As we saw when discussing Gauss’ law for magnetism, we can rewrite the fraction in the integrand:

So this electric field is conservative, and so its integral around the closed circuit is automatically zero. Thus there is no electromotive force around the circuit, and no current flows.

And yet, that’s not actually what we see. Specifically, if we wave a magnet around near such a circuit, a current will indeed flow! Indeed, this is exactly how the simplest electric generators and motors work.

To put some quantitative meat on these qualitative observational bones, we have Faraday’s law of induction. This says that the electromotive force around a circuit is equal to the rate of change of the magnetic flux through any surface bounded by that circuit. What? maybe a formula will help:

where is any surface with . Why can we pick any such surface? Because if is another one then:

We can calculate the boundary of this combined surface:

Since our space is contractible, this means that our surface is itself the boundary of some region .

But Gauss’ law for magnetism tells us that this is automatically zero. That is, every surface has the same flux, and so it doesn’t matter which one we use in Faraday’s law.

Now, we can couple this with our original definition of electromotive force:

But this works no matter what surface we consider, so we come up with the differential form of Faraday’s law:

The answer is summed up in something that is named — very confusingly — “electromotive force”. The word “force” is just a word here, so try to keep from thinking of it as a force. In fact, it’s more analogous to work, in the same way the electric field is analogous to force.

We calculate the work done by a force in moving a particle along a path is given by the line integral

If is conservative, this amounts to the difference in “potential energy” between the start and end of the path. We often interpret a work integral as an energy difference even in a more general setting. Colloquially, we sometimes say that particles “want to move” from high-energy states to low-energy ones.

Similarly, we define the electromotive force along a path to be the line integral

If the electric field is conservative — the gradient of some potential function — then the electromotive force over a path is the difference between the potential at the end and at the start of the path. But, we may ask, why the negative sign? Well, it’s conventional, but I like to think of it in terms of electrons, which have negative charge; given electric field “pushes” an electron in the opposite direction, thus we should take the negative to point the other way.

In any event, just like the electric field measures force per unit of charge, electromotive force measures work per unit of charge, and is measured in units of energy per unit of charge. In the SI system, there is is a unit called a volt — with symbol — which is given by

And often electromotive force is called “voltage”. For example, a battery — through chemical processes — induces a certain difference in the electric potential between its two terminals. In a nine-volt battery this difference is, predictably enough, , and the same difference is induced along the path of a wire leading from one terminal, through some electric appliance, and back to the other terminal. Just like the potential energy difference “pushes” particles along from high-energy states to low-energy ones, so the voltage difference “pushes” charged particles along the wire.

As a first step, though, I want to finally get a good definition of “current density”: it’s a vector field that consists of a charge density and a velocity vector , each of which is a function of space. In our example of an infinite line current, this density was concentrated along the -axis, where the velocity was vertical. But it could exist along a surface, or throughout space; a single particle of charge moving with velocity is a current density concentrated at a single point.

Anyway, so the Biot-Savart law says that the differential contribution to the magnetic field at a point from the current density at point is

So, as for the electric field, we want to integrate over :

Last time we spent a while noting that the fraction here is secretly a closed -form in disguise, so its divergence is zero. This time, I say it’s actually a conservative vector field:

Indeed, this is pretty straightforward to check by rote calculation of derivatives, and I’d rather not get into it. The upshot is we can write:

where the extra term on the second line is automatically zero because the curl is in terms of and the current density depends only on . I write it in this form because now it looks like the other end of a product rule:

Indeed, this is clearer if we write it in terms of differential forms; since the exterior derivative is a derivation we can write

for a function and a -form . If we flip over to a vector field this looks like

Okay, so now we see that is the curl of some vector field, and so the divergence of a curl is automatically zero:

Coupling this with the divergence theorem like last time, we conclude that there is no magnetic equivalent of “charge”, or else the outward flow of through a closed surface would be the integral on the inside of such a charge. But instead we find

and we replace our point charge with a charge distribution over some region of . This may be concentrated on some surfaces, or on curves, or at points, or even some combination of the these; it doesn’t matter. What does matter is that we can write the amount contributed to the electric field at by the charge a point as

So to get the whole electric field, we integrate over all of space!

Now we want to take the divergence of each side with respect to . On the right we can pull the divergence inside the integral, since the integral is over rather than . But we’ve still got a hangup.

Let’s consider this divergence:

Away from this is pretty straightforward to calculate. In fact, you can do it by hand with partial derivatives, but I know a sneakier way to see it.

If you remember our nontrivial homology classes, this is closely related to the one we built on — the case where . In that case we got a -form, not a vector field, but remember that we’re working in our standard with the standard metric, which lets us use the Hodge star to flip a -form into a -form, and a -form into a vector field! The result is exactly the field we’re taking the divergence of; and luckily enough the divergence of this vector field is exactly what corresponds to the exterior derivative on the -form, which we spent so much time proving was zero in the first place!

So this divergence is automatically zero for any , while at zero it’s not really well-defined. Still, in the best tradition of physicists we’ll fail the math and calculate anyway; what if it was well-defined, enough to take the integral inside the unit sphere at least? Then the divergence theorem tells us that the integral of the divergence through the ball is the same as the integral of the vector field itself through the surface of the sphere:

since the field is just the unit radial vector field on the sphere, which integrates to give the surface area of the sphere: . Remember that the fact that this is not zero is exactly why we said the -form cannot be exact.

So what we’re saying is that this divergence doesn’t really work in the way we usually think of it, but we can pretend it’s something that integrates to give us whenever our region of integration contains the point . We’ll call this something , where the is known as the “Dirac delta-function”, despite not actually being a function. Incidentally, it’s actually very closely related to the Kronecker delta

So anyway, that means we can calculate

This integrand is zero wherever , so the only point that can contribute at all is . We may as well consider it a constant and pull it outside the integral:

where we have integrated away the delta function to get . Notice how this is like we usually use the Kronecker delta to sum over one variable and only get a nonzero term where it equals the set value of the other variable.

The result is known as Gauss’ law:

and, incidentally, shows why we wrote the proportionality constant the way we did when defining Coulomb’s law. The meaning is that the divergence of the electric field at a point is proportional to the amount of charge distributed at that point, and the constant of proportionality is exactly .

If we integrate both sides over some region we can rewrite the law in “integral form”:

That is: the outward flow of the electric field through a closed surface is equal to the integral of the charge contained within the surface. The second step here is, of course, the divergence theorem, but this is such a popular application that people often call this “Gauss’ theorem”. Of course, there are two very different statements here: one is the physical identification of electrical divergence with charge distribution, and the other is the geometric special case of Stokes’ theorem. Properly speaking, only the first is named for Gauss.

First of all, symmetry tells us that we may as well just consider the points of the form for , and we will first specialize to the points . Along this line, the field generated by a little piece of charge on one side of the circle points across to the other side of the circle and out further along the axis; the piece on the other side cancels the horizontal contribution, but adds to the outward push. This outward direction along the axis is all that we need to calculate in this case.

The Coulomb law tells us that the differential element of charge at is . The displacement vector is , and its length is . And so the integral is

So there’s some extra push near the origin, but when gets much bigger than , this is effectively the Coulomb law again for a point source with charge $latex — the total charge on the ring.

Pushing off of the center line is a bit rougher. The differential element of charge is the same, of course, but now the displacement vector is . Its magnitude is , leading to the integral

This is, not to put too fine a point on it, really ugly, and it would take us way too far afield to go into it.

However, we can use what we know to determine the electric field generated by a charged plane with a charge density of , measured in charge per unit area. At any point we can drop a perpendicular to the plane. We cut the plane into charged rings of radius and width , which gives us a (linear) charge density of . The result above says that the ring of radius has only a component, which is

So we integrate this out over all radii:

So the field points out from the plane, and it does not fall off with distance at all!

Anyway, just like the electric field the magnetic field obeys a superposition principle, so we can add up the contributions to the magnetic field from all the tiny differential bits of a current-carrying curve by taking an integral. The differential element of charge is again . The velocity is in the direction of the curve — unit vector — and has length . Thus the term near a point is . We will take the charge density as charge per unit of distance and the speed as distance per unit of time, and combine them into the current as the charge flowing through this point on the curve per unit of time. For a curve we have the integral:

As an example, let’s take the infinite line of charge and set it in motion with a speed along the -axis. The charge density makes for a current up the line. Of course, if the current is negative then the charge is just moving in the opposite direction, down the line. The obvious parameterization is , so we have and . Plugging in we find:

where I’ve used a couple convenient substitutions to put the integral into exactly the same form as last time. We can reuse all that work to continue:

We find again that the magnetic field now falls off as the first power of the distance from the line current. As for the direction, it wraps around the line in accordance with the famous “right hand rule”; if you place the thumb of your right hand along the -axis, the field curls around the line in the same direction as your fingers.

As it happens, despite how popular the rule is it’s purely conventional, with no actual physical significance. It’s hard to explain just why that is right now, but it will become clear later. For now, I can justify that it makes no difference in the effect of currents on moving charges, since the Biot-Savart law involves a triple vector product which can be rewritten:

which formula involves no cross products and no choice of right-hand or left-hand rules.

For example, let’s say we have a curve , and along this curve we have a charge. It makes sense to measure the charge in units per unit of distance, like coulombs per meter. We can even let it vary from point to point, getting a function describing the charge per unit length near the point with parameter . To be a little more explicit, if is the “line element” that measures a tiny bit of distance near the point , then measures a little bit of charge near that point.

We can now use the Coulomb law to see what electric field this tiny bit of charge creates at a point with position vector :

since the displacement vector from to is . Now we can take this “differential electric field” and integrate it over the curve, adding up all the tiny contributions to the field at made by all the tiny bits of charge along the curve.

As an example, let’s consider an infinite line of charge along the axis with a constant charge density of ; a piece of the line of length will have charge . Admittedly, this is not a finite-length curve like above, but the same principle applies. We set , so and .

Geometric considerations tell us that the electric field generated by the line at a point is contained in the same plane that contains the line and the point. We can also tell that the field points directly perpendicular to the line; if then the vertical component induced by the chunk at is cancelled out by the component induced by the chunk at . Indeed, we can check that the first gives us

while the second gives us

and the vertical components of these two exactly cancel each other out.

All that remains is to calculate the horizontal component. Without loss of generality we can consider the point , and we must calculate the -component of the electric field by taking the integral

We need an antiderivative of the integrand , and it turns out that fits the bill. Indeed, we check:

as asserted. Thus we continue the integration:

which is a nice, tidy value. More generally, we find

pointing directly away from the (positively) charged line, neither up nor down, and calling off in magnitude as the first power of the distance from the line.

while the second exerts a force

As usual, we just add the forces together to get the resultant

Of course, as we add more particles we just add more terms to the sum. But always we find that the force on the “test” particle with charge times some vector field: . Coulomb’s law tells us that a single point of charge at the origin gives rise to a vector field whose value at the point with position vector is

That is, it’s sort of like the radial vector field only instead of getting larger as we move away from the origin, it gets smaller, falling off as the square of the distance.

As we’ve just seen, the superposition principle for forces leads to a superposition principle for the electric field: the field generated by two or more sources is the (vector) sum of the fields generated by each source separately.

If we have two charged particles that are both moving, then they also feel a different force than the electric one. We call the excess the magnetic force. In magnitude it’s proportional to both the magnitudes of the charges and their speeds, and inversely proportional to the square of the distance between them, and then it gets complicated. It’s probably going to be easier to write this down as a formula first.

So many cross products! Like last time, this is the force exerted on the first particle by the second; is the vector pointing from the second particle to the first, and is the unit vector pointing in the same direction. From this formula, we see that the force is perpendicular to the direction the first particle is moving — the magnetic force can only turn a particle, not speed it up or slow it down — and in the plane spanned by the direction the second particle is moving and the displacement vector between them.

Again, we’ve written the constant of proportionality in a weird way. The “magnetic constant” now appears in the numerator, so its units are almost like the inverse of those on the electric constant. But we’ve also got two velocities to contend with; these lead to a factor of time squared over area, resulting in mass times distance over charge squared. In the SI system we have another convenience unit called the “henry”, with symbol , defined by

which lets us write in units of henries per meter. Specifically, the SI units give it a value of . Yes, I know that this makes the proportionality constant look that much weirder, since it just works out to , but still.

But why didn’t I talk about classical Newtonian mechanics when discussing calculus? As it happens, the application of calculus to Newtonian mechanics is pretty straightforward and boring; the first-pass coverage is pretty much all there is. Electromagnatism, however, is another story. The first-pass treatment is basically all about vector calculus, and that’s great; we’ll go over that a bit, which may be review for some people. But there’s a much deeper story to even classical electromagnetism that uses all this stuff I’ve been saying about differential geometry lately. But for now everything will take place in regular three-dimensional space.

Anyway, we start with Coulomb’s law. This is something that can be experimentally determined, but we’ll take it as an assertion — another axiom — and build from there. When we have two charged particles, they exert a force on each other. The magnitude of the force is proportional to the magnitude of the charge on each particle, and inversely proportional to the square of the distance between them. The direction of the force exerted on the first particle by the second is in the direction of the vector pointing from the second to the first if their charges have the same sign, and in the opposite direction if they have different signs. That is, charges of the same sign push each other apart, while charges of the opposite sign pull each other together.

Let’s write this out in a formula: if the charge on the two particles are and — measured as a positive or negative multiple of some unit — and if the displacment vector from the second particle to the first is , then we have the following formula for the magnitude of the force exerted on the first particle by the second:

since the distance between the particles is given by the length of . To get the direction, we use the unit vector that points from the second particle to the first. It turns out that we also just need to drop the absolute value signs on the charges:

Now, I haven’t explained why the constant of proportionality is written in the weird form , and I’m not going to quite yet. I’ll just say that that’s all this is: a constant that gets the scaling right, not to mention the units. On the right, we’ve got (other than the constant) units of charge squared over area, while on the left we’ve got force, which is mass times distance over time squared. The “electric constant” , thus, must carry units of time squared times charge squared over mass times volume.

In the common SI (metric) system we measure charge in coulombs — after Coulomb’s law — with symbol and we have a convenience unit called the “farad” with symbol , which is given by

Using these units, we can write the electric constant with units of farads per meter. Incidentally, it has the measured value of approximately , but the exact value will be largely irrelevant to us.