# Self-inductance, mutual inductance, and the power and energy of inductors

As Feynman puts it, studying physics is not always about the ‘the great and esoteric heights’. In fact, you usually have to come down from then fairly quickly – studying physics is similar to mountain climbing in that regard 🙂 – and study ‘relatively low-level subjects’, such as electrical circuits, which is what we’ll do in this and the next post.

As I’ve introduced some key concepts in a previous post already, let me recapitulate the basics, which include the concept of the electromotive force, which is basically the voltage, i.e. the potential difference, that’s produced in a loop or coil of wire as the magnetic flux changes. I also talked about the impedance in a AC circuit. Finally, we also discussed the power and energies involved. Important results from this previous discussion include (but are not limited to):

1. A constant speed AC generator will create an alternating current with the emf, i.e. the voltage, varying as V0·sin(ωt).
2. If we only have resistors as circuit elements, and the resistance in the circuit adds up to R, then the electric current in the circuit will be equal to I = Ɛ/R = V/R = (V0/R)·sin(ωt). So that’s Ohm’s Law, basically.
3. The power that’s produced and consumed in an AC circuit is the product of the voltage and the current, so P = Ɛ·I = V·I. We also showed this electrical power is equal to the mechanical power dW/dt that makes the generator run.
4. Finally, we explained the concept of impedance (denoted by Z) using Euler’s formula: Z = |Z|eiθ, mentioning that, if other circuit elements than resistors are involved, such as inductors, then it’s quite likely that the current signal will lag the voltage signal, with the phase factor θ telling us by how much.

It’s now time to introduce those ‘other’ circuit elements. So let’s start with inductors here, and the concept of inductance itself. There’s a lot of stuff to them, and so let’s go over it point by point.

The concept of self-inductance

In its simplest form, an inductor is just a coil, but they come in all sizes and shapes. If you want to see how they might look like, just google some images of micro-electronic inductors, and then, just to see the range of applications, some images of inductors used for large-scale industrial applications. If you do so, you’re likely to see images of transformers too, because transformers work on the principle of mutual inductance, and so they involve two coils, i.e. two inductors.

Contrary to what you might expect, the concept of self-inductance (or inductance tout court) is quite simply: a changing current will cause a changing magnetic field and, hence, some emf. Now, it turns out that the induced emf is proportional to the change in current. So we’ve got another constant of proportionality here, so it’s like how we defined resistance, or capacitance. So, in many ways, the inductance is just another proportionality coefficient. If we denote it by L – the symbol is said to honor the Russian phyicist Heinrich Lenz, whom you know from Lenz’ Law – then we define it as:

L = −Ɛ/(dI/dt)

The dI/dt factor is, obviously, the time rate of change of the current, and the negative sign indicates that the emf opposes the change in current, so it will tend to cause an opposing current. That’s why the emf involved is often referred to as a ‘back emf’. So that’s Lenz’ Law basically. As you might expect, the physicists came up with yet another derived unit, the Henry, to honor yet another physicist, Joseph Henry, an American scientist who was a contemporary of Michael Faraday and independently discovered pretty much the same as Faraday: one henry (H) equals one volt·second per ampere: 1 H = 1 V·s/A.

The concept of mutual inductance

Feynman introduces the topic of inductance with a two-coil set-up, as shown below, noting that a current in coil 1 will induce some emf in coil 2, which he denotes by M12. Conversely, a current in coil 2 will induce some emf in coil 1, which he denoted by M21. M12 and M21 are also constants: they depend on the geometry of the coils, including the length of the solenoid (l), its surface area (S) and the number of loop turns of the coils (N1 and N2). The next step in the analysis is then to acknowledge that each coil should also produce a ‘back emf’ in itself, which we can denote by M11 and M22 respectively, but then these constants are, of course, equal to the self-inductance of the coils so, taking into account the convention in regard to the sign of the self-inductance, we write:

L1 = − M11 and L1 = − M22

You will now wonder: what’s the total emf in each coil, taking into account that we do not only have mutual inductance but also self-inductance? Frankly, when I was a kid, and my father tried to tell me one or two things about this, it confused me very much. I could not imagine what happened in one coil, let alone in two coils. I had this vision of a current producing some ‘back-current’, and then the ‘back-current’ producing ‘back-current’ again, and so I could not imagine how one could solve this problem. So the image in my head was very much like that baking powder box which Feynman evokes when talking about the method of images to find the electric fields in situations with an easy geometry, so that’s the picture of a baking powder box which has on its label a picture of a baking powder box which has… Well… Etcetera. Unfortunately, my father didn’t push us to study math and, therefore, I knew that one could solve such problems mathematically – we’re talking a converging series here – but I did not know how, and that’s why I found it all very confusing.

Now I understand there’s one current only, and one potential difference only, and that the formulas do not involve some infinite series of terms. But… Well… I am not ashamed to say these problems are still testing the (limited) agility of my mind. The first thing to ‘get’ is that we’re talking a back emf, and so that’s not a current but a potential difference. In fact, as I explained in my post on the electromotive force, the term ‘force’ in emf is actually misleading, and may lead to that same erroneous vision that I had as a kid: forces generating counter-forces, that generate counter-forces, that generate counter-forces, etcetera. It’s not like that: we have some current – one current – in a coil and we’ll have some voltage – one voltage – across the coil. If the coil would be a resistor instead of a coil, we’d find that the ratio of this voltage and the current would be some constant R = V/I. Now here we’re talking a coil indeed, so that’s a different circuit element, and we find some other ratio, L = −V/(dI/dt) = −Ɛ/(dI/dt). Why the minus sign? Well… As said, the induced emf will be such that it will tend to counter the current, and current flows from positive to negative as per our convention.

But… Yes? So how does it work when we put this coil in some circuit, and how does the resistance of the inductor come into play? Relax. We’ve just been talking ideal circuit elements so far, and we’ve discussed only two: the resistor and the inductor. We’ll talk about voltage sources (or generators) and capacitors too, and then we’ll link all of these ideal circuit elements. In short, we’ll analyze some real-life electrical circuit soon, but first you need to understand the basics. Let me just note that an ideal inductor appears a zero-resistance conductor in a direct current (DC) circuit, so it’s a short-circuit really! Please try to mentally separate out those ‘ideal’ circuit components. Otherwise you’ll never be able to make sense of it all!

In fact, there’s a good reason why Feynman starts with explaining mutual inductance before discussing a little circuit like the one below, which has an inductor and a voltage source. The two-coil situation above is effectively easier to understand, although it may not look like that at first. So let’s analyze that two-coil situation in more detail first. In other words, let me try to understand the situation that I didn’t understand as a kid. 🙂 Because of the law of superposition, we should add fluxes and changes in fluxes and, hence, we should also add the electromotive forces, i.e. the induced voltages. So, what we have here is that the total emf in coil 2 should be written as:

Ɛ2 = M21·(dI1/dt) + M22·(dI2/dt) = M21·(dI1/dt) – L2·(dI2/dt)

What we’re saying here is that the emf, i.e. the voltage across the coil, will indeed depend on the change in current in the other coil, but also on the change in current of the coil itself. Likewise, the total emf in coil 1 should be written as:

Ɛ1 = M12·(dI2/dt) + M11·(dI1/dt) = M12·(dI2/dt) – L1·(dI1/dt)

Of course, this does reduce to the simple L = −Ɛ/(dI/dt) if there’s one coil only. But so you see where it comes from and, while we do not have some infinite series 🙂 we do have a system of two equations here, and so let me say one or two things about it.

The first thing to note is that it is not so difficult to show that M21 is equal to M12, so we can simplify and write that M21 = M12 = M. Now, while I said ‘not so difficult’, I didn’t mean it’s easy and, because I don’t want this post to become too long, I’ll refer you to Feynman for the proof of this M21 = M12 = M equation. It’s a general proof for any two coils or ‘circuits’ of arbitrary shape and it’s really worth the read. However, I have to move on.

The second thing to note is that this coefficient M, which is referred to as the mutual inductance now (so singular instead of plural) depends on the ‘circuit geometry’ indeed. For a simple solenoid, Feynman calculates it as

M = −(1/ε0c2)·(N1·N2)·S/l,

with the length of the solenoid, S its surface area (S), and N1 and Nthe respective number of loop turns of the two coils. So, yes, only ‘geometry’ comes into play. [Note that’s quite obvious from the formula because a switch of the subscripts of Nand Nmakes no difference, of course!]. Now, it’s interesting to note that M is the same for, let’s say, N= 100 and N2 = 10 and for N= 20 and N2 = 50. In fact, because you’re familiar with what transformers do, i.e. transforming voltages, you may think that’s counter-intuitive. It’s not. The thing with the number of coils does not imply that Ɛ1 and Ɛ2 remain the same. Our set of equations is Ɛ1 = M·(dI2/dt) – L1·(dI1/dt) and Ɛ= M·(dI1/dt) – L2·(dI2/dt), and so Land Lclearly do vary as Nand N2 vary! So… Well… Yes. We’ve got a set of two equations with two independent variables (Iand I2) and two dependent variables (Ɛ1 and Ɛ2). Of course, we could also phrase the problem the other way around: given two voltages, what are the currents? 🙂

Of course, that makes us think of the power that goes in and out of a transformer. Indeed, you’ll remember that power is voltage times current. So what’s going on here in regard to that?

Well… There’s a thing with transformers, or with two-coil systems like this in general, that is referred to as coupling. The geometry of the situation will determine how much flux from one coil is linked with the flux of the other coil. If most, or all of it, is linked, we say the two coils are ‘tightly coupled’ or, in the limit, that they are fully coupled. There’s a measure for that, and it’s called the coefficient of coupling. Let’s first explore that concept of power once more.

Inductance, energy and electric power

It’s easy to see that we need electric power to get some current going. Now, as we pointed out in our previous post, the power is equal to the voltage times the current. It’s also equal, of course, to the amount of work done per second, i.e. the time rate of change of the energy W, so we write:

dW/dt = Ɛ·I

Now, we defined the self-inductance as L = −Ɛ/(dI/dt) and, therefore, we know that Ɛ = −L·(dI/dt), so we have:

dW/dt = −L·I·(dI/dt)

What is this? A differential equation? Yes and no. We’ve got not one but two functions of time here, W and I, and, while their derivatives with respect to time do appear in the equation, what we need to do is just integrate the two sides over time. We get: W = −(1/2)·L·I2. Just check it by taking the time derivative of both sides. Of course, we can add any constant, to both sides in fact, but that’s just a matter of choosing some reference point. We’ll chose our constant to be zero, and also think about the energy that’s stored in the coil, i.e. U, which we define as:

U = −W = −(1/2)·L·I

Huh? What’s going on here? Well… It’s not an easy discussion, but let’s try to make sense of it. We have some changing current in the coil here but, obviously, some kind of inertia also: the coil itself opposes the change in current through the ‘back emf’. It requires energy, or power, to overcome the inertia. We may think of applying some voltage to offset the ‘back emf’, so we may effectively think of that little circuit with an inductor and a voltage source. The voltage V we’d need to apply to offset the inertia would, obviously, be equal to the ‘back emf’, but with its sign reversed, so we have:

V = − Ɛ = L·(dI/dt)

Now, it helps to think of what a current really is: it’s about electric charges that are moving at some velocity v because some force is being applied to them. As in any system, the power that’s being delivered is the dot product of the force and the velocity vectors (that ensures we only take the tangential component of the force into account), so if we have moving charges, the power that is being delivered to the circuit is (F·v)·n. What is F? It’s obviously, qE, as the electric field is the force per unit charge, so E = F/q. But so we’re talking some circuit here and we need to think of the power being delivered to some infinitesimal element ds in the coil, and so that’s (F·v)·n·ds, which can be written as: (F·ds)·n·v. And then we integrate over the whole coil to find: Now, you may or may not remember that the emf (Ɛ) is actually defined as the line integral ∫ E·ds line, taken around the entire coil and, hence, noting that = F/q, and that the current I is equal to I = q·n·v, we got our power equation. Indeed, the integrand or kernel of our integral becomes F·n·v·d= q·E·n·v·d= I·E·ds. Hence, we get our power formula indeed: P = V·I, with V the potential difference, i.e. the voltage across the coil.

I am getting too much into the weeds here. The point is: we’ve got a full and complete analog to the concept of inertia in mechanics here: instead of some force F causing some mass m to change its velocity according to Newton’s Law, i.e. F = m·a = m·(dv/dt), we here have a potential difference V causing some current I to change according to the V = L·(dI/dt) law.

This is very confusing but, remember, the same equations must have the same solutions! So, in an electric circuit, the inductance is really like what the mass is in a mechanics. Now, in mechanics, we’ll say that our mass has some momentum p = m·v, and we’ll also say that its kinetic energy is equal to (1/2)m·v2. We can do the same for our circuit: potential energy is continuously being converted into kinetic energy which, for our inductor, we write as U = (1/2)·L·I2.

Just think about by playing with one of the many online graphing tools. The graph below, for example, assumes the current builds up to some maximum. As it reaches its maximum, the stored energy will also max out. Now, you should not worry about the units here, or the scale of the graphs. The assumption is that I builds up from 0 to 1, and that L = 1, so that makes U what it is. Using a different constant for L, and/or different units for I, will change the scale of U too, but not its general shape, and that shape gives you the general idea. The example above obviously assumes some direct current, so it’s a DC circuit: the current builds up, but then stabilizes at some maximum that we can find by applying Ohm’s Law to the resistance of the circuit: I = V/R. Resistance? But we were talking an ideal inductor? We are. If there’s no other resistance in the circuit, we’ll have a short-circuit, so the assumption is that we do have some resistance in the circuit and, therefore, we should also think of some energy loss to heat from the current in the resistance, but that’s not our worry here.

The illustration below is, perhaps, more interesting. Here we are, obviously, applying an alternating current, and so the current goes in one and then in the other direction, so I > 0, and then I < 0, etcetera. We’re assuming some nice sinusoidal curve for the current here (i.e. the blue curve), and so we get what we get for U (i.e. the red curve): the energy goes up and down between zero and some maximum amplitude that’s determined by the maximum current. So, yes, it is, after all, quite intuitive: building up a current does require energy from some external source, which is used to overcome the ‘back emf’ in the inductor, and that energy is stored in the inductor itself. [If you still wonder why it’s stored in the inductor, think about the other question: where else would it be stored?] How is stored? Look at the graph and think: it’s stored as kinetic energy of the charges, obviously. That explains why the energy is zero when the current is zero, and why the energy maxes out when the current maxes out. So, yes, it all makes sense! 🙂

Let’s now get back to that coupling constant.

The coupling constant

We can apply our reasoning to two coils. Indeed, we know that Ɛ1 = M·(dI2/dt) – L1·(dI1/dt) and Ɛ= M·(dI1/dt) – L2·(dI2/dt). So the power in the two-coils system is dW/dt = Ɛ1·I1 + Ɛ2·I2, so we have:

dW/dt = M·I1(dI2/dt) – L1·I1·(dI1/dt) + M·I2·(dI1/dt) – L2·I2·(dI2/dt)

= – L1·I1·(dI1/dt) – L2·I2·(dI2/dt) + M·I1(dI2/dt)·I2·(dI1/dt)

Integrating both sides, and equating U with −W once more, yields:

U = (1/2)·L1·I1+ (1/2)·L2·I2+ M·I1·I2

[Again, you should just take the time derivative to verify this. If you don’t forget to apply the product rule for the M·I1·I2 term, you’ll see I am not writing too much nonsense here.] Now, there’s an interesting algebraic transformation of this expression, and an equally interesting explanation why we’d re-write the expression as we do. Let me copy it from Feynman so I’ll be using his fancier L and M symbols now. 🙂 So what? Well… Taking into account that inequality above, we can write the relation between M and the self-inductances L1 and L2 using some constant k, which varies between 0 and 1 and which we’ll refer to as the coupling constant: We refer to k as the coupling constant, for rather obvious reasons: if it’s near zero, the mutual inductance will be very small, and if it’s near one, then the coils are said to be ‘tightly coupled’, and the ‘mutual flux linkage’ is then maximized. As you can imagine, there’s a whole body of literature out there relating this coupling constant to the behavior of transformers or other circuits where mutual inductance plays a role.

The formula for self-inductance

We gave the formula for the mutual inductance of two coils that are arranged as one solenoid on top of the other (cf. the illustration I started with):

M = −(1/ε0c2)·(N1·N2)·S/l

It’s a very easy calculation, so let me quickly copy it from Feynman: You’ll say: where is the M here? This is a formula for the emf! It is, but M is the constant of proportionality in front, remember? So there you go. 🙂

Now, you would think that getting a formula for the self-inductance L of some solenoid would be equally straightforward. It turns out that that is not the case. Feynman needs two full pages and… Well… By now, you should now how ‘dense’ his writing really is: if it weren’t so dense, you’d be reading Feynman yourself, rather than my ‘explanations’ of him. 🙂 So… Well.. If you want to see how it works, just click on the link here and scroll down to the last two pages of his exposé on self-inductance. I’ll limit myself to just jotting down the formula he does obtain when he’s through the whole argument: See why he uses a fancier L than ‘my’ L? ‘His’ L is the length of the solenoid. 🙂 And, yes, r is the radius of the coil and n the number of turns per unit length in the winding. Also note this formula is valid only if L >> R, so the effects at the end of the solenoid can be neglected. OK. Done. 🙂

Well… That’s it for today! I am sorry to say but the next post promises to be as boring as this one because… Well… It’s on electric circuits again. 😦

# 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: 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: 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. 🙂 To 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. 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).

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.  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. 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: Now, that’s just a nice sinusoidal function indeed, which will look like the graph below. 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: So what happens when we do connect the wires to the circuit, so we’ve got that closed circuit depicted above (and below)? 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: [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. 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.: 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: 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 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. 🙂