电动力学每日一题 2021/10/12
電動力學每日一題 2021/10/12
(a) To make the EM field trapped inside a perfectly electric conducting cavity, the tangential component of the E-field must vanish on the internal surface:
E0J0(w0R/c)sin?(w0t)=0J0(w0R/c)=0E_0J_0(w_0 R/c)\sin(w_0 t)=0 \\ J_0(w_0R/c)=0E0?J0?(w0?R/c)sin(w0?t)=0J0?(w0?R/c)=0
This means the radius of cylinder must be chosen such that w0R/cw_0 R/cw0?R/c is a zero of Bessel function J0(?)J_0(\cdot)J0?(?).
(b) Maxwell 1st equation delievers the boundary condition of E-field
D⊥(r2,t)?D⊥(r1,t)=σS??0E0J0(w0r∣∣/c)sin?(w0t)=σS(r∣∣,?,±L/2,t)\textbf D_{\perp}(\textbf r_2,t)-\textbf D_{\perp}(\textbf r_1,t)=\sigma_S \\ \mp \epsilon_0 E_0 J_0(w_0 r_{||}/c)\sin(w_0t)=\sigma_S(r_{||},\phi,\pm L/2,t)D⊥?(r2?,t)?D⊥?(r1?,t)=σS???0?E0?J0?(w0?r∣∣?/c)sin(w0?t)=σS?(r∣∣?,?,±L/2,t)
(c) Maxwell 2nd equation delievers the boundary condition of H-field
H∣∣(r2,t)?H∣∣(r1)=Jfree(r0,t)×n^JS(R,?,z,t)=?(E0/Z0)J1(w0R/c)cos?(w0t)z^JS(r∣∣,?,±L/2,t)=±(E0/Z0)J1(w0r∣∣/c)cos?(w0t)z^\textbf H_{||}(\textbf r_2,t)-\textbf H_{||}(\textbf r_1) = \textbf J_{free}(\textbf r_0,t) \times \hat n \\ \textbf J_S(R,\phi,z,t)=-(E_0/Z_0)J_1(w_0R/c)\cos(w_0t)\hat z \\ \textbf J_S(r_{||},\phi,\pm L/2,t)=\pm (E_0/Z_0)J_1(w_0r_{||}/c)\cos(w_0 t)\hat zH∣∣?(r2?,t)?H∣∣?(r1?)=Jfree?(r0?,t)×n^JS?(R,?,z,t)=?(E0?/Z0?)J1?(w0?R/c)cos(w0?t)z^JS?(r∣∣?,?,±L/2,t)=±(E0?/Z0?)J1?(w0?r∣∣?/c)cos(w0?t)z^
(d) The charge-currrency continuinity equation is
??JS(r∣∣,?,±L/2,t)+??tσS(r∣∣,?,±L/2,t)=0\nabla \cdot \textbf J_S(r_{||},\phi,\pm L/2,t) + \frac{\partial}{\partial t}\sigma_S(r_{||},\phi,\pm L/2,t)=0??JS?(r∣∣?,?,±L/2,t)+?t??σS?(r∣∣?,?,±L/2,t)=0
Evaluate
??JS=±(E0/Z0)?r∣∣?r∣∣r∣∣J1(w0r∣∣/c)cos?(w0t)=±(E0/Z0)J1(w0r∣∣/c)+(w0r∣∣/c)J1′(w0r∣∣/c)r∣∣cos?(w0t)=±(E0/Z0)(w0r∣∣/c)J0(w0r∣∣/c)r∣∣cos?(w0t)=±?0E0w0J0(w0r∣∣/c)cos?(w0t)\begin{aligned}\nabla \cdot \textbf J_S & = \pm (E_0/Z_0) \frac{\partial }{r_{||} \partial r_{||}}r_{||}J_1(w_0 r_{||}/c)\cos(w_0 t) \\ & = \pm (E_0/Z_0) \frac{J_1(w_0 r_{||}/c)+(w_0 r_{||}/c)J_1'(w_0 r_{||}/c)}{r_{||}}\cos(w_0 t) \\ & = \pm (E_0/Z_0) \frac{(w_0 r_{||}/c)J_0(w_0 r_{||}/c)}{r_{||}}\cos(w_0 t) \\ & = \pm \epsilon_0 E_0 w_0 J_0(w_0 r_{||}/c)\cos(w_0 t)\end{aligned}??JS??=±(E0?/Z0?)r∣∣??r∣∣???r∣∣?J1?(w0?r∣∣?/c)cos(w0?t)=±(E0?/Z0?)r∣∣?J1?(w0?r∣∣?/c)+(w0?r∣∣?/c)J1′?(w0?r∣∣?/c)?cos(w0?t)=±(E0?/Z0?)r∣∣?(w0?r∣∣?/c)J0?(w0?r∣∣?/c)?cos(w0?t)=±?0?E0?w0?J0?(w0?r∣∣?/c)cos(w0?t)?
??tσS=??t??0E0J0(w0r∣∣/c)sin?(w0t)=??0E0J0(w0r∣∣/c)cos?(w0t)\begin{aligned} \frac{\partial}{\partial t}\sigma_S & = \frac{\partial}{\partial t} \mp \epsilon_0 E_0 J_0(w_0 r_{||}/c)\sin(w_0t) \\ & = \mp \epsilon_0 E_0 J_0(w_0 r_{||}/c)\cos(w_0t)\end{aligned}?t??σS??=?t????0?E0?J0?(w0?r∣∣?/c)sin(w0?t)=??0?E0?J0?(w0?r∣∣?/c)cos(w0?t)?
Above,
??JS+??tσS=0\nabla \cdot \textbf J_S+\frac{\partial}{\partial t}\sigma_S =0??JS?+?t??σS?=0
總結
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