Irodov 3.250: Magnetism as a Small Correction – Amit & Deepali Sinha Foundation

Problem Irodov 3.250

Two protons travel along parallel lines with the same velocity, of magnitude $v = 300$ km/s. Find the ratio of the magnetic force to the electric force with which they act on each other.

The two protons p moving with the same velocity v, a distance r apart, drawn side by side (assumed). — The two protons $p$ moving with the same velocity $\vect v$ , a distance $r$ apart, drawn side by side (assumed).

Given

$v = 300$ km/s, the same for both
the protons abreast, the line joining them perpendicular to $\vect v$ (assumed)

Find

F_m/F_e

Solution

Step 1The fields of the moving proton

Take proton $2$ as the source and proton $1$ , a distance $r$ away, as the one acted on; the protons are taken abreast, so the line joining them is perpendicular to $\vect v$ . With $v \ll c$ , the electric field of proton $2$ at proton $1$ is the Coulomb field, and its magnetic field is that of a moving point charge, $\vect r$ running from $2$ to $1$ :

E = \frac{1}{4\pi\varepsilon_0}\frac{e}{r^2}, \qquad \vect B = \frac{\mu_0}{4\pi}\,\frac{e\,\vect v\times\vect r}{r^3}

Since $\vect v \perp \vect r$ , the magnitude of the magnetic field is

B = \frac{\mu_0}{4\pi}\,\frac{e v}{r^2} \tag{1}

Step 2The two forces

The forces on proton 1 from proton 2, drawn schematically: the electric repulsion F_e and the magnetic attraction F_m, both along the line joining them. — The forces on proton $1$ from proton $2$ , drawn schematically: the electric repulsion $\vect F_e$ and the magnetic attraction $\vect F_m$ , both along the line joining them.

The electric force on proton $1$ pushes it away from $2$ . The magnetic force $e\,\vect v\times\vect B$ is perpendicular to $\vect v$ and, by (1), has the magnitude

F_m = e v B = \frac{\mu_0}{4\pi}\,\frac{e^2 v^2}{r^2}

Its direction follows from the double cross product, since $\vect B$ at $1$ is along $\vect v\times\vect r$ ; expanding it with $\vect v\cdot\vect r = 0$ :

\vect v\times(\vect v\times\vect r) = \vect v\,(\vect v\cdot\vect r) - \vect r\,v^2 = -v^2\,\vect r

So $\vect F_m$ points from $1$ back toward $2$ : the magnetic force attracts. The electric force has the magnitude

F_e = eE = \frac{1}{4\pi\varepsilon_0}\,\frac{e^2}{r^2}

Step 3The ratio

Dividing the two, the charges and the distance cancel:

\frac{F_m}{F_e} = \frac{\mu_0 e^2 v^2/(4\pi r^2)}{e^2/(4\pi\varepsilon_0 r^2)} = \mu_0\varepsilon_0 v^2

With $\mu_0\varepsilon_0 = 1/c^2$ this is $(v/c)^2$ . Numerically,

\frac{F_m}{F_e} = \left(\frac{3.00\times 10^5\text{ m/s}}{3.00\times 10^8\text{ m/s}}\right)^2

\frac{F_m}{F_e} = \left(\frac{v}{c}\right)^2 = 1.00\times 10^{-6}

(2)

Check

The ratio is dimensionless, as it must be, and it vanishes for $v \to 0$ , where only the Coulomb force remains. For protons that are not abreast, with $\vect r$ at the angle $\alpha$ to $\vect v$ , the double cross product $\vect v\times(\vect v\times\vect r)$ has the magnitude $v^2 r\sin\alpha$ , so the ratio of magnitudes becomes $(v/c)^2\sin\alpha$ : the abreast case is the largest.

Remark

With the exact fields of a charge in uniform motion the ratio is still $(v/c)^2$ for protons abreast: the electric force grows to $F_e = \gamma F_0$ , with $F_0$ the Coulomb force in the protons’ common rest frame and $\gamma = (1 - v^2/c^2)^{-1/2}$ , and the net repulsion is $F_e\,(1 - v^2/c^2)$ , which is $F_0/\gamma$ . That is exactly how a force transverse to the motion transforms between frames, so the magnetic force is the relativistic correction that keeps electrostatics consistent in a frame where the charges move.