The magnetic field of an electromagnetic wave travelling in a...

From the given magnetic field equation, $B_y = 2 \times 10^{-8} \sin\left(10^3 z - 2 \times 10^{11} t\right) \text{ T}$, we can extract:
Wave number, $k = 10^3 \text{ rad/m}$
Angular frequency, $\omega = 2 \times 10^{11} \text{ rad/s}$
Magnetic field amplitude, $B_0 = 2 \times 10^{-8} \text{ T}$

The speed of the wave in the medium is:
$v = \dfrac{\omega}{k} = \dfrac{2 \times 10^{11}}{10^3} = 2 \times 10^8 \text{ m/s}$

The amplitude of the electric field is:
$E_0 = v B_0 = (2 \times 10^8) \times (2 \times 10^{-8}) = 4 \text{ V/m}$

The phase argument $(10^3 z - 2 \times 10^{11} t)$ indicates that the wave propagates in the $+z$ direction ($\hat{v} = \hat{k}$).
The magnetic field is in the $+y$ direction ($\hat{B} = \hat{j}$).
The direction of propagation is given by the cross product $\hat{E} \times \hat{B} = \hat{v}$.
We need $\hat{E} \times \hat{j} = \hat{k}$.
Since $\hat{i} \times \hat{j} = \hat{k}$, the electric field must oscillate in the $+x$ direction ($\hat{i}$).

Therefore, the electric field vector is:
$E_x = 4 \sin\left(10^3 z - 2 \times 10^{11} t\right) \text{ V/m}$

Answer: $E_x = 4 \sin\left(10^3 z - 2 \times 10^{11} t\right) \text{ V/m}$