An introduction to electric circuits

In my previous post,I introduced electric motors, generators and transformers. They all work because of Faraday’s flux rule: a changing magnetic flux will produce some circulation of the electric field. The formula for the flux rule is given below:

emf

It is a wonderful thing, really, but not easy to grasp intuitively. It’s one of these equations where I should quote Feynman’s introduction to electromagnetism: “The laws of Newton were very simple to write down, but they had a lot of complicated consequences and it took us a long time to learn about them all. The laws of electromagnetism are not nearly as simple to write down, which means that the consequences are going to be more elaborate and it will take us quite a lot of time to figure them all out.”

Now, among Maxwell’s Laws, this is surely the most complicated one! However, that shouldn’t deter us. 🙂 Recalling Stokes’ Theorem helps to appreciate what the integral on the left-hand side represents:

Stokes theorem

We’ve got a line integral around some closed loop Γ on the left and, on the right, we’ve got a surface integral over some surface S whose boundary is Γ. The illustration below depicts the geometry of the situation. You know what it all means. If not, I am afraid I have to send you back to square one, i.e. my posts on vector analysis. Yep. Sorry. Can’t keep copying stuff and make my posts longer and longer. 🙂

Diagram stokesTo understand the flux rule, you should imagine that the loop Γ is some loop of electric wire, and then you just replace C by E, the electric field vector. The circulation of E, which is caused by the change in magnetic flux, is referred to as the electromotive force (emf), and it’s the tangential force (E·ds) per unit charge in the wire integrated over its entire length around the loop, which is denoted by Γ here, and which encloses a surface S.

Now, you can go from the line integral to the surface integral by noting Maxwell’s Law: −∂B/∂t = ×E. In fact, it’s the same flux rule really, but in differential form. As for (×E)n, i.e. the component of ×E that is normal to the surface, you know that any vector multiplied with the normal unit vector will yield its normal component. In any case, if you’re reading this, you should already be acquainted with all of this. Let’s explore the concept of the electromotive force, and then apply it our first electric circuit. 🙂

Indeed, it’s now time for a small series on circuits, and so we’ll start right here and right now, but… Well… First things first. 🙂

The electromotive force: concept and units

The term ‘force’ in ‘electromotive force’ is actually somewhat misleading. There is a force involved, of course, but the emf is not a force. The emf is expressed in volts. That’s consistent with its definition as the circulation of E: a force times a distance amounts to work, or energy (one joule is one newton·meter), and because E is the force on a unit charge, the circulation of E is expressed in joule per coulomb, so that’s a voltage: 1 volt = 1 joule/coulomb. Hence, on the left-hand side of Faraday’s equation, we don’t have any dimension of time: it’s energy per unit charge, so it’s x joule per coulomb . Full stop.

On the right-hand side, however, we have the time rate of change of the magnetic flux. through the surface S. The magnetic flux is a surface integral, and so it’s a quantity expressed in [B]·m2, with [B] the measurement unit for the magnetic field strength. The time rate of change of the flux is then, of course, expressed in [B]·mper second, i.e. [B]·m2/s. Now what is the unit for the magnetic field strength B, which we denoted by [B]?

Well… [B] is a bit of a special unit: it is not measured as some force per unit charge, i.e. in newton per coulomb, like the electric field strength E. No. [B] is measured in (N/C)/(m/s). Why? Because the magnetic force is not F = qE but F = qv×B. Hence, so as to make the units come out alright, we need to express B in (N·s)/(C·m), which is a unit known as the tesla (1 T = N·s/C·m), so as to honor the Serbian-American genius Nikola Tesla. [I know it’s a bit of short and dumb answer, but the complete answer is quite complicated: it’s got to do with the relativity of the magnetic force, which I explained in another post: both the v in F = qv×B equation as well as the m/s unit in [B] should make you think: whose velocity? In which reference frame? But that’s something I can’t summarize in two lines, so just click the link if you want to know more. I need to get back to the lesson.]

Now that we’re talking units, I should note that the unit of flux also got a special name, the weber, so as to honor one of Germany’s most famous physicists, Wilhelm Eduard Weber: as you might expect, 1 Wb = 1 T·m2. But don’t worry about these strange names. Besides the units you know, like the joule and the newton, I’ll only use the volt, which got its name to honor some other physicist, Alessandro Volta, the inventor of the electrical battery. Or… Well… I might mention the watt as well at some point… 🙂

So how does it work? On one side, we have something expressed per second – so that’s per unit time – and on the other we have something that’s expressed per coulomb – so that’s per unit charge. The link between the two is the power, so that’s the time rate of doing work. It’s expressed in joule per second. So… Well… Yes. Here we go: in honor of yet another genius, James Watt, the unit of power got its own special name too: the watt. 🙂 In the argument below, I’ll show that the power that is being generated by a generator, and that is being consumed in the circuit (through resistive heating, for example, or whatever else taking energy out of the circuit) is equal to the emf times the current. For the moment, however, I’ll just assume you believe me. 🙂

We need to look at the whole circuit now, indeed, in which our little generator (i.e. our loop or coil of wire) is just one of the circuit elements. The units come out alright: the poweremf·current product is expressed in volt·coulomb/second = (joule/coulomb)·(coulomb/second) = joule/second. So, yes, it looks OK. But what’s going on really? How does it work, literally?

A short digression: on Ohm’s Law and electric power

Well… Let me first recall the basic concepts involved which, believe it or not, are probably easiest to explain by briefly recalling Ohm’s Law, which you’ll surely remember from your high-school physics classes. It’s quite simple really: we have some resistance in a little circuit, so that’s something that resists the passage of electric current, and then we also have a voltage source. Now, Ohm’s Law tells us that the ratio of (i) the voltage V across the resistance (so that’s between the two points marked as + and −) and (ii) the current I will be some constant. It’s the same as saying that V and I are inversely proportional to each other.  The constant of proportionality is referred to as the resistance itself and, while it’s often looked at as a property of the circuit itself, we may embody it in a circuit element itself: a resistor, as shown below.

120px-OhmsLaw

So we write R = V/I, and the brief presentation above should remind you of the capacity of a capacitor, which was just another constant of proportionality. Indeed, instead of feeding a resistor (so all energy gets dissipated away), we could charge a capacitor with a voltage source, so that’s a energy storage device, and then we find that the ratio between (i) the charge on the capacitor and (ii) the voltage across the capacitor was a constant too, which we defined as the capacity of the capacitor, and so we wrote C = Q/V. So, yes, another constant of proportionality (there are many in electricity!).

In any case, the point is: to increase the current in the circuit above, you need to increase the voltage, but increasing both amounts to increasing the power that’s being consumed in the circuit, because the power is voltage times current indeed, so P = V·I (or v·i, if I use the small letters that are used in the two animations below). For example, if we’d want to double the current, we’d need to double the voltage, and so we’re quadrupling the power: (2·V)·(2·I) = 22·V·I. So we have a square-cube law for the power, which we get by substituting V for R·I or by substituting I for V/R, so we can write the power P as P = V2/R = I2·R. This square-cube law says exactly the same: if you want to double the voltage or the current, you’ll actually have to double both and, hence, you’ll quadruple the power. Now let’s look at the animations below (for which credit must go to Wikipedia).

Electric_power_source_animation_1 Electric_load_animation_2

They show how energy is being used in an electric circuit in  terms of power. [Note that the little moving pluses are in line with the convention that a current is defined as the movement of positive charges, so we write I = dQ/dt instead of I = −dQ/dt. That also explains the direction of the field line E, which has been added to show that the power source effectively moves charges against the field and, hence, against the electric force.] What we have here is that, on one side of the circuit, some generator or voltage source will create an emf pushing the charges, and then some load will consume their energy, so they lose their push. So power, i.e. energy per unit time, is supplied, and is then consumed.

Back to the emf…

Now, I mentioned that the emf is a ratio of two terms: the numerator is expressed in joule, and the denominator is expressed in coulomb. So you might think we’ve got some trade-off here—something like: if we double the energy of half of the individual charges, then we still get the same emf. Or vice versa: we could, perhaps, double the number of charges and load them with only half the energy. One thing is for sure: we can’t both.

Hmm… Well… Let’s have a look at this line of reasoning by writing it down more formally.

  1. The time rate of change of the magnetic flux generates some emf, which we can and should think of as a property of the loop or the coil of wire in which it is being generated. Indeed, the magnetic flux through it depends on its orientation, its size, and its shape. So it’s really very much like the capacity of a capacitor or the resistance of a conductor. So we write: emf = Δ(flux)/Δt. [In fact, the induced emf tries to oppose the change in flux, so I should add the minus sign, but you get the idea.]
  2. For a uniform magnetic field, the flux is equal to the field strength B times the surface area S. [To be precise, we need to take the normal component of B, so the flux is B·S = B·S·cosθ.]  So the flux can change because of a change in B or because of a change in S, or because of both.
  3. The emf = Δ(flux)/Δt formula makes it clear that a very slow change in flux (i.e. the same Δ(flux) over a much larger Δt) will generate little emf. In contrast, a very fast change (i.e. the the same Δ(flux) over a much smaller Δt) will produce a lot of emf. So, in that sense, emf is not like the capacity or resistance, because it’s variable: it depends on Δ(flux), as well as on Δt. However, you should still think of it as a property of the loop or the ‘generator’ we’re talking about here.
  4. Now, the power that is being produced or consumed in the circuit in which our ‘generator’ is just one of the elements, is equal to the emf times the current. The power is the time rate of change of the energy, and the energy is the work that’s being done in the circuit (which I’ll denote by ΔU), so we write: emf·current = ΔU/Δt.
  5. Now, the current is equal to the time rate of change of the charge, so I = ΔQ/Δt. Hence, the emf is equal to emf = (ΔU/Δt)/I = (ΔU/Δt)/(ΔQ/Δt) = ΔU/ΔQ. From this, it follows that: emf = Δ(flux)/Δt = ΔU/ΔQ, which we can re-write as:

Δ(flux) = ΔU·Δt/ΔQ

What this says is the following. For a given amount of change in the magnetic flux (so we treat Δ(flux) as constant in the equation above), we could do more work on the same charge (ΔQ) – we could double ΔU by moving the same charge over a potential difference that’s twice as large, for example – but then Δt must be cut in half. So the same change in magnetic flux can do twice as much work if the change happens in half of the time.

Now, does that mean the current is being doubled? We’re talking the same ΔQ and half the Δt, so… Well? No. The Δt here measures the time of the flux change, so it’s not the dt in I = dQ/dt. For the current to change, we’d need to move the same charge faster, i.e. over a larger distance over the same time. We didn’t say we’d do that above: we only said we’d move the charge across a larger potential difference: we didn’t say we’d change the distance over which they are moved.

OK. That makes sense. But we’re not quite finished. Let’s first try something else, to then come back to where we are right now via some other way. 🙂 Can we change ΔQ? Here we need to look at the physics behind. What’s happening really is that the change in magnetic flux causes an induced current which consists of the free electrons in the Γ loop. So we have electrons moving in and out of our loop, and through the whole circuit really, but so there’s only so many free electrons per unit length in the wire. However, if we would effectively double the voltage, then their speed will effectively increase proportionally, so we’ll have more of them passing through per second. Now that effect surely impacts the current. It’s what we wrote above: all other things being the same, including the resistance, then we’ll also double the current as we double the voltage.

So where is that effect in the flux rule? The answer is: it isn’t there. The circulation of E around the loop is what it is: it’s some energy per unit charge. Not per unit time. So our flux rule gives us a voltage, which tells us that we’re going to have some push on the charges in the wire, but it doesn’t tell us anything about the current. To know the current, we must know the velocity of the moving charges, which we can calculate from the push if we also get some other information (such as the resistance involved, for instance), but so it’s not there in the formula of the flux rule. You’ll protest: there is a Δt on the right-hand side! Yes, that’s true. But it’s not the Δt in the v = Δs/Δt equation for our charges. Full stop.

Hmm… I may have lost you by now. If not, please continue reading. Let me drive the point home by asking another question. Think about the following: we can re-write that Δ(flux) = ΔU·Δt/ΔQ equation above as Δ(flux) = (ΔU/ΔQ)·Δt equation. Now, does that imply that, with the same change in flux, i.e. the same Δ(flux), and, importantly, for the same Δt, we could double both ΔU as well as ΔQ? I mean: (2·ΔU)/(2·ΔQ) = ΔU/ΔQ and so the equation holds, mathematically that is. […] Think about it.

You should shake your head now, and rightly so, because, while the Δ(flux) = (ΔU/ΔQ)·Δt equation suggests that would be possible, it’s totally counter-intuitive. We’re changing nothing in the real world (what happens there is the same change of flux in the same amount of time), but so we’d get twice the energy and twice the charge ?! Of course, we could also put a 3 there, or 20,000, or minus a million. So who decides on what we get? You get the point: it is, indeed, not possible. Again, what we can change is the speed of the free electrons, but not their number, and to change their speed, you’ll need to do more work, and so the reality is that we’re always looking at the same ΔQ, so if we want a larger ΔU, then we’ll need a larger change in flux, or we a shorter Δt during which that change in flux is happening.

So what can we do? We can change the physics of the situation. We can do so in many ways, like we could change the length of the loop, or its shape. One particularly interesting thing to do would be to increase the number of loops, so instead of one loop, we could have some coil with, say, N turns, so that’s N of these Γ loops. So what happens then? In fact, contrary to what you might expect, the ΔQ still doesn’t change as it moves into the coil and then from loop to loop to get out and then through the circuit: it’s still the same ΔQ. But the work that can be done by this current becomes much larger. In fact, two loops give us twice the emf of one loop, and N loops give us N times the emf of one loop. So then we can make the free electrons move faster, so they cover more distance in the same time (and you know work is force times distance), or we can move them across a larger potential difference over the same distance (and so then we move them against a larger force, so it also implies we’re doing more work). The first case is a larger current, while the second is a larger voltage. So what is it going to be?

Think about the physics of the situation once more: to make the charges move faster, you’ll need a larger force, so you’ll have a larger potential difference, i.e. a larger voltage. As for what happens to the current, I’ll explain that below. Before I do, let me talk some more basics.

In the exposé below, we’ll talk about power again, and also about load. What is load? Think about what it is in real life: when buying a battery for a big car, we’ll want a big battery, so we don’t look at the voltage only (they’re all 12-volt anyway). We’ll look at how many ampères it can deliver, and for how long. The starter motor in the car, for example, can suck up like 200 A, but for a very short time only, of course, as the car engine itself should kick in. So that’s why the capacity of batteries is expressed in ampère-hours.

Now, how do we get such large currents, such large loads? Well… Use Ohm’s Law: to get 200 A at 12 V, the resistance of the starter motor will have to as low as 0.06 ohm. So large currents are associated with very low resistance. Think practical: a 240-volt 60 watt light-bulb will suck in 0.25 A, and hence, its internal resistance, is about 960 Ω. Also think of what goes on in your house: we’ve got a lot of resistors in parallel consuming power there. The formula for the total resistance is 1/Rtotal = 1/R+ 1/R+ 1/R+ … So more appliances is less resistance, so that’s what draws in the larger current.

The point is: when looking at circuits, emf is one thing, but energy and power, i.e. the work done per second, are all that matters really. And so then we’re talking currents, but our flux rule does not say how much current our generator will produce: that depends on the load. OK. We really need to get back to the lesson now.

A circuit with an AC generator

The situation is depicted below. We’ve got a coil of wire of, let’s say, N turns of wire, and we’ll use it to generate an alternating current (AC) in a circuit.

AC generatorCircuit

The coil is really like the loop of wire in that primitive electric motor I introduced in my previous post, but so now we use the motor as a generator. To simplify the analysis, we assume we’ll rotate our coil of wire in a uniform magnetic field, as shown by the field lines B.

motor

Now, our coil is not a loop, of course: the two ends of the coil are brought to external connections through some kind of sliding contacts, but that doesn’t change the flux rule: a changing magnetic flux will produce some emf and, therefore, some current in the coil.

OK. That’s clear enough. Let’s see what’s happening really. When we rotate our coil of wire, we change the magnetic flux through it. If S is the area of the coil, and θ is the angle between the magnetic field and the normal to the plane of the coil, then the flux through the coil will be equal to B·S·cosθ. Now, if we rotate the coil at a uniform angular velocity ω, then θ varies with time as θ = ω·t. Now, each turn of the coil will have an emf equal to the rate of change of the flux, i.e. d(B·S·cosθ)/dt. We’ve got N turns of wire, and so the total emf, which we’ll denote by Ɛ (yep, a new symbol), will be equal to:

Formula emfNow, that’s just a nice sinusoidal function indeed, which will look like the graph below.

graph (1)

When no current is being drawn from the wire, this Ɛ will effectively be the potential difference between the two wires. What happens really is that the emf produces a current in the coil which pushes some charges out to the wire, and so then they’re stuck there for a while, and so there’s a potential difference between them, which we’ll denote by V, and that potential difference will be equal to Ɛ. It has to be equal to Ɛ because, if it were any different, we’d have an equalizing counter-current, of course. [It’s a fine point, so you should think about it.] So we can write:

formula VSo what happens when we do connect the wires to the circuit, so we’ve got that closed circuit depicted above (and below)?

Circuit

Then we’ll have a current I going through the circuit, and Ohm’s Law then tells us that the ratio between (i) the voltage across the resistance in this circuit (we assume the connections between the generator and the resistor itself are perfect conductors) and (ii) the current will be some constant, so we have R = V/I and, therefore:

Formula AC generator

[To be fully complete, I should note that, when other circuit elements than resistors are involved, like capacitors and inductors, we’ll have a phase difference between the voltage and current functions, and so we should look at the impedance of the circuit, rather than its resistance. For more detail, see the addendum below this post.]

OK. Let’s now look at the power and energy involved.

Energy and power in the AC circuit

You’ll probably have many questions about the analysis above. You should. I do. The most remarkable thing, perhaps, is that this analysis suggests that the voltage doesn’t drop as we connect the generator to the circuit. It should. Why not? Why do the charges at both ends of the wire simply discharge through the circuit? In real life, there surely is such tendencysudden large changes in loading will effectively produce temporary changes in the voltage. But then it’s like Feynman writes: “The emf will continue to provide charge to the wires as current is drawn from them, attempting to keep the wires always at the same potential difference.”

So how much current is drawn from them? As I explained above, that depends not on the generator but on the circuit, and more in particular on the load, so that’s the resistor in this case. Again, the resistance is the (constant) ratio of the voltage and the current: R = V/I. So think about increasing or decreasing the resistance. If the voltage remains the same, it implies the current must decrease or increase accordingly, because R = V/I implies that I = V/R. So the current is inversely proportional to R, as I explained above when discussing car batteries and lamps and loads. 🙂

Now, I still have to prove that the power provided by our generator is effectively equal to P = Ɛ·I but, if it is, it implies the power that’s being delivered will be inversely proportional to R. Indeed, when Ɛ and/or V remain what they are as we insert a larger resistance in the circuit, then P = Ɛ·I = Ɛ2/R, and so the power that’s being delivered would be inversely proportional to R. To be clear, we’d have a relation between P and R like the one below.

Capture

This is somewhat weird. Why? Well… I also have to show you that the power that goes into moving our coil in the magnetic field, i.e. the rate of mechanical work required to rotate the coil against the magnetic forces, is equal to the electric power Ɛ·I, i.e. the rate at which electrical energy is being delivered by the emf of the generator. However, I’ll postpone that for a while and, hence, I’ll just ask you, once again, to take me on my word. 🙂 Now, if that’s true, so if the mechanical power equals the electric power, then that implies that a larger resistance will reduce the mechanical power we need to maintain the angular velocity ω. Think of a practical example: if we’d double the resistance (i.e. we halve the load), and if the voltage stays the same, then the current would be halved, and the power would also be halved. And let’s think about the limit situations: as the resistance goes to infinity, the power that’s being delivered goes to zero, as the current goes to zero, while if the resistance goes to zero, both the current as well as the power would go to infinity!

Well… We actually know that’s also true in real-life: actual generators consume more fuel when the load increases, so when they deliver more power, and much less fuel, so less power, when there’s no load at all. You’ll know that, at least when you’re living in a developing country with a lot of load shedding! 🙂 And the difference is huge: no or just a little load will only consume 10% of what you need when fully loading it. It’s totally in line with what I wrote on the relationship between the resistance and the current that it draws in. So, yes, it does make sense:

An emf does produce more current if the resistance in the circuit is low (so i.e. when the load is high), and the stronger currents do represent greater mechanical forces.

That’s a very remarkable thing. It means that, if we’d put a larger load on our little AC generator, it should require more mechanical work to keep the coil rotating at the same angular velocity ω. But… What changes? The change in flux is the same, the Δt is the same, and so what changes really? What changes is the current going through the coil, and it’s not a change in that ΔQ factor above, but a change in its velocity v.

Hmm… That all looks quite complicated, doesn’t it? It does, so let’s get back to the analysis of what we have here, so we’ll simply assume that we have some dynamic equilibrium obeying that formula above, and so I and R are what they are, and we relate them to Ɛ according to that equation above, i.e.:

Formula AC generator

Now let me prove those formulas on the power of our generator and in the circuit. We have all these charges in our coil that are receiving some energy. Now, the rate at which they receive energy is F·v.

Huh? Yes. Let me explain: the work that’s being done on a charge along some path is the line integral ∫ F·ds along this path. But the infinitesimal distance ds is equal to v·dt, as ds/dt = v (note that we write s and v as vectors, so the dot product with F gives us the component of F that is tangential to the path). So ∫ F·ds = ∫ (F·v)dt. So the time rate of change of the energy, which is the power, is F·v. Just take the time derivative of the integral. 🙂

Now let’s assume we have n moving charges per unit length of our coil (so that’s in line with what I wrote about ΔQ above), then the power being delivered to any element ds of the coil is (F·v)·n·ds, which can be written as: (F·ds)·n·v. [Why? Because v and ds have the same direction: the direction of both vectors is tangential to the wire, always.] Now all we need to do to find out how much power is being delivered to the circuit by our AC generator is integrate this expression over the coil, so we need to find:

power

However, the emf (Ɛ) is defined as the line integral ∫ E·ds line, taken around the entire coil, and = F/q, and the current I is equal to I = q·n·v. So the power from our little AC generator is indeed equal to:

Power = Ɛ·I

So that’s done. Now I need to make good on my other promise, and that is to show that Ɛ·I product is equal to the mechanical power that’s required to rotate the coil in the magnetic field. So how do we do that?

We know there’s going to be some torque because of the current in the coil. It’s formula is given by τ = μ×B. What magnetic field? Well… Let me refer you to my post on the magnetic dipole and its torque: it’s not the magnetic field caused by the current, but the external magnetic field, so that’s the B we’ve been talking about here all along. So… Well… I am not trying to fool you here. 🙂 However, the magnetic moment μ was not defined by that external field, but by the current in the coil and its area. Indeed, μ‘s magnitude was the current times the area, so that’s N·I·S in this case. Of course, we need to watch out because μ is a vector itself and so we need the angle between μ and B to calculate that vector cross product τ = μ×B. However, if you check how we defined the direction of μ, you’ll see it’s normal to the plane of the coil and, hence, the angle between μ and B is the very same θ = ω·t that we started our analysis with. So, to make a long story short, the magnitude of the torque τ is equal to:

τ = (N·I·S)·B·sinθ

Now, we know the torque is also equal to the work done per unit of distance traveled (around the axis of rotation, that is), so τ = dW/dθ. Now dθ = d(ω·t) = ω·dt. So we can now find the work done per unit of time, so that’s the power once more:

dW/dt = ω·τ = ω·(N·I·S)·B·sinθ

But so we found that Ɛ = N·S·B·ω·sinθ, so… Well… We find that:

dW/dt = Ɛ·I

Now, this equation doesn’t sort out our question as to how much power actually goes in and out of the circuit as we put some load on it, but it is what we promised to do: I showed that the mechanical work we’re doing on the coil is equal to the electric energy that’s being delivered to the circuit. 🙂

It’s all quite mysterious, isn’t it? It is. And we didn’t include other stuff that’s relevant here, such as the phenomenon of self-inductance: the varying current in the coil will actually produce its own magnetic field and, hence, in practice, we’d get some “back emf” in the circuit. This “back emf” is opposite to the current when it is increasing, and it is in the direction of the current when it is decreasing. In short, the self-inductance effect causes a current to have ‘inertia’: the inductive effects try to keep the flow constant, just as mechanical inertia tries to keep the velocity of an object constant. But… Well… I left that out. I’ll take about next time because…

[…] Well… It’s getting late in the day, and so I must assume this is sort of ‘OK enough’ as an introduction to what we’ll be busying ourselves with over the coming week. You take care, and I’ll talk to you again some day soon. 🙂

Perhaps one little note, on a question that might have popped up when you were reading all of the above: so how do actual generators keep the voltage up? Well… Most AC generators are, indeed, so-called constant speed devices. You can download some manuals from the Web, and you’ll find things like this: don’t operate at speeds above 4% of the rated speed, or more than 1% below the rated speed. Fortunately, the so-called engine governor will take car of that. 🙂

Addendum: The concept of impedance

In one of my posts on oscillators, I explain the concept of impedance, which is the equivalent of resistance, but for AC circuits. Just like resistance, impedance also sort of measures the ‘opposition’ that a circuit presents to a current when a voltage is applied, but it’s a complex ratio, as opposed to R = V/I. It’s literally a complex ratio because the impedance has a magnitude and a direction, or a phase as it’s usually referred to. Hence, one will often write the impedance (denoted by Z) using Euler’s formula:

Z = |Z|eiθ

The illustration below (credit goes to Wikipedia, once again) explains what’s going on. It’s a pretty generic view of the same AC circuit. The truth is: if we apply an alternating current, then the current and the voltage will both go up and down, but the current signal will usually lag the voltage signal, and the phase factor θ tells us by how much. Hence, using complex-number notation, we write:

V = IZ = I∗|Z|eiθ

General_AC_circuit

Now, while that resembles the V = R·I formula, you should note the bold-face type for V and I, and the ∗ symbol I am using here for multiplication. First the ∗ symbol: that’s to make it clear we’re not talking a vector cross product A×B here, but a product of two complex numbers. The bold-face for V and I implies they’re like vectors, or like complex numbers: so they have a phase too and, hence, we can write them as:

  • = |V|ei(ωt + θV)
  • = |I|ei(ωt + θI)

To be fully complete – you may skip all of this if you want, but it’s not that difficult, nor very long – it all works out as follows. We write:

IZ = |I|ei(ωt + θI)∗|Z|eiθ = |I||Z|ei(ωt + θ+ θ) = |V|ei(ωt + θV)

Now, this equation must hold for all t, so we can equate the magnitudes and phases and, hence, we get: |V| = |I||Z| and so we get the formula we need, i.e. the phase difference between our function for the voltage and our function for the current.

θ= θI + θ

Of course, you’ll say: voltage and current are something real, isn’t it? So what’s this about complex numbers? You’re right. I’ve used the complex notation only to simplify the calculus, so it’s only the real part of those complex-valued functions that counts.

Oh… And also note that, as mentioned above, we do not have such lag or phase difference when only resistors are involved. So we don’t need the concept of impedance in the analysis above. With this addendum, I just wanted to be as complete as I can be. 🙂

Induced currents

In my two previous posts, I presented all of the ingredients of the meal we’re going to cook now, most notably:

  1. The formula for the torque on a loop of a current in a magnetic field, and its energy: (i) τ = μ×B, and (ii) Umech = −μ·B.
  2. The Biot-Savart Law, which gives you the magnetic field that’s produced by wires carrying currents:

B formula 2

Both ingredients are, obviously, relevant to the design of an electromagnetic motor, i.e. an ‘engine that can do some work’, as Feynman calls it. 🙂 Its principle is illustrated below.

motor

The two formulas above explain how and why the coil go around, and the coil can be made to keep going by arranging that the connections to the coil are reversed each half-turn by contacts mounted on the shaft. Then the torque is always in the same direction. That’s how a small direct current (DC) motor is made. My father made me make a couple of these thirty years ago, with a magnet, a big nail and some copper coil. I used sliding contacts, and they were the most difficult thing in the whole design. But now I found a very nice demo on YouTube of a guy whose system to ‘reverse’ the connections is wonderfully simple: he doesn’t use any sliding contacts. He just removes half of the insulation on the wire of the coil on one side. It works like a charm, but I think it’s not so sustainable, as it spins so fast that the insulation on the other side will probably come off after a while! 🙂

Now, to make this motor run, you need current and, hence, 19th century physicists and mechanical engineers also wondered how one could produce currents by changing the magnetic field. Indeed, they could use Alessandro Volta’s ‘voltaic pile‘ to produce currents but it was not very handy: it consisted of alternating zinc and copper discs, with pieces of cloth soaked in salt water in-between!

Now, while the Biot-Savart Law goes back to 1820, it took another decade to find out how that could be done. Initially, people thought magnetic fields should just cause some current, but that didn’t work. Finally, Faraday unequivocally established the fundamental principle that electric effects are only there when something is changingSo you’ll get a current in a wire by moving it in a magnetic field, or by moving the magnet or, if the magnetic field is caused by some other current, by changing the current in that wire. It’s referred to as the ‘flux rule’, or Faraday’s Law. Remember: we’ve seen Gauss’ Law, then Ampère’s Law, and then that Biot-Savart Law, and so now it’s time for Faraday’s Law. 🙂 Faraday’s Law is Maxwell’s third equation really, aka as the Maxwell-Faraday Law of Induction:

×E = −∂B/∂t

Now you’ll wonder: what’s flux got to do with this formula? ×E is about circulation, not about flux! Well… Let me copy Feynman’s answer:

Faraday's law

So… There you go. And, yes, you’re right, instead of writing Faraday’s Law as ×E = −∂B/∂t, we should write it as:

emf

That’s a easier to understand, and it’s also easier to work with, as we’ll see in a moment. So the point is: whenever the magnetic flux changes, there’s a push on the electrons in the wire. That push is referred to as the electromotive force, abbreviated as emf or EMF, and so it’s that line and/or surface integral above indeed. Let me paraphrase Feynman so you fully understand what we’re talking about here:

When we move our wire in a magnetic field, or when we move a magnet near the wire, or when we change the current in a nearby wire, there will be some net push on the electrons in the wire in one direction along the wire. There may be pushes in different directions at different places, but there will be more push in one direction than another. What counts is the push integrated around the complete circuit. We call this net integrated push the electromotive force (abbreviated emf) in the circuit. More precisely, the emf is defined as the tangential force per unit charge in the wire integrated over length, once around the complete circuit.

So that’s the integral. 🙂 And that’s how we can turn that motor above into a generator: instead of putting a current through the wire to make it turn, we can turn the loop, by hand or by a waterwheel or by whatever. Now, when the coil rotates, its wires will be moving in the magnetic field and so we will find an emf in the circuit of the coil, and so that’s how the motor becomes a generator.

Now, let me quickly interject something here: when I say ‘a push on the electrons in the wire’, what electrons are we talking about? How many? Well… I’ll answer that question in very much detail in a moment but, as for now, just note that the emf is some quantity expressed per coulomb or, as Feynman puts it above, per unit charge. So we’ll need to multiply it with the current in the circuit to get the power of our little generator.

OK. Let’s move on. Indeed, all I can do here is mention just a few basics, so we can move on to the next thing. If you really want to know all of the nitty-gritty, then you should just read Feynman’s Lecture on induced currents. That’s got everything. And, no, don’t worry: contrary to what you might expect, my ‘basics’ do not amount to a terrible pile of formulas. In fact, it’s all easy and quite amusing stuff, and I should probably include a lot more. But then… Well… I always need to move on… If not, I’ll never get to the stuff that I really want to understand. 😦

The electromotive force

We defined the electromotive force above, including its formula:

emf

What are the units? Let’s see… We know B was measured not in newton per coulomb, like the electric field E, but in N·s/C·m, because we had to multiply the magnetic field strength with the velocity of the charge to find the force per unit charge, cf. the F/q = v×equation. Now what’s the unit in which we’d express that surface integral? We must multiply with m2, so we get N·m·s/C. Now let’s simplify that by noting that one volt is equal to 1 N·m/C. [The volt has a number of definitions, but the one that applies here is that it’s the potential difference between two points that will impart one joule (i.e. 1 N·m) of energy to a unit of charge (i.e. 1 C) that passes between them.] So we can measure the magnetic flux in volt-seconds, i.e. V·s. And then we take the derivative in regard to time, so we divide by s, and so we get… Volt! The emf is measured in volt!

Does that make sense? I guess so: the emf causes a current, just like a potential difference, i.e. a voltage, and, therefore, we can and should look at the emf as a voltage too!

But let’s think about it some more, though. In differential form, Faraday’s Law, is just that ×E = −∂B/∂t equation, so that’s just one of Maxwell’s four equations, and so we prefer to write it as the “flux rule”. Now, the “flux rule” says that the electromotive force (abbreviated as emf or EMF) on the electrons in a closed circuit is equal to the time rate of change of the magnetic flux it encloses. As mentioned above, we measure magnetic flux in volt-seconds (i.e. V·s), so its time rate of change is measured in volt (because the time rate of change is a quantity expressed per second), and so the emf is measured in volt, i.e. joule per coulomb, as 1 V = 1 N·m/C = 1 J/C. What does it mean?

The time rate of change of the magnetic flux can change because the surface covered by our loop changes, or because the field itself changes, or by both. Whatever the cause, it will change the emf, or the voltage, and so it will make the electrons move. So let’s suppose we have some generator generating some emf. The emf can be used to do some work. We can charge a capacitor, for example. So how would that work?

More charge on the capacitor will increase the voltage V of the capacitor, i.e. the potential difference V = Φ1 − Φ2 between the two plates. Now, we know that the increase of the voltage V will be proportional to the increase of the charge Q, and that the constant of proportionality is defined by the capacity C of the capacitor: C = Q/V. [How do we know that? Well… Have a look at my post on capacitors.] Now, if our capacitor has an enormous capacity, then its voltage won’t increase very rapidly. However, it’s clear that, no matter how large the capacity, its voltage will increase. It’s just a matter of time. Now, its voltage cannot be higher than the emf provided by our ‘generator’, because it will then want to discharge through the same circuit!

So we’re talking power and energy here, and so we need to put some load on our generator. Power is the rate of doing work, so it’s the time rate of change of energy, and it’s expressed in joule per second. The energy of our capacitor is U = (1/2)·Q2/C = (1/2)·C·V2. [How do we know that? Well… Have a look at my post on capacitors once again. :-)] So let’s take the time derivative of U assuming some constant voltage V. We get: dU/dt = d[(1/2)·Q2/C]/dt = (Q/C)·dQ/dt = V·dQ/dt. So that’s the power that the generator would need to supply to charge the generator. As I’ll show in a moment, the power supplied by a generator is, indeed, equal to the emf times the current, and the current is the time rate of change of the charge, so I = dQ/dt.

So, yes, it all works out: the power that’s being supplied by our generator will be used to charge our capacitor. Now, you may wonder: what about the current? Where is the current in Faraday’s Law? The answer is: Faraday’s Law doesn’t have the current. It’s just not there. The emf is expressed in volt, and so that’s energy per coulomb, so it’s per unit charge. How much power an generator can and will deliver depends on its design, and the circuit and load that we will be putting on it. So we can’t say how many coulomb we will have. It all depends. But you can imagine that, if the loop would be bigger, or if we’d have a coil with many loops, then our generator would be able to produce more power, i.e. it would be able to move more electrons, so the mentioned power = (emf)×(current) product would be larger. 🙂

Finally, to conclude, note Feynman’s definition of the emf: the tangential force per unit charge in the wire integrated over length around the complete circuit. So we’ve got force times distance here, but per unit charge. Now, force times distance is work, or energy, and so… Yes, emf is joule per coulomb, definitely! 🙂

[…] Don’t worry too much if you don’t quite ‘get’ this. I’ll come back to it when discussing electric circuits, which I’ll do in my next posts.

Self-inductance and Lenz’s rule

We talked about motors and generators above. We also have transformers, like the one below. What’s going on here is that an alternating current (AC) produces a continuously varying magnetic field, which generates an alternating emf in the second coil, which produces enough power to light an electric bulb.

transformer

Now, the total emf in coil (b) is the sum of the emf’s of the separate turns of coil, so if we wind (b) with many turns, we’ll get a larger emf, so we can ‘transform’ the voltage to some other voltage. From your high-school classes, you should know how that works.

The thing I want to talk about here is something else, though. There is an induction effect in coil (a) itself. Indeed, the varying current in coil (a) produces a varying magnetic field inside itself, and the flux of this field is continually changing, so there is a self-induced emf in coil (a). The effect is called self-inductance, and so it’s the emf acting on a current itself when it is building up a magnetic field or, in general, when its field is changing in any way. It’s a most remarkable phenomenon, and so let me paraphrase Feynman as he describes it:

“When we gave “the flux rule” that the emf is equal to the rate of change of the flux linkage, we didn’t specify the direction of the emf. There is a simple rule, called Lenz’s rule, for figuring out which way the emf goes: the emf tries to oppose any flux change. That is, the direction of an induced emf is always such that if a current were to flow in the direction of the emf, it would produce a flux of B that opposes the change in B that produces the emf. In particular, if there is a changing current in a single coil (or in any wire), there is a “back” emf in the circuit. This emf acts on the charges flowing in the coil to oppose the change in magnetic field, and so in the direction to oppose the change in current. It tries to keep the current constant; it is opposite to the current when the current is increasing, and it is in the direction of the current when it is decreasing. A current in a self-inductance has “inertia,” because the inductive effects try to keep the flow constant, just as mechanical inertia tries to keep the velocity of an object constant.”

Hmm… That’s something you need to read a couple of times to fully digest it. There’s a nice demo on YouTube, showing an MIT physics video demonstrating this effect with a metal ring placed on the end of an electromagnet. You’ve probably seen it before: the electromagnet is connected to a current, and the ring flies into the air. The explanation is that the induced currents in the ring create a magnetic field opposing the change of field through it. So the ring and the coil repel just like two magnets with opposite poles. The effect is no longer there when a thin radial cut is made in the ring, because then there can be no current. The nice thing about the video is that it shows how the effect gets much more dramatic when an alternating current is applied, rather than a DC current. And it also shows what happens when you first cool the ring in liquid nitrogen. 🙂

You may also notice the sparks when the electromagnet is being turned on. Believe it or not, that’s also related to a “back emf”. Indeed, when we disconnect a large electromagnet by opening a switch, the current is supposed to immediately go to zero but, in trying to do so, it generates a large “back emf”: large enough to develop an arc across the opening contacts of the switch. The high voltage is also not good for the insulation of the coil, as it might damage it. So that’s why large electromagnets usually include some extra circuit, which allows the “back current” to discharge less dramatically. But I’ll refer you to Feynman for more details, as any illustration here would clutter the exposé.

Eddy currents

I like educational videos, and so I should give you a few references here, but there’s so many of this that I’ll let you google a few yourself. The most spectacular demonstration of eddy currents is those that appear in a superconductor: even back in the 1970s, when Feynman wrote his Lectures, the effect of magnetic levitation was well known. Feynman illustrates the effect with the simple diagram below: when bringing a magnet near to a perfect conductor, such as tin below 3.8°K, eddy currents will create opposing fields, so that no magnetic flux enters the superconducting material. The effect is also referred to as the Meisner effect, after the German physicist Walther Meisner, although it was discovered much earlier (in 1911) by a Dutch physicist in Leiden, Heike Kamerlingh Onnes, who got a Nobel Prize for it.

superconductor

Of course, we have eddy currents in less dramatic situations as well. The phenomenon of eddy currents is usually demonstrated by the braking of a sheet of metal as it swings back and forth between the poles of an electromagnet, as illustrated below (left). The illustration on the right shows how eddy-current effect can be drastically reduced by cutting slots in the plate, so that’s like making a radial cut in our jumping ring. 🙂

eddy currentseddy currents 2

The Faraday disc

The Faraday disc is interesting, not only from a historical point of view – the illustration below is a 19th century model, so Michael Faraday may have used himself – but also because it seems to contradict the “flux of rule”: as the disc rotates through a steady magnetic field, it will produce some emf, but so there’s no change in the flux. How is that possible?

Faraday_disk_generatorFaraday disk

The answer, of course, is that we are ‘cheating’ here: the material is moving, so we’re actually moving the ‘wire’, or the circuit if you want, so here we need to combine two equations:

two laws

If we do that, you’ll see it all makes sense. 🙂 Oh… That Faraday disc is referred to as a homopolar generator, and it’s quite interesting. You should check out what happened to the concept in the Wikipedia article on it. The Faraday disc was apparently used as a source for power pulses in the 1950s. The thing below could store 500 mega-joules and deliver currents up to 2 mega-ampère, i.e. 2 million amps! Fascinating, isn’t it? 🙂800px-Homopolar_anu-MJC

Maxwell, Lorentz, gauges and gauge transformations

I’ve done quite a few posts already on electromagnetism. They were all focused on the math one needs to understand Maxwell’s equations. Maxwell’s equations are a set of (four) differential equations, so they relate some function with its derivatives. To be specific, they relate E and B, i.e. the electric and magnetic field vector respectively, with their derivatives in space and in time. [Let me be explicit here: E and B have three components, but depend on both space as well as time, so we have three dependent and four independent variables for each function: E = (Ex, Ey, Ez) = E(x, y, z, t) and B = (Bx, By, Bz) = B(x, y, z, t).] That’s simple enough to understand, but the dynamics involved are quite complicated, as illustrated below.

Maxwell interactionI now want to do a series on the more interesting stuff, including an exploration of the concept of gauge in field theory, and I also want to show how one can derive the wave equation for electromagnetic radiation from Maxwell’s equations. Before I start, let’s recall the basic concept of a field.

The reality of fields

I said a couple of time already that (electromagnetic) fields are real. They’re more than just a mathematical structure. Let me show you why. Remember the formula for the electrostatic potential caused by some charge q at the origin:

E 4

We know that the (negative) gradient of this function, at any point in space, gives us the electric field vector at that point: E = –Φ. [The minus sign is there because of convention: we take the reference point Φ = 0 at infinity.] Now, the electric field vector gives us the force on a unit charge (i.e. the charge of a proton) at that point. If q is some positive charge, the force will be repulsive, and the unit charge will accelerate away from our q charge at the origin. Hence, energy will be expended, as force over distance implies work is being done: as the charges separate, potential energy is converted into kinetic energy. Where does the energy come from? The energy conservation law tells us that it must come from somewhere.

It does: the energy comes from the field itself. Bringing in more or bigger charges (from infinity, or just from further away) requires more energy. So the new charges change the field and, therefore, its energy. How exactly? That’s given by Gauss’ Law: the total flux out of a closed surface is equal to:

Gauss Law

You’ll say: flux and energy are two different things. Well… Yes and no. The energy in the field depends on E. Indeed, the formula for the energy density in space (i.e. the energy per unit volume) is

D 6

Getting the energy over a larger space is just another integral, with the energy density as the integral kernel:

energy integral

Feynman’s illustration below is not very sophisticated but, as usual, enlightening. 🙂

energy in the field

Gauss’ Theorem connects both the math as well as the physics of the situation and, as such, underscores the reality of fields: the energy is not in the electric charges. The energy is in the fields they produce. Everything else is just the principle of superposition of fields –  i.e. E = E+ E– coming into play. I’ll explain Gauss’ Theorem in a moment. Let me first make some additional remarks.

First, the formulas are valid for electrostatics only (so E and B only vary in space, not in time), so they’re just a piece of the larger puzzle. 🙂 As for now, however, note that, if a field is real (or, to be precise, if its energy is real), then the flux is equally real.

Second, let me say something about the units. Field strength (E or, in this case, its normal component En = E·n) is measured in newton (N) per coulomb (C), so in N/C. The integral above implies that flux is measured in (N/C)·m2. It’s a weird unit because one associates flux with flow and, therefore, one would expect flux is some quantity per unit time and per unit area, so we’d have the m2 unit (and the second) in the denominator, not in the numerator. But so that’s true for heat transfer, for mass transfer, for fluid dynamics (e.g. the amount of water flowing through some cross-section) and many other physical phenomena. But for electric flux, it’s different. You can do a dimensional analysis of the expression above: the sum of the charges is expressed in coulomb (C), and the electric constant (i.e. the vacuum permittivity) is expressed in C2/(N·m2), so, yes, it works: C/[C2/(N·m2)] = (N/C)·m2. To make sense of the units, you should think of the flux as the total flow, and of the field strength as a surface density, so that’s the flux divided by the total area, so (field strength) = (flux)/(area). Conversely, (flux) = (field strength)×(area). Hence, the unit of flux is [flux] = [field strength]×[area] = (N/C)·m2.

OK. Now we’re ready for Gauss’ Theorem. 🙂 I’ll also say something about its corollary, Stokes’ Theorem. It’s a bit of a mathematical digression but necessary, I think, for a better understanding of all those operators we’re going to use.

Gauss’ Theorem

The concept of flux is related to the divergence of a vector field through Gauss’ Theorem. Gauss’s Theorem has nothing to do with Gauss’ Law, except that both are associated with the same genius. Gauss’ Theorem is:

Gauss Theorem

The ·C in the integral on the right-hand side is the divergence of a vector field. It’s the volume density of the outward flux of a vector field from an infinitesimal volume around a given point.

Huh? What’s a volume density? Good question. Just substitute C for E in the surface and volume integral above (the integral on the left is a surface integral, and the one on the right is a volume integral), and think about the meaning of what’s written. To help you, let me also include the concept of linear density, so we have (1) linear, (2) surface and (3) volume density. Look at that representation of a vector field once again: we said the density of lines represented the magnitude of E. But what density? The representation hereunder is flat, so we can think of a linear density indeed, measured along the blue line: so the flux would be six (that’s the number of lines), and the linear density (i.e. the field strength) is six divided by the length of the blue line.

Linear density

However, we defined field strength as a surface density above, so that’s the flux (i.e. the number of field lines) divided by the surface area (i.e. the area of a cross-section): think of the square of the blue line, and field lines going through that square. That’s simple enough. But what’s volume density? How do we count the number of lines inside of a box? The answer is: mathematicians actually define it for an infinitesimally small cube by adding the fluxes out of the six individual faces of an infinitesimally small cube:

Volume density for small cube 1

So, the truth is: volume density is actually defined as a surface density, but for an infinitesimally small volume element. That, in turn, gives us the meaning of the divergence of a vector field. Indeed, the sum of the derivatives above is just ·C (i.e. the divergence of C), and ΔxΔyΔz is the volume of our infinitesimal cube, so the divergence of some field vector C at some point P is the flux – i.e. the outgoing ‘flow’ of Cper unit volume, in the neighborhood of P, as evidenced by writing

Volume density for small cube

Indeed, just bring ΔV to the other side of the equation to check the ‘per unit volume’ aspect of what I wrote above. The whole idea is to determine whether the small volume is like a sink or like a source, and to what extent. Think of the field near a point charge, as illustrated below. Look at the black lines: they are the field lines (the dashed lines are equipotential lines) and note how the positive charge is a source of flux, obviously, while the negative charge is a sink.

equipotential

Now, the next step is to acknowledge that the total flux from a volume is the sum of the fluxes out of each part. Indeed, the flux through the part of the surfaces common to two parts will cancel each other out. Feynman illustrates that with a rough drawing (below) and I’ll refer you to his Lecture on it for more detail.

cut

So… Combining all of the gymnastics above – and integrating the divergence over an entire volume, indeed –  we get Gauss’ Theorem:

Gauss Theorem

Stokes’ Theorem

There is a similar theorem involving the circulation of a vector, rather than its flux. It’s referred to as Stokes’ Theorem. Let me jot it down:

Stokes Theorem

We have a contour integral here (left) and a surface integral (right). The reasoning behind is quite similar: a surface bounded by some loop Γ is divided into infinitesimally small squares, and the circulation around Γ is the sum of the circulations around the little loops. We should take care though: the surface integral takes the normal component of ×C, so that’s (×C)n = (×Cn. The illustrations below should help you to understand what’s going on.

Stokes Theorem 1Stokes Theorem 2

The electric versus the magnetic force

There’s more than just the electric force: we also have the magnetic force. The so-called Lorentz force is the combination of both. The formula, for some charge q in an electromagnetic field, is equal to:

Lorentz force

Hence, if the velocity vector v is not equal to zero, we need to look at the magnetic field vector B too! The simplest situation is magnetostatics, so let’s first have a look at that.

Magnetostatics imply that that the flux of E doesn’t change, so Maxwell’s third equation reduces to c2×B = j0. So we just have a steady electric current (j): no accelerating charges. Maxwell’s fourth equation, B = 0, remains what is was: there’s no such thing as a magnetic charge. The Lorentz force also remains what it is, of course: F = q(E+v×B) = qE +qv×B. Also note that the v, j and the lack of a magnetic charge all point to the same: magnetism is just a relativistic effect of electricity.

What about units? Well… While the unit of E, i.e. the electric field strength, is pretty obvious from the F = qE term  – hence, E = F/q, and so the unit of E must be [force]/[charge] = N/C – the unit of the magnetic field strength is more complicated. Indeed, the F = qv×B identity tells us it must be (N·s)/(m·C), because 1 N = 1C·(m/s)·(N·s)/(m·C). Phew! That’s as horrendous as it looks, and that’s why it’s usually expressed using its shorthand, i.e. the tesla: 1 T = 1 (N·s)/(m·C). Magnetic flux is the same concept as electric flux, so it’s (field strength)×(area). However, now we’re talking magnetic field strength, so its unit is T·m= (N·s·m)/(m·C) = (N·s·m)/C, which is referred to as the weber (Wb). Remembering that 1 volt = 1 N·m/C, it’s easy to see that a weber is also equal to 1 Wb = 1 V·s. In any case, it’s a unit that is not so easy to interpret.

Magnetostatics is a bit of a weird situation. It assumes steady fields, so the ∂E/∂t and ∂B/∂t terms in Maxwell’s equations can be dropped. In fact, c2×B = j0 implies that ·(c2×B ·(j0) and, therefore, that ·= 0. Now, ·= –∂ρ/∂t and, therefore, magnetostatics is a situation which assumes ∂ρ/∂t = 0. So we have electric currents but no change in charge densities. To put it simply, we’re not looking at a condenser that is charging or discharging, although that condenser may act like the battery or generator that keeps the charges flowing! But let’s go along with the magnetostatics assumption. What can we say about it? Well… First, we have the equivalent of Gauss’ Law, i.e. Ampère’s Law:

Ampere Law

We have a line integral here around a closed curve, instead of a surface integral over a closed surface (Gauss’ Law), but it’s pretty similar: instead of the sum of the charges inside the volume, we have the current through the loop, and then an extra c2 factor in the denominator, of course. Combined with the B = 0 equation, this equation allows us to solve practical problems. But I am not interested in practical problems. What’s the theory behind?

The magnetic vector potential

TheB = 0 equation is true, always, unlike the ×E = 0 expression, which is true for electrostatics only (no moving charges). It says the divergence of B is zero, always, and, hence, it means we can represent B as the curl of another vector field, always. That vector field is referred to as the magnetic vector potential, and we write:

·B = ·(×A) = 0 and, hence, B×A

In electrostatics, we had the other theorem: if the curl of a vector field is zero (everywhere), then the vector field can be represented as the gradient of some scalar function, so if ×= 0, then there is some Ψ for which CΨ. Substituting C for E, and taking into account our conventions on charge and the direction of flow, we get E = –Φ. Substituting E in Maxwell’s first equation (E = ρ/ε0) then gave us the so-called Poisson equation: ∇2Φ = ρ/ε0, which sums up the whole subject of electrostatics really! It’s all in there!

Except magnetostatics, of course. Using the (magnetic) vector potential A, all of magnetostatics is reduced to another expression:

2A= −j0, with ·A = 0

Note the qualifier: ·A = 0. Why should the divergence of A be equal to zero? You’re right. It doesn’t have to be that way. We know that ·(×C) = 0, for any vector field C, and always (it’s a mathematical identity, in fact, so it’s got nothing to do with physics), but choosing A such that ·A = 0 is just a choice. In fact, as I’ll explain in a moment, it’s referred to as choosing a gauge. The·A = 0 choice is a very convenient choice, however, as it simplifies our equations. Indeed, c2×B = j0 = c2×(×A), and – from our vector calculus classes – we know that ×(×C) = (·C) – ∇2C. Combining that with our choice of A (which is such that ·A = 0, indeed), we get the ∇2A= −j0 expression indeed, which sums up the whole subject of magnetostatics!

The point is: if the time derivatives in Maxwell’s equations, i.e. ∂E/∂t and ∂B/∂t, are zero, then Maxwell’s four equations can be nicely separated into two pairs: the electric and magnetic field are not interconnected. Hence, as long as charges and currents are static, electricity and magnetism appear as distinct phenomena, and the interdependence of E and B does not appear. So we re-write Maxwell’s set of four equations as:

  1. ElectrostaticsE = ρ/ε0 and ×E = 0
  2. Magnetostatics: ×B = j/c2ε0 and B = 0

Note that electrostatics is a neat example of a vector field with zero curl and a given divergence (ρ/ε0), while magnetostatics is a neat example of a vector field with zero divergence and a given curl (j/c2ε0).

Electrodynamics

But reality is usually not so simple. With time-varying fields, Maxwell’s equations are what they are, and so there is interdependence, as illustrated in the introduction of this post. Note, however, that the magnetic field remains divergence-free in dynamics too! That’s because there is no such thing as a magnetic charge: we only have electric charges. So ·B = 0 and we can define a magnetic vector potential A and re-write B as B×A, indeed.

I am writing a vector potential field because, as I mentioned a couple of times already, we can choose A. Indeed, as long as ·A = 0, it’s fine, so we can add curl-free components to the magnetic potential: it won’t make a difference. This condition is referred to as gauge invariance. I’ll come back to that, and also show why this is what it is.

While we can easily get B from A because of the B×A, getting E from some potential is a different matter altogether. It turns out we can get E using the following expression, which involves both Φ (i.e. the electric or electrostatic potential) as well as A (i.e. the magnetic vector potential):

E = –Φ – ∂A/∂t

Likewise, one can show that Maxwell’s equations can be re-written in terms of Φ and A, rather than in terms of E and B. The expression looks rather formidable, but don’t panic:

Equations 2

Just look at it. We have two ‘variables’ here (Φ and A) and two equations, so the system is fully defined. [Of course, the second equation is three equations really: one for each component x, y and z.] What’s the point? Why would we want to re-write Maxwell’s equations? The first equation makes it clear that the scalar potential (i.e. the electric potential) is a time-varying quantity, so things are not, somehow, simpler. The answer is twofold. First, re-writing Maxwell’s equations in terms of the scalar and vector potential makes sense because we have (fairly) easy expressions for their value in time and in space as a function of the charges and currents. For statics, these expressions are:

Integrals staticsSo it is, effectively, easier to first calculate the scalar and vector potential, and then get E and B from them. For dynamics, the expressions are similar:

Integrals dynamics

Indeed, they are like the integrals for statics, but with “a small and physically appealing modification”, as Feynman notes: when doing the integrals, we must use the so-called retarded time t′ = t − r12/ct’. The illustration below shows how it works: the influences propagate from point (2) to point (1) at the speed c, so we must use the values of ρ and j at the time t′ = t − r12/ct’ indeed!

Retarded timeThe second aspect of the answer to the question of why we’d be interested in Φ and A has to do with the topic I wanted to write about here: the concept of a gauge and a gauge transformation.

Gauges and gauge transformations in electromagnetics

Let’s see what we’re doing really. We calculate some A and then solve for B by writing: B = ×A. Now, I say some A because any A‘ = AΨ, with Ψ any scalar field really. Why? Because the curl of the gradient of Ψ – i.e. curl(gradΨ) = ×(Ψ) – is equal to 0. Hence, ×(AΨ) = ×A×Ψ = ×A.

So we have B, and now we need E. So the next step is to take Faraday’s Law, which is Maxwell’s second equation: ×E = –∂B/∂t. Why this one? It’s a simple one, as it does not involve currents or charges. So we combine this equation and our B = ×A expression and write:

×E = –∂(∇×A)/∂t

Now, these operators are tricky but you can verify this can be re-written as:

×(E + ∂A/∂t) = 0

Looking carefully, we see this expression says that E + ∂A/∂t is some vector whose curl is equal to zero. Hence, this vector must be the gradient of something. When doing electrostatics, When we worked on electrostatics, we only had E, not the ∂A/∂t bit, and we said that E tout court was the gradient of something, so we wrote E = −Φ. We now do the same thing for E + ∂A/∂t, so we write:

E + ∂A/∂t = −Φ

So we use the same symbol Φ but it’s a bit of a different animal, obviously. However, it’s easy to see that, if the ∂A/∂t would disappear (as it does in electrostatics, where nothing changes with time), we’d get our ‘old’ −Φ. Now, E + ∂A/∂t = −Φ can be written as:

E = −Φ – ∂A/∂t

So, what’s the big deal? We wrote B and E as a function of Φ and A. Well, we said we could replace A by any A‘ = AΨ but, obviously, such substitution would not yield the same E. To get the same E, we need some substitution rule for Φ as well. Now, you can verify we will get the same E if we’d substitute Φ for Φ’ = Φ – ∂Ψ/∂t. You should check it by writing it all out:

E = −Φ’–∂A’/∂t = −(Φ–∂Ψ/∂t)–∂(A+Ψ)/∂t

= −Φ+(∂Ψ/∂t)–∂A/∂t–∂(Ψ)/∂t = −Φ – ∂A/∂t = E

Again, the operators are a bit tricky, but the +(∂Ψ/∂t) and –∂(Ψ)/∂t terms do cancel out. Where are we heading to? When everything is said and done, we do need to relate it all to the currents and the charges, because that’s the real stuff out there. So let’s take Maxwell’s E = ρ/ε0 equation, which has the charges in it, and let’s substitute E for E = −Φ – ∂A/∂t. We get:

Capture

That equation can be re-written as:

equation 1

So we have one equation here relating Φ and A to the sources. We need another one, and we also need to separate Φ and A somehow. How do we do that?

Maxwell’s fourth equation, i.e. c2×B = j+ ∂E/∂t can, obviously, be written as c2×− E/∂t = j0. Substituting both E and B yields the following monstrosity:

equation 3

We can now apply the general ∇×(×C) = (·C) – ∇2C identity to the first term to get:

equation 4

It’s equally monstrous, obviously, but we can simplify the whole thing by choosing Φ and A in a clever way. For the magnetostatic case, we chose A such that ·A = 0. We could have chosen something else. Indeed, it’s not because B is divergence-free, that A has to be divergence-free too! For example, I’ll leave it to you to show that choosing ·A such that

equation 5also respects the general condition that any A and Φ we choose must respect the A‘ = AΨ and Φ’ = Φ – ∂Ψ/∂t equalities. Now, if we choose ·A such that ·A = −c–2·∂Φ/∂t indeed, then the two middle terms in our monstrosity cancel out, and we’re left with a much simpler equation for A:

equation 6

In addition, doing the substitution in our other equation relating Φ and A to the sources yields an equation for Φ that has the same form:

equation 7

What’s the big deal here? Well… Let’s write it all out. The equation above becomes:

wave equation

That’s a wave equation in three dimensions. In case you wonder, just check one of my posts on wave equations. The one-dimensional equivalent for a wave propagating in the x direction at speed c (like a sound wave, for example) is ∂2Φ/∂xc–2·∂2Φ/∂t2, indeed. The equation for A yields above yields similar wave functions for A‘s components Ax, Ay, and Az.

So, yes, it is a big deal. We’ve written Maxwell’s equations in terms of the scalar (Φ) and vector (A) potential and in a form that makes immediately apparent that we’re talking electromagnetic waves moving out at the speed c. Let me copy them again:

Equations 2

You may, of course, say that you’d rather have a wave equation for E and B, rather than for A and Φ. Well… That can be done. Feynman gives us two derivations that do so. The first derivation is relatively simple and assumes the source our electromagnetic wave moves in one direction only. The second derivation is much more complicated and gives an equation for E that, if you’ve read the first volume of Feynman’s Lectures, you’ll surely remember:

equation for E

The links are there, and so I’ll let you have fun with those Lectures yourself. I am finished here, indeed, in terms of what I wanted to do in this post, and that is to say a few words about gauges in field theory. It’s nothing much, really, and so we’ll surely have to discuss the topic again, but at least you now know what a gauge actually is in classical electromagnetic theory. Let’s quickly go over the concepts:

  1. Choosing the ·A is choosing a gauge, or a gauge potential (because we’re talking scalar and vector potential here). The particular choice is also referred to as gauge fixing.
  2. Changing A by adding ψ is called a gauge transformation, and the scalar function Ψ is referred to as a gauge function. The fact that we can add curl-free components to the magnetic potential without them making any difference is referred to as gauge invariance.
  3. Finally, the ·A = −c–2·∂Φ/∂t gauge is referred to as a Lorentz gauge.

Just to make sure you understand: why is that Lorentz gauge so special? Well… Look at the whole argument once more: isn’t it amazing we get such beautiful (wave) equations if we stick it in? Also look at the functional shape of the gauge itself: it looks like a wave equation itself! […] Well… No… It doesn’t. I am a bit too enthusiastic here. We do have the same 1/c2 and a time derivative, but it’s not a wave equation. 🙂 In any case, it all confirms, once again, that physics is all about beautiful mathematical structures. But, again, it’s not math only. There’s something real out there. In this case, that ‘something’ is a traveling electromagnetic field. 🙂

But why do we call it a gauge? That should be equally obvious. It’s really like choosing a gauge in another context, such as measuring the pressure of a tyre, as shown below. 🙂

abs-gauge-press

Gauges and group theory

You’ll usually see gauges mentioned with some reference to group theory. For example, you will see or hear phrases like: “The existence of arbitrary numbers of gauge functions ψ(r, t) corresponds to the U(1) gauge freedom of the electromagnetic theory.” The U(1) notation stands for a unitary group of degree n = 1. It is also known as the circle group. Let me copy the introduction to the unitary group from the Wikipedia article on it:

In mathematics, the unitary group of degree n, denoted U(n), is the group of n × n unitary matrices, with the group operation that of matrix multiplication. The unitary group is a subgroup of the general linear group GL(n, C). In the simple case n = 1, the group U(1) corresponds to the circle group, consisting of all complex numbers with absolute value 1 under multiplication. All the unitary groups contain copies of this group.

The unitary group U(n) is a real Lie group of of dimension n2. The Lie algebra of U(n) consists of n × n skew-Hermitian matrices, with the Lie bracket given by the commutator. The general unitary group (also called the group of unitary similitudes) consists of all matrices A such that A*A is a nonzero multiple of the identity matrix, and is just the product of the unitary group with the group of all positive multiples of the identity matrix.

Phew! Does this make you any wiser? If anything, it makes me realize I’ve still got a long way to go. 🙂 The Wikipedia article on gauge fixing notes something that’s more interesting (if only because I more or less understand what it says):

Although classical electromagnetism is now often spoken of as a gauge theory, it was not originally conceived in these terms. The motion of a classical point charge is affected only by the electric and magnetic field strengths at that point, and the potentials can be treated as a mere mathematical device for simplifying some proofs and calculations. Not until the advent of quantum field theory could it be said that the potentials themselves are part of the physical configuration of a system. The earliest consequence to be accurately predicted and experimentally verified was the Aharonov–Bohm effect, which has no classical counterpart.

This confirms, once again, that the fields are real. In fact, what this says is that the potentials are real: they have a meaningful physical interpretation. I’ll leave it to you to expore that Aharanov-Bohm effect. In the meanwhile, I’ll study what Feynman writes on potentials and all that as used in quantum physics. It will probably take a while before I’ll get into group theory though. :-/

Indeed, it’s probably best to study physics at a somewhat less abstract level first, before getting into the more sophisticated stuff.