STEM Problem Detail

Problem Statement

Prompt

Give step by step solution for Problem 11.28

Solution Rendering...

# Problem 11.28 — Give step by step solution for Problem 11.28

**(a)**

$$
A(x,t)
= \dfrac{\mu_0}{4\pi}\int \dfrac{K(t_r)}{r}\, da
= \dfrac{\mu_0}{4\pi}\int \dfrac{K(t_r)}{\sqrt{r^2 + x^2}}\, 2\pi r\, dr
= \dfrac{\mu_0}{2}\int
\dfrac{K \left(t - \sqrt{r^2 + x^2}/c\right)}
{\sqrt{r^2 + x^2}}\, r\, dr.
$$

The maximum range is given by
$$t - \sqrt{r^2 + x^2}/c = 0;$$

$$r_{\max} = \sqrt{c^2t^2 - x^2}$$
(since $$K(t)=0$$ for $$t<0$$).

**(i)**

$$A(x,t)
= \dfrac{\mu_0 K_0}{2}
\int_0^{r_{\max}} \dfrac{r}{\sqrt{r^2 + x^2}}\,dr
= \dfrac{\mu_0 K_0}{2}\sqrt{r^2 + x^2}\Big|_0^{r_{\max}}$$

$$= \dfrac{\mu_0 K_0}{2}
\left(\sqrt{r_{\max}^2 + x^2} - x\right)
= \dfrac{\mu_0 K_0(ct - x)}{2}\,\hat{\mathbf{z}}.$$

$$\mathbf{E}(x,t)
= -\dfrac{\partial \mathbf{A}}{\partial t}
= -\dfrac{\mu_0 K_0}{2}\,\hat{\mathbf{z}},
\quad \text{for } ct > x,
\text{ and } 0,\ \text{for } ct < x.$$

$$\mathbf{B}(x,t)
= \nabla \times \mathbf{A}
= -\dfrac{\partial A_z}{\partial x}\,\hat{\mathbf{y}}
= \dfrac{\mu_0 K_0}{2}\,\hat{\mathbf{y}},
\quad \text{for } ct > x,
\text{ and } 0,\ \text{for } ct < x.$$

**(ii)**

$$A(x,t)
= \dfrac{\mu_0\alpha}{2}
\int_0^{r_{\max}}
\dfrac{t - \sqrt{r^2 + x^2}/c}{\sqrt{r^2 + x^2}}\,r\,dr$$

$$= \dfrac{\mu_0\alpha}{2}
\left[
t\int_0^{r_{\max}}\dfrac{r}{\sqrt{r^2 + x^2}}\,dr
- \dfrac{1}{c}\int_0^{r_{\max}} r\,dr
\right]$$

$$= \dfrac{\mu_0\alpha}{2}
\left[
t(ct - x)
- \dfrac{1}{2c}(c^2t^2 - x^2)
\right]$$

$$= \dfrac{\mu_0\alpha}{4c}
(x^2 - 2ctx + c^2t^2)\,\hat{\mathbf{z}}
= \dfrac{\mu_0\alpha(x - ct)^2}{4c}\,\hat{\mathbf{z}}.$$

$$\mathbf{E}(x,t)
= -\dfrac{\partial \mathbf{A}}{\partial t}
= \dfrac{\mu_0\alpha(x - ct)}{2}\,\hat{\mathbf{z}},
\quad \text{for } ct > x,
\text{ and } 0,\ \text{for } ct < x.$$

$$\mathbf{B}(x,t)
= \nabla \times \mathbf{A}
= -\dfrac{\partial A_z}{\partial x}\,\hat{\mathbf{y}}
= -\dfrac{\mu_0\alpha(x - ct)}{2c}\,\hat{\mathbf{y}},
\quad \text{for } ct > x,
\text{ and } 0,\ \text{for } ct < x.$$

---

**(b)** Let
$$u = \dfrac{1}{c}(\sqrt{r^2 + x^2} - x),$$
so

$$du
= \dfrac{1}{c}\left[\dfrac{1}{2\sqrt{r^2 + x^2}}\,2r\,dr\right]
= \dfrac{r}{\sqrt{r^2 + x^2}}\,dr,$$

and

$$t - \dfrac{\sqrt{r^2 + x^2}}{c}
= t - \dfrac{x}{c} - u.$$

As $$r : 0 \to \infty,\ u : 0 \to \infty.$$
Then

$$A(x,t)
= \dfrac{\mu_0 c}{2}
\int_0^{\infty} K\!\left(t - \dfrac{x}{c} - u\right)du.$$

$$\mathbf{E}(x,t)
= -\dfrac{\partial \mathbf{A}}{\partial t}
= -\dfrac{\mu_0 c}{2}
\int_0^{\infty}
\dfrac{\partial}{\partial t}
K\!\left(t - \dfrac{x}{c} - u\right)du.$$

But

$$\dfrac{\partial}{\partial t}
K\!\left(t - \dfrac{x}{c} - u\right)
= -\dfrac{\partial}{\partial u}
K\!\left(t - \dfrac{x}{c} - u\right).$$

Hence

$$\mathbf{E}(x,t)
= -\dfrac{\mu_0 c}{2}
\int_0^{\infty}
\left[-\dfrac{\partial}{\partial u}
K\!\left(t - \dfrac{x}{c} - u\right)\right]du$$

$$= \dfrac{\mu_0 c}{2}
\left[K\!\left(t - \dfrac{x}{c}\right) - K(-\infty)\right]\hat{\mathbf{z}}.$$

If $$K(-\infty)=0$$,

$$\boxed{
\mathbf{E}(x,t)
= \dfrac{\mu_0 c}{2}\,
K\!\left(t - \dfrac{x}{c}\right)\hat{\mathbf{z}}.
}$$

Meanwhile,

$$\mathbf{B}(x,t)
= -\dfrac{\partial A_z}{\partial x}\,\hat{\mathbf{y}}
= -\dfrac{\mu_0 c}{2}
\int_0^{\infty}
\dfrac{\partial}{\partial x}
K\!\left(t - \dfrac{x}{c} - u\right)du.$$

But

$$\dfrac{\partial}{\partial x}
K\!\left(t - \dfrac{x}{c} - u\right)
= -\dfrac{1}{c}\,\dot{K}\!\left(t - \dfrac{x}{c} - u\right)
= \dfrac{1}{c}\,
\dfrac{\partial}{\partial u}
K\!\left(t - \dfrac{x}{c} - u\right).$$

Therefore

$$\mathbf{B}(x,t)
= -\dfrac{\mu_0}{2}
\int_0^{\infty}
\dfrac{\partial}{\partial u}
K\!\left(t - \dfrac{x}{c} - u\right)du$$

$$= \dfrac{\mu_0}{2}
\left[K\!\left(t - \dfrac{x}{c}\right) - K(-\infty)\right]\hat{\mathbf{y}}.$$

If $$K(-\infty)=0$$,

$$\boxed{
\mathbf{B}(x,t)
= \dfrac{\mu_0}{2}\,
K\!\left(t - \dfrac{x}{c}\right)\hat{\mathbf{y}}.
}$$

Finally,

$$\mathbf{S}
= \dfrac{1}{\mu_0}(\mathbf{E}\times\mathbf{B})
= \dfrac{1}{\mu_0}
\left(\dfrac{\mu_0 c}{2}\right)
\left(\dfrac{\mu_0}{2}\right)
K^2\!\left(t - \dfrac{x}{c}\right)
(-\hat{\mathbf{z}}\times\hat{\mathbf{y}})$$

$$= \dfrac{\mu_0 c}{4}
\left[K\!\left(t - \dfrac{x}{c}\right)\right]^2
\hat{\mathbf{x}}.$$

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

$$\boxed{\dfrac{\mu_0 c}{2}\,[K(t)]^2.}$$

Loading solution...

Problem #27