An electron is constrained to move along the $$y$$-axis with a speed of $$0.1\,c$$ ($$c$$ is the speed of light) in the presence of electromagnetic wave, whose electric field is $$\vec{E} = 30\hat{j}\sin(1.5 \times 10^7 t - 5 \times 10^{-2}x)\,\text{V m}^{-1}$$, where $$t$$ in in seconds and $$x$$ is in meters. The maximum magnetic force experienced by the electron will be: (given $$c = 3 \times 10^8\,\text{m s}^{-1}$$ and electron charge $$= 1.6 \times 10^{-19}\,\text{Coulombs}$$)

We start with the general formula for the magnetic force on a moving charge, which is

$$\vec F_B = q \, \vec v \times \vec B.$$

Here $$q$$ is the charge of the particle, $$\vec v$$ is its velocity vector and $$\vec B$$ is the magnetic‐field vector. The magnitude of this force is therefore

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

where $$\theta$$ is the angle between $$\vec v$$ and $$\vec B$$.

From the statement of the problem:

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

$$v = 0.1\,c = 0.1 \times 3 \times 10^{8}\;\text{m s}^{-1} = 3 \times 10^{7}\;\text{m s}^{-1}.$$

The electric field of the electromagnetic wave is given as

$$\vec E = 30\,\hat{\jmath}\,\sin\bigl(1.5 \times 10^{7} t - 5 \times 10^{-2} x\bigr)\;\text{V m}^{-1}.$$

This field points along the $$\hat{\jmath}$$ (positive $$y$$) direction and contains the term $$-k x$$, showing that the wave is propagating in the $$+x$$ direction. For a plane electromagnetic wave in vacuum we always have the relation

$$B_0 = \frac{E_0}{c},$$

where $$E_0$$ and $$B_0$$ are the peak (amplitude) values of the electric and magnetic fields and $$c$$ is the speed of light.

Reading off the amplitude of the electric field from the given expression, we have

$$E_0 = 30\;\text{V m}^{-1}.$$

Substituting this into the wave relation, the peak magnetic field becomes

$$B_0 = \frac{E_0}{c} = \frac{30}{3 \times 10^{8}}\;\text{T} = 1.0 \times 10^{-7}\;\text{T}.$$

The velocity vector $$\vec v$$ of the electron lies along $$\hat{\jmath}$$ (the $$y$$-axis) and the magnetic field $$\vec B$$ for a wave travelling in the $$+x$$ direction lies along $$\hat{k}$$ (the $$z$$-axis). These two directions are perpendicular, so

$$\theta = 90^\circ \quad\Longrightarrow\quad \sin\theta = 1.$$

Therefore the maximum magnetic force experienced by the electron is obtained simply by inserting the peak values into the magnitude formula:

$$\begin{aligned} F_{\text{max}} &= q\,v\,B_0 \\ &= \bigl(1.6 \times 10^{-19}\bigr)\, \bigl(3 \times 10^{7}\bigr)\, \bigl(1.0 \times 10^{-7}\bigr)\;\text{N}. \end{aligned}$$

Multiplying out step by step,

$$3 \times 10^{7} \times 1.0 \times 10^{-7} = 3 \times 10^{0} = 3,$$

and then

$$F_{\text{max}} = 1.6 \times 10^{-19} \times 3 = 4.8 \times 10^{-19}\;\text{N}.$$

Hence, the correct answer is Option C.