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(B) The total energy density $u$ of an electromagnetic wave is given by $u = \frac{B_0^2}{2 \mu_0}$.Given $u = 1.02 	imes 10^{-8} \ J/m^3$ and $\mu_0

Vedclass · Accepted Answer

$160$

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(B) The total energy density $u$ of an electromagnetic wave is given by $u = \frac{B_0^2}{2 \mu_0}$.Given $u = 1.02 	imes 10^{-8} \ J/m^3$ and $\mu_0 = 4 \pi 	imes 10^{-7} \ T \cdot m/A$.Rearranging the formula for the magnetic field amplitude $B_0$:$B_0^2 = 2 \mu_0 u$$B_0^2 = 2 	imes (4 \pi 	imes 10^{-7}) 	imes (1.02 	imes 10^{-8})$$B_0^2 = 8 \pi 	imes 1.02 	imes 10^{-15}$$B_0^2 \approx 25.13 	imes 1.02 	imes 10^{-15} \approx 25.63 	imes 10^{-15} \approx 256.3 	imes 10^{-16}$Taking the square root:$B_0 \approx 16 	imes 10^{-8} \ T$Since $1 \ nT = 10^{-9} \ T$,we have $B_0 = 160 	imes 10^{-9} \ T = 160 \ nT$.

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