A) 0.014N/C, 36m
B) 0.14N/C, 36m
C) 0.14N/C, 360m
D) 0.014N/C, 360m
Correct Answer: D
Solution :
[d] Frequency of EM wave \[\upsilon =830\,KHz\] \[=830\times {{10}^{3}}Hz.\] |
Magnetic field, \[B=4.82\times {{10}^{-11}}T\] |
As we know, frequency, \[\upsilon =\frac{c}{\lambda }\] |
or\[\lambda =\frac{c}{v}=\frac{3\times {{10}^{8}}}{830\times {{10}^{3}}}\]\[\lambda \simeq 360m\] |
And, \[E=BC=4.82\times {{10}^{-11}}\times 3\times {{10}^{8}}\] |
= 0.014 N/C |
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