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JEE Main 201910 JanuaryMorning Shift
Exam
JEE Main
Year
2019
Date
10 January
Shift
Morning Shift
Subject
Physics
Chapter
Electromagnetic Waves

JEE Main 2019 · 10 January · Morning Shift · Physics · Electromagnetic Waves

If the magnetic field of a plane electromagnetic wave is given by...

If the magnetic field of a plane electromagnetic wave is given by (The speed of light =3×108m/s= 3 \times {10}^{8}m / s)

B=100×10−6sinB = 100 \times {10}^{- 6}sin\left[2\pi \times 2 \times {10}^{15}\left(t - \dfrac{x}{c}\right)\right] then the maximum electric field associated with it is:

Options

  1. A3×104NC−13 \times {10}^{4}N{C}^{- 1}
  2. B4×104NC−14 \times {10}^{4}N{C}^{- 1}
  3. C4.5×104NC−14.5 \times {10}^{4}N{C}^{- 1}
  4. D6×104NC−16 \times {10}^{4}N{C}^{- 1}

Correct answer

A. 3×104NC−13 \times {10}^{4}N{C}^{- 1}

Step-by-step solution

We know that relation between electric field E and magnetic field B is given by:

EB=c⇒E=Bc=100×10−6×3×108Vm−1=3×104Vm−1\dfrac{E}{B} = c ⇒ E = Bc = 100 \times {10}^{- 6} \times 3 \times {10}^{8}V{m}^{- 1} = 3 \times {10}^{4}V{m}^{- 1}

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