The electric field of a plane electromagnetic wave propagating in...

Given:
Electric field amplitude, $E_0 = 30\sqrt{2} \text{ V m}^{-1}$
Speed of light, $c = 3 \times 10^8 \text{ m s}^{-1}$

First, find the amplitude (peak value) of the magnetic field using the relation $c = \dfrac{E_0}{B_0}$:
$B_0 = \dfrac{E_0}{c} = \dfrac{30\sqrt{2}}{3 \times 10^8} = 10\sqrt{2} \times 10^{-8} \text{ T} = \sqrt{2} \times 10^{-7} \text{ T}$

Next, calculate the rms value of the magnetic field using $B_{rms} = \dfrac{B_0}{\sqrt{2}}$:
$B_{rms} = \dfrac{\sqrt{2} \times 10^{-7}}{\sqrt{2}} = 1.0 \times 10^{-7} \text{ T}$

Answer: $1.0 \times 10^{-7} \text{ T}$