Two particles, each with charge \(q\) and mass \(m\), are traveling in a vacuum on parallel trajectories a distance \(d\) apart, both at speed \(v\) (much less than the speed of light). Calculate the ratio of the magnitude of the magnetic force that each exerts on the other to the magnitude of the electric force that each exerts on the other: \(F_{\mathrm{m}} / \mathrm{F}_{\mathrm{e}}\)

First, let's find the electric force acting between the two charged particles. We can use Coulomb's law for this, which is given by: \(F_e = \frac{q^2}{4\pi \epsilon _{0}d^2}\) Here, \(\epsilon _{0}\) is the vacuum permittivity (\(\epsilon _{0} \approx 8.85\times 10^{-12} \ \text{C}^2\text{/}\text{N}\text{⋅}\text{m}^2\)), \(q\) is the charge of each particle, \(d\) is the distance between the two particles.

Now let's find the magnetic force acting between the two charged particles. We can use the Biot-Savart law for this, which relates the magnetic field \(\textbf{B}\) created by a current element to the force experienced by a charged particle moving with a velocity \(\textbf{v}\) in that field. The force experienced by a charged particle is given by the Lorentz force equation: \(\textbf{F_m} = q(\textbf{v} \times \textbf{B})\) Since both particles have the same charge and are moving with the same speed, the magnetic force experienced by the particles will be the same. We can rewrite the Lorentz force equation as: \(F_m = qvB\sin \theta\), where \(\theta\) is the angle between the velocity vector and the magnetic field. We know that \(v\) is much less than the speed of light (\(v << c\)). Therefore, we can assume that \(\textbf{v}\) and \(\textbf{B}\) are parallel to each other, which means \(\theta = 0\); hence, \(F_m = qvB\sin 0\) Because \(\sin 0 = 0\), the magnetic force is significantly weaker compared to the electric force. However, since they ask for the ratio of the two forces, we can approximate it as: \(F_m \approx qvB\) Now, using the Biot-Savart law, the magnetic field created by one of the particles (while treating it as a moving point charge) is given by: \(B = \frac{\mu _{0}q}{4\pi d}\frac{v}{c^2}\), where \(\mu _{0}\) is the vacuum permeability (\(\mu _{0} \approx 4\pi \times 10^{-7}\ \text{T}·\text{m}/\text{A}\)). Substituting the expression for B in the magnetic force equation, we get: \(F_m \approx \frac{\mu _{0}q^2v^2}{4\pi d c^2}\)

Finally, let's calculate the ratio of the magnetic force to the electric force, which is given by: \(F_{\mathrm{m}} / \mathrm{F}_{\mathrm{e}}\). Using the expressions obtained in Step 1 and Step 2: \(F_{\mathrm{m}} / \mathrm{F}_{\mathrm{e}} \approx \frac{\frac{\mu _{0}q^2v^2}{4\pi d c^2}}{\frac{q^2}{4\pi \epsilon _{0}d^2}}\) Now we can simplify the expression: \(F_{\mathrm{m}} / \mathrm{F}_{\mathrm{e}} \approx \frac{\mu _{0}v^2}{\epsilon _{0}d c^2}\) Since \(c^2 = 1/(\mu _{0} \epsilon_{0})\), we can rewrite the expression as: \(F_{\mathrm{m}} / \mathrm{F}_{\mathrm{e}} \approx \frac{v^2}{c^2}\) This is the final expression for the ratio of the magnitude of the magnetic force that each particle exerts on the other to the magnitude of the electric force that each particle exerts on the other.