If a 7.5 gram copper-jacketed lead bullet (say, a 9x19mm Parabellum) was travelling at 360 m/s, how much power would it take to diamagnetically stop it in the space of one meter?
This question comes from discussion about this question on the SF&F SE, about the scene in the X-Men movie where Magneto stops a bullet before it hits someone in the face. I'm aware that a strong enough magnet to cause this effect is going to have a whole heap of other side effects. I'm mainly interested in the ideal case of simply stopping a bullet, rather than the complication of doing it within an inch of someone's face.
Answer
The kinetic energy of the bullet is $\frac12 mv^2 \approx 1$ kilojoule.
If the deceleration is continuous over $x=1$ meter, energy conservation gives us an acceleration $a = v_\text{initial}^2/2x \approx 65,000\,\mathrm{m/s^2} \approx 6600\,g$, and the stopping time is $t = v_\text{initial}/a \approx 5.6$ milliseconds.
Spreading the bullet's energy over the stopping time gives an average power of 175 kilowatts. If you make some hand-waving assumptions that the mechanism for stopping the bullet is inefficient you might multiply this power by a factor of 10–100.
This is a lot of power! But the time interval is very brief. And it's certainly not prima facie unphysical—after all, the gunpowder explosion that launched the bullet involved the same energy transfer and an acceleration length of much less than a meter.
After some thinking, and a silly mistake, I can make an order-of-magnitude estimate of the magnetic field that would have to be involved.
I would expect that the main effect involved in rapidly stopping a bullet would not be diamagnetism, a small effect where the magnetic field strength inside a "non-magnetic" material is changed in its fourth or fifth decimal place (and thus the energy density of the field $E\propto B^2$ is changed in its eighth or tenth decimal place).
The predominant factor on introduction of a strong magnetic field to a bullet would be eddy currents in the material. Wikipedia gives me a formula for energy loss due to eddy currents in a material, $$ P = \frac{\pi^2 B^2 d^2 f^2}{6k\rho D} $$ where $P$ is the power in watts per kilogram, $B$ is the peak field, $d$ is the thickness of the conductor, $f$ is the frequency, $k$ is a dimensionless constant which depends on the geometry, $\rho$ is the resistivity, and $D$ is the mass density. Wiki gives $k=1$ for a thin plane and $k=2$ for a thin wire, so I wild-guess $k=3$ for a zero-dimensional bullet. Using values for the lead core of the bullet, we find the rate of field change \begin{align*} (Bf)^2 &= \mathrm{ \frac{18}{\pi^2} \frac{2\times10^{-7}\,\Omega\,m \cdot 10^4\,kg/m^3}{(10^{-2}\,m)^2} \frac{2\times10^5\,W}{8\times10^{-3}\,kg} } = \mathrm{10^{9} \frac{N^2}{C^2\,m^2} }\\ {}\\ Bf &= \mathrm{ \pi\times10^4\,T/s } \end{align*}
The simplest assumption about the frequency is that the field is being ramped up to its maximum while the bullet stops, so we've seen a quarter-oscillation and $1/f = 20\,\mathrm{ms}$. This gives us a peak field of 600 tesla, which is large, but not absurdly large.
On the other hand, if Magneto is actually an FM radio broadcaster at 100 MHz, he'd need only a field of $$\mathrm{ \frac{ \pi\times10^4\,T/s }{ 10^8\,Hz } = \pi\times10^{-4}\,T. }$$ I don't think that radio engineers ordinarily think about local peak magnetic field strengths, but this isn't outrageous either. My college NPR station has a 100 kW transmitter. However, their antenna isn't shaped correctly to put that entire power into a one-cc volume.
No comments:
Post a Comment