Question 10.24: Suppose you take a plastic ring of radius a and glue charge ...

\lambda(\phi, t)=\lambda_{0}|\sin (\theta / 2)|, \text { where } \theta=\phi-\omega t . So the (retarded) scalar potential at the center is (Eq. 10.26)

V( r , t)=\frac{1}{4 \pi \epsilon_{0}} \int \frac{\rho\left( r ^{\prime}, t_{r}\right)}{ᴫ} d \tau^{\prime}, \quad A ( r , t)=\frac{\mu_{0}}{4 \pi} \int \frac{ J \left( r ^{\prime}, t_{r}\right)}{ᴫ} d \tau^{\prime}                            (10.26)

=\frac{\lambda_{0}}{4 \pi \epsilon_{0}}[2-(-2)]=\frac{\lambda_{0}}{\pi \epsilon_{0}}.

(Note: at fixed t_{r}, d \phi=d \theta , and it goes through one full cycle of  \phi \text { or } \theta .)

Meanwhile I (\phi, t)=\lambda v =\lambda_{0} \omega a|\sin [(\phi-\omega t) / 2]| \hat{ \phi } . From Eq. 10.26 (again)

A (t)=\frac{\mu_{0}}{4 \pi} \int \frac{ I }{ᴫ} d l^{\prime}=\frac{\mu_{0}}{4 \pi} \int_{0}^{2 \pi} \frac{\lambda_{0} \omega a\left|\sin \left[\left(\phi-\omega t_{r}\right) / 2\right]\right| \hat{\phi}}{a} a d \phi.

\text { But } t_{r}=t-a / c \text { is again constant, for the } \phi \text { integration, and } \hat{\phi}=-\sin \phi \hat{ x }+\cos \phi \hat{ y }.

=\frac{\mu_{0} \lambda_{0} \omega a}{4 \pi} \int_{0}^{2 \pi}\left|\sin \left[\left(\phi-\omega t_{r}\right) / 2\right]\right|(-\sin \phi \hat{ x }+\cos \phi \hat{ y }) d \phi \text {. Again, switch variables to } \theta=\phi-\omega t_{r},

\text { and integrate from } \theta=0 \text { to } \theta=2 \pi \text { (so we don't have to worry about the absolute value) }.

=\frac{\mu_{0} \lambda_{0} \omega a}{4 \pi} \int_{0}^{2 \pi} \sin (\theta / 2)\left[-\sin \left(\theta+\omega t_{r}\right) \hat{ x }+\cos \left(\theta+\omega t_{r}\right) \hat{ y }\right] d \theta . Now

=-2 \sin \left(\omega t_{r}\right)+\frac{2}{3} \sin \left(\omega t_{r}\right)=-\frac{4}{3} \sin \left(\omega t_{r}\right).

=-2 \cos \left(\omega t_{r}\right)+\frac{2}{3} \cos \left(\omega t_{r}\right)=-\frac{4}{3} \cos \left(\omega t_{r}\right) . So

A (t)=\frac{\mu_{0} \lambda_{0} \omega a}{4 \pi}\left(\frac{4}{3}\right)\left[\sin \left(\omega t_{r}\right) \hat{ x }-\cos \left(\omega t_{r}\right) \hat{ y }\right]=\frac{\mu_{0} \lambda_{0} \omega a}{3 \pi}\{\sin [\omega(t-a / c)] \hat{ x }-\cos [\omega(t-a / c)] \hat{ y }\}.