An electric field of strength E\~ of charges moving with velocity v creates a magnetic field whose induction \vec{B} = K[\vec{v}\times \vec{E}]. The coefficient K is equal to \varepsilon_0 \mu_0in SI and 1/c in CGS, where c is the speed of light. Prove that the magnetic interaction of two moving charges is weaker than their electrical interaction.

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    <h3 id="back-link"><a href="/#9.2">$\leftarrow$Back</a></h3>

    <h3> Statement </h3>
    <p>
        $9.2.1.$ An electric field of strength $E\~$ of charges moving with velocity $v$ creates a magnetic field whose induction $\vec{B} = K[\vec{v}\times \vec{E}]$. The coefficient $K$ is equal to $\varepsilon_0 \mu_0$

in SI and $1/c$ in CGS, where $c$ is the speed of light. Prove that the magnetic interaction of two moving charges is weaker than their electrical interaction.

    <h3>Proof</h3>
    <p>
        Magnetic field, the induction of which

$$|\vec{B}| = \varepsilon_0 \mu_0\left[\vec{v}\times \vec{E}\right] \le \varepsilon_0 \mu_0vE$$
then the force of magnetic interaction (Lorentz force) of charges
$$F_B = |q[\vec{v} \times \vec{B}]| = qvB \le \varepsilon_0 \mu_0 qv^2E$$
taking into account the force of electrostatic interaction $F_E = qE$, we obtain
$$\frac{F_B}{F_E} \le \varepsilon_0 \mu_0 v^2 = \frac{v^2}{c^2}$$
Considering the smallness of the speed $v \ll c$, then
$$\boxed{\frac{F_B}{F_E} \approx 0}$$

Andrei Yersh

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