Answer
$F = 3.2\times 10^{-16}~N$
Work Step by Step
We can write the expression for the magnetic field produced by a current in a straight wire:
$B = \frac{\mu_0~i}{2~\pi~R}$
We can find the magnitude of the net magnetic field $5.0~cm$ from the wire:
$B = \frac{\mu_0~i}{2~\pi~R}$
$B = \frac{(4\pi\times 10^{-7}~H/m)~(50~A)}{(2\pi)~(0.050~m)}$
$B = 2.0\times 10^{-4}~T$
By the right hand rule, this magnetic field is directed perpendicular to the direction of the current. For example, if the current is moving in the +x direction, the magnetic field is directed in the +z direction.
We can find the magnitude of the force on the electron:
$F = \vert q \vert v\times B$
$F = (1.6\times 10^{-19}~C)(1.0\times 10^7~m/s)(2.0\times 10^{-4}~T)$
$F = 3.2\times 10^{-16}~N$
