A standard problem in elementary EM goes something like this:

An infinite straight wire conducts a stationary current $I$. A conducting rod, perpendicular to the wire, moves with constant velocity $v$, parallel to the wire; its length is $l$ and its minimum distance from the wire is $d$. Determine the potential difference $\Delta V$ between the ends of the rod.

Now, the reasoning behind the problem should be simple. The wire produces a magnetic field $B_{wire}$, which causes the charges inside the rod to move due to the Lorentz force, until the static electric field generated by the separation of charge inside the rod balances the magnetic Lorentz force: $E=-(v\wedge B_{wire})$.

My doubt is this: are we sure that the total electric field is $B_{wire}$, the one due to the current? I know that moving charges produce magnetic fields; the rod overall is neutral, but by what was said above, there will be a charge density on the rod that is not null everywhere, and since the rod is moving, another magnetic field $B_{rod}$ should be produced.

Considering that the negatively charged electrons of the rod pile up at one end, leaving an equal amount of positive charge piled up at the other end, couldn't we model the rod as a sort of electric dipole? And won't a moving electric dipole generate a magnetic field which should be added to $B_{wire}$?

Another consideration that could be made is that the electric field inside the rod is "moving" together with the rod itself, making $\frac{\partial E}{\partial t}\neq 0$, another reason why the total magnetic field shouldn't be just $B_{wire}$!

In conclusion, the question is this: is there, or is there not, and in both cases why, a magnetic field $B_{rod}$ that ought to be added to $B_{wire}$ in this problem?

  • $\begingroup$ "a magnetic field $B_{rod}$ that ought to be added to $B_{wire}$ in this problem" - an ought implies an end. What end will adding $B_{rod}$ to the problem serve? $\endgroup$ Mar 30, 2018 at 11:52
  • $\begingroup$ @AlfredCentauri well, finding $\Delta V$ requires the total magnetic field, doesn't it? But besides that, I just want to know if $B_{rod}\neq 0$ or not. $\endgroup$
    – Adrian
    Mar 30, 2018 at 12:25
  • $\begingroup$ I still have a problem with the use of "stationary current" Stationary current? How is this something we still find in books? What, "constant" is too droll? $\endgroup$ Mar 30, 2018 at 15:12
  • $\begingroup$ Moving charge generate magnetic field. So the moving rod does generates magnetic field which can be calculated using point charge Biot-Savart law. The net magnetic field will be the superposition of the two when the rod moves in steady state. $\endgroup$
    – user115350
    Mar 30, 2018 at 16:00

3 Answers 3


You are right, the rod will have its own magnetic field. This is usually neglected to simplify the discussion. It is possible to do so because effect of the neglected magnetic field is much smaller than effect of the external magnetic field.

  • $\begingroup$ Antenna designers would have a problem with this statement :-) $\endgroup$ Mar 30, 2018 at 15:12
  • $\begingroup$ @MauryMarkowitz Interesting. How so? $\endgroup$
    – Adrian
    Mar 30, 2018 at 15:27
  • $\begingroup$ Antennas typically consist of a series of metal rods arranged near each other. Although they are not moving relative to each other, the field is moving relative to them. The induced field causes the signal to be "rebroadcast", and by positioning them property you can get it to add up at a selected point. So, no ignoring the field in that case! $\endgroup$ Mar 30, 2018 at 15:33
  • $\begingroup$ @MauryMarkowitz the question was not about antennas, but about rod slowly moving with respect to a current carrying wire. In this case, the magnetic field due to motion of the rod will be much weaker than the field that induced the charge separation. $\endgroup$ Mar 30, 2018 at 18:45

For simplicity, model the charge distribution of the rod as two point charges of equal and opposite charge $Q$ separated by a distance $l$.

Stipulate that these two charges are aligned on the $y$-axis and move with constant speed $v$ parallel to the $x$-axis.

There is both an electric force and magnetic force on each charge due to the other charge. It's straightforward to show that the magnetic force is proportional to the electric force

$$\vec F_B = -\frac{v^2}{c^2}\vec F_E$$

and that the total force is attractive.

In the problem you describe, the magnetic force $F_{B_{wire}}$ due to the magnetic field of the current carrying wire balances the total force above.

For the typical non-relativistic case $(v \ll c)$ and so

$$F_{B_{wire}} \ggg F_B$$

which justifies not including $F_B$ in this type of problem.


The moving positive and negative charges induced at the ends of the rod constitute currents which should produce magnetic fields which are superimposed on the magnetic field of the current carrying wire. These charges correspond to a moving electrical dipole which experiences a distinct force in the magnetic field of the wire. Here is a related question concerning the magnetic force on a moving electrical dipole.


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