\text { (a) } A (x, t)=\frac{\mu_{0}}{4 \pi} \int \frac{ K \left(t_{r}\right)}{ ᴫ } d a
=\frac{\mu_{0} \hat{ Z }}{4 \pi} \int \frac{K\left(t_{r}\right)}{\sqrt{r^{2}+x^{2}}} 2 \pi r d r
=\frac{\mu_{0} \hat{ z }}{2} \int \frac{K\left(t-\sqrt{r^{2}+x^{2}} / c\right)}{\sqrt{r^{2}+x^{2}}} r d r.
The maximum r is given by t-\sqrt{r^{2}+x^{2}} / c=0;
r_{\max }=\sqrt{c^{2} t^{2}-x^{2}} (since K(t) = 0 for t < 0).
(i)
A (x, t)=\frac{\mu_{0} K_{0} \hat{ z }}{2} \int_{0}^{r_{m}} \frac{r}{\sqrt{r^{2}+x^{2}}} d r=\left.\frac{\mu_{0} K_{0} \hat{ z }}{2} \sqrt{r^{2}+x^{2}}\right|_{0} ^{r_{m}}=\frac{\mu_{0} K_{0} \hat{ z }}{2}\left(\sqrt{r_{m}^{2}-x^{2}}-x\right)=\frac{\mu_{0} K_{0}(c t-x)}{2} \hat{ z }.
E (x, t)=-\frac{\partial A }{\partial t}=\frac{\mu_{0} K_{0} c}{2} \hat{ z } , for ct > x, and 0, for ct < x.
B (x, t)= \nabla \times A =-\frac{\partial A_{z}}{\partial x} \hat{ y }=\frac{\mu_{0} K_{0}}{2} \hat{ y } , for ct > x, and 0, for ct < x.
(ii)
A (x, t)=\frac{\mu_{0} \alpha \hat{ z }}{2} \int_{0}^{r_{m}} \frac{\left(t-\sqrt{r^{2}+x^{2}} / c\right)}{\sqrt{r^{2}+x^{2}}} r d r=\frac{\mu_{0} \alpha \hat{ z }}{2}\left[t \int_{0}^{r_{m}} \frac{r}{\sqrt{r^{2}+x^{2}}} d r-\frac{1}{c} \int_{0}^{r_{m}} r d r\right]
=\frac{\mu_{0} \alpha \hat{ z }}{2}\left[t(c t-x)-\frac{1}{2 c}\left(c^{2} t^{2}-x^{2}\right)\right]=\frac{\mu_{0} \alpha \hat{ z }}{4 c}\left(x^{2}-2 c t x+c^{2} t^{2}\right)=\frac{\mu_{0} \alpha(x-c t)^{2}}{4 c} \hat{ z }.
E (x, t)=-\frac{\partial A }{\partial t}=\frac{\mu_{0} \alpha(x-c t)}{2} \hat{ z } , for ct > x, and 0, for ct < x.
B (x, t)= \nabla \times A =-\frac{\partial A_{z}}{\partial x} \hat{ y }=-\frac{\mu_{0} \alpha}{2 c}(x-c t) \hat{ y } , for ct > x, and 0, for ct < x.
\text { (b) Let } u \equiv \frac{1}{c}\left(\sqrt{r^{2}+x^{2}}-x\right), \text { so } d u=\frac{1}{c}\left[\frac{1}{2} \frac{1}{\sqrt{r^{2}+x^{2}}} 2 r d r\right]=\frac{1}{c} \frac{r}{\sqrt{r^{2}+x^{2}}} d r , and
t-\frac{\sqrt{r^{2}+x^{2}}}{c}=t-\frac{x}{c}-u, \text { and as } r: 0 \rightarrow \infty, u: 0 \rightarrow \infty . \text { Then } A (x, t)=\frac{\mu_{0} c \hat{ z }}{2} \int_{0}^{\infty} K\left(t-\frac{x}{c}-u\right) d u \text {. } . qed
E (x, t)=-\frac{\partial A }{\partial t}=-\frac{\mu_{0} c \hat{ z }}{2} \int_{0}^{\infty} \frac{\partial}{\partial t} K\left(t-\frac{x}{c}-u\right) d u . \quad \text { But } \frac{\partial}{\partial t} K\left(t-\frac{x}{c}-u\right)=-\frac{\partial}{\partial u} K\left(t-\frac{x}{c}-u\right).
=\frac{\mu_{0} c}{2} \hat{ z } \int_{0}^{\infty} \frac{\partial}{\partial u} K\left(t-\frac{x}{c}-u\right) d u=\left.\frac{\mu_{0} c}{2} \hat{ z }\left[K\left(t-\frac{x}{c}-u\right)\right]\right|_{0} ^{\infty}=-\frac{\mu_{0} c}{2}[K(t-x / c)-K(-\infty)] \hat{ z }
=-\frac{\mu_{0} c}{2} K(t-x / c) \hat{ z },[\text { if } K(-\infty)=0].
Note that (i) and (ii) are consistent with this result. Meanwhile
B (x, t)=-\frac{\partial A_{z}}{\partial x} \hat{ y }=-\frac{\mu_{0} c}{c} \hat{ y } \int_{0}^{\infty} \frac{\partial}{\partial x} K\left(t-\frac{x}{c}-u\right) d u . \text { But } \frac{\partial}{\partial x} K\left(t-\frac{x}{c}-u\right)=\frac{1}{c} \frac{\partial}{\partial u} K\left(t-\frac{x}{c}-u\right).
=-\frac{\mu_{0}}{2} \hat{ y } \int_{0}^{\infty} \frac{\partial}{\partial u} K\left(t-\frac{x}{c}-u\right) d u=-\left.\frac{\mu_{0}}{2} \hat{ y }\left[K\left(t-\frac{x}{c}-u\right)\right]\right|_{0} ^{\infty}=\frac{\mu_{0}}{2}[K(t-x / c)-K(-\infty)] \hat{ y }
=\frac{\mu_{0}}{2} K(t-x / c) \hat{ y },[\text { if } K(-\infty)=0].
S =\frac{1}{\mu_{0}}( E \times B )=\frac{1}{\mu_{0}}\left(\frac{\mu_{0} c}{2}\right)\left(\frac{\mu_{0}}{2}\right) K(t-x / c)[-\hat{ z } \times \hat{ y }]=\frac{\mu_{0} c}{4}[K(t-x / c)]^{2} \hat{ x }.
This is the power per unit area that reaches x at time t; it left the surface at time (t-x / c). Moreover, an equal amount of energy is radiated downward, so the total power leaving the surface at time t is \frac{\mu_{0} c}{2}[K(t)]^{2} .
