Light wave travelling in air along x-direction is given by Ey = 540sin π× 10⁴(x - ct) V m⁻¹. Then, the peak value of magnetic field of wave will be (Given c = 3 × 108 ms ⁻¹) (a) 18 × 10 - 7 T (b) 54 × 10 - 7 T (c) 54 × 10 - 8 T (d) 18 × 10 - 8 T

Answer: (a). Peak electric field E₀ = 540 V/m. B₀ = E₀/c = 5403× 10⁸ = 1.8× 10⁻⁶ T. 1.8× 10⁻⁶ = 18× 10⁻⁷ T. So the answer is (a).

Light wave travelling in air along x-direction is given by Ey =… | JEE Main 2022 (25 Jul, Shift 2)

Q32JEE Main 2022 · 25 JUL · SHIFT 2Easy

Nature of EM Waves & Wave Equation

Light wave travelling in air along x-direction is given by $E_y = 540\sin\pi\times 10^{4}(x - ct)\text{ V m}^{-1}$ . Then, the peak value of magnetic field of wave will be (Given c = 3 $\times$ 108 ms ${}^{-1}$ )

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Answer: (a)

Peak electric field $E_0 = 540\ \text{V/m}$ .
$B_0 = \frac{E_0}{c} = \frac{540}{3\times 10^{8}} = 1.8\times 10^{-6}\ \text{T}$ .
$1.8\times 10^{-6} = 18\times 10^{-7}\ \text{T}$ .
So the answer is (a).

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Nature of EM Waves & Wave Equation question from JEE Main 2022 (25 Jul, Shift 2)

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