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JEE Main 2022 · 24 Jun · Shift 1, Physics question 43

A plane electromagnetic wave travels in a medium of relative permeability 1.61 and relative permittivity 6.44. If magnitude of magnetic intensity is 4.5×10−2 Am−14.5\times10^{-2}\ \mathrm{Am^{-1}} at a point, what will be the approximate magnitude of electric field intensity at that point ?

(Given : Permeability of free space μ0=4π×10−7 NA−2\mu_0=4\pi\times10^{-7}\ \mathrm{NA^{-2}}, speed of light in vacuum c=3×108 ms−1c=3\times10^8\ \mathrm{ms^{-1}})

Choose one option
Answer and worked solution

Official answer: C

Wave speed. v=cμrεr=3×1081.61×6.44=3×1083.22≈9.32×107v=\dfrac c{\sqrt{\mu_r\varepsilon_r}}=\dfrac{3\times10^8}{\sqrt{1.61\times6.44}}=\dfrac{3\times10^8}{3.22}\approx9.32\times10^7 m s−1^{-1}.

Magnetic field. B=μrμ0H=1.61×4π×10−7×4.5×10−2≈9.10×10−8B=\mu_r\mu_0H=1.61\times4\pi\times10^{-7}\times4.5\times10^{-2}\approx9.10\times10^{-8} T.

Electric field. E=vB≈9.32×107×9.10×10−8≈8.48E=vB\approx9.32\times10^7\times9.10\times10^{-8}\approx8.48 V m−1^{-1}.

(C)

Topics:Properties of EM waves
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