Directions : In the following question more than one of the answers given may be correct. Select the correct answer and mark it according to the code:
In a plane electromagnetic wave, the electric field oscillates sinusoidally at a frequency of \[2\times {{10}^{10}}Hz\] and amplitude 54 V. (1) The amplitude of oscillating magnetic field will be\[18\times {{10}^{-8}}Wb\text{ }{{m}^{-2}}\] (2) The amplitude of oscillating magnetic field will be\[18\times {{10}^{-7}}Wb\text{ }{{m}^{-2}}\] (3) The wavelength of electromagnetic wave is 1.5 cm (4) The wavelength of electromagnetic wave is 1.5 mA) 1, 2 and 3 are correct
B) 1 and 2 are correct
C) 2 and 4 are correct
D) 1 and 3 are correct
Correct Answer: D
Solution :
The amplitude of oscillating magnetic field \[{{B}_{0}}=\frac{{{E}_{0}}}{c}=\frac{54}{3\times {{10}^{8}}}\] \[=18\times {{10}^{-8}}Wb/{{m}^{2}}\] The wavelength of electromagnetic wave \[\lambda =\frac{c}{v}=\frac{3\times {{10}^{8}}}{2\times {{10}^{10}}}\] \[=1.5\times {{10}^{-2}}m\] \[=1.5\text{ }cm\] Hence, option is correct.
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