Ans: We are given the current through the straight wire to be, $I=10 \mathrm{~A}$
Speed of the electron, $\mathrm{v}=10^{7} \mathrm{~m} / \mathrm{s}$
Distance of electron from the wire, $\mathrm{R}=2.0 \mathrm{~cm}=2 \times 10^{-2} \mathrm{~m}$,
Force acting on moving electron would be, $\mathrm{F}=\mathrm{qVB} \sin \theta$
We have the expression for magnetic field as,
$$
\mathrm{B}=\frac{\mu_{0}}{4 \pi} \frac{2 I}{r}
$$
Substituting the given values,
$\mathrm{B}=\frac{10^{-7} \times 2 \times 10}{2 \times 10^{-2}}=10^{-4} \mathrm{~T}$ and it is given to be $\perp$ to the plane of paper and directed towards.
Now, force acting on the electron could be given by,
$$
\begin{array}{l}
\mathrm{F}=4 \mathrm{~V} \mathrm{~B} \sin \theta \\
\Rightarrow \mathrm{F}=1.6 \times 10^{-16} \mathrm{~N}
\end{array}
$$
Therefore, we find the force to be, $\mathrm{F}=1.6 \times 10^{-16} \mathrm{~N}$.