A light beam is described by $$E = 800 \sin\omega\left(t - \frac{x}{c}\right)$$. An electron is allowed to move normal to the propagation of light beam with a speed of $$3 \times 10^7$$ m s$$^{-1}$$. What is the maximum magnetic force exerted on the electron?

We are given the plane-polarised electromagnetic wave

$$E \;=\; 800 \,\sin\!\Bigl[\;\omega\!\left(t-\dfrac{x}{c}\right)\Bigr]$$

The coefficient of the sine function is the peak (maximum) electric-field magnitude, so

$$E_0 = 800 \text{ V m}^{-1}$$

For a plane electromagnetic wave in free space the electric and magnetic amplitudes are related by the well-known relation

$$E_0 = c\,B_0$$

where $$c = 3.0 \times 10^{8}\ \text{m s}^{-1}$$ is the speed of light in vacuum. Solving for the magnetic-field amplitude $$B_0$$ we get

$$$\begin{aligned} B_0 &= \dfrac{E_0}{c} \\ &= \dfrac{800}{3.0 \times 10^{8}} \ \text{T} \\ &= \dfrac{8.00 \times 10^{2}}{3.0 \times 10^{8}} \ \text{T} \\ &= \dfrac{8.00}{3.0}\times 10^{-6}\ \text{T} \\ &= 2.666\dots\times 10^{-6}\ \text{T} \\ &\approx 2.67 \times 10^{-6}\ \text{T} \end{aligned}$$$

An electron is made to move with speed

$$v = 3.0 \times 10^{7}\ \text{m s}^{-1}$$

and, as stated, the motion is normal (perpendicular) to the direction of propagation of the beam. In a plane wave the magnetic field itself is already perpendicular to the propagation direction, so by choosing the electron’s velocity also normal to the beam we can arrange the velocity to be perpendicular to the magnetic field. For the magnetic force this gives the maximum value because the factor $$\sin\theta$$ becomes unity. The magnetic (Lorentz) force formula is first written explicitly:

$$F = q\,v\,B\,\sin\theta$$

For the maximum force $$\theta = 90^\circ$$ and $$\sin\theta = 1$$, hence

$$F_{\max} = q\,v\,B_0$$

For an electron the magnitude of the charge is

$$q = e = 1.6 \times 10^{-19}\ \text{C}$$

Substituting all numerical values step by step,

$$$\begin{aligned} F_{\max} &= (1.6 \times 10^{-19}) \times (3.0 \times 10^{7}) \times (2.67 \times 10^{-6}) \ \text{N} \\[4pt] &= 1.6 \times 3.0 \times 2.67 \times 10^{-19 + 7 - 6}\ \text{N} \\[4pt] &= (1.6 \times 3.0)\times 2.67 \times 10^{-18}\ \text{N} \\[4pt] &= 4.8 \times 2.67 \times 10^{-18}\ \text{N} \\[4pt] &= 12.816 \times 10^{-18}\ \text{N} \\[4pt] &= 1.2816 \times 10^{-17}\ \text{N} \end{aligned}$$$

Keeping only three significant figures (because the given data have at most two significant figures), we state

$$F_{\max} \approx 1.28 \times 10^{-17}\ \text{N}$$

The option list is expressed in the form $$A \times 10^{-18}\ \text{N}$$. Writing our result in the same form:

$$$1.28 \times 10^{-17}\ \text{N} = 12.8 \times 10^{-18}\ \text{N}$$$

This matches Option B.

Hence, the correct answer is Option B.