To solve this problem, we need to calculate the magnetic force experienced by an electron traveling parallel to a current-carrying straight conductor. Here's how we can solve it step-by-step:
The magnetic field \(B\) at a distance \(r\) from a long straight conductor carrying current \(I\) is given by the formula:
\(B = \frac{\mu_0 I}{2 \pi r}\)
where \(\mu_0 = 4\pi \times 10^{-7} \, \mathrm{T \cdot m/A}\) is the permeability of free space.
Substitute the given values into the formula:
\(I = 5 \, \mathrm{A}\), \(r = 0.1 \, \mathrm{m}\)
\(B = \frac{4\pi \times 10^{-7} \times 5}{2 \pi \times 0.1} = 10^{-6} \, \mathrm{T}\)
Next, calculate the magnetic force \(F\) experienced by the electron using:
\(F = qvB \sin \theta\)
The electron is moving parallel to the wire, so \(\theta = 90^\circ\) and \(\sin \theta = 1\).
Given the charge of an electron \(q = 1.6 \times 10^{-19} \, \mathrm{C}\) and its speed \(v = 5 \times 10^{6} \, \mathrm{ms^{-1}}\), substitute the values:
\(F = 1.6 \times 10^{-19} \times 5 \times 10^{6} \times 10^{-6}\)
\(F = 8 \times 10^{-18} \, \mathrm{N}\)
Thus, the force experienced by the electron is \(8 \times 10^{-18} \, \mathrm{N}\).
This matches the correct answer from the given options.
Therefore, the correct answer is \(8 \times 10^{-18} \, \mathrm{N}\).