Irodov 3.341: Magnetic Energy Is Relativistically Small

Problem Irodov 3.341

A thin ring of radius $a = 10$ cm, carrying a uniformly distributed charge, rotates about its own axis at angular velocity $\omega = 100$ rad/s. On the axis, at a distance $l = a$ from the ring’s centre, find the ratio of the energy density of the magnetic field to that of the electric field.

The thin ring of radius a with its uniform charge (drawn positive), rotating at ω about its axis (sense chosen), and the point P on the axis at the distance l from the centre. — The thin ring of radius $a$ with its uniform charge (drawn positive), rotating at $\omega$ about its axis (sense chosen), and the point $P$ on the axis at the distance $l$ from the centre.

Given

$a = 10$ cm
$\omega = 100$ rad/s
$l = a$
uniform charge

Find

w_m/w_e

P

Solution

Step 1The fields on the axis

Let $q$ be the ring’s charge. By symmetry about the axis both fields at $P$ point along it. Every charge element is at the same distance $\sqrt{a^2+l^2}$ from $P$ , and only its axial share, the fraction $l/\sqrt{a^2+l^2}$ , survives the sum, so Coulomb’s law gives the electric field

E = \frac{1}{4\pi\varepsilon_0}\,\frac{ql}{(a^2+l^2)^{3/2}} \tag{1}

The rotating ring is a loop current: its charge passes a fixed point $\omega/2\pi$ times per second, so $I = q\omega/2\pi$ . Each element $I\,dl$ is perpendicular to the line to $P$ , and by the Biot and Savart law only the axial fraction $a/\sqrt{a^2+l^2}$ of each contribution survives:

B = \frac{\mu_0 I a^2}{2(a^2+l^2)^{3/2}} = \frac{\mu_0}{4\pi}\,\frac{q\omega a^2}{(a^2+l^2)^{3/2}} \tag{2}

Step 2The ratio of the fields

Dividing (2) by (1), the charge and the distance factor $(a^2+l^2)^{3/2}$ cancel:

\frac{B}{E} = \frac{\mu_0}{4\pi}\,4\pi\varepsilon_0\,\frac{\omega a^2}{l} = \varepsilon_0\mu_0\,\frac{\omega a^2}{l} \tag{3}

Step 3The ratio of the energy densities

The energy densities in vacuum are $w_m = B^2/2\mu_0$ and $w_e = \varepsilon_0E^2/2$ , so their ratio is $B^2/\varepsilon_0\mu_0E^2$ . Squaring (3):

\frac{w_m}{w_e} = \frac{1}{\varepsilon_0\mu_0}\,\left(\varepsilon_0\mu_0\,\frac{\omega a^2}{l}\right)^2

\frac{w_m}{w_e} = \varepsilon_0\mu_0\,\frac{\omega^2a^4}{l^2}

(4)

Setting $l = a$ in (4), writing $\varepsilon_0\mu_0$ as $1/c^2$ , and substituting the numbers:

\varepsilon_0\mu_0\,\frac{\omega^2a^4}{a^2} = \left(\frac{\omega a}{c}\right)^2 = \left(\frac{(100)(0.10)}{3.00\times10^{8}}\right)^2

\frac{w_m}{w_e} = \left(\frac{\omega a}{c}\right)^2 \approx 1.1\times10^{-15}

(5)

The sign of the charge and the sense of rotation drop out: both enter squared.

Check

Dimensions: $\varepsilon_0\mu_0 = 1/c^2$ , so (4) is $(\omega a^2/lc)^2$ , the square of a speed over $c$ , a pure number as a ratio must be. Limits: on the axis close to the centre, $l \to 0$ , the electric field vanishes by symmetry while $B$ stays finite, and the ratio grows without bound, as (4) says; far away, $l \gg a$ , $E$ falls as $1/l^2$ and $B$ as $1/l^3$ , so the ratio falls as $1/l^2$ .

Remark

The rim moves at $\omega a = 10$ m/s, and the magnetic energy density is of order $(v/c)^2$ of the electric one. The magnetic field of moving charge is a relativistic correction to its electric field; it dominates around a current-carrying wire because there the electric fields of the moving electrons and of the fixed ions cancel and leave the small magnetic part alone.