Question

Intensity of electromagnetic wave will be(a) $I=\frac{c\mu_{0}B_{0}^{2}}{2}$(b) $I=\frac{c\epsilon_{0}B_{0}^{2}}{2}$(c) $I=\frac{B_{0}^{2}}{c\mu_{0}}$(d) $I=\frac{E_{0}^{2}}{2c\epsilon_{0}}$

Answer 1

Answer:

Explanation:

Electrical energy density in electric field, uE = 1/2ϵ0E²

Magnetic energy density in magnetic field, uB = B²/2μ0

In an electromagnetic wave, E and B are related as, E = cB

The speed of light in vacuum is given by the equation = c= 1/√μ0ϵ0

Electromagnetic waves and the electrical and magnetic energy densities are related to each other as -

uE = 1/2ϵ0E² = 1/2ϵ0B²c² = 1/2ϵ0B²1/μ0ϵ0 = uB

Thus, the two energy densities are equal. The total energy density u of the wave - u = uE+uB = ϵ0E²

The intensity I is then

I = U/AΔt

= cϵ0E²

Answer 2

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Intensity of electromagnetic wave will be(a) $I=\frac{c\mu_{0}B_{0}^{2}}{2}$

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