Question

two protons move parallel in each other with an equal velocity v = 300 km/s. find the ratio of forces of magnetic and electrical interaction of the protons

Answer 1

10^{-6}

Answer 2

The problem asks for the ratio of the magnetic force to the electrical force between two protons moving parallel to each other with a given velocity.

1. Electrical Force ($F_e$) The electrical force between two protons, each with charge $e$, separated by a distance $r$, is given by Coulomb's Law: $F_e = \frac{1}{4\pi\epsilon_0} \frac{e^2}{r^2}$ This force is repulsive.

2. Magnetic Force ($F_m$) Each moving proton creates a magnetic field, and this field interacts with the other moving proton. Consider one proton ($P_1$) moving with velocity $\vec{v}$. It produces a magnetic field $\vec{B}$ at the position of the second proton ($P_2$). For a point charge $q$ moving with velocity $\vec{v}$, the magnetic field at a perpendicular distance $r$ is: $B = \frac{\mu_0}{4\pi} \frac{qv}{r^2}$ In this case, $q = e$, so the magnetic field produced by $P_1$ at $P_2$'s location is: $B_1 = \frac{\mu_0}{4\pi} \frac{ev}{r^2}$ The direction of $\vec{B_1}$ is perpendicular to both $\vec{v}$ and $\vec{r}$ (the line connecting $P_1$ to $P_2$). The force experienced by $P_2$ due to this magnetic field is given by the Lorentz force formula: $\vec{F_m} = e (\vec{v} \times \vec{B_1})$ Since the protons are moving parallel, their velocities are parallel. The magnetic field $\vec{B_1}$ created by $P_1$ at $P_2$'s location is perpendicular to $P_2$'s velocity $\vec{v}$. Therefore, the magnitude of the magnetic force is: $F_m = evB_1 \sin(90^\circ) = evB_1$ Substitute the expression for $B_1$: $F_m = ev \left( \frac{\mu_0}{4\pi} \frac{ev}{r^2} \right)$ $F_m = \frac{\mu_0}{4\pi} \frac{e^2v^2}{r^2}$ This force is attractive.

3. Ratio of Magnetic to Electrical Force ($F_m/F_e$) Now, we find the ratio of the magnitudes of the magnetic and electrical forces: $\frac{F_m}{F_e} = \frac{\frac{\mu_0}{4\pi} \frac{e^2v^2}{r^2}}{\frac{1}{4\pi\epsilon_0} \frac{e^2}{r^2}}$ Cancel common terms ($e^2/r^2$ and $4\pi$): $\frac{F_m}{F_e} = \mu_0 \epsilon_0 v^2$ We know that the speed of light in vacuum ($c$) is related to $\mu_0$ and $\epsilon_0$ by the relation: $c = \frac{1}{\sqrt{\mu_0 \epsilon_0}} \implies \mu_0 \epsilon_0 = \frac{1}{c^2}$ Substitute this into the ratio: $\frac{F_m}{F_e} = \frac{v^2}{c^2}$

4. Calculation Given velocity $v = 300 \text{ km/s} = 300 \times 10^3 \text{ m/s} = 3 \times 10^5 \text{ m/s}$. The speed of light $c \approx 3 \times 10^8 \text{ m/s}$. $\frac{F_m}{F_e} = \frac{(3 \times 10^5 \text{ m/s})^2}{(3 \times 10^8 \text{ m/s})^2}$ $\frac{F_m}{F_e} = \frac{3^2 \times (10^5)^2}{3^2 \times (10^8)^2}$ $\frac{F_m}{F_e} = \frac{9 \times 10^{10}}{9 \times 10^{16}}$ $\frac{F_m}{F_e} = 10^{10 - 16}$ $\frac{F_m}{F_e} = 10^{-6}$

The ratio of the forces of magnetic and electrical interaction is $10^{-6}$.