A) \[{{E}_{y}}=2\times {{10}^{-7}}sin(0.5\times {{10}^{-7}}\sin \,(0.5\times {{10}^{3}}z\] \[+1.5\times {{10}^{11}}t)V{{m}^{-1}}\]
B) \[{{E}_{y}}2\times {{10}^{-7}}sin(0.5\times {{10}^{3}}z\] \[+1.5\times {{10}^{11}}t)V{{m}^{-1}}\]
C) \[{{E}_{y}}=60\sin \,(0.5\times {{10}^{3}}z+1.5\times {{10}^{11}}t)V{{m}^{-1}}\]
D) \[{{E}_{x}}=60\sin \,(0.5\times {{10}^{3}}z+1.5\times {{10}^{11}}t)V{{m}^{-1}}\]
Correct Answer: D
Solution :
Given,\[{{B}_{y}}=2\times {{10}^{-7}}\sin \,(0.5\times {{10}^{3}}z+1.5\times {{10}^{11}})\,T\]. As the electric field vector is perpendicular to magnetic field (y-axis) and direction of propagation of electromagnetic wave (z-axis) the electric field is along x-axis. Moreover \[\frac{{{E}_{0}}}{{{B}_{0}}}=c\] \[\therefore \] \[{{E}_{0}}=2\times {{10}^{-7}}\times 3\times {{10}^{8}}\] \[=60\,V{{m}^{-1}}\] \[\therefore \] The corresponding value of electric field \[{{E}_{x}}=60\sin \,(0.5\times {{10}^{3}}z=1.5\times {{10}^{11}}t)\,V{{m}^{-1}}\]
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