Em Waves Question 87
Question: The magnetic field in the plane electromagnetic field is given by: $ B _{y}=2\times {{10}^{-7}}\sin (0.5\times 10^{3}z+1.5\times 10^{11}t)T $ The expression for the electric field may be given by
Options:
A) $ E _{y}=2\times {{10}^{-7}}\sin (0.5\times 10^{3}z+1.5\times 10^{11}t)V/m $
B) $ E _{x}=2\times {{10}^{-7}}\sin (0.5\times 10^{3}z+1.5\times 10^{11}t)V/m $
C) $ E _{y}=60\sin (0.5\times 10^{3}z+1.5\times 10^{11}t)V/m $
D) $ E _{x}=60\sin (0.5\times 10^{3}z+1.5\times 10^{11}t)V/m $
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Answer:
Correct Answer: D
Solution:
[d] $ B _{y}=2\times {{10}^{-7}},\sin ,(0.5\times 10^{3}z+1.5\times 10^{11}t)T $ The electric vector is perpendicular to B as well as direction of propagation of electromagnetic wave.
Therefore $ E _{x} $ has to be taken. Further, $ E _{0}=B _{0}\times c $
$ =2\times {{10}^{-7}}\times 3\times 10^{8},V/m $
$ E _{0}=2\times {{10}^{-7}}\times 3\times 10^{8}=60,V/m $
$ \
Therefore $ The corresponding value of the electric field is $ Ex=60,\sin ,(0.5\times 10^{3}z+1.5\times 10^{11}t)V/m $
BETA
