The rate of energy flow per unit area of a sinusoidal electromagnetic wave has an average value of $0.601 \mathrm {~W} / \mathrm { m } ^ { 2 }$. What is the maximum value of the magnetic field in the wave? $\left( c = 3.00 \times 10 ^ { 8 } \mathrm {~m} / \mathrm { s } , \mu _ { 0 } \right.$ $= 4 \pi \times 10 ^ { - 7 } \mathrm {~T} \cdot \mathrm { m } / \mathrm { A } , \varepsilon _ { 0 } = 8.85 \times 10 ^ { - 12 } \mathrm { C } ^ { 2 } / \mathrm { N } \cdot \mathrm { m } ^ { 2 }$ )

Question

Quizplus · Accepted Answer

The maximum value of the magnetic field in a sinusoidal electromagnetic wave can be found using the relationship between the average power (Poynting vector) and the maximum magnetic field strength. The average power per unit area is given by $S_{avg} = \frac{1}{2} \cdot c \cdot \varepsilon_0 \cdot E_{max}^2$, where $E_{max}$ is the maximum electric field strength. The relationship between the maximum electric and magnetic fields in an electromagnetic wave is $E_{max} = c \cdot B_{max}$. By rearranging and solving for $B_{max}$, and using the given average power, we can find the maximum magnetic field strength. Given the average power per unit area is $0.601 \, \mathrm{W/m^2}$, and using the values for $c$ and $\varepsilon_0$, we can calculate $B_{max}$ to be $7.09 \times 10^{-8} \, \mathrm{T}$, which matches choice A.

The Rate of Energy Flow Per Unit Area of a Sinusoidal