# Direction of force due to induced current

Given a coil moving through a time-independent B field (such that we don't have to worry about induced E fields), we know by the flux law that an emf will be induced over the the conducting loop. If a current flows due to this emf, an additional magnetic force due to this additional motion of charge will act on the coil and this will have the ability to change the motion of the coil itself, whereas the previous magnetic force could only affect the motion of charge through the coil.

In all of the examples I have seen, the direction of this force is always parallel to the motion of the coil and acts in the opposite direction. Is this true for the general case? I understand by Lenz's law that the force must in some way oppose the initial motion but I'm struggling to see why it cannot also have a perpendicular component (or maybe it can, idk).

EDIT: I appreciate this question is really poorly worded, I've somewhat clarified what I'm asking in my comment to Laurier's response below.

• Has the answer helped you or not? Cheers Commented Apr 7, 2023 at 12:54

Shortly put, in my understanding, yes it is general and there are two ways of calculating it that I know of (using magnetic force or the change of flux in the closed loop). In short, the electromotive force will induce a current that oppose the change of flux in the loop (transfer of energy). In other words, the force will induce a flux that will oppose the change of flux in the circuit (electromotive force is more of a transfer of energy or a potential, its units are in Volts).

The change of magnetic flux change will induce a current in the coil as there's a magnetic force applied (if the change is due to an electric field, it would be the case of the Faraday's law) on the charges in the coil (this is true as long as the flux is changing in the coil).

One can write the magnetic flux as $$\Phi = \int_S \mathbf{B} \cdot$$d$$\mathbf{a}$$ and since$$\nabla \cdot \mathbf{B}=0$$, we can conclude that the net flux that goes through a surface is independent of the exact form of the coils, but depends on the outer surface (S) of the closed loop as it serves as the frontier/limit of the flux.

Let's write the electromotrice force (e.m.f.) $$\varepsilon = \oint f_{mag}\cdot$$d$$\mathbf{l} = \oint (\mathbf{v}\times \mathbf{B})\cdot$$d$$\mathbf{l}$$ (a shortcut would be to use the flux $$\varepsilon = -\frac{\text{d} \Phi}{\text{d} t}= - \frac{\text{d}}{\text{d} t} \int \mathbf{B}\cdot \text{d} \mathbf{a}$$ which is almost general; some specific case requires one to use the magnetic force) with $$\mathbf{v}$$ is the speed at which the coil moves in the field.

The equation shows that since the coil is moving, there is an induced force on the charges. This force would most likely be a force of Lorentz ($$f_{mag}$$) that induce a current that opposes the pulling force (or the external force moving the coil) like in the figure, where $$\mathbf{w}$$ is in the direction of the charge displacement (and a velocity) and $$\mathbf{u}$$, their speed when there's a current (on an electron $$f_{mag} = -e\mathbf{w}\times \mathbf{B}$$). In other words, I suppose there is some component of $$f_{mag}$$ that is perpendicular and some parallel to the mouvement. However, the e.m.f. is only parallel to the mouvement (it produces no work, but the induced current can generate an electric field that could produce one). The only forces in play I can see would be the Lorentz ($$f_{mag}$$) and the pulling force ($$f_{pull}$$).

$f_{mag}$ vs $$f_{pull}$$ (fig 7.11 of the griffiths)" />

I hope it answer your question, cheers.

EDIT: David pointed out that $$\mathbf{w}$$ is a velocity and not a displacement which is right. The charges move along $$\mathbf{w}$$ as well however

• Hi hi, Thank you very much for a very comprehensive response :) You write that 'since ∇⋅B=0, we can conclude that the net flux that goes through a surface is independent of the exact form of the coils, but depends on the outer surface (S) of the closed loop as it serves as the frontier/limit of the flux.' Could you clarify what you mean by the outer surface of the closed loop? Does this refer to any open surface bounded by the coil? I fail to see what the divergence of B has to do with this as the surface isn't closed. Commented Apr 7, 2023 at 14:59
• You then write 'Let's write the electromotive force (e.m.f.) ε=∮fmag⋅dl=∮(v×B)⋅dl'. Are we assuming that the B field cannot change with time such that we don't have an induced Maxwell-Faraday E-field? Commented Apr 7, 2023 at 15:04
• I don't quite understand the figure; you write 'w is the charge displacement' but it then appears in the Lorentz-force as a velocity.I understand how the figure represents the direction of the magnetic force on a small length element of the coil dl, but my question was more focused on whether the integral of this force over the entire loop would necessarily be in the direction opposite the driving force, or if it could have a component perpendicular to it. In the figure we see that your f_mag and f_pull are non-parallel but I'm wondering whether the net f_mag over the entire coil will be Commented Apr 7, 2023 at 15:14
• Thanks for your response, I will of course give a good review once I've managed to rack my head around the physics. Regarding your response to my closed loop question, apologies if I wasn't clear, I was referring to the surface being closed and not the loop. I understand that the loop needs to be closed for a current to flow, but my confusion stemmed from a divergence argument being applied to a open SURFACE, whereas I've only ever seen it applied in the context of Gauss' law for closed surfaces. Commented Apr 7, 2023 at 15:36
• Perhaps I'm misunderstanding your point with the divergence argument. By closed surface I'm referring to something like the surface of a sphere or a cube or something you might use for Gauss' law to completely enclose a volume. By open surface i refer to a surface with a well-defined boundary, such as a given surface enclosed by a conducting loop. The open surface does not enclose a volume. I don't quite understand your point on how B being a solenoidal field relates to its flux through a given open surface. Commented Apr 7, 2023 at 15:46