An expression for oscillating electric field in a plane electromagnetic wave is given as Ez = 300 sin(5 π× 103x - 3 π× 1011t) Vm ⁻¹ Then, the value of magnetic field amplitude will be: (Given: speed of light in Vacuum c = 3 × 108 ms ⁻¹) (a) 1 × 10 - 6 T (b) 5 × 10 - 6 T (c) 18 × 109 T (d) 21 × 109 T

Answer: (b). From the equation: k = 5π× 10³, ω = 3π× 10¹¹. Wave speed v = (ω)/(k) = 3× 10¹¹5× 10³ = 6× 10⁷ m/s (so the wave is in a medium). B₀ = E₀/v = 3006× 10⁷ = 5× 10⁻⁶ T. So the answer is (b).

An expression for oscillating electric field in a plane… | JEE Main 2022 (30 Jun, Shift 1)

Q33JEE Main 2022 · 30 JUN · SHIFT 1Easy

Nature of EM Waves & Wave Equation

An expression for oscillating electric field in a plane electromagnetic wave is given as Ez = 300 sin(5 $\pi\times$ 103x $-$ 3 $\pi\times$ 1011t) Vm ${}^{-1}$

Then, the value of magnetic field amplitude will be:

(Given: speed of light in Vacuum c = 3 $\times$ 108 ms ${}^{-1}$ )

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Answer: (b)

From the equation: $k = 5\pi\times 10^{3}$ , $\omega = 3\pi\times 10^{11}$ .
Wave speed $v = \frac{\omega}{k} = \frac{3\times 10^{11}}{5\times 10^{3}} = 6\times 10^{7}\ \text{m/s}$ (so the wave is in a medium).
$B_0 = \frac{E_0}{v} = \frac{300}{6\times 10^{7}} = 5\times 10^{-6}\ \text{T}$ .
So the answer is (b).

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