Question 5.11: Find the magnetic field at point P on the axis of a tightly ...

Use Eq. 5.41 for a ring of width dz, with I → nI dz:

B(z)=\frac{\mu_{0} I}{4 \pi}\left(\frac{\cos \theta}{ᴫ^{2}}\right) 2 \pi R=\frac{\mu_{0} I}{2} \frac{R^{2}}{\left(R^{2}+z^{2}\right)^{3 / 2}}                          (5.41)

B=\frac{\mu_{0} n I}{2} \int \frac{a^{2}}{\left(a^{2}+z^{2}\right)^{3 / 2}} d z . \text { But } z=a \cot \theta ,

so  d z=-\frac{a}{\sin ^{2} \theta} d \theta, \text { and } \frac{1}{\left(a^{2}+z^{2}\right)^{3 / 2}}=\frac{\sin ^{3} \theta}{a^{3}} .

B=\frac{\mu_{0} n I}{2} \int \frac{a^{2} \sin ^{3} \theta}{a^{3} \sin ^{2} \theta}(-a d \theta)=-\frac{\mu_{0} n I}{2} \int \sin \theta d \theta=\left.\frac{\mu_{0} n I}{2} \cos \theta\right|_{\theta_{1}} ^{\theta_{2}}=\frac{\mu_{0} n I}{2}\left(\cos \theta_{2}-\cos \theta_{1}\right) .

For an infinite solenoid, \theta_{2}=0, \theta_{1}=\pi, \text { so }\left(\cos \theta_{2}-\cos \theta_{1}\right)=1-(-1)=2, \text { and } B=\mu_{0} n I .