart law,
`dvecB=mu_(0)/(4pi)(Ivec(dl)xxvecr)/r^(3)`
7. Magnitude of this magnetic field is given by,
`|dvecB|=mu_(0)/(4pi)(|Ivec(dl)xxvecr|)/r^(3)`
= `mu_(0)/(4pi)(Idlrsintheta.)/r^(3)`
where `theta.` = angle between `vec(dl)andvecr" but "vec(dl)_|_vecr`
so, `sintheta.="sin"pi/2=1`
`therefore|dvecB|=(mu_(0)I)/(4pi)(dl)/r^(2)" "...(1)`
8. As shown in diagram `r^(2)=x^(2)+R^(2)`
`|dvecB|=mu_(0)/(4pi)(dl)/((x^(2)+R^(2)))" "...(2)`
9. Direction of `dvecB` is given by perpendicular to plane of `vec(dl)andvecr`.
10. `dvecB` can be divide in two components,
parallel component `dB_(x)=dBsintheta`
perpendicular component `dB_(_|_)=dBcostheta`
11. When the components perpendicular to the ring axis are summed over, they cancel out and we obtain null result. Here, `dB_(_|_)=dBcostheta` component due to `vec(dl)` is cancelled by the contribution due to the diametrically opposite `vec(dl)` element shown in figure. Thus, only the parallel component survives.
12. The net contribution along X-direction can be obtained by integrating `dB_(x)=dBsintheta` over the loop,
`dB_(x)=dBsintheta" "...(3)`
From diagram geometry,
`sintheta=R/((x^(2)+R^(2))^(1//2))" "...(4)`
13. Substitute dB and `costheta` in equation (3)
`dB_(x)=mu_(0)/(4pi)(Idl)/((x^(2)+R^(2)))*R/((x^(2)+R^(2))^(1//2))`
`thereforedB_(x)=mu_(0)/(4pi)(IdlR)/((x^(2)+R^(2))^(3//2))" "...(5)`
14. To integrate this term, we can get total magnetic field in x - direction.
`B=intdB_(x)`
`B=(mu_(0)IR)/(4pi(x^(2)+R^(2))^(3//2))intdl`
`B=(mu_(0)IR)/(4pi(x^(2)+R^(2))^(3//2))(2piR)" "[becauseintdl=2piR]`
`thereforeB=(mu_(0)IR^(2))/(2(x^(2)+R^(2))^(3//2))" "...(6)`
15. Vector form of magnetic field,
`vecB=(mu_(0)IR^(2))/(2(x^(2)+R^(2))^(3//2))hati" "[becausehati" is unit vector of x-axis"]`