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(C) The average power per unit area (intensity) of an electromagnetic wave is given by the formula: $I = \frac{c B_{max}^2}{2 \mu_0}$,where $c$ is the

Vedclass · Accepted Answer

$8.8$

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(C) The average power per unit area (intensity) of an electromagnetic wave is given by the formula: $I = \frac{c B_{max}^2}{2 \mu_0}$,where $c$ is the speed of light in vacuum $(3 	imes 10^8 \ m/s)$,$\mu_0$ is the permeability of free space $(4 \pi 	imes 10^{-7} \ T \ m/A)$,and $B_{max}$ is the amplitude of the oscillating magnetic field.Rearranging the formula to solve for $B_{max}$:$B_{max} = \sqrt{\frac{2 \mu_0 I}{c}}$Substituting the given values:$B_{max} = \sqrt{\frac{2 	imes 4 \pi 	imes 10^{-7} 	imes 9240}{3 	imes 10^8}}$$B_{max} = \sqrt{\frac{8 \pi 	imes 9240 	imes 10^{-15}}{3}}$$B_{max} = \sqrt{\frac{232252.8 	imes 10^{-15}}{3}}$$B_{max} = \sqrt{77417.6 	imes 10^{-15}}$$B_{max} = \sqrt{77.4176 	imes 10^{-12}}$$B_{max} \approx 8.798 	imes 10^{-6} \ T \approx 8.8 \ \mu T$.

If the average power per unit area delivered by an electromagnetic wave is $9240 \ W \ m^{-2}$,then the amplitude of the oscillating magnetic field in the $EM$ wave is: (in $\mu T$)

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