enter image description here

Let’s say we have two conducting rail tracks bridged together by a conducting rod of length d. We drop the rod from height H, and it moves into a magnetic field coming upward (out of the page).

According to Faraday’s law, the induced voltage in the circuit is Bdv volts, where v is the instantaneous velocity of the rod to the right. Considering that the resistor has the entire voltage across it, it appears the bridging rod has voltage Bdv across its length d.

The current moves clockwise as a result of the incoming magnetic field, and the Lorentz force pushes the bridging rod to the left as a result of the current.

However, in addition to the downward velocity of positive charges in the rod (with the flow of the induced current), the charges in the rod also move v to the right. How should one consider the Lorentz force on the charges moving to the right? This force would appear to polarize the rod by pushing its positive charges downward.

How should one consider the Lorentz force due to the rod’s horizontal velocity to the right? How would it affect the voltage across the rod across its distance d?


1 Answer 1


You have already accounted for the Lorentz force on the charges moving to the right.
That is how the voltage across the moving rod is set up.

First just consider the rod moving without the resistor being present so no induced current flows.

The Lorentz force on positive charges moving to the right has a magnitude of $Bqv$ and acts downwards.
This results in positive charges moving downwards and at the same time there is a lack of positive charges at the top with is now negative.
That redistribution of charge sets up an electric field $E$ and produces an upward force on the positive charges of magnitude $qE$.

Eventually the magnitude of the downward Lorentz force is equal to the upward electric field force, ie $Bqv=qE$.
The electric field $E$ is equal to the voltage across the rod, $\mathcal E$, divided by the length of the rod, $d$, so, $Bqv= q\frac {\mathcal E} d\Rightarrow \mathcal E = Bdv$, the very same emf you derived via Faraday's law.

Connecting the resistor to the circuit means that now an induced current can flow as the electric circuit is complete.

  • $\begingroup$ This is absolutely fascinating. The Lorentz force appears to polarize the rod into being a battery, with the most positive end being the bottom and the most negative end being the top. It seems Faraday’s law is a convenient shortcut to the manual derivation of “qvB=qE” and so on. $\endgroup$ Jul 20, 2023 at 15:03
  • $\begingroup$ It seems from here that Faraday's law and Lorentz force are not unrelated. However, Lorentz force law is understood to be independent of Maxwell's equations. Could there be some way that actually the Lorentz force expression is derived from Maxwell's equations? $\endgroup$
    – timur
    Oct 29, 2023 at 1:01

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.