Ratio of Maximum Electric and Magnetic Forces on Electron in EM Wave

To find the ratio of the maximum electric force to the maximum magnetic force acting on the electron, we analyze the forces exerted by the electromagnetic wave fields.

1. Electric Force: The electric field of the electromagnetic wave is given by: $E_y = E_0 \sin \omega \left(t - \frac{x}{c}\right)$ where $E_0 = 300\text{ V/m}$ is the amplitude of the electric field.

The magnitude of the maximum electric force ($F_{e, \text{max}}$) acting on an electron with elementary charge $e$ is: $F_{e, \text{max}} = e E_0$

2. Magnetic Force: The electromagnetic wave travels along the $+x$-direction ($\hat{i}$) and its electric field is polarized along the $y$-direction ($\hat{j}$). Since the direction of wave propagation is given by $\hat{E} \times \hat{B}$, the magnetic field must lie along the $z$-direction ($\hat{k}$).

The amplitude of the magnetic field $B_0$ is related to the electric field amplitude $E_0$ by: $B_0 = \frac{E_0}{c}$

The electron is restricted to move in the $y$-direction with speed $v = 1.5 \times 10^6\text{ m/s}$, so its velocity vector is $\vec{v} = v\hat{j}$. Since the velocity vector $\vec{v}$ and the magnetic field $\vec{B}$ are perpendicular to each other, the magnitude of the maximum magnetic force ($F_{m, \text{max}}$) is: $F_{m, \text{max}} = e v B_0 = e v \left(\frac{E_0}{c}\right)$

3. Ratio of Maximum Electric and Magnetic Forces: Taking the ratio of the maximum electric force to the maximum magnetic force: $\frac{F_{e, \text{max}}}{F_{m, \text{max}}} = \frac{e E_0}{e v \left(\frac{E_0}{c}\right)} = \frac{c}{v}$

Substitute the values of the speed of light $c = 3 \times 10^8\text{ m/s}$ and the velocity of the electron $v = 1.5 \times 10^6\text{ m/s}$: $\frac{F_{e, \text{max}}}{F_{m, \text{max}}} = \frac{3 \times 10^8\text{ m/s}}{1.5 \times 10^6\text{ m/s}} = 200$

Thus, the ratio of the maximum electric and magnetic forces acting on the electron is 200, which corresponds to option A.