A circuit with resistance $R$ is given. The emf $E$ in the circuit is produced by a conducting rod moving with constant velocity in a uniform perpendicular magnetic field. The rod is sliding on metallic rails. Constant velocity is due to force $F_{\text{ext}}$ on the rod. Net magnetic force one is perpendicular to the net velocity (the sum of drift velocity and rod velocity).

  1. How does the component of magnetic force along the rod produce the same work as that of $F_{\text{ext}}$ on the rod?
  2. Why does the current need to flow since it was already static condition $Eq=qvB$?
  • $\begingroup$ Are you familiar with basic vector algebra: vector and scalar products ? $\endgroup$ Aug 13, 2021 at 19:58
  • $\begingroup$ Yes, I know that. $\endgroup$ Aug 14, 2021 at 0:16

1 Answer 1

  1. The 'motor effect' force on the rod is of magnitude $$F=BIl$$ It acts in the opposite direction to that of the velocity, $v_\text{rod}$ with which we are pushing the rod along the rails. So the mechanical power input is $$P_\text{mech}=BIlv_\text{rod}$$ The emf induced in the rod as it cuts flux is easily shown from Faraday's law to be $$\mathscr E=Blv_\text{rod}$$ Therefore the electrical power provided is $$P_\text{elec}=\mathscr E I=Blv_\text{rod}I$$ We see that the electrical power provided is equal to the mechanical power (work per second) put in.

This treatment relies on equations that can be derived from a single vector equation: the magnetic Lorentz force law. The appendix establishes the result using just that law, in other words it gives a more fundamental treatment.

  1. You have quoted the equation $$Eq=qvB$$ Here $E$ is the magnitude of the electric field strength (not the emf !) in the rod due to redistribution of charge. This equation applies exactly if $R=\infty$, that is if the rails are not connected to anything and no current flows once the redistribution has taken place.

If $R$ is finite $E$ will be less than $vB$ (in fact, $E=\left(1-\frac{R_\text{rod}}{R} \right)vB$ in which $R$ is the whole circuit resistance) so there will be a force on an electron in the rod of magnitude $q(Bv -E)$ that will be balanced by the mean resistive force due to collisions, as the free electron moves through the rod.

Appendix: working directly with the magnetic Lorentz force on a free electron

The magnetic force on a free electron in the rod is $$\vec F =q \vec v \times \vec B\ \ =\ -e (\vec v_{\text{drift}}+\vec v_{\text{wire}}) \times \vec B\ \ =\ -e \vec v_{\text{drift}}\times \vec B-e\vec v_{\text{rod}} \times \vec B$$ in which $\vec v_{\text{drift}}$ is the velocity of the electron through the rod and $\vec v_{\text{wire}}$ is the velocity of the rod itself as it moves along the rails.

$-e \vec v_{\text{drift}}\times \vec B$ is a force at right angles to $\vec B$ and to $\vec v_{\text{drift}}$ and therefore to the rod. This is the force (per free electron) that we have to oppose with an external force if we are to move the rod along the rails at a steady speed when free electrons are moving through it. So we may write $$\vec F_{\text{ext}}=e \vec v_{\text{drift}}\times \vec B$$ The work we have to do per unit time per free electron pushing the rod is therefore $$ P_{\text{ext}}=e (\vec v_{\text{drift}}\times \vec B).\vec v_{\text{rod}}$$ Because $\vec v_{\text{drift}}$ and $\vec B$ are at right angles to each other, and $\vec v_{\text{drift}}\times \vec B$ is in the same direction as $v_\text{rod}$, $P_\text{ext}$ can be written simply as a product of magnitudes, that is $$ P_{\text{ext}}=e\ v_{\text{drift}}\ v_{\text{rod}}\ B$$ $-e\vec v_{\text{rod}} \times \vec B$ (see first equation), is the force component acting along the wire on a free electron, that is the force that gives rise to the emf. So we may write $$\vec F_{\text{emf}}=-e \vec v_{\text{rod}}\times \vec B$$ The work done per second per free electron by this force is therefore $$ P_{\text{emf}}=-e (\vec v_{\text{rod}}\times \vec B).\vec v_{\text{drift}}$$ Because $\vec v_{\text{rod}}$ and $\vec B$ are at right angles to each other, and $-\vec v_{\text{rod}}\times \vec B$ is in the same direction as $v_\text{drift}$, $P_\text{emf}$ can be written simply as a product of magnitudes, that is $$ P_{\text{emf}}=e\ v_{\text{rod}}\ v_{\text{drift}}\ B$$ We see that the work per unit time urging the free electron through the rod is equal to the mechanical power input per free electron that has to be put into pushing the rod along the rails!


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