Irodov 3.364: A Wire's Field Seen on the Move – Amit & Deepali Sinha Foundation

Problem Irodov 3.364

An inertial frame $K$ contains only a magnetic field, of induction $\vect B = b\,(y\vect i - x\vect j)/(x^2 + y^2)$ , where $b$ is a constant and $\vect i$ , $\vect j$ are the unit vectors of the $x$ and $y$ axes. A frame $K'$ moves relative to $K$ at the constant velocity $\vect v = v\vect k$ , small compared with the speed of light, where $\vect k$ is the unit vector of the $z$ axis; the $z'$ axis coincides with the $z$ axis. Find the electric field strength $\vect E'$ in $K'$ , and describe the shape of this field.

Given

in $K$ : only $\vect B = b\,(y\vect i - x\vect j)/(x^2+y^2)$
$\vect v = v\vect k$ , $v \ll c$
$z' \equiv z$

Find

\vect E'

K'

· the shape of

\vect E'

Solution

Step 1The low-speed transformation

For $v \ll c$ the electric field in a frame moving at $\vect v$ is $\vect E' = \vect E + \vect v\times\vect B$ (3.359). Frame $K$ has no electric field, so

\vect E' = \vect v\times\vect B \tag{1}

Step 2The cross product

Put $\vect v = v\vect k$ and the given $\vect B$ into (1), and use $\vect k\times\vect i = \vect j$ and $\vect k\times\vect j = -\vect i$ :

\vect E' = \frac{bv}{x^2+y^2}\,\vect k\times\left(y\vect i - x\vect j\right) = \frac{bv}{x^2+y^2}\left(y\vect j + x\vect i\right)

The motion is along $z$ , so $x' = x$ and $y' = y$ . Let $\vect r = x\vect i + y\vect j$ be the vector from the $z'$ axis to the point, perpendicular to that axis, with $r^2 = x^2 + y^2$ :

\vect E' = \frac{bv\,\vect r}{r^2}

(2)

The plane z' = 0 seen from the positive z' axis, for b > 0 and v > 0: the circles of B and the radial lines of E' leaving the z' axis. — The plane $z' = 0$ seen from the positive $z'$ axis, for $b > 0$ and $v > 0$ : the circles of $\vect B$ and the radial lines of $\vect E'$ leaving the $z'$ axis.

So $\vect E'$ is radial, pointing away from the $z'$ axis for $bv > 0$ (toward it for $bv < 0$ ), with the magnitude $|bv|/r$ falling inversely with the distance from the axis: the shape of the field of a uniformly charged straight thread along the $z'$ axis.

Check

Units: the given field has the magnitude $B = |b|/r$ , so $b$ is in $\text{T}\cdot\text{m}$ and $bv/r$ is in $\text{T}\cdot\text{m/s} = \text{V/m}$ , a field strength. As $v \to 0$ , $\vect E' \to 0$ , as it must. Gauss’s law over a cylinder of radius $r$ and length $\ell$ about the axis gives $E'\cdot 2\pi r\ell = \lambda'\ell/\varepsilon_0$ , so the equivalent thread carries $\lambda' = 2\pi\varepsilon_0 bv$ , independent of $r$ , as a thread’s charge must be.

Remark

The given $\vect B$ is the field of a straight wire on the $z$ axis carrying the current $I = -2\pi b/\mu_0$ along $\vect k$ . A wire that is neutral in $K$ therefore looks charged in $K'$ : putting this $I$ into the thread charge of the check, with $\varepsilon_0\mu_0 = 1/c^2$ , gives $\lambda' = -Iv/c^2$ . That is the first-order relativistic charge of a current-carrying wire seen by an observer moving along it, the same effect that, read the other way, makes magnetism the relativistic companion of electrostatics.