電動力學(xué)每日一題 2021/10/14

(a) Define ${\text{r}}_{||}=x\hat{x}+y\hat{y}$, $r_{||}=\sqrt{x^2+y^2}$,
$$\text{M}(\text{r},t)=M_0\hat{z}[Rect(x/L_x)Rect(y/L_y)?Circ(r_{||}/R)]Rect(z/L_z)$$

(b)
$$\begin{array}{rl}{\rho }_{bound}^{(m)}(\text{r},t)& =???\text{M}=?\frac{?M_z}{?z}\\ & =M_0[Rect(x/L_x)Rect(y/L_y)?Circ(r_{||}/R)][\delta (z?L_z/2)?\delta (z+L_z/2)]\end{array}$$

$$\begin{array}{rl}{\text{J}}_{bound}^{(e)}& ={\mu }_0^{?1}?\times \text{M}\\ & ={\mu }_0^{?1}{M}_0[?\times Rect(x/L_x)Rect(y/L_y)Rect(z/L_z)\hat{z}??\times Circ(r_{||}/R)Rect(z/L_z)\hat{z}]\end{array}$$

在直角坐標(biāo)系中計算
$$\begin{gathered} ?\times Rect(x/L_x)Rect(y/L_y)Rect(z/L_z)\hat{z} \\ ={\mu }_0^{?1}{M}_{0}Rect(x/L_x)[\delta (y+L_y/2)?\delta (y?L_y/2)]Rect(z/L_z)\hat{x} \\ ?{\mu }_0^{?1}{M}_0[\delta (x+L_x/2)?\delta (x?L_x/2)]Rect(y/L_y)Rect(z/L_z)\hat{y} \end{gathered}$$

在柱坐標(biāo)系中計算
$$?\times Circ(r_{||}/R)Rect(z/L_z)\hat{z}=?{\mu }_0^{?1}{M}_0\delta (r_{||}?R)Rect(z/L_z)\widehat{?}$$

綜上，
$$\begin{gathered} {\text{J}}_{bound}^{(e)}={\mu }_0^{?1}{M}_{0}Rect(x/L_x)[\delta (y+L_y/2)?\delta (y?L_y/2)]Rect(z/L_z)\hat{x} \\ ?{\mu }_0^{?1}{M}_0[\delta (x+L_x/2)?\delta (x?L_x/2)]Rect(y/L_y)Rect(z/L_z)\hat{y} \\ ?{\mu }_0^{?1}{M}_0\delta (r_{||}?R)Rect(z/L_z)\widehat{?} \end{gathered}$$

(c)
$$\begin{array}{rl}{\rho }_{bounded}^{(m)}(\text{k},w)& ={\int }_{?\infty }^{+\infty }{\rho }_{bound}^{(m)}(\text{r},t)e^{?i(\text{k}?\text{r}?wt)}d\text{r}dt\\ & =2\pi M_0\delta (w)(I_1{I}_2?I_3)I_4\end{array}$$

其中
$$\begin{gathered} I_1=\mathcal{F}[Rect(x/L_x)](k_x)=\frac{\sin (L_x{k}_x/2)}{k_x/2} \\ {I}_2=\mathcal{F}[Rect(y/L_y)](k_y)=\frac{\sin (L_y{k}_y/2)}{k_y/2} \\ {I}_3=\mathcal{F}[Circ(r_{||}/R)](k_{||})=\frac{2\pi R}{k_{||}}{J}_1(K_{||}R) \\ {I}_4=\mathcal{F}[\delta (z?L_z/2)?\delta (z+L_z/2)](k_z)=?2i\sin (L_z{k}_z/2) \end{gathered}$$

綜上
$$\begin{array}{r}{\rho }_{bounded}^{(m)}(\text{k},w)=?i4\pi M_0\delta (w)?L_x{L}_{y}sinc(\frac{L_x{k}_x}{2\pi })sinc(\frac{L_y{k}_y}{2\pi })?2\pi R\frac{J_1(R\sqrt{k_x^2+k_y^2})}{\sqrt{k_x^2+k_y^2}}?\sin (L_z{k}_z/2)\end{array}$$

簡直了，考試應(yīng)該不會考這種題吧，照著公式兩個小時也算不完啊。。。

$${\text{J}}_{bound}^{(e)}(\text{k},w)=i2\pi {\mu }_0^{?1}{M}_0\delta (w)L_{z}sinc(\frac{L_x{k}_x}{2\pi })\left[L_x{L}_{y}sinc(\frac{L_x{k}_x}{2\pi })sinc(\frac{L_y{k}_y}{2\pi })(k_y\hat{x}?k_x\hat{y})+2\pi R{J}_1(R\sqrt{k_x^2+k_y^2})\widehat{?}\right]$$

總結(jié)

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