A solenoid of length \ell and radius a consists of N turns of wire carrying current I. Show that at point P along its axis
H=\frac{nI}{2}(\cos\theta_{2}-\cos\theta_{1})a_{z}
where n=N/\ell, \theta _{1} and \theta_{2} are the angles subtended at P by the end turns as illustrated in Figure 7.9. Also show that if \ell\gg a, at the center of the solenoid
H=nIa_{z}
Consider the cross section of the solenoid as shown in Figure 7.9. Since the solenoid consists of circular loops, we apply the result of Example 7.3. The contribution to the magnetic field H at P by an element of the solenoid of length dz is
dH_{z}=\frac{Idla^{2}}{2[a^{2}+z^{2}]^{3/2}}=\frac{Ia^{2}ndz}{2[a^{2}+z^{2}]^{3/2}}
where dl=ndz=(N/\ell)dz. From Figure 7.9, \tan\theta=a/z; that is,
dz=-a\csc^{2}\theta d\theta =-\frac{[z^{2}+a^{2}]^{3/2}}{a^{2}}\sin\theta d\theta
Hence
dH_{z}=-\frac{nI}{2}\sin\theta d\theta
or
H_{z}=-\frac{nI}{2}\int_{\theta_{1}}^{\theta_{2}}\sin\theta d\theta
Thus
H=\frac{nI}{2}(\cos\theta_{2}-\cos\theta_{1})a_{z}
as required. Substituting n=N/\ell, gives
H=\frac{NI}{2\ell}(\cos\theta_{2}-\cos\theta_{1})a_{z}
At the center of the solenoid
\cos\theta_{2}=\frac{\ell/2}{[a^{2}+\ell^{2}/4]^{1/2}}=-\cos\theta_{1}
and
H=\frac{In\ell}{2[a^{2}+\ell^{2}/4]^{1/2}}a_{z}
If \ell\gg a or \theta_{2}\simeq 0^{\circ}, \theta_{1}\simeq 180^{\circ}
H=nIa_{z}=\frac{NI}{\ell}a_{z}