A plane electromagnetic wave of frequency 20 MHz travels in free space along the $+x$ direction. At a particular point in space and time, the electric field vector of the wave is $\mathrm{E}_y=9.3 \mathrm{Vm}^{-1}$. Then, the magnetic field vector of the wave at that point is

<p>For an electromagnetic wave in free space, the magnitudes of the electric and magnetic fields are related by the speed of light, $c \approx 3 \times 10^8 \, \mathrm{m/s},$ through the equation</p> <p>$E = cB.$</p> <p>Here's how to determine the magnetic field:</p> <p><p>We are given the electric field magnitude: $E=9.3 \, \mathrm{V/m}.$</p></p> <p><p>Use the relationship to find the magnetic field magnitude:</p> <p>$B = \frac{E}{c} = \frac{9.3}{3 \times 10^8} \, \mathrm{T}.$</p></p> <p><p>Calculating the value gives:</p> <p>$B \approx 3.1 \times 10^{-8} \, \mathrm{T}.$</p></p> <p><p>Since the electric field is along the $y$ direction and the wave propagates along the $+x$ direction, by the right-hand rule, the magnetic field is oriented along the $z$ direction.</p></p> <p>Thus, the magnetic field vector is</p> <p>$\mathrm{B}_z = 3.1 \times 10^{-8} \, \mathrm{T},$</p> <p>which corresponds to Option C.</p>