An electromagnetic rail gun uses magnetic forces to launch projec...

Question

An electromagnetic rail gun uses magnetic forces to launch projectiles. FIGURE P29.76 shows a 10-cm-long, 10 g metal wire that can slide without friction along 1.0-m-long horizontal rails. The rails are connected to a 300 V source, and a 0.10 T magnetic field fills the space between the rails. Each rail has linear resistivity ⋋ = 0.10 Ω/m, which means that the resistance is ⋋ multiplied by the length of rail through which current flows. Assume that the sliding wire and the left end, where the voltage source is, have zero resistance. The wire is initially placed at x₀ = 5.0 cm then the switch is closed. What is the wire's speed as it leaves the rails?

Accepted Answer

Hi, everyone. Let's take a look at this practice problem dealing with magnetic forces. In this problem, we have a rail gun that launches a metal projectile. The projectile has a mass of 20 g and rests on two horizontal rails that are 1.5 m long and connected to a 500 volt power supply. A magnetic field of 0.08 Tesla acts between the rails rails have a resistivity of 0.15 ohms per meter and the switch is closed when the projectile is in the middle of the rails, ignoring any friction and resistance at the contacts. Uh We need to find the speed of the projectile when it leaves the rails below the question. We're given a diagram of what was described in the problem. We're also given four choices as are possible answers. Choice A is 16 m per second. Choice B is 21 m per second. Choice C is 33 m per second and choice D is 42 m per second. Now, since we have a switch being closed, we're gonna have a complete circuit. So that means there's gonna be a current running through our loop that goes from the battery through one of the rails through the projectile and back on the second rail and back to the um power supply. So that means I have a current that's gonna be flowing downwards on the page. And since I have a current going through a piece of metal that's inside of a magnetic field, it's gonna have a force acting on it. So let's start there and calculate what the force is. And then once we have the force, we'll be able to use it to calculate the speed of the projectile. So recall the definition of a force of the force due to a magnetic field acting on a current. So that uh equation is, if you recall is F is equal to ilb multiplied by the sine of theta. So F is the four, I is the current L is the length of the wire. In this case, it's the length of our projectile and B is the strength of magnetic field and theta is the angle between the magnetic field and the direction of the current. Now, in our set up here, we have the current going down on the screen and we have the magnetic field coming out of the screen. So the angle between those two directions is 90 degrees. So our, the is gonna equal 90 degrees and the s of the is one sign of 90 degrees is one. So I can rewrite our force and that's gonna be equal to ILB, but I wasn't given anything um about the current in the problem. What I was given is the voltage of the power supply and the resistivity of the rails. So, since I have the resistivity, I could find the resistance and with the voltage and resistance, I could calculate the current using Ohm's Law. That's why I'm due here. Recall that Ohm's law is V is equal to Ir and so if I solve that for I, I is going to be equal to V divided by R, we'll plug that into our force equation. So I'll have F is equal to V divided by R lb. Now I need to calculate what the resistance is given the resistivity. So the total resistance for both of the rails is going to be given by, we have capital R as our resistance. Since I have two rails, I'm gonna have a factor of two, the one for each rail and the resistance of each rail is just the resistivity. We'll call that row multiplied by the length of that rail. Now to define the length, we need to define actually a coordinate system. So we'll have our X equal to zero. Basically our origin be at the right end of the rail. And so that means that um the length that is gonna have uh length of the rail that's gonna have the current range through, it is just gonna be equal to X. So our resistance R is gonna be equal to two row X and we'll have the positive X direction going to the left just to make life simple. So we can now plug this um expression for the resistance into our force equation. So that will give me F equal to VLB divided by two row X. So that is the force that's gonna be acting on the projectile. But we're actually interested in the speed of the projectile that when it leaves. So since I have the force acting on an object, it's going to do some work and that work is going to change the connect energy of the projectile. So what we want to use now is our work energy relationship. So for that recall the work energy theorem that says that the total work done on an object is gonna be equal to its change in energy. So W is equal to delta K. So we call our definition for work. The work is going to be the integral of F dotted with DS and that's gonna be equal to our change in connect energy, which is just my final connect energy minus my initial connect energy. So it's KF minus K I. Now, in order to the stop product, we need to figure out the direction that the force is acting in. So to do that, we need to use the right hand rule. So I'm gonna point my fingers in the direction of the current I'm gonna curl my fingers in the direction of the magnetic field and my thumb is gonna be pointing in the direction of the force, which in this case is pointing to the left. And that happens to be in the same direction as our displacement, the bar is moving to the left as well. So that means that I can get rid of that dot product. And it's just gonna be F multiplied by DS where those are now magnitudes instead of vectors. And so I can actually just replace our f in our um integral there with what we had found previously. So we have um V lb divided by two row X. And for the DS, that's just gonna be a small displacement along the direction of the motion and that's just in the X direction. So that'll just be DX. And that limit of integration is going to go from my initial exposition, we'll call it X I to our final exposition. XF now that integration has to be equal to my final kinetic energy. And if you call our definition of kinetic energy, that is one half MV squared, I'll have one half M little V squared. And our initial connect energy is gonna be zero. It's starting from rest. So it has no speed and so therefore, it has no connect energy. So at this point, I need to do some math to do the integration. So I'm gonna actually pull out all the constants outside the integral. So everything that's a constant is the capital VLB divided by two row. And what I'm left with is the integral from X I to XF of one over X DX. And that's gonna be equal to the one half M little V squared. So I can easily integrate one over X. So I will have V capital VLB divided by two row. The integra of one over X is the natural log of X. And that's gonna be evaluated at the two limits going from X I to XF. And that is still just equal to the one half M little V squared. Now, I just need to plug in my limits of integration. So I still have this constant sitting out front. So I'll have the capital VLB divided by two row. We'll have the natural log of XF minus the natural log of X I. But I can combine those two using the properties of natural logs. So I'll just have the natural log of XF divided by X I that's gonna be equal to the one half um little V squared. So at this point, I can kind uh solve this for my speed, the low V the um before I do that, I noticed that the one half halves will cancel out. And so now solving for the speed will have little V is equal to the square root of capital VLB divided by row M. We had to divide over the M and then that fraction gets multiplied by the natural log of XF divided by X I. So at this point I can plug in values. So how little the is equal to the square root or the voltage? That's gonna be the 500 volts or the length? That was the five centimeters I need to convert that into meters. So it's gonna be 0.05 m. The magnetic field strength was the 0.08 teslas. Then on the denominator, I will have the, the resistivity that's gonna be the zero point 15 ohms per meter. And then that gets multiplied by the mass which is 20 g. But I need to convert that into kilograms by dividing by 1000. That would be 0.02 kg. And that fraction is multiplied by the natural log of my final position is gonna be the 1.5 m the end of the rail. And we were told that it starts in the middle. So the middle point would be the uh 0.75 m. So I can calculate everything on the right hand side in my calculator and keeping two significant figures. This turns out to be 21 m per second. And if I look at the possible answer choices, this corresponds to answer B so just a quick recap of what we did, we started by finding the force that was acting on the current due to the magnetic field. And once we found that force, we could use the work energy theorem to calculate the final speed, the projectile as it left the rails. So I hope that this has been useful and I'll see you in the next video.

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