Question

Assuming that the charges are moving nonrelativistically (v≪cv≪c), what can you say about the relationship between the magnitudes of the magnetic and electrostatic forces?

Answer 1

Answer:

The magnetic force is much smaller than electric force

Explanation:

The electric force and magnetic force experienced by each of the particles, assuming that they move parallel to each other with the same speed and the same charge are given by:

$F_e=\frac{e^2}{4\pi \epsilon_0d^2}\\F_m=\frac{\mu_0 v^2e^2}{4\pi d^2}$

So, the relationship between the magnitudes of the magnetic and electrostatic forces is:

$\frac{F_m}{F_e}=\frac{\frac{\mu_0 v^2e^2}{4\pi d^2}}{\frac{e^2}{4\pi \epsilon_0d^2}}\\\frac{F_m}{F_e}=\epsilon_0 \mu_0v^2$

Recall that $\epsilon_0\mu_0=\frac{1}{c^2}$:

$F_m=\frac{v^2}{c^2}F_e$

If the charges are moving nonrelativistically (v≪c), So $F_m$<<$F_e$, in this case, the magnetic force is much smaller than electric force.