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Why does an EMF generate across a rod moving linearly in a region of magnetic field constant with space and time?

While proving it, we considered the rod to be connected to a conducting rectangular loop (whose one side, say right side, is the rod itself with the top and bottom sides extending to infinity towards right). The derivation uses the fact that as we slide the rod towards right, the magnitude of magnetic flux passing through the area of the loop increases (since the area of the loop increases), to oppose which, a current sets up in the loop such that the direction of the magnetic field it would produce (inside the loop) due to its flow, will oppose the change in the increasing magnetic field.

But now how can we extrapolate this derivation to cases in which the loop (the other 3 sides) is entirely missing? If there is no loop, then there is no region through which the flux passing change, upon moving a rod linearly in a magnetic field constant with space and time. Then why?

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  • $\begingroup$ It's not quite clear to me why your derivation requires a current 'round the loop to explain the emf due to the moving rod. By inserting a resistor with resistance $R$ in the loop, you can make the current arbitrarily small by letting $R\rightarrow\infty$, correct? $\endgroup$ – Alfred Centauri Aug 24 at 22:01
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We can't really use the flux rule validated for cases 1) where conductors are stationary and magnetic field changes in time to cases 2) where conductors move in static magnetic field, without some justification of why it should work also in 2). These are two different kinds of electromagnetic induction. However, it turns out, both experimentally (as Faraday found), and theoretically as explained below, that the formula is the same and the flux rule works in both cases, even though the mechanism is different.

In case 2) of motional EMF, there are actually further two distinct cases.

2a) Let the rod move with velocity $\mathbf v$ in external magnetic field $\mathbf B$, but it is not connected to any circuit. The motional EMF of magnitude $Bvl$ is present in the sense there is magnetic force acting on the charge carriers of magnitude $\mathbf v \times \mathbf B$ per unit charge and hypothetical work of this force along the rod of length $l$, if charge was moved along the rod, would be $Bvl$. This hypothetical work is called motional EMF. As you can see, in this case of sole rod with no current flowing, motional EMF is somewhat technical-theoretical concept; while EMF is present, there is actually no work being done, because magnetic force on charged particle with velocity $\mathbf u$ inside the rod is given by the Lorentz formula $q\mathbf u\times \mathbf B$, so the force is perpendicular to velocity of the charged particle and thus does no work and no net work is being done on the mobile charges (so EMF, despite its name, is not very electromotive de facto). Also there is no induced current so charge does not really move along the rod either. The only thing that happens is that after extremely short time after start of the motion, the charges on the rod surface shift a little so the rod gets polarized until the force of electric field of the surface charges acting on any mobile charge carrier inside the rod cancels out the external magnetic force acting on that same mobile charge carrier. This electric field is quantified by voltage of the same magnitude as the motional EMF has, but the directions of force are opposite, so they cancel each other and no current is flowing. One could perhaps say that in this case, the EMF is present, but it is "latent", purely due to magnetic forces which do nothing but maintain electrically polarized rod with voltage $Bvl$ between the ends of the rod.

2b) The moment the rod is connected to a closed circuit, things change very much. Some excess charge on the rod surface is sucked away along wires, there is no longer full-strength electric field needed to counteract the magnetic forces on the mobile charges. Mobile charge carriers inside the rod move a little along the rod (due to magnetic force being stronger than electric force) and current starts flowing. The external magnetic forces now do more; as the particle has some component of velocity along the rod, the magnetic forces have component perpendicular to the rod and so induce small shift in distribution of the mobile charge in the rod in direction perpendicular to the rod. This shift causes new interaction between the rod and the mobile charge carriers: the charges push on the walls of the rod as they are blocked by the rod walls from jumping out.

A different kind of EMF appears, one which is due to forces from the rod itself and which actually does some work on the mobile charges, pushing them against ohmic resistance in the direction of current. This electromotive force is sum of those internal forces between the rod and the mobile charge carriers that arise as reaction to abovementioned shift of the mobile charge carriers. This "real working EMF" still has magnitude $Bvl$ and this is simply the work of the rod done as it pushes unit charge along the rod of length $l$ from one end to another.

As the rod does work on the mobile charges, it loses kinetic energy, so it either slows down or there has to be some other agent pushing the rod and maintaining or increasing its speed and thus being the supply of energy for work of EMF on the current.

The EMF magnitude is the same in both cases 2a),2b), $Bvl$. This is equal to rate of change of magnetic flux passing through any imaginary closed loop involving the rod (this is easy to see by drawing such loop and writing down how much area changes per unit time), so Faraday's formulation of law of electromagnetic induction in terms of magnetix flux is valid also in this case of motional EMF. But as described above, the mechanism in the case motional EMF is completely different from that where magnetic field changes in time and is associated with induced electric field, which is the force behind EMF in that case.

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    $\begingroup$ @Ján Lalinský I understood your point 2a almost completely except the last few lines, but only half of point 2b. The other half of 2b is quite unclear to get. $\endgroup$ – user233565 Aug 25 at 4:48
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    $\begingroup$ Please explain these lines from 2a: "This electric field is quantified by voltage of same magnitude that the motional EMF has so no current is flowing. One could almost say that there is no "real" EMF present, only latent EMF which is mere magnetic forces which do nothing but maintain electrically polarized rod with voltage Bvl between the ends of the rod." By the first three lines, do you mean that initially the magnetic field created an EMF in one direction but when the charges start accumulating at corners according to that EMF, then they create an electric field which create an EMF of.... $\endgroup$ – user233565 Aug 25 at 4:55
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    $\begingroup$ .... equal magnitude but opposite in direction? Also, in the latent EMF line, I think you would like to edit it so that it seems more as a voltage than a force. $\endgroup$ – user233565 Aug 25 at 4:57
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    $\begingroup$ What do you mean by this in 2b: "The external magnetic forces now do more; they induce small shift in distribution of the mobile charge in the rod in direction perpendicular to the rod and thus cause new interaction between the rod and the mobile charge carriers (the charges are blocked by the rod walls from jumping out). A different kind of EMF appears, one which is due to forces from the rod itself and one which actually does some work on the mobile charges, pushing them against ohmic resistance in the direction of current." Please explain clearly. $\endgroup$ – user233565 Aug 25 at 5:01
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    $\begingroup$ Also, you again might want to reconsider that EMF is a not a force but a potential difference due to that force in these lines in 2b: "This electromotive force is sum of those internal forces between the rod and the mobile charge carriers that arise as reaction to abovementioned shift of the mobile charge carriers" $\endgroup$ – user233565 Aug 25 at 5:05
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Just because the rod isn't connected to a loop doesn't mean you can't view it as being part of a loop. You can picture a curve part of which goes along the rod, and where the rest of the curve is stationary. The flux rule $$ \mathcal{E} = -\frac{d\Phi}{dt} $$ doesn't require that the curve be defined by a physical loop of some conductive material, it also holds for our imaginary loop. For this loop, since only the part of the loop along the rod is moving, the emf along the rest of the loop is zero and the total emf is induced entirely across the rod.

That said, you certainly don't have to take the rod as being part of some loop. Motional emf is due to the magnetic component of the Lorentz force. Assuming the charges have settled so there is no movement of charge along the rod, the motional emf across two ends of the rod (say $A$ and $B$) is given by

$$ \int\limits_A^B{\vec{f}_\vec{B}·d\vec{\ell}} = \int\limits_A^B{(\vec{v}\times\vec{B})·d\vec{\ell}} $$ where $\vec{f}_\vec{B}$ is the magnetic force per unit charge and $\vec{v}$ is the velocity of a small piece $d\ell$ of the rod. Evaluating this integral will give the same result as the flux rule. In fact, the flux rule for motional emf can be derived by evaluating this integral along a loop.

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