Irodov 3.365: A Moving Observer Sees Circles – Amit & Deepali Sinha Foundation

Problem Irodov 3.365

In an inertial frame $K$ there is no magnetic field, only the electric field $\vect E = a\,(x\vect i + y\vect j)/(x^2 + y^2)$ , where $a$ is a constant and $\vect i$ , $\vect j$ , $\vect k$ are the unit vectors of the $x$ , $y$ , $z$ axes. A second inertial frame $K'$ moves relative to $K$ with the constant velocity $\vect v = v\vect k$ , small compared with the speed of light, and its $z'$ axis coincides with the $z$ axis. Find the magnetic induction $\vect B'$ in $K'$ , and describe the shape of its field lines.

Oblique view of the frame K with the field E, radial from the z axis in planes normal to it (drawn for a > 0), and the velocity v of the frame K' along z. — Oblique view of the frame $K$ with the field $\vect E$ , radial from the $z$ axis in planes normal to it (drawn for $a > 0$ ), and the velocity $\vect v$ of the frame $K'$ along $z$ .

Given

$\vect E = a\,(x\vect i + y\vect j)/(x^2+y^2)$ , $\vect B = 0$ in $K$
$\vect v = v\vect k$ , $v \ll c$
$z' = z$

Find

\vect B'

K'

· the shape of its lines

Solution

Step 1The field transformation at low speed

For a frame moving with velocity $\vect v \ll c$ , the fields in $K'$ follow from those in $K$ by the non-relativistic transformation:

\vect E' = \vect E + \vect v\times\vect B, \qquad \vect B' = \vect B - \frac{\vect v\times\vect E}{c^2} \tag{1}

With $\vect B = 0$ the second relation of (1) leaves only the cross term:

\vect B' = -\frac{\vect v\times\vect E}{c^2} \tag{2}

Step 2The cross product

With $\vect v = v\vect k$ , the products $\vect k\times\vect i = \vect j$ and $\vect k\times\vect j = -\vect i$ give

\vect v\times\vect E = \frac{a v}{x^2+y^2}\,\bigl(x\,\vect k\times\vect i + y\,\vect k\times\vect j\bigr) = \frac{a v\,(x\vect j - y\vect i)}{x^2+y^2}

Putting this into (2):

\vect B' = \frac{a v\,(y\vect i - x\vect j)}{c^2\,(x^2+y^2)} \tag{3}

Step 3The vector form and the shape of the lines

Let $\vect r = x\vect i + y\vect j$ be the vector from the $z'$ axis to the field point, at the distance $r$ from that axis. The bracket in (3) is a cross product with $\vect k$ :

\vect r\times\vect k = x\,\vect i\times\vect k + y\,\vect j\times\vect k = y\vect i - x\vect j

Multiplying by $v$ turns $\vect k$ into $\vect v$ , and $x^2 + y^2 = r^2$ , so (3) becomes

\vect B' = \frac{a\,\vect r\times\vect v}{c^2 r^2}

(4)

with the magnitude $B' = a v/(c^2 r)$ . The vector $\vect r\times\vect v$ is normal both to the $z'$ axis and to $\vect r$ , so $\vect B'$ lies in planes normal to the axis and is tangent to circles about it, with a magnitude that is the same all round each circle:

the lines of

\vect B'

are circles centred on the

z'

axis

For $a v > 0$ they run clockwise seen from the $+z$ side; reversing $a$ or $v$ reverses them.

Check

Units: $a$ is a field times a length, in V, so $a v/(c^2 r)$ is $\text{V}\cdot(\text{m/s})/(\text{m}^2\text{/s}^2\cdot\text{m}) = \text{V}\cdot\text{s/m}^2 = \text{T}$ . Independent route: the field is that of a thread on the $z$ axis carrying the charge $\lambda = 2\pi\varepsilon_0 a$ per unit length. In $K'$ the thread moves with $-\vect v$ , a current $I = \lambda v$ along $-z$ , whose field is $\mu_0 I/(2\pi r) = \mu_0\varepsilon_0 a v/r = a v/(c^2 r)$ , circling clockwise seen from $+z$ by the right-hand rule, as in (4).

Remark

The observer in $K'$ calls this field magnetic, the observer in $K$ sees none: the circles of the Biot and Savart field of a current are the electrostatic field of the same charges seen by someone moving past them. The two fields are parts of one object, and how it splits into $\vect E$ and $\vect B$ depends on the frame.