How do I calculate change in flux in an augmented rail accelerator without violating conservation of energy?

Question

Suppose I have two rails and a wire connecting them I'd like to accelerate. Then $\frac{d\phi}{dt}=Bvl$, where $B$ is the magnetic field strength where the wire is, $v$ is the velocity, and $l$ is the length.

The force on the wire is $F=B\times Il$ (and assuming $B$ and $I$ are perpendicular $F=BIl$) Lets assume all other forces on the wire are very small (arbitrarily small). Then the power going into the wire is the force on the wire (or arbitrarily close to that force), Because $E=\frac{1}{2}mv^2$ and the derivative of energy is power $P=ma$, $F=ma$, $F=P$.

Just to check my sanity, I wanted to make sure the power going into the wire's motion was equal to the power coming out of the circuit.

So $P=VI$, and $V=\frac{d\phi}{dt}=Bvl$. $P=\frac{d\phi}{dt}I=BvlI$, and finally $BvlI=BIl$. Now that's obviously wrong. My sanity check has proven I'm insane.

But that's only the beginning of my problems.

Supposing I can figure out how the power out of the circuit equals the power into the wire's motion. Well, I can change the one without changing the other. Consider these two cases:

Case 1: Suppose I change the magnetic permeability of the wire; Perhaps the wire's relative permeability is 100. Then the force on the wire would be 100 times greater! (given the same current). But the voltage due to the motion in the magnetic field is the change in flux. The flux inside the circuit is determined by the permeability of air $\mu_a$, not the permeability of the wire $\mu_w$! I can change the power output by increasing $\mu_w$, without increasing the power input, because flux is determined by $\mu_a$ not $\mu_w$.

Case 2: Suppose I increase the B field by adding an inductor with the wires parallel to my rails. This will also increase the force on the projectile (by increasing the total B field), but not the rate at which area is added to the circuit!

to make case 2 more clear, if the rails are in the center of an inductor with N turns and each turn runs the length of the rails and back, then the magnetic field is $B=b(N+1)$ where b is the magnetic flux density due to one turn around the inductor (plus one because rails will also generate a magnetic field). The force on the projectile is $F=BIl$ but the change in flux is $\frac{d\phi}{dt} = bvl$, not $\frac{d\phi}{dt} = bvl$.

Both of these examples lead me to conclude that I've been lied to. EMF is not the derivative of flux. Instead I think the EMF is equal to the rate at which H-field lines pass through the wire times the permeability, regardless of the flux density on the other side of the wire. Is that the case? If not, where am I going wrong?

Answer 1

Considering only an idealized model where the rails and the projectile are both 1-dimensional, straight wires.

Start with the Lorentz force $\mathbf{F}=q(\mathbf{E}+\mathbf{v}\times \mathbf{B})$. Assuming zero electric field, and rewriting in terms of currents via $q\mathbf{v}=I\boldsymbol{\ell}$, then we have $\mathbf{F}=I\boldsymbol{\ell}\times\mathbf{B}$. For $\mathbf{B}\perp\boldsymbol{\ell}$, we have $F=I\ell B$. Good so far. Where we find our first issue is with your claim that the power going into the wire is the same as the force on the wire, which is not dimensionally consistent. I believe the error you have made is thinking that the following holds:

$$\mathrm{d}_t\left[\frac{1}{2}mv^2\right]=ma$$

However, this is not the case. Instead, we really have (via the chain rule):

$$\mathrm{d}_t\left[\frac{1}{2}mv^2\right]=P=m\mathbf{a}\cdot\mathbf{v}=\mathbf{F}\cdot\mathbf{v}$$

Which for $\mathbf{F}\perp\mathbf{v}$ will give us $P_{mech}=Fv=Bv\ell I$. Now with this corrected form of the mechanical power, let's proceed with your sanity checks. Find the electrical power into the system:

$$\begin{align*}P_{elec}&=\mathcal{E}I\\\mathcal{E}=\mathrm{d}_t\phi=Bv\ell\implies&=Bv\ell I\\\implies P_{elec}&=P_{mech}~~~~~\checkmark\end{align*}$$

So this is all good!

I believe the mistake with your case 1 is thinking that the magnetic permeability of the wire changes the force it feels, which I do not think is true. The only way I can think of for this effect to occur would be from the force on a magnetic moment. However, the magnetic moment is given $\mathbf{m}=\iiint\mathrm{d}V\,\mathbf{M}=\mathbf{M}V$ which reduces to zero in our case (no-volume limit). The Lorentz force in a material depends not on $\mathbf{B}$, but on $\mathbf{H}$ (see Physics LibreTexts). As such, the force on the wire will not increase alongside the permeability.

I am a little unclear as to what you are exactly describing with your case 2. If you are describing a sort of transformation where you take each rail from a straight line to a tight-coiled helix, then the mistake there is thinking that this will increase the $\mathbf{B}$ field in the region bound by the circuit's loop. This is also not true - the field around a helix is the same as around a wire for $d\gg r$.

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Regarding your clarified case 2, we can model this by keeping the current as $I$ in the projectile, but replacing the currents in each of the rails with $2I$ (the other option is to replace currents everywhere with $2I$, but then we can just rewrite it as $I$ and nothing changes). This could perhaps be physically realized by running an additional parallel wire alongside each rail which is part of a separate circuit. This will increase the magnetic field to $2B$, doubling the force on the wire (and also the mechanical power). But you can see that this will also double the emf and with it the electrical power as well, so they remain in balance.