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Origin of non-conservative Faraday

  1. May 13, 2010 #1
    The non-conservative nature of an induced E-field due to a changing magnetic flux has always interested me.

    In the case of a constant magnetic field and a circuit with a changing surface, the changing magnetic flux can actually be written as a line integral of v x B (*), with v the speed of the changing circuit at each point. So basically you're integrating the Lorentz Force qv x B around a closed circuit (or generally: loop). This means the non-conservative nature of this law (or anyway as far as motional emf is concerned) arises out of the concept of Lorentz Force. Why is this, as B is actually defined by the Lorentz Force?

    Maybe you don't understand what my problem is. It is the fact I don't understand why it directly arises out of the Lorentz Force. For example: a free charge undergoing the Lorentz Force is experiencing a conservative force (since no work is ever done), but in the case of a charge in a conductor, you don't even need anything more than the Lorentz Force to make it non-conservative. Maybe it has something to do with the Hall-effect, but I honestly have no clue.

    Thanking all helpers,
    mr. vodka

    (*) int([v x B] .dl) = - int(B. [v x dl]) = - d(phi)/dt (cf. Electromagnetic Fields and Waves, Lorrain, Corson, Lorrain [p413])

    NB: In class when we have to calculate the current in a circuit with an induced E-field, we pretend there is an imaginary voltage source with the same emf as associated with the changing magnetic field and then we ignore the magnetic field. Is it obvious this should work? I do not understand why this little trick is allowed (for one thing: you're making a non-conservative electric field a conservative one)
    Last edited: May 13, 2010
  2. jcsd
  3. May 14, 2010 #2
    Hm, maybe I'm being vague, so I'll post a supershort recap:

    Lorentz' Force is intricately conservative for free charges (it even does no work at all), while it is unconservative for charges in conductors. Huh?
  4. May 14, 2010 #3
    vxB is not really a conservative force. Usually that is reserved for forces which can be written as the gradient of a potential energy.

    [tex]F = -\nabla V[/tex]

    The flux rule cannot be reduced to this form, either from changing B in time, or movement of the conductor.

    The rule of summed voltage around a loop of wire being zero has nothing to do with conservative forces. It is a statement of how free charge behaves in conductors. If you go around a conductor adding up the voltage changes, and get to a non-zero number, then that means there is now a voltage drop over an infinitesimal segment of wire, causing an electric field in the conductor. Charge would tend to flow to stamp out this field.
  5. May 14, 2010 #4
    What's "a circuit with a changing surface"?
  6. May 14, 2010 #5


    Staff: Mentor

    In a changing magnetic flux the E-field is not conservative in the sense that it cannot be written as the gradient of a scalar field, but it is conservative in the sense that energy is conserved. So I am not really sure what you are asking about here.
  7. May 16, 2010 #6
    Hm, I'm not sure what you mean (in bold). Are you implying that it is as if there is full-fledged battery on an infinitesimal segment of the wire? What do you mean to say with this exactly? Are you saying you can view everything as if there is a battery on a random spot? Is that self-evident?

    And why is the rule of summed voltage not related to conservative forces? For fields from charges: V = circle(E.ds) = 0 which is equivalent to E being a conservative field just because F is a conservative force.

    "Circuit with induced motional e.m.f." or something like that, know what I mean? Letting the dA of the flux variate, not the B.

    Are those two things not the same? Well of course every field is conservative somehow due to the conservation of energy, but an induced E-field is not in itself conservative is it? (energy-wise, although I didn't know there was a distinction) Because if you move a charge in a loop and stop where you began, you somehow gained or lost energy, while being at the exact same spot in space and time as far as the field is concerned! Of course other quantities know the difference and somehow "restore" the energy, but I don't understand where the non-conservative nature comes from. Take the Lorentz' force qvxB: for any free charge, no work is done and thus the force is conservative. Place the same charge in a circular conductor (i.e. place some constraints) and the Lorentz' force delivers work and becomes unconservative. Why/how the jump?
  8. May 16, 2010 #7


    Staff: Mentor

  9. May 16, 2010 #8
    Consider the equation you show from your book,

    (*) int([v x B] .dl) = - int(B. [v x dl]) = - d(phi)/dt .

    The confusing point in this problem/equation is that the v variable doesn't discriminate between the v charge in (1.) the charge carriers of the original, unchanging circuit, and (2.) the charge carriers in the moving circuit. In the first case, you still have the same equation, but obviously the RHS is zero and you wouldn't have any complaints. In the second case, you get these weird observations.

    I'll consider the simple case where the circuit is just expanding/contracting in a plane perpendicular to the field B. Then where is the energy coming from? Clearly, it's not coming from the Lorenz force. It's coming from whatever's pushing (doing the work) to expand/contract that circuit. That force has nothing to do with the Lorenz force, and so there should be no cause to think that the Lorenz force is acting nonconservatively in this situation.
  10. May 17, 2010 #9
    Another easier, possibly more correct way to say this is the following. The equation you give above (the same one in my post above) uses v to refer to the loop velocity. That is not the Lorenz force there in that equation! The Lorenz force refers to the velocity of the charges. Now, the loop velocity and the charge velocity are related, but they're not the same thing. You could have zero loop velocity and nonzero charge velocity, for example.
  11. May 17, 2010 #10
    mordechai, thank you for your interesting view. I'm not entirely sure though. Please follow me in the following thought experiment:

    Construct a rectangular, conducting loop, with its right side/bar extendable. Now say we have a constant homogeneous B-field pointing directly down into the surface of the loop. Now pull the right bar away with a constant speed v, enlarging the area A of the loop with dA/dt = Lv with L the length of the moving bar. Faraday says we get an emf V = BLv. But there's also another way of calculating the emf: look at a specific electron in the right bar that is moving with a constant speed v to the right. (-e)*vxB gives us a Lorentz Force on the electron pointing downward along the bar (i.e. the electron is pushed to the bottom of the moving bar). Also consider the (general) equations F = qE and EL = V. This gives us F = qV/L and as F = -evB and q = -e, we have V = BvL.

    Now I'd like to make two seperate remarks on this:
    1) At least in this case, the origin of Faraday's Law is found entirely in Lorentz' Force. Since Faraday's Law produces an inherently non-conservative E-field, the Lorentz' Force must account for this and thus somehow be an unconservative force after all... (?)

    2) You argue that the energy gained (in form of an E-field) is coming from the one deforming the circuit. I think this is a more subtle point... Indeed, as soon as the charges themselves have acquired an equilibrium speed u in the conductor, there is a second Lorentz Force, pointing directly against the vector v and thus the person pulling along the bar at speed v must deliver a constant power against these electrons. But where does this power go to? It goes to the dissipation of energy in the system, not the to creation of the E-field.
    As for the energy which actually set up the E-field, this seems to have come directly from the Lorentz' Force, because when the charges were initially fixed in the conductor, there was no force pushing against the outside bar-mobilizer.
  12. May 17, 2010 #11
    The short answer to your question is that the Lorentz force is perpendicular to the 'total' velocity of the charges. In a conductor, charges have a relative speed with respect to the conductor that is responsible for the electric current flow. However, if the conductor is moving itself, then their velocity with respect to a stationary observer is the sum of these two velocities (for speeds much smaller than the speed of light). The Lorentz force is perpendicular to this total velocity. Let us do some vector algebra. Denote the relative velocity by [itex]\mathbf{v}'[/itex], the velocity of the conductor (or, at least the element of the conductor where the charge presently is) as [itex]\mathbf{u}[/itex] and the external magnetic field with [itex]\mathbf{B}[/itex]. The total velocity of the charge is:

    \mathbf{v} = \mathbf{v}' + \mathbf{u}

    So, the Lorentz force acting on the charge is:

    \mathbf{F}_{\textup{L}} = q (\mathbf{v} \times \mathbf{B})

    The component of the force parallel to the relative velocity [itex]\mathbf{v}'[/itex] is proportional to the dot product:

    \mathbf{v}' \cdot \mathbf{F}_{\textup{L}} = q \mathbf{v}' \cdot (\mathbf{v} \times B) = q \mathbf{B} \cdot (\mathbf{v}' \times (\mathbf{v}' + \mathbf{u})) = q \mathbf{B} \cdot (\mathbf{v}' \times \mathbf{u}) \neq 0

    as long as [itex]\mathbf{v}' \times \mathbf{u} \neq 0[/itex].
  13. May 19, 2010 #12
    Hm, Dickfore, thank you for your post. I'm not sure if I get your point though. Your remark about v = v' + u and taking that in account is very just. But why, if you're looking at the power delivered by the magnetic field, may one take the dot product of F with v' and not v? (as is the general definition, I think).
  14. May 19, 2010 #13
    I will answer with the following question: Is the Lorentz force the only force acting on the charges?
  15. May 19, 2010 #14
    In this problem, the v x B force is generating an electric field which gives you an electromotive force. But that doesn't mean the v x B force acts nonconservatively. See below.

    I'm not at all convinced by your arguments in this claim. The work being done to expand the wire area is the force pushing radially (or generally, perpendicular to B). This force has nothing to do with the Lorentz force -- it's generated by the machine or the fingers or whatever is pushing outwards on the loop. Even if the Lorentz force resists this expansion (which I suspect it doesn't, as in my prior response) that doesn't imply that the Lorentz force is acting nonconservatively. You would have to show that explicitly.

    But this task would inevitably end in failure, as the v x B force doesn't do any work. This follows absolutely immediately from first principles, and it really doesn't even make sense to talk about a force that doesn't do work being conservative or nonconservative. Conservative forces are defined as such in terms of work. Alternatively, one might say that the lorentz force is trivially conservative, since it always does zero work on any path. (That probably should have been stated earlier. )

    By the way -- the Lorentz force means the q[E + v x B] force. In all of this post, we are talking about the v x B force, not the Lorentz force. Perhaps this terminology is causing your confusion. The Lorentz force is most definitely nonconservative -- but the E=0 Lorentz force is not.

    I realize you are attempting to link a logical chain something like:

    v x B induced --> generates E field through Faraday --> E field nonconservative --> v x B nonconservative,

    But this is logically fallacious. The v x B force (may) redirect energy into an E field that may be nonconservative, but that doesn't make the v x B force nonconservative.
    Last edited: May 19, 2010
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