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JEE Main 202226 JulyEvening Shift
Exam
JEE Main
Year
2022
Date
26 July
Shift
Evening Shift
Subject
Physics
Chapter
Electromagnetic Waves

JEE Main 2022 · 26 July · Evening Shift · Physics · Electromagnetic Waves

The oscillating magnetic field in a plane electromagnetic wave is...

The oscillating magnetic field in a plane electromagnetic wave is given by By=5×10−6sin⁡1000π(5x−4×108t)T{B}_{y} = 5 \times {10}^{- 6}\sin 1000\pi\left(5x - 4 \times {10}^{8}t\right)T. The amplitude of electric field will be

Options

  1. A15×102Vm−115 \times {10}^{2}{Vm}^{- 1}
  2. B5×10−6Vm−15 \times {10}^{- 6}{Vm}^{- 1}
  3. C16×1012Vm−116 \times {10}^{12}{Vm}^{- 1}
  4. D4×102Vm−14 \times {10}^{2}{Vm}^{- 1}

Correct answer

D. 4×102Vm−14 \times {10}^{2}{Vm}^{- 1}

Step-by-step solution

Given here, By=5×10−6sin⁡1000π(5x−4×108t)T{B}_{y} = 5 \times {10}^{- 6}\sin 1000\pi\left(5x - 4 \times {10}^{8}t\right)T

Here, peak value of magnetic field, B0=5×10−6T{B}_{0} = 5 \times {10}^{- 6}T

Now the speed of the wave v=ωk=1000π×4×1081000π×5=8×107ms−1v = \dfrac{\omega}{k} = \dfrac{1000\pi \times 4 \times {10}^{8}}{1000\pi \times 5} = 8 \times {10}^{7}m{s}^{- 1}

Using relation between magnetic and electric field,

Peak value of electric field is E0=B0v=5×10−6×8×107=4×102Vm−1{E}_{0} = {B}_{0}v = 5 \times {10}^{- 6} \times 8 \times {10}^{7} = 4 \times {10}^{2}V{m}^{- 1}

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