Calculating the limits of magnetic acceleration

[Mr Bean]

So if I'm reading you right, in order to fire the gun with a 1 lb projection it must be 500 meters long and 50 GW's worth of energy?

I'm think I'm reading you wrong on what sized projectile the secondary calculations account fore.

[starslayer]

Ampère's circuital law is just the enclosed current: cInt[ B·dl ] = µ₀ I_enc. Maxwell's contribution is the addition of the displacement current term: … + µ₀ε₀ dΦ_E/dt

Quite true. I guess I'm just used to how most physicists around here phrase it. Most of the ones I know just drop "and Maxwell's correction term" when discussing Ampere's Law.

The energy of a dipole μ in a magnetic field B is U = -μ·B, if we assume the dipole moment to be neglible. The force would just be the gradient of the same: F = ∇U(x). (Expressing this in terms of μ and B gets funky.)

Also true. I had the opportunity to ask someone about the force between permanent magnets and why they move. He said that the force between them will be F = (µ * ∇) * B. As for why they move, he said that B-fields effectively trick E-fields into doing their work for them. As the electrons orbit the nuclei in the magnets, they produce a dipole moment (and also contribute one due to their own spin). In a magnet, obviously, these are mostly all aligned in the same direction. The B-field exerts a Lorentz force on them, and they begin to move towards the one of the ends of the magnet. They do so because the B-field isn't completely aligned along the magnet's axis. If it were, the magnet wouldn't move at all. Anyways, as they move, the protons in the nuclei exert a force on them, trying to pull them back. In effect, the electrons bang into one end of the magnet, moving it towards or away from the other (attraction or repulsion).

So if I'm reading you right, in order to fire the gun with a 1 lb projection it must be 500 meters long and 50 GW's worth of energy?

I'm think I'm reading you wrong on what sized projectile the secondary calculations account fore.

What my calculations say is that for a 1 pound projectile, if we want the gun to fire in a hundredth of a second, the gun must be 500 meters long and requires a constant power input of 50 GW over that time (power is defined as work/energy per time). This translates into about 500 MJ of total energy input in said hundredth of a second at 5% efficiency.

[Wyrm]

Quite true. I guess I'm just used to how most physicists around here phrase it. Most of the ones I know just drop "and Maxwell's correction term" when discussing Ampere's Law.

The proper name of the law would be "Ampère-Maxwell Law" in this case, but which Ampère law is usually grokked from the context (like Kirchhoff's laws). To wit, if I speak about "Ampère's law" with regards to force, you're supposed to think "Ampère's force law"; otherwise, I mean "Ampère's law" to mean "Ampère's circuital law" or equivalently "Ampère-Maxwell Law" (The Ampère's law that was modified and included with Maxwell's laws).

Also true. I had the opportunity to ask someone about the force between permanent magnets and why they move. He said that the force between them will be F = (µ * ∇) * B. As for why they move, he said that B-fields effectively trick E-fields into doing their work for them. As the electrons orbit the nuclei in the magnets, they produce a dipole moment (and also contribute one due to their own spin). In a magnet, obviously, these are mostly all aligned in the same direction. The B-field exerts a Lorentz force on them, and they begin to move towards the one of the ends of the magnet. They do so because the B-field isn't completely aligned along the magnet's axis. If it were, the magnet wouldn't move at all. Anyways, as they move, the protons in the nuclei exert a force on them, trying to pull them back. In effect, the electrons bang into one end of the magnet, moving it towards or away from the other (attraction or repulsion).

Of course, the only place where a magnet could be aligned with a magnetic field completely is if it's in a unidirectional field, which will be uniform if there are no currents about. Of course, in this case, the gradient of the potential vanishes, and therefore, there is no force.

[Mr Bean]

What my calculations say is that for a 1 pound projectile, if we want the gun to fire in a hundredth of a second, the gun must be 500 meters long and requires a constant power input of 50 GW over that time (power is defined as work/energy per time). This translates into about 500 MJ of total energy input in said hundredth of a second at 5% efficiency.

Is there a serious effect of increasing the length of firing time? IE can a massive decrease be had by increasing the firing time? Or would the speed of the projectile negate that?

[Wyrm]

What my calculations say is that for a 1 pound projectile, if we want the gun to fire in a hundredth of a second, the gun must be 500 meters long and requires a constant power input of 50 GW over that time (power is defined as work/energy per time). This translates into about 500 MJ of total energy input in said hundredth of a second at 5% efficiency.

Is there a serious effect of increasing the length of firing time? IE can a massive decrease be had by increasing the firing time? Or would the speed of the projectile negate that?

Increasing the time of firing solves several issues, including saturation of the round, and timing issues, at the expense of a lengthening barrel. Increasing time allows you to decrease peak B, meaning that the acceleration more easily remains in the B² regime, and the round doesn't have to relax as much (magnetically speaking) between kicks. Furthermore, increasing time allows more accurate knowledge of the round's trajectory, allowing you to more accurately time the next kick. Of course, the point is to get this round to high speed, and we're already pushing 500 meters. At 500 MJ expended power at 5% efficiency (energy that is not kinetic energy of the projectile) for a 1 lb round equals 10.5 km/s, so increasing the time by another hundreth of a second increases the barrel length by around 50 m.

Forget mounting this sucker on a warship, but a mass-driver to orbit would be fine.

[starslayer]

I've noticed that we seem to be thinking that saturating the round limits the effectiveness of the gun. It doesn't (or shouldn't, anyways). When I asked about bar magnets, I also asked him about how magnetic saturation will affect the gun. He said that it shouldn't, insofar as increasing the field will still increase the force on the projectile, as the force equation for bar magnets holds for our projectile (the coil will look like a bar magnet at the start from the standpoint of its field). I.e., mu will max out, but grad B won't.

A few other notes: Would making a multi-stage gun help solve our length problem? How much could we increase the efficiency by using room-temperature superconductor and such? And how much can we "realistically" decrease the firing time as well?

[Ariphaos]

Well, it could be--that depends on the magnetizing field. If it matches the induction field by the same permeability factor as the surrounding medium, the force should vanish completely. The above expression is an upper limit in the idealization of constant source currents with projectile far from saturation and high permeability. Let's refine the bound a bit. In the same situation as before,
W = (1/8π)Int[ (H₀·B - H·B₀) d³x ] = (1/8π)Int[ (1/μ₀-1/μ)B·B₀ d³x ], μ₀ = 1 free space (cgs).
The corresponding force, F_x = ∂W/∂x, should now be obvious; SI units gain a 4π, etc.

Still not quite getting this - in particular, what is d? Or did I forget some basic calculus?

I'll try picking something simpler for now - taking Iron, μ = .011 for 91 amp-meters (~1.14 oerstads) for H, B and B₀ = 1 (I assume the value for the field strength in free space should be calculated separately, but skipping that for now...)

(1/8π)(1/μ₀-1/.011)Int[ d³x ]

I think I've got it crazy wrong here.

As you can see, when saturated (internal B maximal but constant), the force on the projectile will be linear in original field B₀ (and thus also linear in current), rather than quadratic as before.

That part I understand well enough. Just the math getting there.

[Beowulf]

So if I'm reading you right, in order to fire the gun with a 1 lb projection it must be 500 meters long and 50 GW's worth of energy?

I'm think I'm reading you wrong on what sized projectile the secondary calculations account fore.

The fly in the soup is that your output power is pulsed, therefore your input power is dependant on how long you want to charge before you do the shot.

Of course, he still did his math wrong, because power scales with the square of the velocity, not linearly with velocity. Also, there's a confusion between Joules and Watts, which you've also done.

[Wyrm]

I note that, once the projectile reaches the midpoint of the coil, you want to turn the coil off. Otherwise, the projectile will demagnetize down the hysteresis and your coil will rob some of the hard work you just did, or worse, the moment interaction with the coil will cause the projectile to torque around to the same effect, and that torquing cannot do anything good for your barrel.

Of course, taking down a magnetic field requires energy to be dissipated or reabsorbed (as per Kuroneko's eqn).

[starslayer]

I note that, once the projectile reaches the midpoint of the coil, you want to turn the coil off. Otherwise, the projectile will demagnetize down the hysteresis and your coil will rob some of the hard work you just did, or worse, the moment interaction with the coil will cause the projectile to torque around to the same effect, and that torquing cannot do anything good for your barrel.

Well, you might also end up simply causing the projectile to oscillate back and forth inside the gun (this certainly happened when I built one using a car battery and was firing paper clips across the room). The projectile probably will not have a significant torque on it, as all the forces that would cause it to do so balance out, with the exception of the force gradient between both ends, but that gradient is along the axis of the projectile and the gun.

Of course, he still did his math wrong, because power scales with the square of the velocity, not linearly with velocity. Also, there's a confusion between Joules and Watts, which you've also done.

Not only that, (and I messed up too, though it doesn't affect my calculations: 100 * 80^2 should be 640,000, not 160,000), but as I explained earlier, Deathdealer messed up his units such that they became meaningless; he got mass times energy, not power, which is energy or work divided by time. Therefore, as I explained, we must pick either our desired power input/output, or the time in which we want the gun to fire. I chose to specify the time.

[Mr Bean]

Which we keep coming around to the same situation, this gun is 500 meter long(Possible for a solid land instillation) but impossible for anything mobile, and it only fires a single 1 pound projectile, at a rather high speed. Were one to simply strap a rocket on the back of said 1 pound projectile the speed might be much easier to achieve.

As I initial suspected, such a weapon is impossibly impractical at this time.

[Wyrm]

Well, you might also end up simply causing the projectile to oscillate back and forth inside the gun (this certainly happened when I built one using a car battery and was firing paper clips across the room). The projectile probably will not have a significant torque on it, as all the forces that would cause it to do so balance out, with the exception of the force gradient between both ends, but that gradient is along the axis of the projectile and the gun.

The gradient of the B field reverses once you get past the halfway point. The B field remains in the same direction, but it weakens, so the gradient points the other way, as does the force. That changes the behavior of the lateral equilibrium: going in, the equilibrium is stable; going out, the equilibrium is unstable.

[starslayer]

The gradient of the B field reverses once you get past the halfway point. The B field remains in the same direction, but it weakens, so the gradient points the other way, as does the force. That changes the behavior of the lateral equilibrium: going in, the equilibrium is stable; going out, the equilibrium is unstable.

I agree that the direction of the force reverses; this is easily seen because of the problem of oscillation caused by leaving the power on too long. Perhaps I simply don't know enough about EM (a very good possibility), but I'm having trouble understanding why reversing the gradient's direction changes the equilibrium here. Shouldn't it simply retard the projectile's motion? I guess I'm just not seeing where the torque is going to come from. I'm assuming that we've cut the power at the optimum time, so that we provide maximum acceleration for the projectile, and allow time for the coil's inductance issues to resolve themselves as much as possible.

We also have another, unstated problem. In any real coilgun, as the projectile goes on its merry way, the coil and projectile then repel each other quite strongly if we've cut the power soon enough. We may have a problem even with using this thing as mass driver to reach orbit. If we tried, the coil may destroy itself on each firing due to the compression forces.

[Kuroneko]

Is B₀ the magnetic field absent the slug, and B the magnetic field with the slug, or what?

Yes.

When you say difference, do you mean taking the work equation for setting up the magnetic field around the coil without the projectile (found in your W = ... equation above) and subtracting the work equation with the projectile?

Yes, of course. One can show that the integral of (B+B₀)·(H-H₀) vanishes, which is all one needs to go from H·B-H₀·B₀ to the stated result. Alternatively, with some foresight, one can simply work backwards from the magnetization version of the formula (independent discussion of which is below), assuming a linear medium: M = (μ/μ₀-1)H. Now, two things should might be emphasized. One is that the above calculation assumes a magnetostatic case, and that it assumed a linear medium. But since the problem was to find an upper limit and the force degrades with projectile velocity, the former isn't really an issue, although we have some cause for concern over the latter in highly saturated environments.

If I recall, in general, B = μ₀(H + M) where M is magnetization.

In SI units, yes.

Given that ferromagnets have hysterisis, I suspect that the derivitive wrt time for magnetization is (d/dt)M = f(M, H, (d/dt)H) for some f. Correct?

Yes. A fully general case would have something of the nature of (∂/∂t)M = f(γ,η), where γ,η are in C^1((-∞,0),ℝ³). On the other hand, here the approximation of the magnetization being parallel to the field is quite natural, so we may take μ as μ(B,H) or equivalently μ(M,H). Then differentiating M = B/μ - H with the latter interpretation of μ gives an equation implicitly equivalent to yours.

---

As for why they move, he said that B-fields effectively trick E-fields into doing their work for them. As the electrons orbit the nuclei in the magnets, they produce a dipole moment (and also contribute one due to their own spin). ...

That's true, but for the purposes of calculation, it is easier to think of force as simply the spatial derivative of work, while the latter can be interpreted in the following manner: the energy of a magnetic field is the energy it takes to establish it from the zero state, so that the changing magnetic field induces, by Faraday's law, an electric field, which is what actually does the work.

I had the opportunity to ask someone about the force between permanent magnets and why they move. He said that the force between them will be F = (µ * ∇) * B.

Perhaps some clarification is in order. If M is the magnetic dipole density (i.e., magnetization) of the magnet, the force density on a permanent magnet in an external field is indeed (M×∇)×B₀ = ∇(M·B₀), the equality following from the fact that B is always solenoidal (∇·B₀ = 0). Thus we have a well-defined potential energy density -M·B₀.

Here is a direct connection to the work integral of the last page, W = (1/2)Int[ M·B d³x ], the differences being that the internal dipoles are not permanent, thus the factor of 1/2, and with an opposite sign because of contribution of the work done by the sources against the back-emf in establishing the field. The force components still have the same signs; one is determined by a negative partial and the other by a positive one. This halving isn't something peculiar to magnetism--the same thing happens for the potential difference of a linear dielectric in, say, a parallel-plate capacitor connected to a battery, and work done by the battery, because of the work required to keep a constant voltage. Mathematically, δw = H·δB = H·δ(μH) = (μ/2)δ(H²) = (1/2)δ(H·B).

In general, the energy density of a magnetostatic field being B²/2μ₀ (equivalent to the above dot product in linear media) is something found in almost every book on electromagnetism, so I won't bother with a justification. Again, for those that don't want to dredge through integral differences (since I did receive PMs about this), substituting M = (μ/μ₀-1)H (also only valid for linear media) gets us the result on the previous page.

I've noticed that we seem to be thinking that saturating the round limits the effectiveness of the gun. It doesn't (or shouldn't, anyways).

I've said previously that saturation changes the the regime from being quadratic in B to being linear in B, and that if the projectile is saturated in such a way as to give the same permeability as the surrounding medium (full diffusion), then the force on the projectile actually vanishes. As far as I know, those statements are correct, although in practice, even if this perfect diffusion of the magnetic field were to be obtained, it wouldn't be but for a moment. The only caveat is that the actual formula is no longer applicable there.

A few other notes: Would making a multi-stage gun help solve our length problem? How much could we increase the efficiency by using room-temperature superconductor and such? And how much can we "realistically" decrease the firing time as well?

Well, superconductors have effectively infinite conductivity. In regions of the magnetic field as the sole source of current, the field diffuses according to ∂B/∂t = -∇×E = -∇×(J/σ) = -∇×(∇×B)/(μσ) = ∇²B/(μσ), where σ is the conductivity of the material and μ is assumed to be independent of frequency (edit: the equation in terms of electromagnetic vector potential is actually identical in form, contrary to the comment that resided here). For σμ≫1, the field does does not diffuse, which means that the surface Eddy currents must cancel the field completely. In other words, a superconducting material that does not have an initial magnetic field itself acts as a diamagnet of zero permeability. In such a case, M = -H.

Hmm... perhaps some assumptions about efficiency would simplify this to be tractable. I'll think on it later.

---

Still not quite getting this - in particular, what is d? Or did I forget some basic calculus?

Differential. The usage of d³x is common among physicists to indicate volume integrals. A very common alternative is dV. If the parameters are actually constant across the entire volume, then one can factor them out, and Int[ d³x ] becomes simply the volume of the projectile.

I'll try picking something simpler for now - taking Iron, μ = .011 for 91 amp-meters (~1.14 oerstads) for H, B and B₀ = 1 (I assume the value for the field strength in free space should be calculated separately, but skipping that for now...)
(1/8π)(1/μ₀-1/.011)Int[ d³x ]
I think I've got it crazy wrong here.

There are some complications. For the force, calculating a single value does would not tell you the force--you need a partial derivative for that. However, what you could do is calculate two values separated by some distance h along the trajectory; then (W-W')/h is the average force over that trajectory. Thus, if the value did not change, you have no force, no matter how strong the fields are. This is applicable only over small timeframes (and thus small h), because coilguns are not magnetostatic. Alternatively, if you're not worried about the timeframe other than assuming that it's small, you could calculate the difference in potentials between no magnetization and full magnetization. This would tell you the kinetic energy of the projectile at the end of this operation, although it tells you nothing of the force because the timeframe is unknown.

[Ariphaos]

Differential. The usage of d³x is common among physicists to indicate volume integrals. A very common alternative is dV. If the parameters are actually constant across the entire volume, then one can factor them out, and Int[ d³x ] becomes simply the volume of the projectile.

Ah, thanks.

There are some complications. For the force, calculating a single value does would not tell you the force--you need a partial derivative for that. However, what you could do is calculate two values separated by some distance h along the trajectory; then (W-W')/h is the average force over that trajectory. Thus, if the value did not change, you have no force, no matter how strong the fields are. This is applicable only over small timeframes (and thus small h), because coilguns are not magnetostatic. Alternatively, if you're not worried about the timeframe other than assuming that it's small, you could calculate the difference in potentials between no magnetization and full magnetization. This would tell you the kinetic energy of the projectile at the end of this operation, although it tells you nothing of the force because the timeframe is unknown.

The timeframe is part of the point of the exercise (determining the feasibility of a relativistic kill vehicle - or rather the lack thereof). And h would have to either be large or calculated many times over the course of the exercise for some projectiles, so it would seem the partial derivative is the only way to go, if I needed to be specific.

But I don't. Even with the upper bound you gave, it's easy to show just how absurd it is to get any sort of large projectile up to .99+ of c.

[Wyrm]

I'm having trouble understanding why reversing the gradient's direction changes the equilibrium here. Shouldn't it simply retard the projectile's motion? I guess I'm just not seeing where the torque is going to come from.

There will be no torque if the magnetized projectile exits the coil dead center. Of course, this is rarely going to be the case — the projectile is almost always going to be a little bit off-center. Since this magnetized projectile is going to line up with the field, and the field lines spread out at the poles (since they then must curl back towards the other end), at the other end there will be a torque to make the projectile line up with this spreading out field, and the projectile will be rotated out of true.

I've changed my mind that this will be an unstable equilibrium if the coil's field was to remain on. An unstable equilibrium would require small perturbations to be such that the magnet will be pushed away from the equilibrium point, which is not the case, but see below.

I'm assuming that we've cut the power at the optimum time, so that we provide maximum acceleration for the projectile, and allow time for the coil's inductance issues to resolve themselves as much as possible.

Here's where a problem is introduced. If we had a constant current, then there will be not one, but three oscillations in a projectile. The oscillation you identified is the oscillation along the axis. But there will also be an oscillation transverse to the axis, due to the fact that the projectile can never really be dead center, and an oscillation in the orientation of the projectile, due to the fact that the projectile can never be perfectly aligned with the field initially either (especially if it enters off true).

This would pose no problem if the coil's field was constant, but it isn't. The fields turn off, so when they do, there is likely to be residual transverse momentum and angular momentum. Further, because of the nature of oscillations, the more the projectile's been corrected to true when the field is turned off, the worse the residual momentum tends to be. So while the field is off, the projectile will tend to drift. If your gun is designed poorly, these drifts will not tend to cancel out, but reinforce, and you could be one unhappy coilgunner.

[starslayer]

Thanks, Wyrm. What you say makes perfect sense, and I can easily see these effects becoming significant problem in a coil of this size. Railguns are so much simpler... .

[Ariphaos]

I think for a coilgun of any size you might want to fire a sphere. The first debate I had about RKVs here, someone mentioned firing long, thin projectiles. Every coil it passes through is going to put its own little bit of stress on the projectile. Not just in terms of torque but also because the amount of force is not constant throughout the projectile.

A sphere would seem to minimize this issue.

[Kuroneko]

Well, I did take a look at this again. However, it's not quite finished, and I'm going to have to cut this short for the moment, as I'm going to be busy this weekend. Right now, the coilgun and projectile are modeled as infinitely thin cylindrical sheets. It is certainly possible to extend them to have finite, positive thickness, so if anyone wants to take this up and run with it, they're quite welcome to.

---

Perhaps a completely different approach is in order. The main assumption below is that constant current densities are maintained where appropriate, at some maximal value dictated by other considerations (more on that later). This and other idealizations (such as perfect timing) are actually very unrealistic, but they are unproblematic if our objective is to find an upper bound. Current densities are considered because the following discussion is also intentended to be in the limit as the number of coils and turns per coil goes to infinity; if the current densities is kept constant, this does not mean the fields diverge, as eventually loops can be added to a coil of fixed size only by making the thickness of each smaller, but the maximum current each loop can support decreases accordingly.

For two simple, closed, differentiable curves of fixed orientation carrying respective currents I₀ and I₁, with x₀ and x₁ being positions of line elements along those curves in an arbitrary parameterization, the field of the latter at x₀ is given by (with x = x₀-x₁)
[1] dB = (μ₀/4π)I₁(dl₁×x)/x³,
and since the force element on the former is dF = (μ₀/4π)(dl₀×B), the force is given by a double line integral of
[2] (μ₀/4π)(I₀I₁)(dl₀×(dl₁×x))/x³ = (μ₀/4π)(I₀I₁)[dl₁(dl₀·x/x³) - (dl₀·dl₁)x/x³].
In the first term in the square brackets of right-hand side, the parentheses contain a gradient of a potential, so by the gradient theorem, it is non-contributing over closed curves. If we allow the curves to undergo rigid motion, i.e., letting x₀ and x₁ be fixed parameterizations relative to some point identified as a center, and then vary only the separation between centers z (from the latter loop to the former), then we can write a mutual inductance
[3] M(z) = (μ₀/4π)IntInt[ dl₀·dl₁/|x+z| ]
and F = (I₀I₁)∇M(z). This is obvious because once again the integrand takes the form of the usual inverse-square potential, with z as the varying position. Note that using the Neumann formula (derived typically by considering fluxes through the loops) one could also skip almost directly to [3], but I think relying on Biot-Savart instead makes it clearer why it is thus.

Suppose that each curve lies in a plane perpendicular to a common axis along which all motion is constrained, and that the two curves are similar and scaled, i.e., that there are parameterizations x₀,x₁ with x₁ = ηx₀ for some η>1. Then F = (I₀I₁)(dM/dz). An obvious case is that of coaxial parallel circles, parameterizing each by angles α,β around the common axes, we can write dl₀·dl₁ = r₀r₁cos(α-β)dαdβ, and the mutual inductance as
[4] M(z) = M(r₀,r₁;z) = (μ₀/4π)IntInt[ dαdβ (r₀r₁cos(α-β))/sqrt(r₀² - 2r₀r₁cos(α-β) + r₁² + z²) ].
However, we don't actually need to evaluate this integral in its full generality; in each case, because of geometrical impositions on the curves, the function is monotonic on non-negative z, and vanishes at infinity. Therefore, taking |M(0)| = M₀, the work Int[ F dz ] = Int[ (I₀I₁)(dM/dz) dz ] has a magnitude that is bounded above by
[5] W₊ = (I₀I₁)M₀,
if both currents are positive. This is true regardless of the geometry of the curves involved, so long as they have proportional parameterizations in planes perpendicular to the common axis. Although we'll assume that the curves a circular loops, the following is therefore generalizable to any such pairs of curves (with M₀ calculated via [3] with z = 0). For the circular case, the argument α-β is redundant, as doing the integral in either α or β, the argument ranges linearly from 0 to 2π. MATLAB cranked this integral:
mu0*(2*r0^2*EllipticK(2/(r0+r1)*r0^(1/2)*r1^(1/2))+2*r0*EllipticK(2/(r0+r1)*r0^(1/2)*r1^(1/2))*(r1-r0)+
(r1-r0)^2*EllipticK(2/(r0+r1)*r0^(1/2)*r1^(1/2))-4*EllipticE(2/(r0+r1)*r0^(1/2)*r1^(1/2))*r0^2-
4*EllipticE(2/(r0+r1)*r0^(1/2)*r1^(1/2))*r0*(r1-r0)-EllipticE(2/(r0+r1)*r0^(1/2)*r1^(1/2))*(r1-r0)^2)/(r0+r1),
a nasty mess which can be somewhat simplified by defining A = (r₀+r₁)/2, G = sqrt(r₀r₁), the arithmetic and geometric means of the radii. Then:
[6] M₀ = M₀(r₀,r₁) = μ₀[(2A-G²/A) K(G/A) - 2A E(G/A)],
where K and E are complete elliptic integrals of the first and second kind, respectively.

Now, we extend this result axially. Suppose the we have an idealized coil of infinitely many turns and length L, always energized on one side of the projectile. Then for linear current densities λ₀,λ₁ of the projectile and coils, respectively, the work done is bounded by
[7] W₊ = IntInt[ (λ₀dx₀)(λ₁dx₁)M₀ ] = (λ₀λ₁)(lL)M₀,
where the l≪L (if not, the above still an upper bound, but an even worse one than usual). If the coils are energized in such a way that coils one one side always push and on the other always pull, then the resulting work is always less than double that of the above equation.

TODO: Extend this azimuthally for volume current densities J₀,J₁.
TODO: Figure out the radial stresses involved. This should be B²/2μ₀, given some appropriate model of the a realistic (non-infinitesimal thickness, after azimuthal extension) coils, and gives us a bound for the current in the coils. For the projectile, bound current Jb = ∇×M gives an upper bound for J₀ given some magnetization state, if ferromagnetic.