Force on projectile along rails (Railgun)

[EmBista]

Homework Statement

see attachment

Projectile P sits between two wide rails of circular cross section; a source of current sends current through the rails and through the (conducting) projectile...

Homework Equations

F=μ₀Li_ai_b/2∏d

The Attempt at a Solution

Direction is away from source because direction of the currents are opposite.

i_a=i_b
so, F=μ₀Li²/2∏d
F=i²μ₀/2∏ \frac{L}{d}

Here I'm stuck. Am I using the right equation or am I totally wrong.
otherwise how does \frac{L}{d}=ln\frac{w+R}{R}

 

[collinsmark]

Homework Statement

see attachment

Projectile P sits between two wide rails of circular cross section; a source of current sends current through the rails and through the (conducting) projectile...

Homework Equations

F=μ₀Li_ai_b/2∏d

That equation is for calculating the force between two parallel wires. Hypothetically, you could use that to determine the force of one rail on the other, but that doesn't help. You need to determine the force from the rails onto the projectile, not the force of the rails on themselves.

The force on a wire in a uniform magnetic field is F = iL x B. But the magnetic [Edit: in this problem -- i.e. the magnetic field isn't uniform in this problem] field on the projectile isn't constant all the way across it. It varies depending on distance from the center of either rail. So you'll need to represent the differential force as
dF = idL x B

(dL is the infinitesimally small length of a piece of the projectile.)

What's the magnetic field B, of an infinitely long wire (as a function of the distance from the center of the wire)?

I'm pretty sure you are supposed to approximate the rails as being infinite in this calculation. But there is something you must consider: the current only flows through the rails on the left side of the conducting projectile (no current flows to the right of the projectile). So the magnetic field strength is only half of what it would be for an otherwise infinite rail. (But still don't forget that there are two rails.)

 

[EmBista]

So I use dF=idL x B

B=μi/2∏r .. where r=w/2 + R.. still not getting anywhere really.
then B=μi/(2∏(w/2 + R))

 

[collinsmark]

So I use dF=idL x B

B=μi/2∏r .. where r=w/2 + R.. still not getting anywhere really.
then B=μi/(2∏(w/2 + R))

That's only the magnetic field at the point at the center of the projectile. What about all the points on the projectile that are closer/farther to/from the rails?

For an infinitely long wire (or rail), the magnetic field is given by

B = \frac{\mu_0 i}{2 \pi r},

as you implied. For this problem, the magnetic field varies over the distance from r = R to r = R + w.

And thus the force varies across the projectile, The force at a particular point of infinitesimal length dr on the projectile is

dF = idr \times B

Hint: you need to integrate. (and as I mentioned before, the magnetic field caused by a single rail is only half of what it would be of an infinitely long current carrying wire. And also there are two rails.)

[Edit: Oh, and welcome to Physics Forums, btw! ]

 

[EmBista]

Is this right then?
B = \frac{\mu_0 i}{2 \pi r},
dF = idr * B
dF = \frac{\mu_0 i^2}{2 \pi r} dr = \frac{\mu_0 i^2}{2 \pi} \frac{1}{r} dr
F = \frac{\mu_0 i^2}{2 \pi} \int^{w+R}_{R}\frac{1}{r} dr
F = \frac{\mu_0 i^2}{2 \pi} ln \frac{w+R}{R}
direction of the force is to the right as only the top rail has a current flowing through it directed to the right.

How do I go about the 2nd part of the quesition? what formula do I use.

Thank you for all the help so far, I think I'll be coming back to this site a lot. The homework help is really great :)

 

[collinsmark]

Is this right then?

'Looks right to me.

direction of the force is to the right as only the top rail has a current flowing through it directed to the right.

I'm not quite sure I follow your logic there. Yes, the direction of the force is to the right, as you said. But you're going to have to think in three dimensions and use the right hand rule to justify the direction. Which direction does the magnetic field point?

How do I go about the 2nd part of the quesition? what formula do I use.

I'm just guessing here, but I'm guessing you should approximate it using one of your kinematics equations for uniform acceleration. 'That and Newton's second law of motion.

 

[EmBista]

I'm not sure how to apply the Right Hand rule on this, would you be able to explain it to me clearly. I had a look at here but that doesn't help me, (I'm just not getting my head around it)
Also how would I write the answer if I'm using the Right Hand rule?

edit: tried to find out how to use it.. here it goes:
put my thumb in direction of the current on P (down). Then my fingers 'should' (please explain why/'if' magnetic field is pointing down) point down into the page in the direction of the magnetic field. And then my palm points towards the direction of the Force (right).
Still how would I answer this in an exam.Part two was actually easy, i wanted to use F=ma before I asked about the question but I didn't see that they gave the mass of the projectile in the question. (maybe because it was late for me)

I ended up using..
F=ma
v=at
d=vt^2/2
then found 't'
v=at=2290.15m/s

 

[collinsmark]

I'm not sure how to apply the Right Hand rule on this, would you be able to explain it to me clearly. I had a look at here but that doesn't help me, (I'm just not getting my head around it)
Also how would I write the answer if I'm using the Right Hand rule?

There are a number of different techniques out there involving he right hand rule. You'll need to pick a couple that you like and stick with them.

I say a couple, because you need to use the right hand rule differently in different situations. You'll need to use the right hand rule to do the following, and maybe a different method for each of these two scenarios:

• (a) Given a known current (or velocity of charged particle), find the direction of the magnetic field that it produces.
• (b) Given a known magnetic field, find the direction of the force on current carrying wires or moving particles.

The above reasons are different steps. You can't do both in a single step.

So let's get back to the railgun. It is also a two step process. You can't find the direction of the force in a single step. First find the direction of the magnetic field. Then find the direction of the force.

edit: tried to find out how to use it.. here it goes:
put my thumb in direction of the current on P (down). Then my fingers 'should' (please explain why/'if' magnetic field is pointing down) point down into the page in the direction of the magnetic field.

Yes, that right!

The first step is to figure out the direction of the magnetic field. Put your thumb in the direction of the current, and your fingers curl around in the direction of the magnetic field. In this case, we need to find the magnetic field caused by the rails. Don't bother with the projectile for now. Using the right hand rule, you can see that both rails produce magnetic fields that point into the page. Even though the currents are in opposite directions from each other, the magnetic fields produced in the middle (where the projectile is) are always into the page [Edit: for this particular problem. Other problems might be different].

The same idea even works with charged particles. But be careful here: it assumes that the charged particle is positive. Put your thumb in the direction of the positive charged particle's velocity. Your fingers curl around in the direction of the magnetic field.

If the moving charged particle is negative, it's the reverse. Put your thumb in the opposite direction of motion for things like electrons. Or alternately, you could use your left hand, and keep you thumb pointed in the velocity direction. How you handle it is up to you.

But all that is only the first part.

And then my palm points towards the direction of the Force (right).

Hold on! don't try to do everything in one step. Once you find the direction of the magnetic field, you need to start over to find the force. Remember, it's a two step process.

There's more than one way to use the right hand rule to find the force. I'll tell you the method that has always worked for me (but if you find a method you like better, use that one):

Hold your hand out with your fingers straight. Put your fingers in the direction of the magnetic field. Put your thumb in the direction of the current in the wire that you want to find the force on. The direction of the force is out of your palm.

Back to the railgun. We've already found the direction of the magnetic field, so we're not concerned with the rails anymore. Forget about the rails for this step. We're only concerned with the projectile. Put your fingers in the direction of the magnetic field (in this case, into the page). Put your thumb in the direction of the current flowing through the projectile. The direction of the force on the projectile is the direction out of your palm.

The same idea can be used for particles traveling through a magnetic field. Put your fingers in the direction of the field. For positive charges, put your thumb in the same direction as the particle's velocity, and the direction of the force on that particle is out of your palm.

For negative particles (such as electrons) the force is into your palm. Or alternately, you can use your left hand, or perform some other reversal that makes sense.

However you go about doing it, it's a two step process: the first step is to find the direction of the field created by object A. The second step is to find the force that the field exerts on object B.

Still how would I answer this in an exam.

You can use the right hand rule in exams. Trust me, everybody will be doing weird hand things when the exam takes place. Instructors expect this.

Part two was actually easy, i wanted to use F=ma before I asked about the question but I didn't see that they gave the mass of the projectile in the question. (maybe because it was late for me)

I ended up using..
F=ma
v=at
d=vt^2/2
then found 't'
v=at=2290.15m/s

I ended up with something pretty close to what you found. But you might want to redo your calculations and use more significant figures (when doing your calculations) to eliminate rounding errors. (Also, one of your kinematics formulas for uniform motion can be applied directly, without having to find t first.)

 

[EmBista]

You can use the right hand rule in exams. Trust me, everybody will be doing weird hand things when the exam takes place. Instructors expect this.

What I meant by the question is how would I write down the answer in an exam. Do I just describe how I used the RH rule?
ie. I used the RH rule over the current i with my thumb pointing in the direction of the current and thus find the magnetic field pointing into the page between the two rails. Then using the RH rule over the projectile, with my thumb pointing in the direction of the current and my fingers pointing into the page in the direction of the magnetic field, I found the direction of the force F to be to the right.Ok so I get the right hand rule.. first find the magnetic field using the RH rule over the rail. Then use the RH rule for the projectile to find the direction of the force.

for part 2, I made an error because i made m=10 (g) but it should have been m=0.01 (kg) because force is in kg.
I ended up getting 7.242 * 10^-4 m/s (4 sig fig)

Thank you very much for the help.

 

[collinsmark]

What I meant by the question is how would I write down the answer in an exam. Do I just describe how I used the RH rule?
ie. I used the RH rule over the current i with my thumb pointing in the direction of the current and thus find the magnetic field pointing into the page between the two rails. Then using the RH rule over the projectile, with my thumb pointing in the direction of the current and my fingers pointing into the page in the direction of the magnetic field, I found the direction of the force F to be to the right.


Ok so I get the right hand rule.. first find the magnetic field using the RH rule over the rail. Then use the RH rule for the projectile to find the direction of the force.

In my experience, the student is simply asked to give the direction of the force and/or the direction of the magnetic field.

It's not common to have to explicitly explain the right hand rule on a test. But as long as you know how to use it yourself, you can determine the correct direction.

for part 2, I made an error because i made m=10 (g) but it should have been m=0.01 (kg) because force is in kg.
I ended up getting 7.242 * 10^-4 m/s (4 sig fig)

Actually, the answer that I got was much closer to your original answer. My answer agrees with your first significant figure, and almost the second. My answer was in the form of 2xxx m/s.

 

[EmBista]

I handed the problem in for marking yesterday so I'll see what the right answer was when I get it back.

Thank you for all the help.