I'm trying to build a homemade rail accelerator. I understand that the force on a wire is proportional to the cross product of the magnetic field vector and the current vector. I asked my professor how this didn't violate the conservation of energy, seeing as I can subject the wire to a large permanent magnet, which will increase the force. I asked if perhaps the magnetic field increased the resistance in the wire. He replied that it would not increase the resistance in the wire, but that the wire's movement through the field will cause "back flow", which I guess is technically "impedance" not "resistance". And I take it the backflow is proportional to the velocity of the wire through the magnetic field.

Now, if I've got this summary of the science right, my question is what voltage is required to overcome that backflow? Can you tell me the voltage I need to produce I amps of current through a wire going V meters per second through a magnetic field of t teslas?

  • $\begingroup$ You can subject wire or any other body to howsoever large a force. This does not violate conservation of energy. If you got the body to acquire arbitrarily large kinetic energy then it would be interesting, but this is exactly what does not happen. Trams and electric trains show that you need to supply electric energy to get an increase of kinetic energy of the vehicle. So violation of conservation of energy happens; only conversion from electric energy to mechanical energy and heat. $\endgroup$ Dec 9, 2023 at 11:50
  • $\begingroup$ I understand that force is not energy. But forces can cause kinetic energy, and that's the point. $\endgroup$ Dec 10, 2023 at 5:37
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    $\begingroup$ I'm not sure what they meant with "backflow", but there will be motional EMF acting on the current, which can be also called back-emf, since it acts against the EMF of the source exciting the current, and thus current won't get arbitrarily high, not even if the wire has zero resistance. $\endgroup$ Dec 10, 2023 at 22:28

1 Answer 1


If we imagine a circuit made of the moving wire, rails and a voltage source, we can apply to it Kirchhoff's 2nd circuital law, which states that sum of all EMFs acting on the current in a conductive loop equals sum of all terms $R_kI_k$ due to all conductive elements in the current loop.

Thus if internal resistance of the voltage source is $R_i$, resistance of the rails $R_r$, resistance of the piece of wire $R_w$, the Kirchhoff law states that current $I$ obeys the equation

$$ \mathscr{E}_0 + \mathscr{E}_{motional} = (R_i + R_r + R_w)I $$ where $\mathscr{E}_0$ is EMF of the voltage source, $\mathscr{E}_{motional} = -VBL$ is the motional EMF due to motion of wire of length $L$ with speed $V$ in magnetic field $B$ (this emf acts against the emf of the voltage source, thus the minus sign).

To answer your question, the EMF (of the voltage source) needed to have current $I$ while the wire is moving with speed $V$ is

$$ \mathscr{E}_0 = (R_i + R_r + R_w)I + VBL\tag{*} $$

Thus the needed voltage source depends not only on current $I$ we want to achieve, but also on velocity $V$ it currently has. Initially, $V=0$ and

$$ \mathscr{E}_0 = (R_i + R_r + R_w)I. $$ With increasing velocity $V$, the required EMF of the source is getting higher and higher. In reality, simple voltage source does not increase its EMF from its initial value (for zero current), it is more usual that EMF drops while current rises. The above formula (*) means that with increasing velocity, $I$ will decrease. Macroscopic magnetic force on the wire is proportional to $I$, so the wire element will still be accelerated, but its acceleration will decrease in time.


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