We all know the expression for the potential difference across a conducting rod moving through a region with a uniform magnetic field:

motional emf $\epsilon = B l v$

Now my question is different. Is it possible to find the charge separation which the motion has induced?

A (maybe) naïve attempt which seems obvious to me is to find the capacitance $C$ of the rod.

Once we find the capacitance $C$, we already know the potential difference $\epsilon$ across it.

So we can use:

$C = \frac QV$

However, I have no idea how to proceed further. How can you define capacitance for what is essentially a battery?

Any answers or insight will be highly appreciated.

  • $\begingroup$ I don't see any charge buildup. If the rod on the right is moving to the right at a constant velocity v, the rod has a constant length l, and the magnetic field is uniformly B, then there will be a current I=Blv/R (where R is the resistance of the rod), if you remove the rod there is nothing to have velocity v. Are you removing a small section of the rod to replace with a capacitor? $\endgroup$
    – Timaeus
    Commented Jan 11, 2015 at 20:56
  • $\begingroup$ You should provide us with a circuit representation of your device. Usually, this is modeled with resistors, EMF and eventually inductors. No capacitor involved. $\endgroup$
    – TZDZ
    Commented Jan 12, 2015 at 9:07
  • $\begingroup$ I am sorry for the confusion. There was a slight conceptual doubt involved. There will of course, be no charge accumulation if you have a closed circuit. I have made edits to the original question. $\endgroup$ Commented Jan 12, 2015 at 11:07
  • $\begingroup$ Lenz' law requires a loop and a magnetic flux variation though it. Where would be the magnetic flux variation in this case ? $\endgroup$
    – TZDZ
    Commented Jan 13, 2015 at 8:06
  • 1
    $\begingroup$ Look at Hall the Hall effect. It should give you part of the answer. $\endgroup$
    – WalyKu
    Commented Jan 16, 2015 at 11:08

3 Answers 3


Capacitance is a purely geometric constant. Let $\gamma(\lambda)$ be a curve that traverses the capacitor across the voltage drop (say from hi to low voltage), with end points at $\lambda_1$ and $\lambda_2$, and let $A(\lambda)$ be the section of the capacitor as you move along $\gamma$. Then the capacitance is given by $$\frac1C=\int\limits_{\lambda_1}^{\lambda_2}\frac\epsilon{A(\lambda)}\text d\lambda.$$ The picture above seems to show a constant section $A$ across the voltage drop, so assuming $\epsilon$ to be homogeneous, $$\frac1C = \epsilon\frac lA,$$ whence $$C = \epsilon\frac Al,$$ i.e. a parallel plate capacitor. The charge is then $$Q = \epsilon ABv.$$

  • $\begingroup$ So, sir, you're telling me that the rod is equivalent to placing two plates at the same distance of separation from each other? $\endgroup$ Commented Jan 16, 2015 at 13:23
  • $\begingroup$ What the above computation shows is that your system with the moving rod and the magnetic field is equivalent to a DC circuit with a generator of f.e.m. $Blv$ connected to a parallel plate capacitor of separation $l$ and surface $A$. $\endgroup$
    – Phoenix87
    Commented Jan 16, 2015 at 14:19
  • $\begingroup$ Please do not type out the entire working of the solution. It is against site policy : meta.physics.stackexchange.com/a/715/57075 $\endgroup$
    – Gaurav
    Commented Jan 16, 2015 at 16:09
  • 1
    $\begingroup$ Well this doesn't quite seem a homework question tbh. $\endgroup$
    – Phoenix87
    Commented Jan 16, 2015 at 16:11
  • $\begingroup$ In the first paragraph: "This includes not just questions from actual homework assignments, but also self-study problems, puzzles, etc." This is a homework-related question, so the guidline applies. $\endgroup$
    – Gaurav
    Commented Jan 16, 2015 at 16:24

The problem is exactly equivalent to one with a stationary rod immersed in an external electrostatic field $\overrightarrow{E}=-\overrightarrow{v}\times\overrightarrow{B}$ (this can be demonstrated by considering an observer inside the rod, seeing charges experiment a null Lorentz force after electrostatic equilibrium is reached). Therefore, if the rod is a cylindrical conductor, a superficial charge density $\rho_s$ will be induced at the tips of the rod. However, $\rho_s$ is not uniform along the circular tips, and in general, must be obtained by computational methods such as the Method of Moments. Of course, you could also compute the capacitance $C$ between the two circular parallel plates in order to obtain $Q$, but this has not simple analytical solution neither and aproximations such as $C=\varepsilon\frac{A}{\ell}$ cannot be made since plates are separated by a distance longer than their diameter.

Consequently, you can only estimate the accumulated charge on the rod tips accurately, by computational methods.


You need to calculate that from point where, limitations are in copper and they would be density off current that wire could withstand in storage environment, such as; metal storages or direct storage of the magnetic field in cupper. To much density reproduce overheating. That would be a copper limitation capacity. Density isn't same for every diameter and diamagnetic material.

So, because limitation depends from density off the current that he can withstand, that's copper capacity, use: length, Siemens and diameter, stand for all diamagnets in magnetic field- that's amount off eletricity from battery point of view.

And it depend from interval change that have created that amount, where and how that change is stored.

 Difference would be only in way off creating it:

  1. Where - in iron, nickel... ferromagnetic environment or directly in copper.
  2. How - from mechanical point of view machine need to have standard construction possibility to make 50-60 repetition in sec, the amount that need to be changed in Hz. Note that it's not same to make AC and DC in ferromagnetic environment, magnetic field density or limitations that ferromagnetic environment could withstand depends off the material and rate off the change-Hz. Also magnetization capacity off iron elements is 2,2 to max 3 Tesla, depending off material and environment.

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