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If a conductive loop is partial or incomplete(wrt. $A$), is Faraday's law of induction still valid?

$$\varepsilon = -\frac{\delta \Phi_B}{\delta t}$$

Intuitively it seems possible to define the flux $\Phi_B$, however, I must take into account the geometric changes wrt. the area since it's partial, if it half as diagrammed, then its logical to adjust the area to $\frac{A}{2}$ from a simple symmetry.

Edit Note: The question is aimed mostly for a partial conducting loop.

Geodesic
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    I don't have time for a proper answer, but you may be interested to the concept of partial inductance, introduced by Clayton Paul, see e.g. this. There is also a book by Paul on the topic. – Massimo Ortolano Feb 07 '19 at 10:54
  • Make your question more precise: You need to specify, if the field B(in) varies in time. – flaudemus Feb 07 '19 at 11:44
  • There is no such thing as a 'partial' loop, this is just a wire. The whole closed loop thing is a kind of mathematical trick to get the emf along the wire using Stoke's Theorum. – crobar Feb 07 '19 at 13:55
  • @Crobar the question is more like: "Is it possible to induce and emf & have current flow in the semicircular loop" which is an oversimplified yes. I kept confusing that point with the "loop" in Faraday's law(i.e the mathematical trick you mentioned). – Geodesic Feb 07 '19 at 15:23

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(a) Faraday's law applies to a complete loop. No parts of the loop need to be conducting. To say that there'e an emf along a path in the form of a loop means that if a charge goes round (or is taken round) that loop (in the sense given by Lenz's law), work is done on it by the non-conservative electric field associated with the changing magnetic field.

(b) In general it is meaningless to talk about the emf induced in an isolated segment of conductor in a changing magnetic field, because we are free to complete the loop in any way we please, and different loops will have different emfs induced in them. For example, in your left hand diagram the boundary of the grey area will have twice the emf around it as an area bounded by your conducting path and a diameter of the grey area.

(c) But suppose the loop has been defined, eg as the boundary of the grey circular patch in the left hand diagram. It would be hard not to agree that, by symmetry, the emf in the conducting portion is half that induced in the whole loop. Likewise, I want to say that the emf in each side of the perimeter of the square is a quarter that induced in the whole perimeter. But that doesn't entitle us to say that we've now found the emf in the conducting part of either loop, viewed as an isolated conductor, because we're back to (b) again!

[This answer has been extensively rewritten. The material now replaced is below...

So in your diagrams we can talk about the emfs along the whole perimeters of the grey areas. Any partitioning of emfs into parts along the conducting bits and the non-conducting bits would be quite arbitrary. Take your left hand diagram; we could perfectly well complete the brown conducting path by going underneath the conducting part rather than above it. Then, supposing that the changing magnetic field extended below the conducting path, we'd have an emf in the opposite sense around the loop! This makes nonsense of attempts to assign an emf to an isolated segment of a conductor.]

Philip Wood
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  • I didn't make that distinction clear in the question, but I did; *A partial conducting loop. – Geodesic Feb 07 '19 at 10:51
  • Also, I believe I can create a configuration where the emf would be half, using the same diagram you discussed, but rather than having it underneath I'll wire the ends using connections going into the page such that their partial area is parallel to $B$. What do you think? – Geodesic Feb 07 '19 at 10:52
  • Diagram: https://imgur.com/a/2Ec4CfV – Geodesic Feb 07 '19 at 10:56
  • Not quite with you: at some stage the wires going into the page would have to join up again to make a loop. Can't see that taking them out of the plane of the page makes much difference if $\vec B$ is uniform at at right angles to page. The emf in a semicircular loop and the non-planar loop shown in your supplementary diagram would both be half the emf in the circular loop. – Philip Wood Feb 07 '19 at 11:01
  • The diagram shows how the wires would connect to the semicircular loop, however I purposefully showed it in that manner to get an "idea" of how the conductive loop is complete. The semicircular segment is pepandicular to the flux, and as stated by you should produced half of the emf because it perfectly half :) – Geodesic Feb 07 '19 at 11:06
  • Agreed. So are we in disagreement about $anything$? My point is that it makes no sense to talk about the emf induced in an isolated segment of conductor because the choice of loop in which to include that segment is arbitrary, and different loops have different emfs induced in them. – Philip Wood Feb 07 '19 at 11:11
  • @PhilipWood I am asking because I do not know the answer! "It would be hard not to agree that, by symmetry, the emf in the conducting portion is half that induced in the whole loop." Does that mean that a piece of wire bent in the form of a semicircle, in air with a changing magnetic field, would have a redistribution of charge in it so one end has an excess of positive charges and the other end would have an excess of negative charges? – Farcher Feb 07 '19 at 15:43
  • @Farcher Brilliant question, and I'm not confident of the answer. My feeling is that there $wouldn't$ be separation of charge because – can you see me waving my hands? – the semicircle wouldn't know what loop it's supposed to be a part of! I'll think on. [The situation is quite different for a moving conductor cutting magnetic flux, for which the emf, due to the magnetic Lorentz force, is truly localised.] – Philip Wood Feb 07 '19 at 16:10
  • @PhilipWood If there is an electric field inside a conductor then mobile charges will move so they must in this case if there is an emf created? – Farcher Feb 07 '19 at 17:13
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    By "in this case" I assume you mean the OP's original case of the stationary conductor in a changing magnetic field. I'm afraid that I don't know how to apply the fundamental equation, $curl \vec E =-\frac{d \vec B}{dt}$ to an isolated segment of conductor; only to a complete loop. – Philip Wood Feb 07 '19 at 17:19
  • Adding to the discussion, our agreement (on a $\varepsilon \neq 0$ ,for my post) is true if the change of magnetic field was due to it's weakness in strength, or simply moving it away from the loop. Right @PhilipWood? – Geodesic Mar 05 '19 at 19:14
  • @Farcher I've just come across Manishearth's (6 or 7 year old) answer to this question "Will an emf be induced across the ends of a stationary metal rod placed in a time-varying magnetic field?" I found it interesting. [Sorry but I don't know how to do links.] – Philip Wood Jun 11 '19 at 23:20
  • @PhilipWood The link to Manishearth's answer is here. – Farcher Jun 11 '19 at 23:44
  • @PhilipWood To produce a link [Put information about link in square brackets](Put the address of the link in round brackets) – Farcher Jun 11 '19 at 23:46
  • @Farcher Many thanks. I'll try to do this next time I have the urge to provide a link. – Philip Wood Jun 12 '19 at 08:05
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If a conductive loop is partial or incomplete(wrt. A ), is Faraday's law of induction still valid?

No, a 'partial loop' does not define (as a boundary defines) an area, so the area integral of $$\Phi = \int {\int_A{ B \cdot dA }}$$ is not defined. Only a whole loop encloses the area A (and it must be a directed loop, so that the B direction and dA orientation determines the sign of the flux).

Whit3rd
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  • If the boundary loop does not have to be conducting, and I placed a partial conducting loop in the region, and varied $B$ over time, wouldn't it produce an half of which a complete loop would? Better diagram: https://imgur.com/a/2Ec4CfV – Geodesic Feb 07 '19 at 10:57
  • I didn't make the distinction clear in the question, I did now. – Geodesic Feb 07 '19 at 10:58
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    @Geodesic - the emf that is calculated is not dependent on conductivity of any objects. To measure voltage between two points, with a meter, you will always be making a complete loop (defined by the meter test leads) to make a complete circuit, and THAT depends on a conductive path. It also completes the loop. – Whit3rd Feb 07 '19 at 11:01
  • I'm a bit confused here. 1) So we both agree that there could be an EMF and current flow(if the loop is complete) due to the semicircular loop being placed on the boundary lines 2) The area is defined by our "imaginary" loop, and the semicircular loop lies on half of that portion, why can't the flux law be used again? You response seems straightforward but I'm confused because the "closed loop" does not have to be fully encompassed by a conductor. – Geodesic Feb 07 '19 at 18:30
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    @Geodesic: The 'flux law' requires the defined area. You can split the area and have multiple part calculations, but splitting the boundary does NOT split the area of the integral, it makes that area undefined. You need to at least define a dummy boundary to do the calculation. – Whit3rd Feb 08 '19 at 06:44
  • I reviewed Feynman's lectures it would've been useful if I differentiated between the mathematical closed curve $\Gamma$, and the fixed surface $S$, where $S$ can be semicircular, or circular. But $\Gamma$ remains the same for both :) – Geodesic Mar 05 '19 at 19:18
  • While reading the topic again, I remembered this question, so I thought adding the comment could help someone else later on. – Geodesic Mar 05 '19 at 19:19
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It is perhaps best to look at Faraday's law, which is one of Maxwell's equations, in this form ?

$$\oint_{\text {closed loop L} }\vec E \cdot d\vec l = - \frac{d}{dt}\iint_{\text{open surface S}} \vec B \cdot d\vec S$$

where $\vec E$ is the electric field and $\vec B$ the magnetic flux density.

The closed loop and the open surface are defined in the "butterfly net" diagram below.

enter image description here

The integral on the left hand side of the equation (emf) is for a closed loop and that closed loop defines the edge of the open surface which in your case is a plane.

There is no mention of what the loop is made, which then means that the loop can be part electrical conductor and part electrical insulator but the loop must be closed.

The problem is that you can evaluate the right hand side and then this gives you the left hand side (emf) but it does not give you the distribution of the electric field, $\vec E$, within the loop.

Farcher
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