The oscillating magnetic field in a plane electromagnetic wave is given by By = 5× 10⁻⁶sin 1000π(5x - 4× 10⁸t) T. The amplitude of electric field will be: (a) 15× 10² V m⁻¹ (b) 5× 10⁻⁶ V m⁻¹ (c) 16× 10¹² V m⁻¹ (d) 4× 10² V m⁻¹

Answer: (d). From the equation: k = 1000π× 5, ω = 1000π× 4× 10⁸. Wave speed v = (ω)/(k) = 4× 10⁸5 = 8× 10⁷ m/s (a medium, not vacuum). E₀ = vB₀ = 8× 10⁷× 5× 10⁻⁶ = 400 V/m. So the answer is (d) 4× 10² V m⁻¹.

The oscillating magnetic field in a plane electromagnetic wave is… | JEE Main 2022 (26 Jul, Shift 2)

Q31JEE Main 2022 · 26 JUL · SHIFT 2Easy

Nature of EM Waves & Wave Equation

The oscillating magnetic field in a plane electromagnetic wave is given by

$B_y = 5\times 10^{-6}\sin 1000\pi\left(5x - 4\times 10^{8}t\right)\text{ T}$ . The amplitude of electric field will be:

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Answer: (d)

From the equation: $k = 1000\pi\times 5$ , $\omega = 1000\pi\times 4\times 10^{8}$ .
Wave speed $v = \frac{\omega}{k} = \frac{4\times 10^{8}}{5} = 8\times 10^{7}\ \text{m/s}$ (a medium, not vacuum).
$E_0 = vB_0 = 8\times 10^{7}\times 5\times 10^{-6} = 400\ \text{V/m}$ .
So the answer is (d) $4\times 10^{2}\ \text{V m}^{-1}$ .

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Nature of EM Waves & Wave Equation question from JEE Main 2022 (26 Jul, Shift 2)

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