FlyVirgin744 From United States of America, joined Jul 1999, 1313 posts, RR: 1 Posted (9 years 6 months 4 weeks 20 hours ago) and read 6604 times:
If someone can help me with this and maybe some questions in the future, I'd really appreciate it. I know some of you out there love physics!
This has to do with a Rail Gun. I'm stumped.
This problem explores how a current-carrying wire can be accelerated by a magnetic field. You will use the ideas of magnetic flux and the EMF due to change of flux through a loop.
A conducting rod is free to slide on two parallel rails with negligible friction. At the right end of the rails, a voltage source of strength V in series with a resistor of resistance R makes a closed circuit together with the rails and the rod. The rails and the rod are taken to be perfect conductors. The rails extend to infinity on the left. The arrangement is shown in the figure.
There is a uniform magnetic field of magnitude B, pervading all space, perpendicular to the plane of rod and rails. The rod is released from rest, and it is observed that it accelerates to the left.
What is the acceleration a(t)(sub r) of the rod? Take m to be the mass of the rod. Express your answer as a function of V, B, the velocity of the rod v(t)sub r, L, R, and the mass of the rod m.
I've tried fiddleling around with the units in the 6 variables but can't come up with acceleration. Thanks in advanced,
Matt
Sometimes I go about in pity for myself and all the while a great wind carries me across the sky.
JetService From United States of America, joined Feb 2000, 4798 posts, RR: 13 Reply 1, posted (9 years 6 months 4 weeks 20 hours ago) and read 6594 times:
David b. should be able to help you. He's an expert at dealing with friction from accelerating rods and fiddling with units.
David b. From United States of America, joined Jun 2001, 3148 posts, RR: 6 Reply 2, posted (9 years 6 months 4 weeks 19 hours ago) and read 6589 times:
David b. should be able to help you. He's an expert at dealing with friction from accelerating rods and fiddling with units
Do I know................well, there an other member that can help you more then I can. He/she has much more experience. Doesn't he/she JS?
FlyVirgin744 From United States of America, joined Jul 1999, 1313 posts, RR: 1 Reply 3, posted (9 years 6 months 4 weeks 19 hours ago) and read 6575 times:
Awesome, I needed that laugh
Sometimes I go about in pity for myself and all the while a great wind carries me across the sky.
David b. From United States of America, joined Jun 2001, 3148 posts, RR: 6 Reply 4, posted (9 years 6 months 4 weeks 18 hours ago) and read 6575 times:
Bobrayner From United Kingdom, joined Apr 2003, 2227 posts, RR: 7 Reply 6, posted (9 years 6 months 4 weeks 8 hours ago) and read 6530 times:
Think of the railgun setup as an inductor, in which one element is free to move.
I'm not in the habit of giving people verbatim answers for their homework, and you haven't given all the information anyway, but we can do a walkthrough.
You already know the current though the loop (Ohm's law). Use this to find the induced magnetic field. You could form a Biot-Savart integral in 4 parts (both rails, the rod, and the power connection) but this might be too deep. Whichever method you choose, be careful with vector directions.
In real life, resistance wouldn't be zero; you could apply Faraday's law to this function, and get the induced EMF. It's actually a very simple inductor, so the EMF effectively causes a "counter-current" such that the actual current flow through the rod is less than you would expect from V=IR.
Anyway, back to the question. You have a function for the induced magnetic field; add this to the existing ('background') magnetic field B to get a function for the total magnetic field. Use Lorentz to find the force that this field creates on the rod as current flows through it.
Once you have the force on the rod, the rest is basic calculus - just use F=ma &c. It's not linear because the force is a function of the rod's position.
If you want to be really thorough, go through the same solution relativistically - after all, the rails are infinitely long, the rod will eventually be moving pretty fast...
FlyVirgin744 From United States of America, joined Jul 1999, 1313 posts, RR: 1 Reply 8, posted (9 years 6 months 4 weeks 5 hours ago) and read 6508 times:
While I understand somewhat better now, no help.
I'm getting the mu_0 in my answer when applying the B field, and that is not in the answer.
If you don't mind, I'm passing the 12 hour mark, tell what you think the answer is, if it is correct I'd appreciate you explaining it to me in great detail, cause right now, I don't seem to be getting it.
Thanks
Matt
Sometimes I go about in pity for myself and all the while a great wind carries me across the sky.
FlyVirgin744 From United States of America, joined Jul 1999, 1313 posts, RR: 1 Reply 9, posted (9 years 6 months 3 weeks 6 days 15 hours ago) and read 6477 times:
If this helps anyone else trying to help:
Thanks
Sometimes I go about in pity for myself and all the while a great wind carries me across the sky.
There's a lot of railgun junk on the web, but I think this one is most similar to your example. Equation 34 is pretty close to what you need? It gives you a function for F, but you can change that to displacement [r] thanks to Newton; F=m(d²r/dt²).
Try that. If it still doesn't work, I'll be back in a few hours...
FlyVirgin744 From United States of America, joined Jul 1999, 1313 posts, RR: 1 Reply 12, posted (9 years 6 months 3 weeks 5 days 15 hours ago) and read 6441 times:
When I try putting mu_0 in the answer, it says the solution is not dependant on mu_0.
I hate these stupid online homework assignments.
Thanks for that file though, I hadn't come across that one yet.
Sometimes I go about in pity for myself and all the while a great wind carries me across the sky.
FlyVirgin744 From United States of America, joined Jul 1999, 1313 posts, RR: 1 Reply 14, posted (9 years 6 months 3 weeks 3 days 20 hours ago) and read 6418 times:
I finally was able to get the answer from a friend who gave up and pushed "Give answer."
Here it is for all of you interested:
(V-B*L*v_r(t))(R*B*L)/(R*m)
Thanks,
Matt
Sometimes I go about in pity for myself and all the while a great wind carries me across the sky.
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