NEET Physics Properties of EM Waves MCQ Question
An electromagnetic wave travels in free space with an electric field amplitude of 6.3 V/m. If the speed of light is 3 x 10^8 m/s, what is the magnitude of the corresponding magnetic field?
1 x 10^-8 T
1 x 10^-8 T
5 x 10^-8 T
5 x 10^-8 T
Correct Answer
Detailed Explanation
The magnitude of the magnetic field (B) is calculated using the relation B = E/c, where E is the electric field amplitude and c is the speed of light. Thus, B = 6.3 V/m / 3 x 10^8 m/s = 2.1 x 10^-8 T.
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