If the magnetic field of a plane electromagnetic wave is given by (the speed of light = 3 × 108 m/s) B = 100 × 10–6 sin [2π× 2× 10¹⁵(t - x/c)] then the maximum electric field associated with it is - (a) 4.5 × 104 N/C (b) 4 × 104 N/C (c) 6 × 104 N/C (d) 3 × 104 N/C

Answer: (d). E₀ = cB₀ E₀ = 3×10⁸×100×10⁻⁶ = 3×10⁴ N/C So the answer is (d) 3×10⁴ N/C.

If the magnetic field of a plane electromagnetic wave is given by… | JEE Main 2019 (10 Jan, Shift 1)

Q63JEE Main 2019 · 10 JAN · SHIFT 1Easy

Nature of EM Waves & Wave Equation

If the magnetic field of a plane electromagnetic wave is given by (the speed of light = 3 × 108 m/s) B = 100 × 10–6 sin $\left[2\pi\times 2\times 10^{15}\left(t - \frac{x}{c}\right)\right]$ then the maximum electric field associated with it is -

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Answer: (d)

$E_0 = cB_0$
$E_0 = 3\times10^{8}\times100\times10^{-6} = 3\times10^{4}\text{ N/C}$
So the answer is (d) $3\times10^{4}$ N/C.

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