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JEE Mains · Physics · STD 12 - Electromagnetic waves
  • JEE Mains 2019
  • Easy
  • MCQ

If the magnetic field of a plane electromagnetic wave is given by (The speed of light \( = 3 \times {10^8}\,m/s\) ) \(B = 100 \times {10^{ - 6}}\,\sin \,\left[ {2\pi  \times 2 \times {{10}^{15}}\,\left( {t - \frac{x}{c}} \right)} \right]\)  then the maximum electric field associated with it is

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Answer & Solution

Correct Answer

(B) \(3 \times {10^4}\,N/C\)

Step-by-step Solution

Detailed explanation

\(E_{0}=B_{0} \times C\)

\(=100 \times 10^{-6} \times 3 \times 10^{8}\)

\(=3 \times 10^{4} \,N / C\)

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