| MadSci Network: Physics |
This is a question that involves a lot of physics. I would urge you to conduct some experiments because things are complex enough that you will probably quickly find out things that I missed. I will sketch out the formulas you need and try to point out things you should be aware of. I don’t think I have made any major mistakes here, but as I don’t really know magnetics very well, beware. One problem is that to really understand the physics involved you need to understand calculus a bit. I’ll try to frame my answer so that this is not necessary, but if you have not taken a calculus course yet you may want to turn to a teacher or friend who can help you. A word on units. I’ll use SI units here. SI uses meters, kilograms and seconds, so we will use these units for length, mass and time, respectively. The SI unit for magnetic field B is the Tesla, the Earth’s magnetic field is about (1/20,000) of a Tesla, but getting fields as strong as a Tesla is not out of the question if you use lots of turns in your solenoid and keep the radius small. I can’t bring myself to call B the magnetic induction, but that is really the term favored by most users of SI units. By the way, MRI scanners used in medical imaging use magnetic fields that are anywhere from .01 Tesla to 1.5 Tesla, with most scanners being in the .5 to 1.5 Tesla range. The unit used for current is Amperes, and one ampere of current is readily accessible from most power supplies. I think, but am not sure, that lantern batteries can supply about 10 amps of current for short times, say one second. Inductance of coils is measured in Henrys, and the symbol L is usually used for inductance. Real world inductances that you can make easily are of the order of microhenries (one millionth of a Henry) to millihenries (one thousandth of a Henry). We will start out by assuming the simplest solenoid we can, which is just a circular loop with current I0 and radius r0 with N0 turns. By N0 turns I mean that the wire is looped around N0 times, so that if you made a solenoid with the wire looping around 100 times, then N0 = 100. We will assume that the wire is thin enough that the 100 turns of wire can be modeled as lying right on top of one another. This is not true, of course, but as long as the fractional variation in wire radius is not large you can take the average radius and the formulas should be pretty close. You do not want the turns to be spread out along the solenoid axis either, as this will cut the maximum force you can supply to the BB. More about this later. We will take the loop to be centered in an x-y-z coordinate system with the loop centered at the origin and lying in the x-y plane. The axis of the loop is then along the z direction. Now, the magnetic field along the z axis is B = Mu0*N0*I0*(r0^2)/(2*[(r0)^2 + (z)^2 ]^3/2 ) I’m afraid the typography is not so good here. By (r0^2), I mean raise r0 to the second power, that is, square it. By [ ]^3/2, I mean raise the quantity inside the brackets to the 3/2 power. The variable Mu0 is the permeability of free space, which equals 4*pi*([10]^(-7)) newton/(ampere^2) = .000001257 newton/(ampere^2). Let’s put in some numbers here to see what sorts of fields are possible. Suppose the coil had 100 turns and was 1 cm in radius ( =.01 meters) and you ran 10 Amps through it. Then the field in the middle of the solenoid, where z=0, would be just Bmax= Mu0*N0*I0/(2*r0)= .00000126*100*10/(2*.01) = .063 Tesla To get a large B field use lots of turns, keep the radius small and use a lot of current. Now I’m going to assume that the current in the solenoid is fixed and that there are no eddy currents in the ferrous material making up the projectile. With these assumptions the force on the BB is (in magnitude) (d/dz)(m*B), where m is the magnetization of the BB and B is the field of the solenoid. The (d/dz) operator means take the derivative of whatever is to its right. The derivative is a concept from calculus, and it tells you how fast something is varying, that is, the function’s slope. If the product m*B doesn’t vary with position, then its derivative is zero. Now if you made a long solenoid the field inside the solenoid would be constant and there would be no force at those points in the BB’s path. The largest force would occur when the BB was about half the radius of the coil away from the origin, for the case we considered earlier. I will also assume that the material that the BB is made from is magnetically soft. This is a questionable assumption, but it will get us in the right ballpark. In that case the magnetization m = V*((Mur-1)/Mu0)*B, where V is the volume of the BB, Mur is the relative permeability of the material the BB is made from, and B is the applied magnetic field. Mur varies a lot for materials, and can range from at least 500 to 2000 for ferrous materials. Mur equals (approximately) 1 for nonmagnetic materials. If the applied field is high enough, the material is guaranteed to saturate and the linear relation does not hold. Saturation B fields vary widely also, but if the field is below .5 Tesla the relation between m and B should be approximately linear. At low fields a ferrous material that has been exposed to high fields retains some remnant magnetization, which also violates the assumed linearity between m and B. But it is o.k. for rough estimates. Next we have to add up the effect of the force that varies with position to figure out how much energy goes into accelerating the BB. This involves integration, another calculus concept. Your integral is particularly easy because you are just integrating the derivative of a function, which winds up being the difference of the function at the stop and start points. Far away from the solenoid the field is zero, and then the energy pumped into the BB when it gets to the center of the solenoid, where the field is at Bmax, winds up being just V*((Mur-1)/Mu0)*(Bmax)^2, where Bmax is just the maximum of the B field which occurs at z=0, assuming we stay on the z axis. Now the mass of the BB is equal to the density, rho, of the BB, times its volume, V. The kinetic energy of the BB is .5*Mass*(Velocity^2). Setting the kinetic energy equal to the energy the magnetic field has pumped in gives .5*rho*V*(Velocity^2) = V *((Mur-1)/Mu0)*((Bmax)^2). Solving for the velocity we get Velocity = Bmax * Sqrt[(2*(Mur-1)/(rho*Mu0)] , where Sqrt means square root. To get a high velocity you want a large magnetic field, high permeability, and a low density. Note that the volume, V, of the BB dropped out. Putting in numbers for iron and taking Mur = 1000 and the density, rho= 8,000 kg/meter^3, and Bmax = .063 Tesla we get Velocity = 28 meters/second (approx. 60 miles per hour). Not impressive for a gun, where it is not unusual for rifle bullets to go faster than the speed of sound in air, which is about 300 meters/second. Things to think about: 1) You mention turning the field off when the BB is in the middle of the solenoid so that it doesn’t slow down on the other side. I agree that this is desirable, but a coil that has current running through it is tough to turn off quickly. This principle is used in car ignition systems. A large current is run through a coil (solenoid) and then a switch is opened that interrupts the current through the coil. A large voltage is generated in this way that is further stepped up by a transformer and then used to generate a spark at the spark plug. Basically, if you open up a circuit that has current running through a coil, you will usually cause a spark to short out the switch for a brief time. This is not great for the switch, by the way. The discharge time (in seconds) for an inductor in series with a coil is L/R, where L is the coil inductance in Henries and R is the resistance in the circuit. You should measure the coil inductance and resistance and compute L/R to find out how long it will take the field to ramp up and ramp down. 2) Stability of the BB trajectory. I am not sure that the BB will drop straight through the solenoid. It may tend to veer to one side and wind up sticking to the solenoid. I have never tried this, but I know that if you get too close to the edge of the solenoid it will definitely not make it through. Worst case, you might need to string the BB on a wire so it can’t drift off the central axis. 3) We have assumed that the current through the solenoid is constant. I think that for anything you build, this assumption is probably a good one. However, if you managed to make a very efficient launcher I think you would notice an effect whereby the current would dip in the solenoid as the BB approached it. There is a version of one of Maxwell’s equations known as Lenz’s law that states that if the magnetic flux looping through a conductor changes with time, an electric field will be set up that sets up a current that sets up a B whose changing flux is opposite to the changing flux of the first B field. [Don’t worry if you found that confusing, it confuses everyone at first.] If you drop a permanent magnet through your solenoid with only an oscilloscope or a voltmeter across the solenoid you should see a significant voltage spike as the magnet drops through the solenoid. This effect is the basis for how a generator works. The same effect will tend to set up a current (known as an eddy current) in the BB that will partially cancel the magnetization of the BB until the eddy current dies out. If you use a material known as ferrite, you can have fairly high permeability and almost zero conductivity. In that case, the eddy currents will die out very quickly. Good luck with trying to build something to test this out. I hope this has been of some help to you.
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