Apparent ambiguity in the E field direction in Faraday's Law

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I'm confused by an apparent ambiguity in the direction the E field in Faraday's law:

∫ E°dl = - ∂/∂t ∫ B°da

Faraday's law says the change in magnetic flux through an open surface gives rise to an emf equal to E°dl taken around the closed loop which is the boundary of the open surface.
And also says the induced E field around the loop exists even if there is not an actual wire loop to support current flow.

My understanding is that E is always parallel to dl around the closed loop. This is supported by:

(1) The Walter Lewin MIT lecture at

at time 20:22 into the lecture where he says "E and dl are always in the same direction if you stay in the wire."

(2) Also Example 7.7, page 306 of "Introduction to Electrodynamics, 3th ed." by David Griffiths which implies E is parallel to dl.

Here is a simple example of the ambiguity involving two intersecting loops which are circles in the xy plane and a B(t) field perpendicular to the xy plane:

(1) Let a time variable uniform B(t) field be everywhere perpendicular to the xy plane.
(2) Let C1 be a circle in the xy plane where the associated flux surface is the area enclosed by C1 in the xy plane.
(3) Likewise, let C2 be another circle in the xy plane where the associated flux surface is the area enclosed by C2 in the xy plane.
(4) In addition, C1 and C2 intersect where P is one point of intersection.
(5) Applying Faraday's law to both circles we see an E tangent to C1 at P and also a different E tangent to C2 at P. Thus two E vectors at the same point in the xy plane, each pointing in a different directions in the xy plane. How can this be?

Please point out my error in this analysis.
 

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  • #2
Charles Link
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The EMF that you calculate around a loop of wire only takes into account the component of ## E ## along ## dl ##. The calculation of the EMF gives you zero information about any component of ## E ## that may exist perpendicular to ## dl ##. ## \\ ## In general, any time you choose a closed loop path over which you are applying Stokes theorem to Faraday's law in differential form, so that ## \mathcal{E}= \oint E \cdot dl =\iint \nabla \times E \cdot dA=-\iint \frac{\partial{B}}{\partial{t}} \cdot dA ##, you get zero info on any component of ## E ## that might exist perpendicular to the path of the closed loop integral. There could be a very large component of ## E ## perpendicular to the path of integration, but it will not contribute to the EMF.
 
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  • #3
sophiecentaur
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each pointing in a different directions in the xy plane.
What is worrying you about this? In your additional experiment, all that's happening is that the emf's could cancel along the common parts of the two circles (if they are touching over the whole length of the common section - or if they just cross at two points the area between the two overlapping (non-touching) wire sections will have a current circulating round it in the same sense as the current around the outer loop.
It strikes me that it's like a resistive (figure of eight) network with two loops and two batteries. It is quite possible to arrange the resistors so that no current flows through the middle leg. Current will just flow round the outer, big loop.
 
  • #4
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The EMF that you calculate around a loop of wire only takes into account the component of ## E ## along ## dl ##. The calculation of the EMF gives you zero information about any component of ## E ## that may exist perpendicular to ## dl ##. ## \\ ## In general, any time you choose a closed loop path over which you are applying Stokes theorem to Faraday's law in differential form, so that ## \mathcal{E}= \oint E \cdot dl =\iint \nabla \times E \cdot dA=-\iint \frac{\partial{B}}{\partial{t}} \cdot dA ##, you get zero info on any component of ## E ## that might exist perpendicular to the path of the closed loop integral. There could be a very large component of ## E ## perpendicular to the path of integration, but it will not contribute to the EMF.
Thank you for your response which I understand. However, this leads to other questions: (1) is there a perpendicular component of E due to the changing flux and (2) if so, how would I find it using Faraday's law?
 
  • #5
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What is worrying you about this? In your additional experiment, all that's happening is that the emf's could cancel along the common parts of the two circles (if they are touching over the whole length of the common section - or if they just cross at two points the area between the two overlapping (non-touching) wire sections will have a current circulating round it in the same sense as the current around the outer loop.
It strikes me that it's like a resistive (figure of eight) network with two loops and two batteries. It is quite possible to arrange the resistors so that no current flows through the middle leg. Current will just flow round the outer, big loop.
Thank you for the response. In my thought experiment there are only two imaginary loops which intersect at two points (no wires, thus no currents) thus only E fields in space. Basic question was how can here be two different E fields at the common point P where the circles intersect.
 
  • #6
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Thank you for your response which I understand. However, this leads to other questions: (1) is there a perpendicular component of E due to the changing flux and (2) if so, how would I find it using Faraday's law?
I'm assuming there are only E fields due to the changing flux.
 
  • #7
sophiecentaur
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Thank you for the response. In my thought experiment there are only two imaginary loops which intersect at two points (no wires, thus no currents) thus only E fields in space. Basic question was how can here be two different E fields at the common point P where the circles intersect.
So the result is what the maths says it is, It can only be worrying you because you are, in fact, assuming there is some physically connected equipment. Those fields aren't really there. They are fictitious until you measure the effects of them on a wire.
 
  • #8
Charles Link
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Thank you for your response which I understand. However, this leads to other questions: (1) is there a perpendicular component of E due to the changing flux and (2) if so, how would I find it using Faraday's law?
If the symmetry of the problem allows you to construct a circular loop, with the tangential component of ## E ## the same everywhere on the loop by symmetry, oftentimes, Faraday's law can be used to compute this component of ## E ##. In those cases, no info is obtained about any possible ## E ## that may be perpendicular to the loop. Sometimes, I believe Gauss' law can be used if there is sufficient symmetry, and in some cases will allow you to compute a perpendicular component of ## E ## and/or show that it is non-zero. Sometimes, the luxury of some symmetry is present, other times the calculations can be quite difficult. ## \\ ## It is also worth pointing out, that any electric field that is due to static electrical charges has ## \nabla \times E=0 ##, so that Faraday's law gives zero info on the contribution of static electric charges to the electric field at any given point. You can often compute an electric field ## E_{induced} ## that is caused by Faraday's law, but you still need to consider what can be essentially solutions to the homogeneous equation ## \nabla \times E=0 ##, (from static electrical charges, if any are present), to get the complete solution to ## E ##. ## \\ ## And I believe this can get even more advanced, where it is necessary to consider Lienard-Wiechert type solutions. That is a rather advanced topic, but you might find it of interest, and you may want to google it.
 
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If the symmetry of the problem allows you to construct a circular loop, with the tangential component of ## E ## the same everywhere on the loop by symmetry, oftentimes, Faraday's law can be used to compute this component of ## E ##. In those cases, no info is obtained about any possible ## E ## that may be perpendicular to the loop. Sometimes, I believe Gauss' law can be used if there is sufficient symmetry, and in some cases will allow you to compute a perpendicular component of ## E ## and/or show that it is non-zero. Sometimes, the luxury of some symmetry is present, other times the calculations can be quite difficult. ## \\ ## It is also worth pointing out, that any electric field that is due to static electrical charges has ## \nabla \times E=0 ##, so that Faraday's law gives zero info on the contribution of static electric charges to the electric field at any given point. You can often compute an electric field ## E_{induced} ## that is caused by Faraday's law, but you still need to consider what can be essentially solutions to the homogeneous equation ## \nabla \times E=0 ##, (from static electrical charges, if any are present), to get the complete solution to ## E ##. ## \\ ## And I believe this can get even more advanced, where it is necessary to consider Lienard-Wiechert type solutions. That is a rather advanced topic, but you might find it of interest, and you may want to google it.

Thank you Mr. Link for your patience. I'm sorry to say I'm still a bit confused and perhaps it can be cleared up by an answer to the following question:
With a single uniform, time dependent B(t) field point in the z direction (i.e., the B(t) is everywhere perpendicular to the xy plane). And with an imaginary loop in the xy plane (i.e., no conducting elements make up the loop). And with no other static or moving charges present. Is the E field in Faraday's Law the result of the changing flux and only that changing flux? That is, the E on the left hand side of the following equation completely determined by the the changing flux on the right hand side?
∫ E°dl = - ∂/∂t ∫ B°da
Another way of asking the question might be, does the choice of where you place the loop and its associated area have any effect on the E, or does E depend only on B(t) and thus could be calculated independent of the choice of loops and Faraday's Law?
 
  • #10
Charles Link
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And I believe what I posted in response so far is correct, but there can apparently be other complicating factors when considering applications of Faraday's law=and in the presence of circuit loops with resistors, these apparently can actually be simplifying factors. See e.g. this recent posting on PF: https://www.physicsforums.com/threads/induced-emf-and-current.941717/#post-5956204 Electrical currents for the circuit are computed, and it is assumed the currents are small enough that they don't generate strong magnetic fields of their own to further complicate the calculation. For the final solution for the electric field that results in this problem, it can be computed in each of the resistors using the formula for current density ## J=\sigma E ##, with ## \sigma=\frac{1}{\rho} ## where ## \rho ## is the resistivity of the resistor material. (The physical dimensions and other parameters for the resistors would need to be included). What the electric field is in regions where there is no material of the physical circuit is of little consequence. Meanwhile, if the circuit with the resistors was not present, the same changing magnetic field over the same area would no doubt generate different results for the electric field resulting from Faraday's law. You could also throw in capacitive and inductive components in the circuit, and these will affect the computation of the electric field. ## \\ ## In general, it seems fair to conclude when materials and circuit components are present in a changing magnetic field, they can clearly affect the solutions that you get when applying Faraday's law. You can't simply compute the electric field as if these materials weren't present.
 
  • #11
Charles Link
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To answer your latest question,(post 9), I think the answer is that Faraday's law by itself, which consists of doing a computation using the changing magnetic flux over a plane, does not uniquely determine the electric field in that plane. ## \\ ## I believe the complete induced electric field can be computed from the changing magnetic field with no other materials present, e.g. if we were to insert a single test charge and see how it responds, etc. It would experience the Lorentz force given by ## F=q(E+v \times B ) ## where the ## E ## is the induced ## E ## computed from Faraday's law. To compute the induced ## E ## field at a given location from Faraday's law, it can help if the changing magnetic field has some symmetry that can be utilized.
 
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  • #12
sophiecentaur
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I think the answer is that Faraday's law by itself, which consists of doing a computation using the changing magnetic flux over a plane, does not uniquely determine the electric field in that plane. \\
+1
There are many incidences of this sort of thing and they only introduce a cognitive dissonance if you forget that the calculated 'effect' assumes a particular reference frame.
 
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  • #13
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What is worrying you about this? In your additional experiment, all that's happening is that the emf's could cancel along the common parts of the two circles (if they are touching over the whole length of the common section - or if they just cross at two points the area between the two overlapping (non-touching) wire sections will have a current circulating round it in the same sense as the current around the outer loop.
It strikes me that it's like a resistive (figure of eight) network with two loops and two batteries. It is quite possible to arrange the resistors so that no current flows through the middle leg. Current will just flow round the outer, big loop.
Thank you for your reply.
 
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  • #14
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To answer your latest question,(post 9), I think the answer is that Faraday's law by itself, which consists of doing a computation using the changing magnetic flux over a plane, does not uniquely determine the electric field in that plane. ## \\ ## I believe the complete induced electric field can be computed from the changing magnetic field with no other materials present, e.g. if we were to insert a single test charge and see how it responds, etc. It would experience the Lorentz force given by ## F=q(E+v \times B ) ## where the ## E ## is the induced ## E ## computed from Faraday's law. To compute the induced ## E ## field at a given location from Faraday's law, it can help if the changing magnetic field has some symmetry that can be utilized.
Thank you for your response.
 
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After further work and thinking about your responses with respect to finding the E field given a uniform B(t) field in the z-direction using Faraday's Law, I have come up with following:

∇×E=−∂B(t)/∂t (Faraday's Law in differential form)

One solution is E = <−yβ/2,xβ/2,0> where β = −∂B/∂t in the z-direction. Essentially this is the solution given in Example 7.7, page 306 of "Introduction to Electrodynamics, 3th ed." by David Griffiths. (Griffith "cheats" by tacitly assuming the E field's direction and proceeds to find |E| ). E is a field in the xy plane which "rotates" about the z-axis. The problem with this solution is that it depends on where I choose to locate the z-axis. I start with a uniform B field perpendicular to an infinite plane and in that plane I choose an origin for xyz with z in the direction of the B field. Once I pick the origin with z parallel to B the E = <−yβ/2,xβ/2,0> solution works and the E solution rotates about that particular z. But this implies the solution depends on where I plunk down the z-axis. Surely there must be one "correct" E field answer that doesn't depend on my choice of where to place a z-axis in the infinite B field.
 
  • #16
Charles Link
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The form ## \nabla \times A=B ##, where ## B ## is a constant vector, does have a well known solution of ## A(\vec{r}) =-(\frac{1}{2}) \vec{r} \times B ##. This, I think, is basically what Griffiths is using there, with ## A ## as ## E ##, and ## B ## on the right side being ## -(\frac{\partial{B}}{\partial{t}})=\beta \hat{k} ##. ## \\ ## Editing: A different gauge, essentially the addition of a homogeneous solution, would result by letting ## \vec{r}=\vec{r}-\vec{r}_o ##, for any constant vector ## \vec{r}_o ##. It would still have ## \nabla \times A=B ##. ## \\ ## Also, for ## \nabla \times A=\vec{f}(r) ##, there is a Biot-Savart type integral solution: ## A(r)=(\frac{1}{4 \pi}) \int \frac{\vec{f}(r') \times (r-r')}{|r-r'|^3} \, d^3 r' ##, (where ## r ## and ## r' ##, and ## A ## and ## B ## are vectors, and the vector symbol is understood). In addition, any solution to the homogeneous equation ## \nabla \times A=0 ## can be added to this ## A ##. This Biot-Savart type integral approach sometimes works as an alternative to doing a Stokes theorem (see post 2) type integral approach. (In the Biot-Savart problem, ## A ## is replaced by ## B ##, and ##\vec{f}(r)=\mu_o J(r) ##). ## \\ ## It should be noted, if ## \vec{f} ## has a complex time dependence, in general, solutions to this equation will be of the Lienard-Wiechert form mentioned in post 8, but that topic is quite advanced. ## \\ ## We have been showing the solution for ## E ## to the differential equation ## \nabla \times E=-\frac{\partial{B}}{\partial{t}} ## is not unique. The question then is, what is the correct solution? I think the Lienard-Wiechert method will always get you the correct result, and some of these other methods to solve the differential equation will not, in general, give you the correct result for ## E ## unless you can select the correct homogeneous solution. For some of the simpler cases, the correct solution emerges right away, and no additional homogeneous solution is required. ## \\ ## Additional editing: When you introduce a changing magnetic field with infinite extent, you might start encountering problems of the type that this recent thread was trying to address: https://www.physicsforums.com/threads/breakdown-of-gauss-law.926835/ I'm not sure they ever reached a final definitive answer.
 
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The form ## \nabla \times A=B ##, where ## B ## is a constant vector, does have a well known solution of ## A(\vec{r}) =-(\frac{1}{2}) \vec{r} \times B ##. This, I think, is basically what Griffiths is using there, with ## A ## as ## E ##, and ## B ## on the right side being ## -(\frac{\partial{B}}{\partial{t}})=\beta \hat{k} ##. ## \\ ## Editing: A different gauge, essentially the addition of a homogeneous solution, would result by letting ## \vec{r}=\vec{r}-\vec{r}_o ##, for any constant vector ## \vec{r}_o ##. It would still have ## \nabla \times A=B ##. ## \\ ## Also, for ## \nabla \times A=\vec{f}(r) ##, there is a Biot-Savart type integral solution: ## A(r)=(\frac{1}{4 \pi}) \int \frac{\vec{f}(r') \times (r-r')}{|r-r'|^3} \, d^3 r' ##, (where ## r ## and ## r' ##, and ## A ## and ## B ## are vectors, and the vector symbol is understood). In addition, any solution to the homogeneous equation ## \nabla \times A=0 ## can be added to this ## A ##. This Biot-Savart type integral approach sometimes works as an alternative to doing a Stokes theorem (see post 2) type integral approach. (In the Biot-Savart problem, ## A ## is replaced by ## B ##, and ##\vec{f}(r)=\mu_o J(r) ##). ## \\ ## It should be noted, if ## \vec{f} ## has a complex time dependence, in general, solutions to this equation will be of the Lienard-Wiechert form mentioned in post 8, but that topic is quite advanced. ## \\ ## We have been showing the solution for ## E ## to the differential equation ## \nabla \times E=-\frac{\partial{B}}{\partial{t}} ## is not unique. The question then is, what is the correct solution? I think the Lienard-Wiechert method will always get you the correct result, and some of these other methods to solve the differential equation will not, in general, give you the correct result for ## E ## unless you can select the correct homogeneous solution. For some of the simpler cases, the correct solution emerges right away, and no additional homogeneous solution is required. ## \\ ## Additional editing: When you introduce a changing magnetic field with infinite extent, you might start encountering problems of the type that this recent thread was trying to address: https://www.physicsforums.com/threads/breakdown-of-gauss-law.926835/ I'm not sure they ever reached a final definitive answer.
Thank you, I will definitely follow up on this information.
 
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