How Is Faraday's Law Applied in Time-Varying Loops?

[Eye_in_the_Sky]

Faraday's law of induction is

emf = - (d/dt) ∫_S B∙da .

When the closed loop (serving as the boundary of the surface S) is independent of time, the above relation is equivalent to the Maxwell equation

curl E = - ∂B/∂t .

However, when the closed loop C (i.e. the boundary of S) is itself a function of time, the following two questions seem relevant to ask:

(i) How is the first equation above to be applied?

(ii) Is this method consistent with Galilean (or Lorentz) invariance?

 

[reilly]

Great question. Requires subtle and sophisticated arguments to indicate that Faraday's Law works for moving loops/field sources. One of the best discussions of this issue, I think, is given by Purcell in his Electricity and Magnetism, Vol. 2 of the Berkeley Physics Course. (When I taught this subject, it took one or two lecture sessions to cover the topic, both at the intermediate undergraduate level(Purcell), and the graduate level (Jackson or Panofsky and Phillips). A good part of the issue is calculating the flux through a moving loop. And, again, Purcell's discussion is superb.

So, I suggest some reading is in order. But, perhaps someone has a better intuitive approach, which can be explained more briefly than several pages of argument.

Regards,
Reilly Atkinson

 

[Crosson]

The answer is straightfoward calculus. If boundary S is a function of time, then the limits of the integrand are functions of time, and you just carry the derivative through according to what is called liebniz's rule.

(look it up www.mathworld.com)

If you are worrying about lorentz invariance, setup the integrand in the unprimed frame and differentiate wrt unprimed time, do it again in some other primed frame with respect to that prime time. Then it remains for you to calculate an invariant, transform it, and satisfy your curiosity.

For the record, the invariants are E*B (dot product) and (E^2)-(B^2).

It is funny, the title of your post is "Faraday" (the name of the em field tensor) but you phrased your question in terms of maxwells equations. If you are taking a lorentz transformation, or doing anything I suggested, use the field tensor.

 

[Eye_in_the_Sky]

The answer is straightfoward calculus. If boundary S is a function of time, then the limits of the integrand are functions of time, and you just carry the derivative through according to what is called liebniz's rule.

Yes. Working out the right-hand side of Faraday's law is just straightforward calculus.

On the right-hand side, we have

RHS ≡ - (d/dt) ∫_S(t) B∙da

= - ∫_S(t) ∂B/∂t ∙ da - lim_∆t→0(1/∆t)[ ∫_S(t+∆t) - ∫_S(t) ] B∙da

= ∫_S(t) curl E ∙ da - ∫_C(t) (B x v_C) ∙ dl

= ∫_C(t) (E + v_C x B) ∙ dl ,

where v_C denotes the "velocity" at time t of the line-element dl on C(t).

Note that, in the above derivation, the following two relations have been employed:

(i) div B = 0 ;

(ii) curl E = - ∂B/∂t .

This is all straightforward enough.
______________

The difficulty, however, lies in the left-hand side of Faraday's law. How is the emf at time t associated with the closed loop C(t) to be defined? According to the usual definition, the emf is given by the work done on a unit charge moved once around the loop. This suggests using the following definition:

emf(t) ≡ ∫_C(t) (E + v x B) ∙ dl .

But what is the v here? It obviously includes a component parallel to dl corresponding to the (hypothetical) motion of the unit charge along the curve C(t), and this component, of course, gives no contribution to the emf. It is only a component of v perpendicular to dl which can give some contribution ... yet, the (hypothetical) motion of the unit charge along C(t) would not include such a component.

So, perhaps the answer is to simply use the above definition in a 'conceptual way' with v set equal to the component of the velocity of the line-element dl perpendicular to the dl-direction. In that case, we can then just as well set v = v_C , because any component of v_C parallel to dl makes no contribution anyways. Doing this yields

emf(t) ≡ ∫_C(t) (E + v_C x B) ∙ dl ,

which is precisely the expression for the RHS of Faraday's law as it was derived above.
______________

One of the best discussions of this issue, I think, is given by Purcell in his Electricity and Magnetism, Vol. 2 of the Berkeley Physics Course. (When I taught this subject, it took one or two lecture sessions to cover the topic, both at the intermediate undergraduate level(Purcell), and the graduate level (Jackson or Panofsky and Phillips). A good part of the issue is calculating the flux through a moving loop. And, again, Purcell's discussion is superb.

I don't have Purcell (although it does seem worthwhile to attempt to procure a copy).

You wrote: "A good part of the issue is calculating the flux through a moving loop." This hurdle, I think, I have been able to overcome; and as Crosson points out, it requires no more than straightforward calculus. The result is to be found in what I recorded above, namely,

(d/dt) ∫_S(t) B∙da = ∫_S(t) ∂B/∂t ∙ da + ∫_C(t) (B x v_C) ∙ dl ,

in which div B = 0 has been utilized. (Also, remember that v_C is the velocity at time t of the line-element dl on C(t); thus, in general, v_C will vary from point to point along C(t) depending upon how the curve changes over time at the given point.)

So, it appears to me that my only real difficulty lies in finding the 'proper' definition of: emf for a time-dependent closed loop.

Upon achieving that, I would then like to probe the question of where/when/how (?) either of Galilean or Lorentz invariance comes into the picture.

... I do have Jackson's book. I'll take a look there and see what it has to offer.

 

[Eye_in_the_Sky]

three questions

J. D. Jackson, Classical Electrodynamics

Section 6.1, Faraday's Law of Induction
----------------------------------------------

The changing [magnetic] flux [linking a circuit] induces an electric field around the circuit, the line integral of which is called the electromotive force, emf. The electromotive force causes a current flow, according to Ohm's law.

... The magnetic flux linking the circuit is defined by

F = ∫_S B∙n da –––– (6.1)

... The electromotive force around the circuit is

emf = ∫_C E'∙dl –––– (6.2)

where E' is the electric field at the line element dl of the circuit C. Faraday's observations are summed up in the mathematical law,

emf = - k dF/dt –––– (6.3)

... The constant of proportionality k depends on the choice of units for the electric and magnetic field quantities. It is not, as might at first be supposed, an independent empirical constant to be determined from experiment. As we will see immediately, once the units and dimensions in Ampere's law have been chosen, the magnitude and dimensions of k follow from the assumption of Galilean invariance for Faraday's law.

... Okay, but why write E'[/color] and not just plain E without a prime?

Let us now consider Faraday's law for a moving circuit and ... Galilean invariance. Expressing (6.3) in terms of the integrals over E' and B, we have

∫_C E'∙dl = - k (d/dt) ∫_S B∙n da –––– (6.4)

The induced electromotive force is proportional to the total time derivative of the flux—the flux can be changed by changing the magnetic induction or by changing the shape or orientation or position of the circuit. In form (6.4) we have a far-reaching generalization of Faraday's law. The circuit C can be thought of as any closed geometrical path in space, not necessarily coincident with an electric circuit. Then (6.4) becomes a relation between the fields themselves. It is important to note, however, that the electric field, E' is the electric field at dl in the coordinate system or medium in which dl is at rest, since it is that field which causes current to flow if a circuit is actually present.

1) Does the above prescription for defining the emf in terms of E' also apply to a Lorentz invariant scenario?

2) If so, then just how, specifically?

3) Then, in terms of the E and B fields in the laboratory frame, what does the emf look like?[/color]


Next, the relevant principle of relativity happens to be:

In particular, consider Faraday's observations. It is expected and experimentally verified that the same current is induced in a secondary circuit whether it is moved while the primary circuit through which current is flowing is stationary or it is held fixed while the primary circuit is moved in the same relative manner.

And finally, Faraday's law is said to be precisely valid in the context of special relativity:

Because we considered a Galilean transformation, the result [we obtained for E' in terms of E and B] ... is an approximation only valid for speeds small compared to the speed of light. ... Faraday's law is no approximation, however.

... Are there any comments on and/or answers out there to my three questions above?[/color]