An electromagnetic wave travelling in x-direction is described by field equation $$E_y = 300 \sin \omega \left(t - \frac{x}{c}\right)$$. If the electron is restricted to move in y-direction only with speed of $$1.5 \times 10^6$$ m/s then ratio of maximum electric and magnetic forces acting on the electron is __________.

The given plane electromagnetic wave is moving along the positive $$x$$-axis.
For such a wave, if the electric field vector is along $$y$$, the magnetic field vector is along $$z$$ and their magnitudes are related by

$$B = \frac{E}{c} \qquad -(1)$$

Maximum value of the electric field from the equation
$$E_y = 300 \sin\!\left[\omega\!\left(t-\frac{x}{c}\right)\right]$$
is

$$E_{\text{max}} = 300 \ \text{V m}^{-1}$$

Therefore, using $$(1)$$, the maximum magnetic field is

$$B_{\text{max}} = \frac{E_{\text{max}}}{c} = \frac{300}{3 \times 10^{8}} = 1 \times 10^{-6} \ \text{T}$$

For an electron (charge $$q = -e$$) moving only along the $$y$$-direction with speed
$$v = 1.5 \times 10^{6}\ \text{m s}^{-1}$$, the forces are

Electric force: $$F_E = q\,E$$ (along $$\pm y$$)
Magnetic force: $$F_B = q\,vB$$ (magnitude only)

Hence the ratio of the maximum magnitudes is

$$\frac{F_{E,\text{max}}}{F_{B,\text{max}}} = \frac{qE_{\text{max}}}{q\,v\,B_{\text{max}}} = \frac{E_{\text{max}}}{v\,\left(E_{\text{max}}/c\right)} = \frac{c}{v}$$

Substituting $$c = 3 \times 10^{8}\ \text{m s}^{-1}$$ and $$v = 1.5 \times 10^{6}\ \text{m s}^{-1}$$:

$$\frac{F_{E,\text{max}}}{F_{B,\text{max}}} = \frac{3 \times 10^{8}}{1.5 \times 10^{6}} = \frac{3}{1.5}\times 10^{2} = 2 \times 10^{2} = 200$$

Therefore, the required ratio is $$200$$.

Option A which is: $$200$$