A) \[\frac{2{{E}_{0}}}{c}\hat{j}\,\sin kz\,\sin \omega t\]
B) \[\frac{2{{E}_{0}}}{c}\hat{j}\cos kz\,\cos \omega t\]
C) \[\frac{2{{E}_{0}}}{c}\hat{j}\sin kz\,\cos \omega t\]
D) \[-\frac{2{{E}_{0}}}{c}\hat{j}\sin kz\,\sin \omega t\]
Correct Answer: A
Solution :
\[\frac{dE}{dz}=-\,\frac{dE}{dt}\] \[\frac{dE}{dz}=-\,2{{E}_{0}}\,K\,\sin \,kz\,\cos \omega t\,=-\frac{dB}{dt}\] \[dB=+2{{E}_{0}}K\,\sin kz\,\cos \omega t\,dt\] \[B=+2{{E}_{0}}K\,\sin kz\,\int_{{}}^{{}}{\cos \,\omega tdt}\] \[=+2{{E}_{0}}\,\frac{k}{\omega }\,\sin \,kz\,\sin \omega t\] \[\frac{{{E}_{0}}}{{{B}_{0}}}=\frac{\omega }{k}=C\] \[B=\frac{2{{E}_{0}}}{C}\sin \,kz\,\sin \omega t\]
You need to login to perform this action.
You will be redirected in
3 sec
