- #1
yungman
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The is example 10.1 in page 417. The example is the find the current density from given condition:
[tex] V=0,\;\;\hbox{ and } \;\;\vec A=\frac{\mu_0 k}{4c}(ct-|x|)^2 \hat z \;\hbox { for x = +ve and }\;\; \vec A=0 \;\;\hbox { for x = -ve.}[/tex]
[tex] c=\frac 1 {\sqrt{\mu_0 \epsilon_0}}[/tex]
From this, you get:
[tex]\vec E= -\frac {\partial \vec A}{\partial t} \;=\; -\frac {\mu_0 k}{2} (ct-|x|)\hat z \;\;\;\hbox { and }\;\;\; \vec B = \nabla X \vec A = ^+_- \hat y \frac{\mu_0 k}{2c}(ct-|x|) [/tex]
[tex]\nabla X \vec B = -\hat z \frac {\mu_0 k}{2c} = \mu_0 \vec J + \mu_0 \epsilon_0 \frac{\partial \vec E}{\partial t} = \mu_0 \vec J -\hat z \mu \epsilon_0 \frac{\partial^2 \vec A}{\partial t^2} = \mu_0 \vec J - \hat z \frac{\mu_0 k c}{2}[/tex]
[tex] \Rightarrow \vec J = -\hat z \frac {\mu_0 k}{2c} + \hat z \frac{\mu_0 k c}{2}[/tex]
The book claimed [itex] \vec J = 0 [/itex] without going through the steps.
I cannot get that, please help.
Thanks
Alan
[tex] V=0,\;\;\hbox{ and } \;\;\vec A=\frac{\mu_0 k}{4c}(ct-|x|)^2 \hat z \;\hbox { for x = +ve and }\;\; \vec A=0 \;\;\hbox { for x = -ve.}[/tex]
[tex] c=\frac 1 {\sqrt{\mu_0 \epsilon_0}}[/tex]
From this, you get:
[tex]\vec E= -\frac {\partial \vec A}{\partial t} \;=\; -\frac {\mu_0 k}{2} (ct-|x|)\hat z \;\;\;\hbox { and }\;\;\; \vec B = \nabla X \vec A = ^+_- \hat y \frac{\mu_0 k}{2c}(ct-|x|) [/tex]
[tex]\nabla X \vec B = -\hat z \frac {\mu_0 k}{2c} = \mu_0 \vec J + \mu_0 \epsilon_0 \frac{\partial \vec E}{\partial t} = \mu_0 \vec J -\hat z \mu \epsilon_0 \frac{\partial^2 \vec A}{\partial t^2} = \mu_0 \vec J - \hat z \frac{\mu_0 k c}{2}[/tex]
[tex] \Rightarrow \vec J = -\hat z \frac {\mu_0 k}{2c} + \hat z \frac{\mu_0 k c}{2}[/tex]
The book claimed [itex] \vec J = 0 [/itex] without going through the steps.
I cannot get that, please help.
Thanks
Alan
Last edited: