1. Given data

• Electron speed along $y$-axis: $v = 0.1c = 0.1 \times 3 \times 10^8 = 3 \times 10^7\ \text{m/s}$
• Electric field of the EM wave: $\vec E = 30\,\hat j\,\sin(1.5\times10^7 t - 5\times10^{-2}x)\ \text{V/m}$

So the electric field is along $\hat j$ (i.e. along $y$-axis), and the wave propagates in the $+x$ direction.

2. Find the magnetic field amplitude

For an electromagnetic wave, $E_0 = cB_0$ Hence, $B_0 = \frac{E_0}{c} = \frac{30}{3\times10^8} = 1\times10^{-7}\ \text{T}$

3. Direction of magnetic field

In an EM wave, propagation direction is along $\vec E \times \vec B$.

• $\vec E$ is along $\hat j$
• Propagation is along $\hat i$

Therefore $\vec B$ must be along $\hat k$ (or $z$-axis), since $\hat j \times \hat k = \hat i$

Thus the electron moves along $y$ while magnetic field is along $z$, so $\vec v \perp \vec B$.

4. Magnetic force on the electron

Magnitude of magnetic force is $F_B = |q|vB\sin\theta$ Since $\theta = 90^\circ$, $F_B = |q|vB$ Maximum force occurs when $B = B_0$: $F_{B,\max} = e v B_0$ Substitute values: $F_{B,\max} = (1.6\times10^{-19})(3\times10^7)(1\times10^{-7})$ $F_{B,\max} = 4.8\times10^{-19}\ \text{N}$

5. Check options

• A: $4.8\times10^{-19}\,\text{N}$ ✅
• B: $2.4\times10^{-18}\,\text{N}$ ❌
• C: $3.2\times10^{-18}\,\text{N}$ ❌
• D: $1.6\times10^{-18}\,\text{N}$ ❌

Therefore, the correct option is A.