A) \[6\,V{{m}^{-1}}\]
B) \[60\text{ }V{{m}^{-1}}\]
C) \[10/6V{{m}^{-1}}\]
D) none of these
Correct Answer: B
Solution :
We can write the electric and magnetic fields as sinusoidal functions of positions and time t. \[E={{E}_{0}}\sin (kx-\omega t)\] \[B={{B}_{0}}\sin (kx-\omega t)\] In this\[{{E}_{0}}\]and\[{{B}_{0}}\]are the amplitudes of the field. Further, \[c=\frac{{{E}_{0}}}{{{B}_{0}}}=speed\text{ }of\text{ }light\] or \[{{E}_{0}}={{B}_{0}}c\] Given, \[{{B}_{0}}=2\times {{10}^{-7}}T,C=3\times {{10}^{8}}m/s\] \[{{E}_{0}}=2\times {{10}^{-7}}\times 3\times {{10}^{8}}=60\,T\,m/s=6\,V{{m}^{-1}}\]
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