电动力学每日一题 2021/10/15 Fourier变换法计算均匀电流密度产生的磁场

无限长均匀电流
无限长圆柱面均匀电流密度

无限长均匀电流

假设z轴上有一根非常细的电线，携带均匀电流I0I_0I0，则空间中的电流密度可以表示为
J(r,t)=I0δ(x)δ(y)z^\textbf J(\textbf r,t)=I_0 \delta(x)\delta(y)\hat zJ(r,t)=I0δ(x)δ(y)z^

它的Fourier变换为
J(k,w)=(2π)2I0δ(kz)δ(w)z^\textbf J(\textbf k,w)=(2 \pi)^{2}I_0 \delta(k_z)\delta(w)\hat zJ(k,w)=(2π)2I0δ(kz)δ(w)z^

则磁场的Fourier变换为
H(k,w)=ik×J(k,w)k2−(w/c)2=i(2π)2I0δ(kz)δ(w)k2−(w/c)2k×z^\textbf H(\textbf k,w)=\frac{i \textbf k \times \textbf J(\textbf k ,w)}{k^2 - (w/c)^2}=\frac{i(2 \pi)^{2}I_0 \delta(k_z)\delta(w)}{k^2-(w/c)^2} \textbf k \times \hat zH(k,w)=k2−(w/c)2ik×J(k,w)=k2−(w/c)2i(2π)2I0δ(kz)δ(w)k×z^

因为k=k∣∣k^∣∣+kzz^\textbf k=k_{||} \hat k_{||}+k_z \hat zk=k∣∣k^∣∣+kzz^，其中k∣∣=k∣∣k^∣∣=kxx^+kyy^=k∣∣cos⁡ϕr^∣∣\textbf k_{||}=k_{||} \hat k_{||}=k_x \hat x+ k_y \hat y=k_{||}\cos \phi \hat r_{||}k∣∣=k∣∣k^∣∣=kxx^+kyy^=k∣∣cosϕr^∣∣

所以
k×z^=(k∣∣k^∣∣+kzz^)×z^=k∣∣cos⁡ϕr^∣∣×z^=k∣∣cos⁡ϕϕ^H(k,w)=i(2π)2I0δ(kz)δ(w)k∣∣2+kz2−(w/c)2k∣∣cos⁡ϕϕ^\textbf k \times \hat z=(k_{||} \hat k_{||}+k_z \hat z) \times \hat z = k_{||} \cos \phi \hat r_{||} \times \hat z = k_{||}\cos \phi \hat \phi \\ \textbf H(\textbf k,w)=\frac{i(2 \pi)^{2}I_0 \delta(k_z)\delta(w)}{k_{||}^2+k_z^2-(w/c)^2} k_{||} \cos \phi \hat \phi k×z^=(k∣∣k^∣∣+kzz^)×z^=k∣∣cosϕr^∣∣×z^=k∣∣cosϕϕ^H(k,w)=k∣∣2+kz2−(w/c)2i(2π)2I0δ(kz)δ(w)k∣∣cosϕϕ^

并且
∫−∞+∞dkxdky=∫02π∫0+∞k∣∣dk∣∣dϕ\int_{-\infty}^{+\infty}dk_xdk_y=\int_{0}^{2 \pi} \int_0^{+\infty} k_{||}d k_{||}d \phi∫−∞+∞dkxdky=∫02π∫0+∞k∣∣dk∣∣dϕ

计算H\textbf HH的Fourier逆变换，
H(r,t)=(2π)−4∫−∞+∞H(k,w)ei(k⋅r−wt)dkdw=(2π)−4∫−∞+∞i(2π)2I0δ(kz)δ(w)k∣∣2+kz2−(w/c)2k∣∣cos⁡ϕϕ^ei(k⋅r−wt)dkdw=iI0ϕ^4π2∫−∞+∞1k∣∣2k∣∣cos⁡ϕei(k∣∣⋅r)dkxdky=iI0ϕ^4π2∫−∞+∞cos⁡ϕei(k∣∣⋅r)dk∣∣dϕ=I0ϕ^2π∫0+∞J1(k∣∣r∣∣)dk∣∣=I0ϕ^2πr∣∣\begin{aligned} \textbf H(\textbf r,t) & = (2 \pi)^{-4}\int_{-\infty}^{+\infty} \textbf H(\textbf k,w)e^{i(\textbf k \cdot \textbf r - wt)}d\textbf k d w \\ & = (2 \pi)^{-4}\int_{-\infty}^{+\infty} \frac{i(2 \pi)^{2}I_0 \delta(k_z)\delta(w)}{k_{||}^2+k_z^2-(w/c)^2} k_{||} \cos \phi \hat \phi e^{i(\textbf k \cdot \textbf r - wt)}d\textbf k d w \\ & = \frac{i I_0 \hat \phi}{4 \pi^2}\int_{-\infty}^{+\infty} \frac{1}{k_{||}^2} k_{||} \cos \phi e^{i(\textbf k_{||} \cdot \textbf r)}dk_xdk_y \\ & = \frac{i I_0 \hat \phi}{4 \pi^2}\int_{-\infty}^{+\infty} \cos \phi e^{i(\textbf k_{||} \cdot \textbf r)}dk_{||}d \phi \\ &= \frac{I_0 \hat \phi}{2\pi}\int_0^{+\infty}J_1(k_{||}r_{||})d k_{||} = \frac{I_0 \hat \phi}{2 \pi r_{||}} \end{aligned}H(r,t)=(2π)−4∫−∞+∞H(k,w)ei(k⋅r−wt)dkdw=(2π)−4∫−∞+∞k∣∣2+kz2−(w/c)2i(2π)2I0δ(kz)δ(w)k∣∣cosϕϕ^ei(k⋅r−wt)dkdw=4π2iI0ϕ^∫−∞+∞k∣∣21k∣∣cosϕei(k∣∣⋅r)dkxdky=4π2iI0ϕ^∫−∞+∞cosϕei(k∣∣⋅r)dk∣∣dϕ=2πI0ϕ^∫0+∞J1(k∣∣r∣∣)dk∣∣=2πr∣∣I0ϕ^

无限长圆柱面均匀电流密度

考虑一个轴在z轴上的无限高薄壳圆柱体，电流密度为
J(r,t)=J0δ(ρ−R)z^\textbf J(\textbf r,t)=J_0 \delta(\rho-R)\hat zJ(r,t)=J0δ(ρ−R)z^

它的Fourier变换为
J(k,w)=∫−∞+∞J(r,t)e−i(k⋅r−wt)drdt=∫−∞+∞J0δ(ρ−R)z^e−i(k⋅r−wt)drdt=4π2δ(kz)δ(w)J0z^∫−∞+∞δ(ρ−R)e−i(kxx+kyy)dxdy=4π2δ(kz)δ(w)J0z^∫02π∫0+∞δ(ρ−R)e−ik∣∣ρcos⁡ϕρdρdϕ=4π2Rδ(kz)δ(w)J0z^∫02πe−ik∣∣Rcos⁡ϕdϕ=8π3RJ0δ(kz)δ(w)J0(k∣∣R)z^\begin{aligned} \textbf J(\textbf k,w) & = \int_{-\infty}^{+\infty} \textbf J(\textbf r,t) e^{-i(\textbf k\cdot \textbf r - wt)}d \textbf r dt \\ & = \int_{-\infty}^{+\infty} J_0 \delta(\rho-R)\hat z e^{-i(\textbf k\cdot \textbf r - wt)}d \textbf r dt \\ & = 4 \pi^2 \delta(k_z)\delta(w)J_0 \hat z \int_{-\infty}^{+\infty}\delta(\rho-R) e^{-i(k_xx+k_yy)} dxdy \\ & = 4 \pi^2 \delta(k_z)\delta(w)J_0 \hat z \int_0^{2 \pi} \int_0^{+\infty} \delta(\rho-R)e^{-ik_{||} \rho \cos \phi} \rho d \rho d \phi \\ & = 4 \pi^2R \delta(k_z)\delta(w)J_0 \hat z \int_0^{2 \pi}e^{-ik_{||} R \cos \phi} d \phi \\ & = 8 \pi^3 RJ_0\delta(k_z)\delta(w) J_0(k_{||}R)\hat z\end{aligned}J(k,w)=∫−∞+∞J(r,t)e−i(k⋅r−wt)drdt=∫−∞+∞J0δ(ρ−R)z^e−i(k⋅r−wt)drdt=4π2δ(kz)δ(w)J0z^∫−∞+∞δ(ρ−R)e−i(kxx+kyy)dxdy=4π2δ(kz)δ(w)J0z^∫02π∫0+∞δ(ρ−R)e−ik∣∣ρcosϕρdρdϕ=4π2Rδ(kz)δ(w)J0z^∫02πe−ik∣∣Rcosϕdϕ=8π3RJ0δ(kz)δ(w)J0(k∣∣R)z^

所以磁场为
H(r,t)=(2π)−4∫−∞+∞ik×J(k,w)k2−(w/c)2ei(k⋅r−wt)dkdw=iJ0Rϕ^2π∫−∞+∞k∣∣cos⁡ϕk∣∣2J0(k∣∣R)eik∣∣⋅r∣∣dk∣∣=iJ0R(−ϕ^)2π∫02π∫0+∞cos⁡ϕJ0(k∣∣R)eik∣∣r∣∣cos⁡ϕdk∣∣dϕ=J0Rϕ^∫0+∞J0(k∣∣R)J1(k∣∣r∣∣)dk∣∣={0,ρ<RJ0Rρ,ρ>R\begin{aligned} \textbf H(\textbf r,t) & = (2 \pi)^{-4} \int_{-\infty}^{+\infty} \frac{i \textbf k \times \textbf J(\textbf k,w)}{k^2-(w/c)^2}e^{i(\textbf k\cdot \textbf r-wt)} d\textbf kdw \\ & = \frac{iJ_0R \hat \phi}{2 \pi} \int_{-\infty}^{+\infty}\frac{k_{||}\cos \phi}{k_{||}^2}J_0(k_{||}R)e^{i \textbf k_{||}\cdot \textbf r_{||}}d \textbf k_{||} \\ & =\frac{iJ_0R (-\hat \phi)}{2 \pi} \int_{0}^{2 \pi} \int_0^{+\infty}\cos \phi J_0(k_{||}R)e^{i k_{||} r_{||} \cos \phi}d k_{||} d \phi \\ & = J_0 R \hat \phi \int_0^{+\infty}J_0(k_{||}R)J_1(k_{||}r_{||})dk_{||} \\ & = \begin{cases} 0, \rho < R \\ \frac{J_0R}{\rho}, \rho>R\end{cases}\end{aligned}H(r,t)=(2π)−4∫−∞+∞k2−(w/c)2ik×J(k,w)ei(k⋅r−wt)dkdw=2πiJ0Rϕ^∫−∞+∞k∣∣2k∣∣cosϕJ0(k∣∣R)eik∣∣⋅r∣∣dk∣∣=2πiJ0R(−ϕ^)∫02π∫0+∞cosϕJ0(k∣∣R)eik∣∣r∣∣cosϕdk∣∣dϕ=J0Rϕ^∫0+∞J0(k∣∣R)J1(k∣∣r∣∣)dk∣∣={0,ρ<RρJ0R,ρ>R