A plane electromagnetic wave of wavelength 3. | vedclass

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(A) The intensity $I$ of an electromagnetic wave is given by the formula $I = \frac{1}{2} c \epsilon_0 E_0^2$. We know that $c = \frac{1}{\sqrt{\mu_0

Vedclass · Accepted Answer

$119.4$

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(A) The intensity $I$ of an electromagnetic wave is given by the formula $I = \frac{1}{2} c \epsilon_0 E_0^2$. We know that $c = \frac{1}{\sqrt{\mu_0 \epsilon_0}}$,which implies $\epsilon_0 = \frac{1}{\mu_0 c^2}$. Substituting this into the intensity formula: $I = \frac{1}{2} c \left(\frac{1}{\mu_0 c^2}ight) E_0^2 = \frac{E_0^2}{2 \mu_0 c}$. Given: $E_0 = 300 \ Vm^{-1}$,$\mu_0 = 4\pi 	imes 10^{-7} \ Hm^{-1}$,and $c = 3 	imes 10^8 \ ms^{-1}$. $I = \frac{(300)^2}{2 	imes (4\pi 	imes 10^{-7}) 	imes (3 	imes 10^8)}$. $I = \frac{90000}{2 	imes 4\pi 	imes 10^{-7} 	imes 3 	imes 10^8} = \frac{90000}{24\pi 	imes 10} = \frac{9000}{24\pi} = \frac{375}{\pi}$. Using $\pi \approx 3.14159$,$I \approx \frac{375}{3.14159} \approx 119.36 \ Wm^{-2}$. Rounding to one decimal place,$I \approx 119.4 \ Wm^{-2}$.

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