Easy MCQ +4 / -1 PYQ · JEE Mains 2020

A plane electromagnetic wave of frequency 25 GHz is propagating in vacuum along the z-direction. At a particular point in space and time, the magnetic field is given by $\overrightarrow B = 5 \times {10^{ - 8}}\widehat jT$. The corresponding electric field $\overrightarrow E$ is (speed of light c = 3 × 108 ms–1)

  1. A 15 $\widehat i$V / m Correct answer
  2. B -15 $\widehat i$V / m
  3. C 1.66 × 10<sup>–16</sup> $\widehat i$V / m
  4. D -1.66 × 10<sup>–16</sup> $\widehat i$V / m

Solution

$\overrightarrow E = \overrightarrow B \times \overrightarrow V$ <br><br>= $$\left( {5 \times {0^{ - 8}}\widehat j} \right) \times \left( {3 \times {{10}^8}\widehat k} \right)$$ <br><br>= ${15\,\widehat i}$ V/m

About this question

Subject: Physics · Chapter: Electromagnetic Waves · Topic: Properties of EM Waves

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