# Magnetic dipoles and their torque and energy

We studied the magnetic dipole in very much detail in one of my previous posts but, while we talked about an awful lot of stuff there, we actually managed to not talk about the torque on a it, when it’s placed in the magnetic field of other currents, or some other magnetic field tout court. Now, that’s what drives electric motors and generators, of course, and so we should talk about it, which is what I’ll do here. Let me first remind you of the concept of torque, and then we’ll apply it to a loop of current. 🙂

The concept of torque

The concept of torque is easy to grasp intuitively, but the math involved is not so easy. Let me sum up the basics (for the detail, I’ll refer you to my posts on spin and angular momentum). In essence, for rotations in space (i.e. rotational motion), the torque is what the force is for linear motion:

1. It’s the torque (τ) that makes an object spin faster or slower around some axis, just like the force would accelerate or decelerate that very same object when it would be moving along some curve.
2. There’s also a similar ‘law of Newton’ for torque: you’ll remember that the force equals the time rate of change of a vector quantity referred to as (linear) momentum: F = dp/dt = d(mv)/dt = ma (the mass times the acceleration). Likewise, we have a vector quantity that is referred to as angular momentum (L), and we can write: τ (i.e. the Greek tau) = dL/dt.
3. Finally, instead of linear velocity, we’ll have an angular velocity ω (omega), which is the time rate of change of the angle θ that defines how far the object has gone around (as opposed to the distance in linear dynamics, describing how far the object has gone along). So we have ω = dθ/dt. This is actually easy to visualize because we know that θ, expressed in radians, is actually the length of the corresponding arc on the unit circle. Hence, the equivalence with the linear distance traveled is easily ascertained.

There are many more similarities, like an angular acceleration: α = dω/dt = d2θ/dt2, and we should also note that, just like the force, the torque is doing work – in its conventional definition as used in physics – as it turns an object instead of just moving it, so we can write:

ΔW = τ·Δθ

So it’s all the same-same but different once more 🙂 and so now we also need to point out some differences. The animation below does that very well, as it relates the ‘new’ concepts – i.e. torque and angular momentum – to the ‘old’ concepts – i.e. force and linear momentum. It does so using the vector cross product, which is really all you need to understand the math involved. Just look carefully at all of the vectors involved, which you can identify by their colors, i.e. red-brown (r), light-blue (τ), dark-blue (F), light-green (L), and dark-green (p).

So what do we have here? We have vector quantities once again, denoted by symbols in bold-face. Having said that, I should note that τ, L and ω are ‘special’ vectors: they are referred to as axial vectors, as opposed to the polar vectors F, p and v. To put it simply: polar vectors represent something physical, and axial vectors are more like mathematical vectors, but that’s a very imprecise and, hence, essential non-correct definition. 🙂 Axial vectors are directed along the axis of spin – so that is, strangely enough, at right angles to the direction of spin, or perpendicular to the ‘plane of the twist’ as Feynman calls it – and the direction of the axial vector is determined by a convention which is referred to as the ‘right-hand screw rule’. 🙂

Now, I know it’s not so easy to visualize vector cross products, so it may help to first think of torque (also known, for some obscure reason, as the moment of the force) as a twist on an object or a plane. Indeed, the torque’s magnitude can be defined in another way: it’s equal to the tangential component of the force, i.e. F·sin(Δθ), times the distance between the object and the axis of rotation (we’ll denote this distance by r). This quantity is also equal to the product of the magnitude of the force itself and the length of the so-called lever arm, i.e. the perpendicular distance from the axis to the line of action of the force (this lever arm length is denoted by r0). So, we can define τ without the use of the vector cross-product, and in not less than three different ways actually. Indeed, the torque is equal to:

1. The product of the tangential component of the force times the distance r: τ = r·Ft= r·F·sin(Δθ);
2. The product of the length of the lever arm times the force: τ = r0·F;
3. The work done per unit of distance traveled: τ = ΔW/Δθ or τ = dW/dθ in the limit.

Phew! Yeah. I know. It’s not so easy… However, I regret to have to inform you that you’ll need to go even further in your understanding of torque. More specifically, you really need to understand why and how we define the torque as a vector cross product, and so please do check out that post of mine on the fundamentals of ‘torque math’. If you don’t want to do that, then just try to remember the definition of torque as an axial vector, which is:

τ = (τyz, τzx, τxy) = (τx, τy, τz) with

τx = τyz = yFz – zFy (i.e. the torque about the x-axis, i.e. in the yz-plane),

τy = τzx = zFx – xFz (i.e. the torque about the y-axis, i.e. in the zx-plane), and

τz = τxy = xFy – yFx (i.e. the torque about the z-axis, i.e. in the xy-plane).

The angular momentum L is defined in the same way:

L = (Lyz, Lzx, Lxy) = (Lx, Ly, Lz) with

Lx = Lyz = ypz – zpy (i.e. the angular momentum about the x-axis),

Ly = Lzx = zpx – xpz (i.e. the angular momentum about the y-axis), and

Lz = Lxy = xpy – ypx (i.e. the angular momentum about the z-axis).

Let’s now apply the concepts to a loop of current.

The forces on a current loop

The geometry of the situation is depicted below. I know it looks messy but let me help you identifying the moving parts, so to speak. 🙂 We’ve got a loop with current and so we’ve got a magnetic dipole with some moment μ. From my post on the magnetic dipole, you know that μ‘s magnitude is equal to |μ| = μ = (current)·(area of the loop) = I·a·b.

Now look at the B vectors, i.e. the magnetic field. Please note that these vectors represent some external magnetic field! So it’s not like what we did in our post on the dipole: we’re not looking at the magnetic field caused by our loop, but at how it behaves in some external magnetic field. Now, because it’s kinda convenient to analyze, we assume that the direction of our external field B is the direction of the z-axis, so that’s what you see in this illustration: the B vectors all point north. Now look at the force vectors, remembering that the magnetic force is equal to:

Fmagnetic = qv×B

So that gives the F1F2F3, and F4 vectors (so that’s the force on the first, second, third and fourth leg of the loop respectively) the magnitude and direction they’re having. Now, it’s easy to see that the opposite forces, i.e. the F1F2 and F3Fpair respectively, create a torque. The torque because of Fand Fis a torque which will tend to rotate the loop about the y-axis, so that’s a torque in the xz-plane, while the torque because of Fand Fwill be some torque about the x-axis and/or the z-axis. As you can see, the torque is such that it will try to line up the moment vector μ with the magnetic field B. In fact, the geometry of the situation above is such that Fand Fhave already done their job, so to speak: the moment vector μ is already lined up with the xz-plane, so there’s not net torque in that plane. However, that’s just because of the specifics of the situation here: the more general situation is that we’d have some torque about all three axes, and so we need to find that vector τ.

If we’d be talking some electric dipole, the analysis would be very straightforward, because the electric force is just Felectric = qE, which we can also write as E = Felectric =/q, so the field is just the force on one unit of electric charge, and so it’s (relatively) easy to see that we’d get the following formula for the torque vector:

τ = p×E

Of course, the p is the electric dipole moment here, not some linear momentum. [And, yes, please do try to check this formula. Sorry I can’t elaborate on it, but the objective of this blog is not substitute for a textbook!]

Now, all of the analogies between the electric and magnetic dipole field, which we explored in the above-mentioned post of mine, would tend to make us think that we can write τ here as:

τ = μ×B

Well… Yes. It works. Now you may want to know why it works 🙂 and so let me give you the following hint. Each charge in a wire feels that Fmagnetic = qv×B force, so the total magnetic force on some volume ΔV, which I’ll denote by ΔF for a while, is the sum of the forces on all of the individual charges. So let’s assume we’ve got N charges per unit volume, then we’ve got N·ΔV charges in our little volume ΔV, so we write: ΔF = N·ΔV·q·v×B. You’re probably confused now: what’s the v here? It’s the (drift) velocity of the (free) electrons that make up our current I. Indeed, the protons don’t move. 🙂 So N·q·v is just the current density j, so we get: ΔF = j×BΔV, which implies that the force per unit volume is equal to j×B. But we need to relate it to the current in our wire, not the current density. Relax. We’re almost there. The ΔV in a wire is just its cross-sectional area A times some length, which I’ll denote by ΔL, so ΔF = j×BΔV becomes ΔF = j×BAΔL. Now, jA is the vector current I, so we get the simple result we need here: ΔF = I×BΔL, i.e.  the magnetic force per unit length on a wire is equal to ΔF/ΔL = I×B.

Let’s now get back to our magnetic dipole and calculate Fand F2. The length of ‘wire’ is the length of the leg of the loop, i.e. b, so we can write:

F= −F2 = b·I×B

So the magnitude of these forces is equal F= F2 = I·B·b. Now, The length of the moment or lever arm is, obviously, equal to a·sinθ, so the magnitude of the torque is equal to the force times the lever arm (cf. the τ = r0·F formula above) and so we can write:

τ = I·B·b·a·sinθ

But I·a·b is the magnitude of the magnetic moment μ, so we get:

τ = μ·B·sinθ

Now that’s consistent with the definition of the vector cross product:

τμ×= |μ|·|B|·sinθ·n = μ·B·sinθ·n

Done! Now, electric motors and generators are all about work and, therefore, we also need to briefly talk about energy here.

The energy of a magnetic dipole

Let me remind you that we could also write the torque as the work done per unit of distance traveled, i.e. as τ = ΔW/Δθ or τ = dW/dθ in the limit. Now, the torque tries to line up the moment with the field, and so the energy will be lowest when μ and B are parallel, so we need to throw in a minus sign when writing:

τ = −dU/dθ ⇔ dU = −τ·dθ

We should now integrate over the [0, θ] interval to find U, also using our τ = μ·B·sinθ formula. That’s easy, because we know that d(cosθ)/dθ = −sinθ, so that integral yields:

U = 1 − μ·B·cosθ + a constant

If we choose the constant to be zero, and if we equate μ·B with 1, we get the blue graph below:

The μ·B in the U = 1 − μ·B·cosθ formula is just a scaling factor, obviously, so it determines the minimum and maximum energy. Now, you may want to limit the relevant range of θ to [0, π], but that’s not necessary: the energy of our loop of current does go up and down as shown in the graph. Just think about it: it all makes perfect sense!

Now, there is, of course, more energy in the loop than this U energy because energy is needed to maintain the current in the loop, and so we didn’t talk about that here. Therefore, we’ll qualify this ‘energy’ and call it the mechanical energy, which we’ll abbreviate by Umech. In addition, we could, and will, choose some other constant of integration, so that amounts to choosing some other reference point for the lowest energy level. Why? Because it then allows us to write Umech as a vector dot product, so we get:

Umech = −μ·B·cosθ = −μ·B

The graph is pretty much the same, but it now goes from −μ·B to +μ·B, as shown by the red graph in the illustration above.

Finally, you should note that the Umech = −μ·B formula is similar to what you’ll usually see written for the energy of an electric dipole: U = −p·E. So that’s all nice and good! However, you should remember that the electrostatic energy of an electric dipole (i.e. two opposite charges separated by some distance d) is all of the energy, as we don’t need to maintain some current to create the dipole moment!

Now, Feynman does all kinds of things with these formulas in his Lectures on electromagnetism but I really think this is all you need to know about it—for the moment, at least. 🙂

# The field of electric and magnetic dipoles

You’l surely remember the electric field of an electric dipole, as depicted below:

We distinguish two components:

1. A z-component along the axis of the dipole itself, and
2. A so-called transverse component, i.e. the component that is perpendicular to the axis of the dipole, or the z-axis.

Pythagoras’ rule then gives us the total field, which is equal to:

I’ll give you the formulas for both components in a moment, but let’s first introduce the concept of a magnetic dipole. Look at the magnetic field of a solenoid below, and imagine we reduce the solenoid to one loop of current only. What would we get?

We get a magnetic field that resembles the electric field from an electric dipole. Of course, it’s a magnetic field, and it’s not the field of an electric dipole but of a magnetic dipole which, in this case, is the field of a small loop of current. Feynman depicts the geometry of the situation, and the relevant variables, as follows:

Now, in my previous post, in which I presented the magnetic vector potential, I pointed out that the equations for the x-, y- and z-component of the vector potential A are Poisson equations and, therefore, they are mathematically identical to the Poisson equation for electrostatics. Now that’s why the x-, y- and z-component of the vector potential arising from a current density j is exactly the same as the electric potential Φ that would be produced by a charge density ρ equal to jx, jy  and jz respectively, divided by c2, and we use that fact to calculate them. Huh? Yes. Let me just jot it down the equations once more, so you can appreciate what I am saying here:

2Φ = –ρ/ε0 (Poisson equation for electrostatics)

2A = –j0c(Poisson equation for magnetostatics)

⇔ Ax = –jx0c2, Ay = –jy0cand Az = –jz0c2

We didn’t go through all of the nitty-gritty of the exercise for a straight wire and a solenoid in our previous post, but we will do so now for our single loop of current. It’s kinda tedious stuff, so just hang in there and try to get through it. 🙂

We’ll start with Ax. To calculate Ax, we need jx, so that’s the component of the current density that’s shown below. [As you can see, jz is zero, and we’ll talk about jy in a minute.]

What are those + and − signs next to those two loop legs? Well… Think about what we wrote above: we need to think of jdivided by cas some charge density, and the associated Φ will then equal Ax. It’s an easy principle: the same equations have the same solutions. We’re only switching symbols. So look at the geometry of the situation: we have a uniform current and, hence, a uniform current density, in each leg of the loop, so jis some constant over the two legs shown here. Therefore, the electrostatic equivalent is that of two charged rods with some charge density λ.

Now, it’s easy to see that λ jx/c2 = I/cand… So… Well… Hmm… […] So how do we calculate the electrostatic potential Φ from two charged rods? Well… We haven’t calculated that in any of our posts, but if R is ‘large enough’ (and it should be, because we’re interested in the fields only at distances that are large compared to the size of the loop), then Φ will be the potential of a dipole with (opposite) charges that are equal to the (opposite) charges of the rods. Now, λ is a charge density, so we need to multiply it with the length of the rods a to get the charge. And then we need to multiply the charge with the distance between the two charges to get the dipole moment, so we write:

p = λab = Iab/c2

OK. That’s a nice result because we can now use that to calculate the potential. Indeed, you may or may not remember the formula for the potential of an electric dipole, but we wrote it as follows in one of my posts:

Φ = −p·φ= −(1/4πε0)p·(1/R) = −(1/4πε0)(p·er)/R2

φ0 is the gradient of φ0, and φis the potential of a unit point of charge: φ0 = 1/4πε0r. In any case, don’t worry too much about this formula right now, because I’ll give you some easier version of it later on. So let’s just move on. To calculate that vector product p·er = |p|·|er |·cosθ = p·cosθ, we need to know the angle θ between them, and the direction of the dipole moment. That’s simple enough: the vector p points in the negative y-direction, so cosθ = –y/R, where y is a coordinate of P. So we have:

Now we do the grand substitution, the hat-trick really: Φ becomes Ax, and λ jx/c2 = I/c2, and so we get:

Huh? Yes. It’s a great trick really. Brilliant! 🙂 You’ll get it. Just think about what happened for a while.

Next thing is jy. So what about it? Well… As you may have suspected, it’s a perfectly symmetrical situation, so we can just repeat the same reasoning but swap x for y and y for x. We get:

Note the minus sign has disappeared, but you can explain that yourself. And, of course, we only have the vector potential here, so we still need to calculate B from it, using the B×A equation, which is three equations involving all possible cross-derivatives really:

Phew! It looks quite horrible indeed! Feynman says: “It’s a little more complicated than electrostatics, but the same idea.” But… Well… I guess it’s a bit hard to not take this statement with a pinch of salt. For starters, there is no intuitive physical interpretation of the magnetic vector potential A, as opposed to the electric potential Φ. More importantly, calculating E = −Φ involves only three derivatives, and that’s a helluva lot easier than calculating not less than six cross-derivatives and keep track of their order and all of that. In any case, nothing much we can do about, so let’s grind through it. And to be fully honest: Az is zero, so the ∂Az/∂y and ∂Az/∂x derivatives above are also zero, so that leaves only four cross-derivatives to be calculated.

Let’s first define the magnetic dipole moment, however. We talked a lot about it, but what is it? We said it’s similar to the electric dipole moment = q·d, whose magnitude was λab = Iab/cin this example. So what have we got here? No charge, but a current, so that’s charge per second. No distance either, but some area a·b. And then the 1/c2 factor that, somehow, always slips in because we’ve got it in Maxwell’s equation too. So, we said the fields look the same, and just like Φ was proportional to p, we now see that A is proportional to I·a·b, so it’s quite similar really and so we’ll just define the magnetic dipole moment as:

μ = (current)·(area of the loop) = I·a·b

Now, the area of the loop doesn’t have to be some square: it could also be some circle or some triangle or whatever. You should just change the formula to calculate the area accordingly. You’ll say: where’s the 1/c2 factor? It’s true: we didn’t put it in. Why? I could say: that’s the convention, but that’s a bit too easy as an answer, I guess. I am not quite sure but the best answer I have is that the 1/c2 factor has nothing to do with the physicality of the situation: it’s part of Maxwell’s equations, or the electromagnetic force equations, if you want. So it’s typical of the magnetic force but it’s got nothing to do with our little loop of current, so to speak. So that’s why we just leave it out in our definition of μ.

OK. That should be it. However, I should also note that we can (and, hence, we will) make a vector quantity out of μ, so then we write it in boldface: μ. The symmetry of the situation implies that μ should be some vector that’s normal to the plane of the loop because, if not, all directions would be equally likely candidates. But up or down? As you may expect, we’ll use the right-hand rule again, as illustrated below: if your fingers point in the direction of current, your thumb will point in μ‘s direction. The illustration below also gives the direction of A, because we can now combine the definition of μ and our formulas for Ax, Ay and Az and write:

Note that the formula for A above is quite similar to the formula for Φ below, but not quite the same: in the formula for A we’ve got a vector cross product – so that’s very different from the vector dot product below – and then we’ve also got that 1/cfactor. So watch out: it’s the “same-same but different”, as they say in Asia. 🙂[…] OK. Sooooo much talk, and still I did not calculate B to effectively show that the magnetic field of a magnetic dipole looks like the electric field of an electric dipole indeed. So let me no longer postpone those tedious calculations and effectively do that B×A cross-product using our new formula for A. Well… Calculating Bx and By is relatively simple because A‘s z-component is zero, so ∂Az/∂x = ∂Az/∂y = 0. So for Bx, we get:

For By and Bz, we get:

The … in the formulas above obviously stands for μ/4πε0c2. Now, let’s compare this with the components of the electric field (E) of an electric dipole. Let’s re-write the formula for the electric potential above as:

Note that this formula assumes the z-axis points in the same direction as the dipole, and that the dipole is centered at the origin of the coordinate system, so it’s the same coordinate system as the one for the magnetic dipole above. Therefore, we have that cosθ = z/r, and so the formula above is equivalent to:

Now, let’s find the electric field by using the E = (Ex, Ey, Ez) = −Φ = (−∂Φ/∂x, Φ, −∂Φ/∂y, −∂Φ/∂z) equation. For Ez, we get:

which we can re-write as:

Huh?  Yes. Note that z also appears in r, because r = (x2+y2+z2)1/2, and so we need to apply the product (or quotient) rule for derivatives: (u·v)’ = u’·v + u·v’. Note that this simplifies to E = (1/4πε0)·(2p/r3) for the field at distance r from the dipole in the direction along its axis.

OK, so we’re done with Ez. For Eand Ey, we get:

So… Well… Yes! We’ve checked it: the formulas for Ex, Ey, and Ez have exactly the same shape as those for Bx, By, and Bz , except for that 1/c2 factor and, of course, the fact that we switched p for μ, so we may have some other number there too. [Oh – before I forget – I promised to give you the formula for the transverse component. The transverse component is, obviously, equal to (Ex2 + Ey2)1/2. So it’s just Pythagoras’ rule once more. Let me refer you to Feynman for the explicit formula, i.e. the formula in terms of r and θ, or in terms of x and y, as I don’t need it here and there’s too much clutter already in this post.]

So… Well… Yes, that’s all very interesting! Indeed, as Feynman notes, it is quite remarkable that, starting with completely different laws, E=ρ/ϵ0 and ×j/ϵ0c2, we end up with the same kind of field.

Is it? Well… Yes and no. You should, of course, note that the sources whose configuration we summarizing here by the dipole moments are physically quite different—in one case, it’s a circulating current; in the other, a pair of charges—one above and one below the plane of the loop for the corresponding field. So… Well… It’s the same-same but different indeed! 🙂

Now, to conclude this post, we should, perhaps, have a look at the units and order of magnitude of what we’re talking about here. The unit of charge and the unit of current are related: 1 ampere is a current of 1 coulomb per second. The coulomb is a rather large unit of charge, as it’s equivalent to the charge of some 6.241×1018 protons. Now, that’s less than a mole, which is about 6×1023, but it’s still a quite respectable number. 🙂 Having said that, a current of one ampere is not so exceptional in everyday life, and so a magnetic dipole moment of 1 A·m2 (i.e. the equivalent of an electric dipole moment of 1 C·m) would not be exceptional either.

The really interesting thing about the magnetic force is that 1/c2 factor, which causes its magnitude to be so much smaller than that of the electric force. Indeed, the electric field at distance r from the dipole in the direction along its axis is given by E = (1/4πε0)·(2p/r3). Assuming p = 1 C·m, and noting that 1/4πε≈ 9×10N·m2/C2 we get E ≈ (18×109)·r−3 N/C. So we’ve got an incredibly large (18×109) factor in front here! [Note how the units come out alright: the electric field is effectively measured in newton per coulomb, indeed.]

What do we get for the magnetic field (along the z-axis once more) from a magnetic dipole with moment μ = 1 A·m2? The formula is now B = (1/4πε0c2)·(2μ/r3). Now, 1/4πε0c2 ≈ (1×10−7) N·s2/C2, so B ≈ (2×10−7)·r−3 N·s/C·m. So we’ve got an incredibly small (2×10−7) factor in front here! [As for the unit, 1 N·s/C·m is the unit of the magnetic field indeed. It’s referred to as the tesla (1 T = 1 N·s/C·m), and it makes the force equation come out alright: the magnetic force on some charge q is, indeed, Fmagnetic = qv×B, with v expressed in m/s, and so that’s why we need to multiply the N/C factor with an s/m factor in the unit for B.]

So look at this! The point to note is the relative strength – I should say weakness – of the magnetic force as compared to the electric force: they differ with a factor equal to c≈ 9×1016. That’s quite incredible, especially because our electric motors and generators work through the magnetic force, as I’ll show in my next posts. In fact, looking at the Fmagnetic = qv×B, and comparing it to the electric force (Felectric = q·E), we may say that the magnitude of the magnetic force, as compared to the magnitude of the electric force, is weaker because of two factors:

1. First, there is the relative velocity of the charge that we are looking, which is equal to β = v/c. The electrostatic Coulomb force has no such factor.
2. Second, there is the 1/c factor, which we also encountered when discussing radiation pressure, or the momentum of light, which is equal to p = E/c, with the symbol p denoting momentum here – not some dipole moment – and with E denoting the energy of the light photons – not the electric field involved!

Both combine to give us the actual relative strength of the magnetic versus the electric force, so we can write that relative strength as:

|Fmagnetic|/|Felectric| = FB/FE = (v/c)·(1/c) = v/c2

Now, I am tempted to now write a lot about that 1/c factor, but I can’t do that here, as this post has become way too long already. Let me just add a small addendum here. It’s, perhaps, a not-so-easy piece 🙂 but I warmly recommend trying to work through it.

Addendum: On the momentum of light and radiation pressure

Look at the illustration below: it represents a beam of light, i.e. electromagnetic radiation, originating at some source S and hitting some charge q. What charge? Whatever charge the electromagnetic radiation is affecting, so you may want to think of an electron or a proton in a mirror or a piece of glass, or whatever other surface that is absorbing a photon fully or partially—so that’s any charge that is absorbing the energy of the light.

I’ve written a post on this before. It concluded a series of posts on electromagnetic radiation by noting that we usually look at the electric field only when discussing radiation. The electric field is given by the following formula for E:

I know. It looks like a monster, but the first term is just the Coulomb effect. For the explanation of the second and term term, I need to refer you to the mentioned section in Feynman’s Lecture on it. [Sorry for that, but I am still struggling somewhat with this equation myself. If and when I fully ‘get it’ myself, I’ll jot down a summary of my understanding of it. Now now, however.]

The point is: when discussing light, like interference and what have you, we usually forget about the B vector, for which the formula is the following:

So we define B by referring to E. Of course, this again shows the electric and magnetic force are one and the same phenomenon, really. The cross-product gives you both (i) the magnitude of B, which is equal to B = E/c (the magnitude of the unit vectors er’ and n in the vector cross-product are obviously equal to one, and sinθ is equal to one too), and (ii) its direction: it’s perpendicular to both E as well as to er’, which is the unit vector from the point P where E is measured to the charge q that is producing the field.

The B = E/c equation tells us why the magnetic field is never looked at when discussing electromagnetic radiation: because it’s so tiny. It’s so tiny because of that 1/c factor, and so the B vector in the illustration above is surely not to scale: if all was drawn to scale, you just wouldn’t see it, because it’s 1/c times smaller than E. However, it’s there, and the illustration shows how the magnetic force resulting from it looks like, if we forget about scale, that is. 🙂

The magnetic force is, obviously, F = qv×B, and you need to apply the usual right-hand screw rule to find the direction of the force. [Please note that the order of magnitude of the force is the same as that of B so… Well… Again I need to warn you the illustration is not to scale and, hence, somewhat misleading. But I can’t think of an alternative to it.] As you can see, the magnetic force – as tiny as it is – is oriented in the direction of propagation, and it is what is responsible for the so-called radiation pressure.

Indeed, there is a ‘pushing momentum’ here, and we can calculate how strong it is. In fact, we should check how strong it is in order to see if it could, potentially, make a space ship like the one below. 🙂

Sorry. That was just to add a lighter note. 🙂 I know posts like these are rather tedious, so I just wanted to lighten it all up. 🙂 So… Well… Back to our F = qv×B equation. Because of that B = –er’×E/c formula, we can substitute B for E/c, and so we get the following formula for the magnitude of the magnetic force:

F = q·v·E/c

Now, the charge q times the electric field is the electric force on the charge, and the force on the charge times the velocity is equal to dW/dt, so that’s the time rate of change of the work that’s being done on the charge.

Huh? I know: not easy to see what’s being done here. Think of it like this: work equals force times distance, so W = qE·s in this case. Hence, dW/dt = d(qE·s)/dt = qE·ds/dt = qE·ds/dt. [Note that, if you’d check Feynman’s analysis, you’ll find that it’s based on average values here for some reason, so he writes 〈F〉 and 〈E〉, but I am not quite sure why, so I simplified here. I’ll try to figure the why of these averages later, and let you know.]

Again, think physical, rather than mathematical: what’s the charge here? It’s whatever charge the electromagnetic radiation is hitting, so you should think of an electron or a proton in a mirror or a piece of glass or whatever other surface. Now, because of the energy conservation principle, dW/dt must also be equal to the energy that is being absorbed from the light per second. So F = (dW/dt)/c.

Now, Newton’s Law (he had many, of course, so let me be precise: I am talking Newton’s Second Law of Motion) tells us that the force is the rate of change of the momentum: F = m·a = m·dv/dt = d(m·v)/dt. Hence, integrating both sides of the F = (dW/dt)/c equation gives us the associated momentum of the light: p = W/c.

In short, we know that light carries energy, but so we also know that light also carries momentum, and that momentum is due to the magnetic force, and it’s equal to 1/c times the energy. To be clear: it’s real pressure, so when light is emitted from a source there is a recoil effect, and when light hits some charge, it’s the same thing: the momentum of the light is being conserved as the charge that’s absorbing the energy picks it up. To be fully complete: we also have Newton’s Third Law of Motion coming into play: for every action, there is an equal and opposite reaction.

So… Well… That’s about it, I think. Just note that you’ll usually see this momentum written as p = E/c, with E denoting the energy of the radiation, not the electric field (sorry for the switch in symbols). In this equation, E is given by the Planck-Einstein relation E = h·f, with h Planck’s constant and the frequency of the light.

Interesting! So it all makes sense! Isn’t it wonderful how all these equations come together in the end? 🙂 However, I shouldn’t digress even more, and so I’ll leave it to you to further reflect on this. 🙂

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