The magnetic field on the axis of a long solenoid having n turns per unit length and carrying a current is

Explanation

(1) A solenoid consists of a large number of circular coils placed close to each other. Each of these coils is not in a single plane, but if the turns are close we can assume them to be in a single plane.
(2) We can apply the formula for the magnetic field due to a single coil we derived just not (Eq. 29-15). Each of the coils has a different distance from the point, but we can find the magnetic field due to a short part of the solenoid, and then add up the contribution due to each.
Calculations : If we consider a coil of width `dx` at a distance x from the point P on the axis of the solenoid as shown in Fig. 29-11, we have
`dB=(mu_(0))/(4pi)(2pi(dN)iR^(2))/(4pi(R^(2)+x^(2))^(3//2))`. (29-16)
If n is the number of turns per unit length, `dN=ndx`.
As in Fig.29-11 `x=Rtanphi`, that is `dx=Rsec^(2)phidphi`, so
`dB=(mu_(0))/(4pi)(2pi(ndx)iR^(2))/((R^(2)+R^(2)tan^(2)phi)^(3//2))=(mu_(0))/(4pi)(2pini)cosphidphi` (29-17)
and hence
`B=(mu_(0))/(4pi)(2pini)int_(-alpha)^(beta)cosphidphi`
That is
`B=(mu_(0))/(4pi)(2pini)(sinalpha+sinbeta)`.

This is the desired result and from this it is clear that
1. If the solenoid is of infinite length and the point is well inside the solenoid,
`alpha=beta=(pi)/(2)`
So,
`beta=(mu_(0))/(4pi)(2pini)(1+1)`
`B=mu_(0)ni`
2. If the solenoid is of infinite length and the point is near one end,
`alpha=0`, `beta=(pi)/(2)`
So,
`beta=(mu_(0))/(4pi)(2pini)(1+0)=(1)/(2)(mu_(0)ni)`