Trying to understand the direction of the induced electric field

[m74z00219]

Hi all,

I'm studying the electricity and magnetism this semester at university and I am very confused by faraday's law of induction.

I understand the following case: an infinitely long wire with a changing current (quasistatic B-field approximation - the frequency is't very high)

**Ampere's law will give you the B field B = [(uo * I)/(2*pi*s)]*phi_hat

**you can then calculate the time rate of change of the flux through an imaginary rectangular area above the infinitely long wire - this is the induced EMF

**you then calculate EMF by using the potential equation. that is,

EMF = integral of electric field over the path bounding the aforementioned surface

Equating these two can lead to a solution for the induced electric field of a infinitely long wire with changing current "I(t)".

The induced E field will be parallel to the wire and has a dependence of natural log of the radial distance from the wire.

Okay, now let's consider a uniform magnetic field. Defined as

B_vec = B*z_hat

Imagine this field is changing in time, but at any given moment still uniform throughout space. You can pick a circular surface at the origin (parallel to the B-Field) and determine the induced electric field - it will be circumferential (like the B-field around a wire).

I find this perplexing because I wouldn't expect the induced electric field to depend on the coordinate system when the B-field is uniform throughout space. Is there anyone that could help me make sense of this conundrum?

Thanks,
M74

[korjik]

Hi all,

I'm studying the electricity and magnetism this semester at university and I am very confused by faraday's law of induction.

I understand the following case: an infinitely long wire with a changing current (quasistatic B-field approximation - the frequency is't very high)

**Ampere's law will give you the B field B = [(uo * I)/(2*pi*s)]*phi_hat

**you can then calculate the time rate of change of the flux through an imaginary rectangular area above the infinitely long wire - this is the induced EMF

**you then calculate EMF by using the potential equation. that is,

EMF = integral of electric field over the path bounding the aforementioned surface

Equating these two can lead to a solution for the induced electric field of a infinitely long wire with changing current "I(t)".

The induced E field will be parallel to the wire and has a dependence of natural log of the radial distance from the wire.

Okay, now let's consider a uniform magnetic field. Defined as

B_vec = B*z_hat

Imagine this field is changing in time, but at any given moment still uniform throughout space. You can pick a circular surface at the origin (parallel to the B-Field) and determine the induced electric field - it will be circumferential (like the B-field around a wire).

I find this perplexing because I wouldn't expect the induced electric field to depend on the coordinate system when the B-field is uniform throughout space. Is there anyone that could help me make sense of this conundrum?

Thanks,
M74

Why do you think that this field is dependent on the coordinate system?

[m74z00219]

I think this because it seems that if i compute the induced electric field at the origin (due to a changing uniform B-field in the z direction and using a surface parallel to the B-field), the induced E-field will be circumferential about the z-axis. Why should the E-field be "circulating" about this location, why should it be circulating about anything?

I know I'm thinking about this in the wrong way, but I haven't figured out how yet.


M74

[korjik]

I think this because it seems that if i compute the induced electric field at the origin (due to a changing uniform B-field in the z direction and using a surface parallel to the B-field), the induced E-field will be circumferential about the z-axis. Why should the E-field be "circulating" about this location, why should it be circulating about anything?

I know I'm thinking about this in the wrong way, but I haven't figured out how yet.


M74

The electric field isnt circulating, it is circular. The why is that the E field induced by a changing B field but be perpendicular to the B field. That is one of the things that Maxwells equations say.

[Ken G]

I think this because it seems that if i compute the induced electric field at the origin (due to a changing uniform B-field in the z direction and using a surface parallel to the B-field), the induced E-field will be circumferential about the z-axis. Why should the E-field be "circulating" about this location, why should it be circulating about anything?

This is a natural confusion, and it stems from the fact that the B field you describe does not actually specify the electric field completely. Maxwell's equations are differential equations, which means they require boundary conditions. So it turns out that just knowing that B field doesn't tell us the electric field, but it does tell us the integral of the electric field around a loop. So there is a little bit of a swindle going on that you have exposed-- since the B field does specify the integral of the E field around a loop (the EMF), we simply choose a particular solution for the E field that circulates around the origin. We have broken the symmetry ourselves, because we know it doesn't matter how we break that symmetry. For example, you could choose a different point and imagine the E field circles around that point (for a different set of boundary conditions), but sure enough, if you calculate the integral around the original loop around the origin, you'll get the same EMF as if the field circled the origin.

[korjik]

This is a natural confusion, and it stems from the fact that the B field you describe does not actually specify the electric field completely. Maxwell's equations are differential equations, which means they require boundary conditions. So it turns out that just knowing that B field doesn't tell us the electric field, but it does tell us the integral of the electric field around a loop. So there is a little bit of a swindle going on that you have exposed-- since the B field does specify the integral of the E field around a loop (the EMF), we simply choose a particular solution for the E field that circulates around the origin. We have broken the symmetry ourselves, because we know it doesn't matter how we break that symmetry. For example, you could choose a different point and imagine the E field circles around that point (for a different set of boundary conditions), but sure enough, if you calculate the integral around the original loop around the origin, you'll get the same EMF as if the field circled the origin.

Well said. Gives a far better answer that I would have.

[Ken G]

Thanks-- it's a tricky business, those pesky boundary conditions. I think they are an important "missing link" in how physics is often conveyed.

[m74z00219]

Thanks-- it's a tricky business, those pesky boundary conditions. I think they are an important "missing link" in how physics is often conveyed.

Thanks Ken, that partially clears it up for me. Perhaps, the boundary conditions are my issue: I'm still trying to understand them. What you're saying I find simultaneously satisfying and puzzling: it seems there no way to globally define this E-field (for the case of the uniform B-field in z direction) whilst only knowing about the B-field. Of course, if you set up a real loop of wire, you could predict exactly the current that would be generated (by virtue of knowing the iEMF).

[Ken G]

Yes, you can know the IEMF without actually knowing the full E field. The IEMF only depends on the rate of change of B flux within the loop of wire, but the full E field needs to know what is happening elsewhere as well. It's a bit like fishing-- if all you care about is the fish that are caught in the sweep of your net, you really don't care about the full distribution of fish everywhere outside the net.

[Jerry]

It's not at all like fishing. Why do you assume there is anything at all in my net?

[Ken G]

There's always something in the net-- the question is, does it taste good with capers?

[publius]

I'll give you another one, one you'll find boils down to the same thing.

What is the electric field of an infinite charge distribution over all 3D space?

From symmetry, the field can't depend on where we are. No matter how we translate, things look the same. Further, it can't depend on rotation, either. No matter how we rotate around, things look the same. Now, the only vector that can sastify that is the zero vector.

But that violates div E = rho.

You can make the E field be whatever you want there, actually. You can adopt spherical coordinates at some point and make the field be radial about the point. But you can shift the origin and get the same thing. Or, if you like cartesians, you can make the field be a constant in a given direction, any one you like depending on how you set up your Gaussian surfaces.


-Richard

[tusenfem]

I'll give you another one, one you'll find boils down to the same thing.

What is the electric field of an infinite charge distribution over all 3D space?

From symmetry, the field can't depend on where we are. No matter how we translate, things look the same. Further, it can't depend on rotation, either. No matter how we rotate around, things look the same. Now, the only vector that can sastify that is the zero vector.

But that violates div E = rho.

You can make the E field be whatever you want there, actually. You can adopt spherical coordinates at some point and make the field be radial about the point. But you can shift the origin and get the same thing. Or, if you like cartesians, you can make the field be a constant in a given direction, any one you like depending on how you set up your Gaussian surfaces.


-Richard

Oh now you're just getting nasty!

[Ken G]

I'll give you another one, one you'll find boils down to the same thing.

Right, it's the way any field equation that is defined in differential form works. Being in differential form makes it look like it's entirely local, but this merely hides the importance of the global issues. If we use integral forms instead (like instead of div E = rho, use that the electric flux through a closed surface equals the enclosed charge), it is much more clear what ambiguities exist, which will need to be resolved by additional information (generally some boundary that breaks translational symmetry).

[DrRocket]

I'll give you another one, one you'll find boils down to the same thing.

What is the electric field of an infinite charge distribution over all 3D space?

From symmetry, the field can't depend on where we are. No matter how we translate, things look the same. Further, it can't depend on rotation, either. No matter how we rotate around, things look the same. Now, the only vector that can sastify that is the zero vector.

But that violates div E = rho.

You can make the E field be whatever you want there, actually. You can adopt spherical coordinates at some point and make the field be radial about the point. But you can shift the origin and get the same thing. Or, if you like cartesians, you can make the field be a constant in a given direction, any one you like depending on how you set up your Gaussian surfaces.


-Richard

I assume that by an "infinite charge distribution" you mean a uniform charge distribution.

I think that as much as anything your analysis demonstrates that it is impossible to have such a distribution, other than a distribution of zero charge. Among other things it would violate the net charge neutrality of the universe.

The example also demonstrates that physics is not mathematics. It is quite possible, as you have done, to set up a hypothetical problem that is mathematically well defined but physically impossible. So, given a physically impossible problem statement, one ought not be very surprised if it results in a violation of physical laws.