Electromotive force computation and induced current

Question

I was reading Griffiths' Electrodynamics (4th ed, p. 303 and subsequent) chapter on electromotive force, and some doubts about the nature of this force per unit charge have arisen. In the first paragraph, the author considers a rectangular circuit partially immersed in a uniform magnetic field. The circuit is being dragged by an external force with a constant velocity $\mathbf{v} \perp \mathbf{B}$. This causes the wire segment perpendicular to the field to experience a force which in turn causes the free charges in the conductor to start moving upward, thus a current in the clockwise direction arises. This current, if I'm not mistaken, tends to oppose the change in flux (in this case, it decreases, since the area involved diminishes). The emf is therefore: $$ \mathcal{E} = \oint \mathbf{f}_{mag} \cdot d\mathbf{l} = vBh $$ where $h$ is the height of the sole segment of the wire where the force is parallel to the direction of $d\mathbf{l}$.
Everything checks out, but then the author makes an interesting remark that I fail to understand:
"Notice that the integral you perform to calculate $\mathcal{E}$ is carried out at one instant of time-take a snapshot of the loop..." I know this is just a very intuitive way of thinking about it, but if you freeze the system then there's no velocity at all, everything stops and thus there's no magnetic force to integrate. Just by looking at the definition of force per unit charge itself[ $\mathbf{f}_{mag} = (\mathbf{v} \times \mathbf{B})$ ], one gets the idea that the latter exists because the free charges in the conductor are moving. Is this approach used only for the sole purpose of making the infinitesimal element of wire $d\mathbf{l}$ point vertically and making the integration trivial? It seems to me that the other approach, namely the one where you consider the path of a single charge in motion, is the only one to be both formally correct and reasonable. Then I went on reading and realized that the situation might be more complicated than I originally believed: I started wondering what is actually $\mathbf{f}_{mag}$? Let's for example consider problem $7.7$ (pag.311, I'll add a picture for calrity). The free charges in the metal bar experience a vertical force ($\mathbf{v \times B}$) that sets them in motion. We can compute the emf via the flux rule and obtain the expression for the induced current by applying: $$\mathcal{E} = R I$$ Now the bar experiences a magnetic force: $$\mathbf{F}_{mag} = \oint (\mathbf{v} \times \mathbf{B})dq$$ which is directed in the opposite direction of $\mathbf{v}$ (aka, towards the left of the bar). Now then, the same force now acts upon the charges in the metal bar but it acts simultaneously in two different directions. Is this actually possible? No, at least to me, therefore I conclude that: $$\mathbf{f}_{mag} \ne \mathbf{F}_{mag}$$ although they are evidently the same thing. What am I misinterpreting? Is Griffiths' explanation clear, or is there some subtleties hidden that I missed while reading? Any help is much appreciated, as always.

Answer 1

Yes, there are very fine subtleties.

Is this approach used only for the sole purpose of making the infinitesimal element of wire $d\mathbf l$ point vertically and making the integration trivial? It seems to me that the other approach, namely the one where you consider the path of a single charge in motion, is the only one to be both formally correct and reasonable.

No, the integration has to be done over a fixed path in space defined by the whole circuit at some time, not over the actual path the current-carrying mobile charge makes in the course of its motion.

If you tried to do the latter, you would have path element $d\mathbf l$ parallel to particle's velocity, and then you would find the integral to be zero. This might look surprising and suspicious, but is correct, because magnetic part of the Lorentz force does no work on the charged particle. That's because this force is always perpendicular to velocity of the charged particle.

The motional emf, for the path from the end A to the end B of the rod, due to its motion in magnetic field, can be expressed as

$$ \mathscr{E}_{motional} = \int_A^B \mathbf v_{conductor} \times \mathbf B_{ext} \cdot d\mathbf l . $$ Notice $\mathbf v_{conductor}$ here is not velocity of the mobile charge, but velocity of the conductor body element. This, together with using the fixed vertical $d\mathbf l$ not affected by charge's actual motion, makes the vector $\mathbf v\times \mathbf B$ have some component along $d\mathbf l$, and thus the integral may be non-zero.

The intensity $\mathbf v_{conductor} \times \mathbf B_{ext}$ pushing on current is not actually the external Lorentz force per unit charge, but the effective EMF intensity due to combined effect of both 1) the external Lorentz force, and 2) the internal forces from the rest of the conductor, acting on the mobile charges. So denoting the motional EMF intensity as $\mathbf f_{mag}$ is very inappropriate and bound to cause confusion. It is better to use something like $\mathbf E^*$, or $\mathbf E_{motional}$.

Now the bar experiences a magnetic force: $$\mathbf{F}_{mag} = \oint (\mathbf{v} \times \mathbf{B})dq$$

Oh no, that's not so. In general, the moving conductor experiences macroscopic magnetic force (a very different concept from the motional EMF above):

$$ \mathbf F_{magnetic} = \int_{conductor} \mathbf j \times \mathbf B~ dV $$ For the moving bar in the example, $\mathbf j$ does not have exactly the direction of the bar motion. It is not exactly parallel to $\mathbf v$, but has a vertical component as well, which makes the magnetic force expression lead to formula

$$ \mathbf F_{magnetic} = -BIL\mathbf e_{right}, $$ so the magnetic force acting on the moving conductor acts against the conductor motion, thus slowing it down.

In energy terms, kinetic energy of the conductor(or work of external force pushing it) transforms into work of motional EMF acting on the current. External magnetic field does no work, it just enables this transfer.

Answer 2

Adding a little to @JánLalinský, the power balancing equation (conservation of energy) goes like this.

$\mathcal E_{\rm motional} = B\, L\,v$ and so the electrical power dissipated in the circuit is $\mathcal E_{\rm motional}\,I= B\,L\,v\,I$.

The external power needed to move the bar at a constant velocity is $F_{\rm external}\,v = B\,I\, L \, v$.

So the external power need to move the bar at constant velocity is equal to the electrical power dissipated in the circuit.