Suppose we have a charge moving at velocity $\mathbf{v}$ in the same plane of a square wire.

If I sit in a reference frame where the square wire is still, since the charge is moving with velocity $\textbf{v}$ in this coordinate system, I will see an induced current in the wire.

$$\textbf{B} = \frac{\textbf{v}}{c^2} \times \textbf{E} $$ $$ \frac{d\phi_B}{dt} \neq 0 $$

Now, what If I choose a reference system where the charged particle is at its origin?

According to this frame, since the the electric charge(and its electric field) is static, $\text{rot}\,\textbf{E}$ will be zero.

$$\nabla \times\mathbf{E} = 0$$

But this means that there is no induced current.
Are my assumptions right? If not, how should I estimate the induced current in a reference system bound with a moving charge at its origin?

  • $\begingroup$ can you explain your notations.... I don't use the same ,I think. $\endgroup$ – ABC May 25 '13 at 11:57
  • $\begingroup$ $\mathbf{rot}$ means $\mathbf{\nabla\times}$ or $\mathbf{curl}$. But what is $\phi$? And what is a square wire? $\endgroup$ – fffred May 26 '13 at 5:09
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    $\begingroup$ Moving charge doesn't induce current in a conducting loop. You need a moving magnet to do that. So I think your question is wrong. $\endgroup$ – QuantumDot May 26 '13 at 5:13
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    $\begingroup$ @QuantumDot "Moving charge" is like a variable current and a variable current produces a variable magnetic field. So IMH it induces current in a conducting loop. $\endgroup$ – antonjs May 27 '13 at 6:43
  • $\begingroup$ I think, the point is that the particle is not moving perpendicular to the square's plane. $\endgroup$ – Ali May 27 '13 at 12:10

There's no contradiction (there never is in relativity problems...): a transient current flows in the square as it passes by the charge.

You're correct that the conservative electric field in the charge's rest frame means that the line integral around a square in that frame is zero, but that's not the appropriate integral to use for a moving wire.

The induced current in the square is determined by the line integral of E around the square at a fixed time in the square's rest frame (the "primed" frame). To calculate this quantity in the charge's rest frame (the "unprimed" frame) the integrand $\boldsymbol{E' \cdot} d\boldsymbol{l'}$ must be transformed:

  • the electro-magnetic field transforms to the static coulomb field of the charge.
  • the path infinitesimals transform according to $dx = \gamma \, dx'$ and $dy = dy'. (t'$ is fixed.)

It's that $\gamma$ that changes the character of the integral to be calculated and gives a non-zero result. This effect is a typically small relativistic correction, but even a small correction matters when the non-relativistic result is 0.


In these type of questions, the following effects usually exist, but are usually missed:

  • The electric potential(and ergo the electric field), that corresponding points of the wire feel, varies. This effect will polarize the wire which is always negligible, and we are going to ignore it as well.
  • in a moving wire, the electric charges will feel the Magnetic Force(which doesn't exist in this case). $$ q \left( \mathbf v \times \mathbf B \right) $$
  • Special Relativity will affect these questions as well. The length contraction is one effect; the other is the relativistic Electromotive Force (relativistic Ohm's law as well). In the latter we will have a term including $ \mathbf v \left(\mathbf v . \mathbf E \right)$(more on this below).

In this case the first two will not have any effect. On the other hand, if you calculate the last one precisely, it should give the same result as the one acquired in the other reference frame.

To calculate the Emf in a moving wire one has to calculate an integral proportional to $$\oint \left(\gamma \left( \mathbf{E} + \mathbf{v} \times \mathbf{B} \right ) - \left ({\gamma-1} \right ) ( \mathbf{E} \cdot \mathbf{\hat{v}} ) \mathbf{\hat{v}} \right).\mathbf{dl}$$, where the integral goes around the wire in an instant of time. This comes from the relativistic field transformations.

Now looking at this formula and getting back to our problem, since $\nabla \times \mathbf{E}=0$ and $\mathbf{B}=0$ the first two terms of the integral are zero; unlike the last term which will give non-zero answer.

  • $\begingroup$ I will slightly update my comment according to the update in the question. $\endgroup$ – Ali May 27 '13 at 12:24

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