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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

發布時間：2025/4/14 编程问答 28 豆豆

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電動力學每日一題 2021/10/15 Fourier變換法計算均勻電流密度產生的磁場

- 無限長均勻電流
- 無限長圓柱面均勻電流密度

無限長均勻電流

假設z軸上有一根非常細的電線，攜帶均勻電流 $I_0$ ，則空間中的電流密度可以表示為
$J(r,t)=I0δ(x)δ(y)z^\textbf J(\textbf r,t)=I_0 \delta(x)\delta(y)\hat z$

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

則磁場的Fourier變換為
$H(k,w)=ik×J(k,w)k2?(w/c)2=i(2π)2I0δ(kz)δ(w)k2?(w/c)2k×z^\textbf H(\textbf k,w)=\frac{i \textbf k \times \textbf J(\textbf k ,w)}{k^2 - (w/c)^2}=\frac{i(2 \pi)^{2}I_0 \delta(k_z)\delta(w)}{k^2-(w/c)^2} \textbf k \times \hat z$

因為 $k=k∣∣k^∣∣+kzz^\textbf k=k_{||} \hat k_{||}+k_z \hat z$ ，其中 $k∣∣=k∣∣k^∣∣=kxx^+kyy^=k∣∣cos??r^∣∣\textbf k_{||}=k_{||} \hat k_{||}=k_x \hat x+ k_y \hat y=k_{||}\cos \phi \hat r_{||}$

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

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

計算 $H\textbf H$ 的Fourier逆變換，
$H(r,t)=(2π)?4∫?∞+∞H(k,w)ei(k?r?wt)dkdw=(2π)?4∫?∞+∞i(2π)2I0δ(kz)δ(w)k∣∣2+kz2?(w/c)2k∣∣cos???^ei(k?r?wt)dkdw=iI0?^4π2∫?∞+∞1k∣∣2k∣∣cos??ei(k∣∣?r)dkxdky=iI0?^4π2∫?∞+∞cos??ei(k∣∣?r)dk∣∣d?=I0?^2π∫0+∞J1(k∣∣r∣∣)dk∣∣=I0?^2πr∣∣\begin{aligned} \textbf H(\textbf r,t) & = (2 \pi)^{-4}\int_{-\infty}^{+\infty} \textbf H(\textbf k,w)e^{i(\textbf k \cdot \textbf r - wt)}d\textbf k d w \\ & = (2 \pi)^{-4}\int_{-\infty}^{+\infty} \frac{i(2 \pi)^{2}I_0 \delta(k_z)\delta(w)}{k_{||}^2+k_z^2-(w/c)^2} k_{||} \cos \phi \hat \phi e^{i(\textbf k \cdot \textbf r - wt)}d\textbf k d w \\ & = \frac{i I_0 \hat \phi}{4 \pi^2}\int_{-\infty}^{+\infty} \frac{1}{k_{||}^2} k_{||} \cos \phi e^{i(\textbf k_{||} \cdot \textbf r)}dk_xdk_y \\ & = \frac{i I_0 \hat \phi}{4 \pi^2}\int_{-\infty}^{+\infty} \cos \phi e^{i(\textbf k_{||} \cdot \textbf r)}dk_{||}d \phi \\ &= \frac{I_0 \hat \phi}{2\pi}\int_0^{+\infty}J_1(k_{||}r_{||})d k_{||} = \frac{I_0 \hat \phi}{2 \pi r_{||}} \end{aligned}$

無限長圓柱面均勻電流密度

考慮一個軸在z軸上的無限高薄殼圓柱體，電流密度為
$J(r,t)=J0δ(ρ?R)z^\textbf J(\textbf r,t)=J_0 \delta(\rho-R)\hat z$

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

所以磁場為
$H(r,t)=(2π)?4∫?∞+∞ik×J(k,w)k2?(w/c)2ei(k?r?wt)dkdw=iJ0R?^2π∫?∞+∞k∣∣cos??k∣∣2J0(k∣∣R)eik∣∣?r∣∣dk∣∣=iJ0R(??^)2π∫02π∫0+∞cos??J0(k∣∣R)eik∣∣r∣∣cos??dk∣∣d?=J0R?^∫0+∞J0(k∣∣R)J1(k∣∣r∣∣)dk∣∣={0,ρ<RJ0Rρ,ρ>R\begin{aligned} \textbf H(\textbf r,t) & = (2 \pi)^{-4} \int_{-\infty}^{+\infty} \frac{i \textbf k \times \textbf J(\textbf k,w)}{k^2-(w/c)^2}e^{i(\textbf k\cdot \textbf r-wt)} d\textbf kdw \\ & = \frac{iJ_0R \hat \phi}{2 \pi} \int_{-\infty}^{+\infty}\frac{k_{||}\cos \phi}{k_{||}^2}J_0(k_{||}R)e^{i \textbf k_{||}\cdot \textbf r_{||}}d \textbf k_{||} \\ & =\frac{iJ_0R (-\hat \phi)}{2 \pi} \int_{0}^{2 \pi} \int_0^{+\infty}\cos \phi J_0(k_{||}R)e^{i k_{||} r_{||} \cos \phi}d k_{||} d \phi \\ & = J_0 R \hat \phi \int_0^{+\infty}J_0(k_{||}R)J_1(k_{||}r_{||})dk_{||} \\ & = \begin{cases} 0, \rho < R \\ \frac{J_0R}{\rho}, \rho>R\end{cases}\end{aligned}$

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