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JEE Main 202224 JuneMorning Shift
Exam
JEE Main
Year
2022
Date
24 June
Shift
Morning Shift
Subject
Physics
Chapter
Electromagnetic Waves

JEE Main 2022 · 24 June · Morning Shift · Physics · Electromagnetic Waves

A plane electromagnetic wave travels in a medium of relative...

A plane electromagnetic wave travels in a medium of relative permeability 1.611 . 61 and relative permittivity 6.446 . 44. If magnitude of magnetic intensity is 4.5×10−2Am−14 . 5 \times {10}^{- 2}A{m}^{- 1} at a point, what will be the approximate magnitude of electric field intensity at that point ?

(Given : Permeability of free space μ0=4π×10−7NA−2{\mu}_{0} = 4\pi \times {10}^{- 7}N{A}^{- 2}, speed of light in vacuum c=3×108ms−1c = 3 \times {10}^{8}{ms}^{- 1})

Options

  1. A16.96Vm−116 . 96V{m}^{- 1}
  2. B2.25×10−2Vm−12 . 25 \times {10}^{- 2}V{m}^{- 1}
  3. C8.48Vm−18 . 48V{m}^{- 1}
  4. D6.75×106Vm−16 . 75 \times {10}^{6}V{m}^{- 1}

Correct answer

C. 8.48Vm−18 . 48V{m}^{- 1}

Step-by-step solution

The speed of light in a given medium is given by,

v=cμrεrv = \dfrac{c}{\sqrt{{\mu}_{r}{ε}_{r}}}

Magnetic field is B=μrμ0HB = {\mu}_{r}{\mu}_{0}H.

Also, the relation between the magnetic field and electric field is given by,

E=vB⇒E=cμrεr×μrμ0H=cεr×μr×μ0H⇒E=3×1086.44×1.61×4π×10−7×4.5×10−2=8.48Vm−1E = vB ⇒ E = \dfrac{c}{\sqrt{{\mu}_{r}{ε}_{r}}} \times {\mu}_{r}{\mu}_{0}H = \dfrac{c}{\sqrt{{ε}_{r}}} \times \sqrt{{\mu}_{r}} \times {\mu}_{0}H ⇒ E = \dfrac{3 \times {10}^{8}}{\sqrt{6 . 44}} \times \sqrt{1 . 61} \times 4\pi \times {10}^{- 7} \times 4 . 5 \times {10}^{- 2} = 8 . 48V{m}^{- 1}

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