Irodov 3.362: The Slow Charge's Field by Transformation

Problem Irodov 3.362

A point charge $q$ moves at a constant velocity $\vect v$ , small compared with the speed of light. Using the transformation formulas for the fields, find the magnetic induction $\vect B$ the charge produces at the point whose position relative to the charge is given by the radius vector $\vect r$ .

The charge q moving with velocity v, and the radius vector r from the charge to the field point P. — The charge $q$ moving with velocity $\vect v$ , and the radius vector $\vect r$ from the charge to the field point $P$ .

Given

$q$
$\vect v = \text{const}$ , $v \ll c$
$\vect r$ from the charge to the point

Find

\vect B

at that point

Solution

Step 1The field in the charge’s own frame

Let $K'$ be the frame that moves with the charge, at the velocity $\vect v$ relative to the laboratory frame $K$ . In $K'$ the charge is at rest, so there is only its Coulomb field and no magnetic field:

\vect E' = \frac{q}{4\pi\varepsilon_0}\,\frac{\vect r}{r^3}, \qquad \vect B' = 0 \tag{1}

At these speeds lengths are the same in both frames, so the radius vector is the same $\vect r$ in $K'$ and in $K$ .

Step 2Back to the laboratory frame

Applied from $K'$ to $K$ , the rule reads

\vect B = \vect B' - (-\vect v)\times\frac{\vect E'}{c^2} = \vect B' + \frac{\vect v\times\vect E'}{c^2}

Putting in (1):

\vect B = \frac{1}{c^2}\,\vect v\times\frac{q\,\vect r}{4\pi\varepsilon_0 r^3} = \frac{1}{4\pi\varepsilon_0 c^2}\,\frac{q\,\vect v\times\vect r}{r^3}

Step 3The induction of the moving charge

With $\varepsilon_0\mu_0 = 1/c^2$ the prefactor is $\mu_0/(4\pi)$ :

\vect B = \frac{\mu_0}{4\pi}\,\frac{q\,\vect v\times\vect r}{r^3}

(2)

Its magnitude is $\mu_0 q v\sin\theta/(4\pi r^2)$ , with $\theta$ the angle between $\vect v$ and $\vect r$ , and it is perpendicular to both; it circles the line of motion.

The induction B at P, along v· r: out of the page here, for q > 0. — The induction $\vect B$ at $P$ , along $\vect v\times\vect r$ : out of the page here, for $q > 0$ .

Check

Units: $\mu_0 q v/r^2$ is $(\text{T}\cdot\text{m/A})\cdot(\text{C}\cdot\text{m/s})/\text{m}^2 = \text{T}$ , since $\text{C/s} = \text{A}$ . On the line of motion ( $\vect r \parallel \vect v$ ) the result vanishes, as the symmetry of the Coulomb field about that line requires; dividing by $\mu_0$ gives $\vect H = q\,\vect v\times\vect r/(4\pi r^3)$ , the same field 3.354 finds from the displacement current.

Remark

Writing $q\vect v$ as $I\,d\vect l$ turns the result into the Biot and Savart law. The magnetic field of a slow charge is therefore nothing but its Coulomb field seen from a frame in which the charge moves, a first-order relativistic effect of size $v/c$ relative to $E$ : $cB = (v/c)E\sin\theta$ .