Question 5.62: A thin glass rod of radius R and length L carries a uniform ...

For a dipole at the origin and a field point in the x z plane (\phi = 0) , we have

=\frac{\mu_{0}}{4 \pi} \frac{m}{r^{3}}\left[3 \sin \theta \cos \theta \hat{ x }+\left(2 \cos ^{2} \theta-\sin ^{2} \theta\right) \hat{ z }\right] .

Here we have a stack of such dipoles, running from z = –L/2 to z = +L/2. Put the field point at s on the x axis. The \hat{ x } components cancel (because of symmetrically placed dipoles above and below z = 0), leaving B = \frac{\mu_{0}}{4 \pi} 2 M \hat{ z } \int_{0}^{L / 2} \frac{\left(3 \cos ^{2} \theta-1\right)}{r^{3}} d z ,

where M is the dipole moment per unit length:

m=I \pi R^{2}=(\sigma v h) \pi R^{2}=\sigma \omega R \pi R^{2} h \Rightarrow M =\frac{m}{h}=\pi \sigma \omega R^{3} .

Now  \sin \theta=\frac{s}{r}, \text { so } \frac{1}{r^{3}}=\frac{\sin ^{3} \theta}{s^{3}} ; z=-s \cot \theta \Rightarrow d z=\frac{s}{\sin ^{2} \theta} d \theta . Therefore 

=\left.\frac{\mu_{0} \sigma \omega R^{3}}{2 s^{2}} \hat{ z }\left(-\cos ^{3} \theta+\cos \theta\right)\right|_{\pi / 2} ^{\theta_{m}}=\frac{\mu_{0} \sigma \omega R^{3}}{2 s^{2}} \cos \theta_{m}\left(1-\cos ^{2} \theta_{m}\right) \hat{ z }=\frac{\mu_{0} \sigma \omega R^{3}}{2 s^{2}} \cos \theta_{m} \sin ^{2} \theta_{m} \hat{ z } .

But  \sin \theta_{m}=\frac{s}{\sqrt{s^{2}+(L / 2)^{2}}}, \text { and } \cos \theta_{m}=\frac{-(L / 2)}{\sqrt{s^{2}+(L / 2)^{2}}}, \text { so } B =-\frac{\mu_{0} \sigma \omega R^{3} L}{4\left[s^{2}+(L / 2)^{2}\right]^{3 / 2}} \hat{ z } .