电动力学每日一题 2021/10/14
電動力學每日一題 2021/10/14
(a) Define r∣∣=xx^+yy^\textbf r_{||}=x\hat x+y\hat yr∣∣?=xx^+yy^?, r∣∣=x2+y2r_{||}=\sqrt{x^2+y^2}r∣∣?=x2+y2?,
M(r,t)=M0z^[Rect(x/Lx)Rect(y/Ly)?Circ(r∣∣/R)]Rect(z/Lz)\textbf M(\textbf r,t)=M_0\hat z[Rect(x/L_x)Rect(y/L_y)-Circ(r_{||}/R)]Rect(z/L_z)M(r,t)=M0?z^[Rect(x/Lx?)Rect(y/Ly?)?Circ(r∣∣?/R)]Rect(z/Lz?)
(b)
ρbound(m)(r,t)=???M=??Mz?z=M0[Rect(x/Lx)Rect(y/Ly)?Circ(r∣∣/R)][δ(z?Lz/2)?δ(z+Lz/2)]\begin{aligned}\rho_{bound}^{(m)}(\textbf r,t) & = -\nabla \cdot \textbf M = -\frac{\partial M_z}{\partial 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{aligned}ρbound(m)?(r,t)?=???M=??z?Mz??=M0?[Rect(x/Lx?)Rect(y/Ly?)?Circ(r∣∣?/R)][δ(z?Lz?/2)?δ(z+Lz?/2)]?
Jbound(e)=μ0?1?×M=μ0?1M0[?×Rect(x/Lx)Rect(y/Ly)Rect(z/Lz)z^??×Circ(r∣∣/R)Rect(z/Lz)z^]\begin{aligned} \textbf J_{bound}^{(e)} &= \mu_0^{-1}\nabla \times \textbf M \\ & =\mu_0^{-1} M_0[\nabla \times Rect(x/L_x)Rect(y/L_y)Rect(z/L_z)\hat z-\nabla \times Circ(r_{||}/R)Rect(z/L_z)\hat z]\end{aligned}Jbound(e)??=μ0?1??×M=μ0?1?M0?[?×Rect(x/Lx?)Rect(y/Ly?)Rect(z/Lz?)z^??×Circ(r∣∣?/R)Rect(z/Lz?)z^]?
在直角坐標系中計算
?×Rect(x/Lx)Rect(y/Ly)Rect(z/Lz)z^=μ0?1M0Rect(x/Lx)[δ(y+Ly/2)?δ(y?Ly/2)]Rect(z/Lz)x^?μ0?1M0[δ(x+Lx/2)?δ(x?Lx/2)]Rect(y/Ly)Rect(z/Lz)y^\nabla \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?×Rect(x/Lx?)Rect(y/Ly?)Rect(z/Lz?)z^=μ0?1?M0?Rect(x/Lx?)[δ(y+Ly?/2)?δ(y?Ly?/2)]Rect(z/Lz?)x^?μ0?1?M0?[δ(x+Lx?/2)?δ(x?Lx?/2)]Rect(y/Ly?)Rect(z/Lz?)y^?
在柱坐標系中計算
?×Circ(r∣∣/R)Rect(z/Lz)z^=?μ0?1M0δ(r∣∣?R)Rect(z/Lz)?^\nabla \times Circ(r_{||}/R)Rect(z/L_z)\hat z =-\mu_0^{-1}M_0\delta(r_{||}-R)Rect(z/L_z) \hat \phi?×Circ(r∣∣?/R)Rect(z/Lz?)z^=?μ0?1?M0?δ(r∣∣??R)Rect(z/Lz?)?^?
綜上,
Jbound(e)=μ0?1M0Rect(x/Lx)[δ(y+Ly/2)?δ(y?Ly/2)]Rect(z/Lz)x^?μ0?1M0[δ(x+Lx/2)?δ(x?Lx/2)]Rect(y/Ly)Rect(z/Lz)y^?μ0?1M0δ(r∣∣?R)Rect(z/Lz)?^\textbf 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) \hat \phiJbound(e)?=μ0?1?M0?Rect(x/Lx?)[δ(y+Ly?/2)?δ(y?Ly?/2)]Rect(z/Lz?)x^?μ0?1?M0?[δ(x+Lx?/2)?δ(x?Lx?/2)]Rect(y/Ly?)Rect(z/Lz?)y^??μ0?1?M0?δ(r∣∣??R)Rect(z/Lz?)?^?
(c)
ρbounded(m)(k,w)=∫?∞+∞ρbound(m)(r,t)e?i(k?r?wt)drdt=2πM0δ(w)(I1I2?I3)I4\begin{aligned} \rho^{(m)}_{bounded}(\textbf k,w)& =\int_{-\infty}^{+\infty} \rho^{(m)}_{bound}(\textbf r,t)e^{-i(\textbf k\cdot \textbf r-wt)}d \textbf rdt \\ & = 2 \pi M_0\delta(w) (I_1I_2-I_3)I_4\end{aligned}ρbounded(m)?(k,w)?=∫?∞+∞?ρbound(m)?(r,t)e?i(k?r?wt)drdt=2πM0?δ(w)(I1?I2??I3?)I4??
其中
I1=F[Rect(x/Lx)](kx)=sin?(Lxkx/2)kx/2I2=F[Rect(y/Ly)](ky)=sin?(Lyky/2)ky/2I3=F[Circ(r∣∣/R)](k∣∣)=2πRk∣∣J1(K∣∣R)I4=F[δ(z?Lz/2)?δ(z+Lz/2)](kz)=?2isin?(Lzkz/2)I_1 = \mathcal{F}[Rect(x/L_x)](k_x)=\frac{\sin(L_xk_x/2)}{k_x/2} \\ I_2 = \mathcal{F}[Rect(y/L_y)](k_y)=\frac{\sin(L_yk_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_zk_z/2)I1?=F[Rect(x/Lx?)](kx?)=kx?/2sin(Lx?kx?/2)?I2?=F[Rect(y/Ly?)](ky?)=ky?/2sin(Ly?ky?/2)?I3?=F[Circ(r∣∣?/R)](k∣∣?)=k∣∣?2πR?J1?(K∣∣?R)I4?=F[δ(z?Lz?/2)?δ(z+Lz?/2)](kz?)=?2isin(Lz?kz?/2)
綜上
ρbounded(m)(k,w)=?i4πM0δ(w)[LxLysinc(Lxkx2π)sinc(Lyky2π)?2πRJ1(Rkx2+ky2)kx2+ky2]sin?(Lzkz/2)\begin{aligned} \rho^{(m)}_{bounded}(\textbf k,w) = -i4 \pi M_0 \delta(w) \left[ L_xL_y sinc(\frac{L_xk_x}{2 \pi})sinc(\frac{L_yk_y}{2 \pi})-2 \pi R \frac{J_1(R\sqrt{k_x^2+k_y^2})}{\sqrt{k_x^2+k_y^2}} \right]\sin(L_zk_z/2) \end{aligned}ρbounded(m)?(k,w)=?i4πM0?δ(w)???Lx?Ly?sinc(2πLx?kx??)sinc(2πLy?ky??)?2πRkx2?+ky2??J1?(Rkx2?+ky2??)????sin(Lz?kz?/2)?
簡直了,考試應該不會考這種題吧,照著公式兩個小時也算不完啊。。。
Jbound(e)(k,w)=i2πμ0?1M0δ(w)Lzsinc(Lxkx2π)[LxLysinc(Lxkx2π)sinc(Lyky2π)(kyx^?kxy^)+2πRJ1(Rkx2+ky2)?^]\textbf J^{(e)}_{bound}(\textbf k,w)=i 2 \pi \mu_0^{-1}M_0\delta(w)L_z sinc(\frac{L_xk_x}{2 \pi})\left[ L_xL_y sinc(\frac{L_xk_x}{2 \pi})sinc(\frac{L_yk_y}{2 \pi})(k_y \hat x - k_x \hat y)+2 \pi R J_1(R\sqrt{k_x^2+k_y^2}) \hat \phi \right]Jbound(e)?(k,w)=i2πμ0?1?M0?δ(w)Lz?sinc(2πLx?kx??)[Lx?Ly?sinc(2πLx?kx??)sinc(2πLy?ky??)(ky?x^?kx?y^?)+2πRJ1?(Rkx2?+ky2??)?^?]
總結
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