For the plane electromagnetic wave given by E = E₀ sin (ω t - kx) and B = B₀ sin (ω t - kx), the ratio of average electric energy density to average magnetic energy density is (a) 1 (b) 4 (c) 2 (d) 1/2

Answer: (a). uE = 1/4ε₀ E₀², uB = (B₀²)/(4μ₀) With B₀ = E₀/c and c² = 1/(μ₀ε₀): (B₀²)/(μ₀) = ε₀ E₀² So uE = uB, ratio = 1. So the answer is (a) 1.

For the plane electromagnetic wave given by E = E₀ sin (ω t - kx) and… | JEE Main 2023 (6 Apr, Shift 1)

Q10JEE Main 2023 · 6 APR · SHIFT 1Easy

Energy Density & Intensity

For the plane electromagnetic wave given by $E = E_0 \sin(\omega t - kx)$ and $B = B_0 \sin(\omega t - kx)$ , the ratio of average electric energy density to average magnetic energy density is

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Answer: (a)

$\langle u_E \rangle = \tfrac{1}{4}\varepsilon_0 E_0^2$ , $\langle u_B \rangle = \dfrac{B_0^2}{4\mu_0}$
With $B_0 = E_0/c$ and $c^2 = 1/(\mu_0\varepsilon_0)$ : $\dfrac{B_0^2}{\mu_0} = \varepsilon_0 E_0^2$
So $\langle u_E \rangle = \langle u_B \rangle$ , ratio $= 1$ .
So the answer is (a) 1.

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Energy Density & Intensity question from JEE Main 2023 (6 Apr, Shift 1)

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