JEE Main 2022 · 26 Jul · Shift 2 · Physics Q36

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PhysicsSingle correct+4 / -1Question 36 of 90

The oscillating magnetic field in a plane electromagnetic wave is given by $B_y=5\times10^{-6}\sin1000\pi\left(5x-4\times10^8t\right)$ T. The amplitude of electric field will be :

15\times10^2

^{-1}

5\times10^{-6}

^{-1}

16\times10^{12}

^{-1}

4\times10^2

^{-1}

Answer and worked solution

Official answer: D

Wave speed $\frac\omega k=\frac{4\times10^8}5=8\times10^7$ m s $^{-1}$ (as written).

$E_0=vB_0$ $=8\times10^7\times5\times10^{-6}=400$ V m $^{-1}$ $=4\times10^2$ V m $^{-1}$ .

(D)

Topics:Properties of EM waves

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JEE Main 2022 · 26 Jul · Shift 2, Physics question 36