EMF in a solenoid

A colleague identified a puzzle concerning Faraday’s law. A circular loop at the center of a long solenoid with time-varying magnetic field will run a current due to the emf, which is the path integral of the (curly) non-Coulomb electric field. Put another loop beside the centered one and the puzzle is that the non-Coulomb electric field must be in the same direction in both loops where the two loops nearly touch, yet current will run in opposite directions in the two loops at the neighboring locations. Here’s a solution to the puzzle.

In the figure, the dashed circle is centered in a solenoid in which there is a time-varying spatially-uniform magnetic field that is increasing with time into the page. By symmetry, the non-Coulomb (NC) electric field is tangent to the dashed circle as indicated by the green arrows. By “non-Coulomb” is meant that this is a curly electric field associated with a time-varying magnetic field, not a non-curly “Coulomb” electric field associated with stationary charges.

At larger radii, $E_{NC}$ increases linearly with increasing radius as indicated by the green arrows to the right of the dashed circle, as can be seen from this calculation, where k is a constant:

$2\pi rE = \pi r^2 dB/dt$

$E = kr$

The emf around the dashed circle, the path integral of the electric field, is

$\text{emf}_1 = (2\pi r_1)(kr_1)$

If we were to put a metal ring with resistance per unit length b where the dashed circle is, there would be a current in the ring of amount

$I_1 = \text{emf}_1/(2\pi r_1 b) = E_1L_1/(2\pi r_1 b) = (k r_1)(2\pi r_1)/(2\pi r_1 b) = kr_1/b$

Next consider the path marked by a blue line, which has been constructed to enclose the same amount of area as is enclosed by the dashed circle:

$\pi r_1^2 = \dfrac{1}{4}(\pi r_2^2 - \pi r_1^2)$

$\dfrac{5}{4}r_1^2 = \dfrac{1}{4}r_2^2$

$r_2 = \sqrt{5}r_1$

Because the two paths enclose the same area, with $dB/dt$ the same everywhere, the emf must be the same around both paths. Moving counterclockwise from the lower left corner of the second path, we can calculate the path integral of the non-Coulomb electric field, and we do indeed find the same emf as that for the circular path:

$\text{emf}_2 = 0 + (kr_2)(2\pi r_2)/4 + 0 + (-kr_1)(2\pi r_1)/4$

$\text{emf}_2 = 2\pi(kr_2^2 - kr_1^2)/4$

$\text{emf}_2 = 2\pi(5kr_1^2 - kr_1^2)/4 = 2\pi kr_1^2$

$\text{emf}_2 = (2\pi r_1)(kr_1) = \text{emf}_1$

If we place a metal wire along the outer path, with the same resistance b per unit length, the current $I_2$ will be less than $I_1$ because the path $L_2$ is longer than the path $L_1$ ($2\pi r_1$):

$L_2 = 2(r_2-r_1) + \dfrac{2\pi r_2}{4} + \dfrac{2\pi r_1}{4} = 1.20L_1$

With the same emf along the longer length of wire, $I_2 = I_1/1.20$.

An important difference between the two configurations of wire is the distribution of polarization charges on the surface of the wire. In the circular case, electrons are accelerated inward along their circular path, so the wire must be polarized transverse to the wire, with the outer part of the wire charged negatively and the inner part charged positively, to provide an outward-pointing Coulomb electric field that accelerates the electrons inward. This transverse polarization is extremely small because in a good metal conductor the drift speed (and therefore the centripetal acceleration $v^2/r$) is extremely small.

The situation is much more complex when a wire follows the outer path. Conservation of charge in the steady state, with a wire of constant cross section and uniform resistivity, means that the current $I_2$ must be the same all along the path. This in turn means that the net electric field, non-Coulomb field plus Coulomb field due to surface charges, must have the same magnitude all along the path, as is indicated by the orange arrows in the diagram. These orange arrows are drawn slightly shorter than the green arrows drawn on the circular path, because the emf is the same but the wire length is 1.20 times as long.

The non-Coulomb electric field $E_{NC}$ polarizes the metal, making the upper left section of the wire positively charged and the lower right section negatively charged. These polarization charges are the source of a Coulomb field $E_{C}$ which, when added to the non-Coulomb field, makes a net electric field that has constant magnitude around the path, corresponding to the constant current around the path. The Coulomb field can be visualized qualitatively by thinking about what Coulomb field is required to add to the non-Coulomb field to give the indicated net field $E_\text{net}$.

Note that the path integral of the Coulomb field $E_\text{C}$ will be zero, because the electric field made by charges has zero curl. That part of the net field that is the non-Coulomb field is solely responsible for the non-zero path integral that is the emf.

For clarity in the diagram the arrows representing electric field are drawn just outside the paths, but in the presence of a wire they represent the electric field inside the wire; surface charges will contribute to non-zero Coulomb field outside the wire with components perpendicular to the wire.

For interactive display of the electric field in the presence of polarization surface charges, in electrostatic and circuit situations including circuit loops inside a solenoid with constant dB/dt, see tinyurl.com/SurfaceCharge.

Bruce Sherwood

This entry was posted in Uncategorized. Bookmark the permalink.

23 Responses to EMF in a solenoid

1. Dusan says:

Hi Professor!

“Conservation of charge in the steady state, with a wire of constant cross section and uniform resistivity, means that the current I_2 must be the same all along the path. This in turn means that the net electric field, non-Coulomb field plus Coulomb field due to surface charges, must have the same magnitude all along the path, as is indicated by the orange arrows in the diagram.”

I understand that, in the steady state, the only component of the resultant electric field is the one that follows the curvature of the wire (since if the perpendicular one were to exist, that would lead to the accumulation of the charge, which would eliminate it).

What is not clear to me, is why that component of the field must have the same magnitude all along the path.

Could you explain to me (qualitatively, physically), in some similar fashion, why electric field must have the same magnitude all along the path? What would happen if that wasn’t the case? How would different magnitude along the path eventually turn into the same magnitude?

• BruceSherwood says:

The current can be expressed as nqAv, where n is the density of mobile charges (coulombs per cubic meter), q is the number of coulombs carried by one mobile charge (-1.6e-19 C for an electron), A is the cross-secional area, and v is the average speed of the mobile charges.

Also, v is proportional to the electric field E; v = uE, where u is called the “mobility”. Putting it all together, the current is nqAuE. If the wire is of uniform composition and constant cross section, then n, q, A, and u are all constant along the wire. If E varied along the wire, the current would vary along the wire, which would mean that for some sections of the wire the current into that section would be different from the current out of that section, which would mean that the surface charge on that section would change, which would mean that the DC steady state had not yet been achieved. The change will be in the direction to make the input and output currents be more nearly equal, and this transient will continue until the two currents (and two field magnitudes) become equal.

Of course if the wire does not have uniform composition or constant cross-sectional area, the electric field would vary along the wire.

2. Dusan says:

______E1_________________x______________E2_________

middle of the wire
____________________________________________________

If this is the wire, and on the left of the “x” is E1, on the right of the “x” is E2, (E1 > E2, electric fields are in the direction of the wire), charge can only be accumulated near the “x”, but how can he affect the component of the field in the direction of the wire?

That accumulated charge near the “x” can only affect the electric field perpendicular to the path of the wire. That’s what’s confusing me.

• BruceSherwood says:

This (temporary) situation will create a ring of positive surface charge around the wire at and near location x. This ring of charge will contribute a pattern of electric field that has large horizontal components. See the following program, that shows the remarkable pattern of electric field inside a series of charged rings (with a gradient of charge along the rings):

http://www.glowscript.org/#/user/matterandinteractions/folder/matterandinteractions/program/18-Erings

• BruceSherwood says:

You might find it instructive to study the patterning of surface charge and electric field in circuits and electrostatic situations shown at tinyurl.com/SurfaceCharge

• Dusan says:

And E_ring will be oriented against E1, and will be in the same orientation as E2? So the charges will accumulate, E_ring will grow, until there is the same electric field left and right from the “x”?

• BruceSherwood says:

Yes, the ring field will reduce E1 and increase E2 until E1 and E2 become equal. We know that in the steady state E1 MUST be equal to E2; if they are not equal, the system has not reached the steady state. It’s a fedback system.

• Dusan says:

Thank you professor. I really appreciate your work!

3. Edin says:

I was reading discussion you had with Dusan, and got little confused.

a) Wouldn’t that ring of charge also force current to occupy smaller area when passing through him (which is something we don’t want, and it’s not something
accumulated charge wanted in the first place)? Ring of charge has horizontal component of electric field (in direction of the wire), but also the vertical one (which I think is not something we would want).

b) Looking at this conversation, if we have a long, straight wire, that has for example a linearly decreasing electric field along her, rings of charge that were accumulated to even-out electric field would all be equally charged (there wouldn’t be any gradient of surface charge, if I’m not mistaken?) How are then those rings with gradient of surface charge made (the ones from your book, and in one of the links you mentioned)?

The second question is the one I would like a little more detailed explanation if it’s possible.

Edin

4. BruceSherwood says:

a) The field inside a line of charged rings, with a gradient of charge along the rings, has no vertical component. See this:

http://www.glowscript.org/#/user/matterandinteractions/folder/matterandinteractions/program/18-Erings

b) In the steady state of a DC circuit a long straight wire of constant composition and constant cross section cannot have a “linearly decreasing electric field”, because that would mean that the current entering one end of the wire would be greater than the current leaving the other end of the wire, since the current is proportional to the density n of mobile charges, the charge q of each mobile charge, the cross-sectional area A of the wire, the mobility u of those charges, and the electric field E: current = nqAuE, where the drift speed v = uE. In order to produce the required pattern of electric field in the long wire, the transient leads to a constant gradient of surface charge.

• Edin says:

I’m sorry to bother, but it’s not obvious to me why would “the transient lead to a constant gradient of surface charge”? What is the essence of it? Is electric field (I’m talking before the steady state) strongest at one end of the wire, so that will make the first ring at that end charged the most? I just don’t understand the mechanism behind it.

When I mentioned linearly decreasing electric field, I was thinking before the steady state also.

5. BruceSherwood says:

Suppose the long wire does indeed have (temporarily) a decreasing electric field along its length. To simplify the discussion, let’s suppose that the mobile charges are positive (“holes”), just to avoid sign issue. The strong electric field at the start of the wire drives a large current down the wire, and since the field in this transient situation has components perpendicular to the wire, some positive mobile charges will be driven onto the downstream surface of the wire. The new rings of positive charge contribute an electric field in the opposite direction to the original field near the start of the wire, so that the net field near the start of the wire will be reduced, and the electric field further down the wire will be increased. Already we see that the electic field in the wire is becoming more uniform along the wire than it was at the start of the process.

At tinyurl.com/SurfaceCharge you can see the results of an offline computation of the steady-state distribution of surface charge, obtained by iteratively determining the transient currents that change the surface charge distribution. The iteration is terminated when additional steps are seen to make little or no difference.

• Edin says:

But why the first ring has to have a bigger charge than the last one? All rings could have equal surface charge along the wire, and still it will be true what you said that “the net field near the start of the wire will be reduced, and the electric field further down the wire will be increased”.

6. BruceSherwood says:

Concerning the steady state: If the rings all have equal charge, the electric field inside the (very long) wire will be zero, which would be the case for a long wire that is charged but not connected to other circuit elements. For a circuit, there exist analytic proofs that the surface charge density for a long straight wire must in fact have a constant gradient, and the VPython program I referenced demonstrates this effect.

I was addressing your question of how this steady state comes into being, and I offered a mechanism, namely that if the electric field at some moment decreases as you move along the wire, the first thing that happens is that the field configuration will necessarily change the surface charge distribution in such a way as to decrease the field at the starting end and increase the field farther down the wire, thereby reducing the variation of the field along the wire.

7. Daniel says:

Hi professor. What about the emf inside a current carrying solenoid (a self-inducting one)?
On page 919 (the inductance section of Chapter 20) of “Matter and Interactions 4th Edition” book, it says that the non-Coulomb electric field polarizes the solenoid and these new surface charges produce a Coulomb field that’s opposite to the non-Coulomb field and that they’re almost equal in magnitude.

“If the current is increasing, the non-Coulomb electric field E⃗
NC curls clockwise
around the solenoid, opposing the increase in the current and polarizing the
solenoid (Figure 22.40). The new surface charges produce a Coulomb electric
field E⃗
C that follows the wire and points opposite to the non-Coulomb field.
If the resistance of the solenoid wire is very small, the magnitude of the two
electric fields is nearly equal (EC ≈ ENC).
Note the similarity to a battery: a non-Coulomb emfind maintains a charge
separation, and there is a potential drop ∆Vsol along the solenoid that is
numerically equal to emfind. If the wire resistance rsol is significant, we have
∆Vsol = emfind − rsolI, just like a battery with internal resistance. You can
think of this self-induced emf as making the solenoid act like a battery that
has been put in the circuit “backwards,” opposing the change in the current ”

But how can there be a potential drop if the net field would be nearly zero? I can’t quite understand how. I’d really appreciate it if you’d answer my question. ^ ^

8. BruceSherwood says:

This is a subtle point. Potential difference is determined by the path integral of the Coulomb field (the field made by charges); the non-Coulomb field does not contribute to the potential difference. A simpler case is that of a “mechanical battery” discussed in the textbook, a device where a charge separation is produced and maintained by turning a crank to move charges along a conveyor belt, charging up the ends of the “battery”. Suppose the battery is connected to a resistive wire. By turning the crank you maintain a gradient of surface charge along the wire, which produces a Coulomb field inside the wire. The potential difference along the wire is equal and opposite to the potential difference across the battery associated with the buildup of charges at the ends of the battery. To repeat, the Coulomb electric field in the battery is in the opposite direction to the Coulomb electric field in the wire. The round-trip integral of the Coulomb electric field is zero, as is always the case with the Coulomb field. In the mechanical battery case your cranking maintains this situation. In the case of the time-varying magnetic field in the solenoid, it is the non-Coulomb electric field that maintains the situation.

9. Daniel says:

So, at first there’s a steady current in the solenoid wire so there will be a charge gradient. Then as the current changes(and therefore the magnetic field) there’s a non-Coulomb electric field that alters the charge gradient by separating the surface charges a bit more but the potential drop contributed by E(NC) is through the E(C) due to the separated charges and therefore there will be a sizeable potential drop through the solenoid.

Is it correct professor?

10. BruceSherwood says:

I wouldn’t say it that way. When you first attach the solenoid to the power supply, the current rises from zero to some nonzero value; there is initially no “steady current”. The rising current is associated with a time-varying magnetic field that is associated with a non-Coulomb curly electric field that polarizes the solenoid, establishing a gradient of surface charge along the solenoid, resulting in a Coulomb electric field in the solenoid wire (the integral of this Coulomb field along the solenoid is the potential difference across the solenoid). After the transient, as long as the power supply can continue to increase its voltage output at a constant rate, there will be a steady current in the solenoid.

• Daniel says:

After the transient, will what i thought be true? Also how does the solenoid have a steady current if the voltage output from power supply is increasing?

• BruceSherwood says:

You are right; I misspoke. For dB/dt to remain constant requires that dI/dt remain constant, which means that I must steadily increase.

11. Daniel says:

Ahh sorry I forgot about the dI/dt part. Thanks professor! The mechanical battery analogy helped a lot. Sorry to bother but one last question. Why is there a curly electric field whenever a magnetic flux changes? It feels like the curly field pops out of nowhere. Is it a fundamental law or is there something I don’t know yet?

• BruceSherwood says:

Faraday’s law is a fundamental law, one of four “Maxwell’s equations” that together encompass the nature of electric and magnetic interactions. It is no more and no less mysterious than Gauss’s law and Ampere’s law (two more of the four “Maxwell’s equations”). It isn’t something to be “explained”; Faraday’s law embodies mathematically the experimental, observational fact that a time-varying magnetic field is accompanied by a curly electric field. See the chapter on Faraday’s law in our textbook, and then see the following chapter on electromagnetic radiation for the full exposition of Maxwell’s equations.

12. Daniel says:

Thanks a lot professor. Thanks for being patient with me too. Have a nice day! ^_^