Electromagnetic Waves

JEE Main level, single correct, medium

70 of 165

Question 70 of 165

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})

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Answer: (C) 8.48Vm−18.48V{m}^{-1}

The speed of light in a given medium is given by,
v=cμrϵrv=\dfrac{c}{\sqrt{{\mu }_{r}{\epsilon }_{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 \Rightarrow E=\dfrac{c}{\sqrt{{\mu }_{r}{\epsilon }_{r}}}\times {\mu }_{r}{\mu }_{0}H=\dfrac{c}{\sqrt{{\epsilon }_{r}}}\times \sqrt{{\mu }_{r}}\times {\mu }_{0}H \Rightarrow 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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