Skip to content
dronacharya.ai
← JEE Main 2022 · 26 Jun · Shift 1
PhysicsSingle correct+4 / -1Question 45 of 90

JEE Main 2022 · 26 Jun · Shift 1, Physics question 45

If Electric field intensity of a uniform plane electro magnetic wave is given as
E=−301.6sin⁡(kz−ωt) a^x+452.4sin⁡(kz−ωt) a^y Vm.E=-301.6\sin(kz-\omega t)\,\hat a_x+452.4\sin(kz-\omega t)\,\hat a_y\ \frac{\mathrm V}{\mathrm m}.
Then, magnetic intensity 'H' of this wave in A m−1^{-1} will be :

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

Choose one option
Answer and worked solution

Official answer: C

Magnitudes. H=Eη0H=\dfrac E{\eta_0} with η0=μ0c=120π≈377 Ω\eta_0=\mu_0c=120\pi\approx377\ \Omega: 301.6377=0.8\dfrac{301.6}{377}=0.8, 452.4377=1.2\dfrac{452.4}{377}=1.2 A m−1^{-1}.

Directions. The wave moves along +z+z, so H⃗=1η0z^×E⃗\vec H=\dfrac1{\eta_0}\hat z\times\vec E: z^×x^=y^\hat z\times\hat x=\hat y and z^×y^=−x^\hat z\times\hat y=-\hat x. Ex=−301.6E_x=-301.6 gives Hy=−0.8H_y=-0.8; Ey=+452.4E_y=+452.4 gives Hx=−1.2H_x=-1.2.

(C)

Topics:Properties of EM waves
← Previous questionNext question →