Special theory of relativity > The transformation of electric and magnetic fields > Problem 14.3.24

Statement

$14.3.24.$

The magnetic moment of a long flat solenoid with current is equal to M. What
electric moment will arise in this solenoid when it moves laterally at a speed
v parallel to flat surfaces?

Solution

We start analysing Rest frame S'

In the system S' the solenoid is at rest:

The solenoid is "long and flat": its turns define a plane; the magnetic moment $\mathbf{M}$
is perpendicular to that plane (normal direction). Let us take
take $\mathbf{M}' =M\, \hat{\mathbf{z}}$

There is no electric moment
$\mathbf{p}' = 0$.

In the laboratory system S, the solenoid moves with velocity $\mathbf{v} = v\, \hat{\mathbf{x}}$
parallel to the flat surfaces (hence, perpendicular to $\mathbf{M})$

Under a boost with velocity
$ \mathbf{v}$
the electric dipole moment $\mathbf{p}$ and magnetic dipole moment $\mathbf{m}$ transform according to:

$\begin{aligned}
\mathbf{p} &= \gamma\left( \mathbf{p}' + \frac{\mathbf{v}}{c} \times \mathbf{m}' \right) - \frac{\gamma^2}{\gamma+1},\frac{\mathbf{v}}{c}\left( \frac{\mathbf{v}}{c} \cdot \mathbf{p}' \right),\
\mathbf{m} &= \gamma\left( \mathbf{m}' - \frac{\mathbf{v}}{c} \times \mathbf{p}' \right) - \frac{\gamma^2}{\gamma+1},\frac{\mathbf{v}}{c}\left( \frac{\mathbf{v}}{c} \cdot \mathbf{m}' \right).
\end{aligned}$

In our case:$ \mathbf{p}' = 0$ and $\mathbf{v} \cdot \mathbf{m}' = 0 $(because they are perpendicular)
Therefore:

$\boxed{\mathbf{p} = \frac{\gamma}{c},\mathbf{v} \times \mathbf{M}}, \qquad
\mathbf{m} = \gamma,\mathbf{M}$

The magnetic moment in S increases by a factor of $\gamma$
and an electric moment appears perpendicular to $\mathbf{v}$ and to $\mathbf{M}$

For $ v \ll c (\gamma \approx 1)$

In CGS: $\displaystyle \mathbf{p} \approx \frac{1}{c}\, \mathbf{v} \times \mathbf{M}$.

In SI: the factor is
$1/c^2$
$\displaystyle \mathbf{p} = \frac{1}{c^2}\, \mathbf{v} \times \mathbf{M} $

(with $\mathbf{M}$ in $A·m²$)

Magnitude (if $\mathbf{v} \perp \mathbf{M})$:
$ p = \frac{v}{c}\, M (CGS)$
or $ p = \frac{v}{c^2}\, M$(SI)

Direction:
perpendicular to$ \mathbf{v} $and to $\mathbf{M}$,
that is, parallel to the flat surfaces but perpendicular to the motion.

Answer

$\boxed{\mathbf{p} = \frac{\gamma}{c},\mathbf{v} \times \mathbf{M} ;;(\text{CGS}), \qquad
\mathbf{p} = \frac{\gamma}{c^2},\mathbf{v} \times \mathbf{M} ;;(\text{SI})}$

In the low-velocity limit $(\gamma \to 1)$:

$p \approx \frac{v}{c}M ;;(\text{CGS}), \qquad
p \approx \frac{v}{c^2}M ;;(\text{SI})$

The electric moment appears because, when in motion, the currents that produce the magnetic moment are no longer simultaneous in the new system, generating an effective dipolar charge distribution