The equation of a light wave,normally inciden | vedclass

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(A) The intensity $I$ of an electromagnetic wave is given by the formula $I = \frac{B_0^2}{2\mu_0} c$,where $B_0$ is the peak magnetic field,$\mu_0$ i

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$1.2$

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(A) The intensity $I$ of an electromagnetic wave is given by the formula $I = \frac{B_0^2}{2\mu_0} c$,where $B_0$ is the peak magnetic field,$\mu_0$ is the permeability of free space $(4\pi 	imes 10^{-7} 	ext{ T m/A})$,and $c$ is the speed of light $(3 	imes 10^8 	ext{ m/s})$.Given $B_0 = 100 	ext{ nT} = 100 	imes 10^{-9} 	ext{ T} = 10^{-7} 	ext{ T}$.Substituting the values into the formula:$I = \frac{(10^{-7})^2}{2 	imes 4\pi 	imes 10^{-7}} 	imes 3 	imes 10^8$$I = \frac{10^{-14}}{8\pi 	imes 10^{-7}} 	imes 3 	imes 10^8$$I = \frac{3 	imes 10^{-14} 	imes 10^8}{8\pi 	imes 10^{-7}}$$I = \frac{3 	imes 10^{-6}}{8\pi 	imes 10^{-7}} = \frac{30}{8\pi} = \frac{15}{4\pi} \approx \frac{15}{12.566} \approx 1.1936 	ext{ W/m}^2$.Rounding to one decimal place,we get $1.2 	ext{ W/m}^2$.

The equation of a light wave,normally incident on a surface,is given by $B = (100 \text{ nT}) \sin(2\pi(10^{15}t - (3 \times 10^{-7})x) + \frac{\pi}{6})$. Find the intensity of light on that surface in $W/m^2$.

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