MadSci Network: Physics

Re: How do I calculate the velocity of a projectile exiting an solenoid?

Date: Wed Jun 3 01:36:43 1998
Posted By: Don Pettibone, Other (pls. specify below), Ph.D. in Applied Physics, Quadlux Inc.
Area of science: Physics
ID: 895504314.Ph

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 
donít think I have made any major mistakes here, but as I donít really know 
magnetics very well, beware.

One problem is that to really understand the physics involved you need
to understand calculus a bit.  Iíll 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.  Iíll 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 Earthís 
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 canít 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 )

Iím 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).  Letís 
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 Iím 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) 


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 functionís slope.  If the product m*B 
doesnít 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 BBís 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 

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 


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 

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 doesnít 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 canít 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 Maxwellís equations known 
as Lenzís 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.  [Donít 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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