1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Flux of Circuit made from Metal Bar on Parallel Rails

  1. Apr 7, 2015 #1
    1. The problem statement, all variables and given/known data

    A simple model of a railgun is a metal bar which runs on two long parallel rails. The two rails are connected to a charged capacitor with capacitance C and a resistor with resistance R. After charging, the capacitor can discharge through the circuit. You may assume that the rails and sliding bar are perfect conductors, and that the bar moves without any friction. The two rails are cylinders with radius r with centers separated by a distance d. Assume that the sliding bar is a distance x along the rails where x>>d>>r.

    (a) At some instant, the circuit has current I running through it. If the sliding bar is fixed in position, what is the total magnetic flux through the middle of the circuit? You may assume that the rails are long enough that fringe effects can be ignored and the field from the rails can be approximated to be that of infinite wires. You may also ignore the contributions from the sliding bar and the capacitor at the end of the circuit.

    2. Relevant equations

    ∫B⋅da = ΦB

    3. The attempt at a solution

    I tried to model the wires as infinite, then use the principle of superposition to add up the magnetic fields in the center of the circuit. I assumed the current flows counter-clockwise through the circuit. I also assigned my coordinate system with the x-axis along the bottom rail, z-axis vertical (across the circuit, in the same direction as the bar), and the y-axis pointing out of the plane of the page. From this approach I found the magnetic field enclosed in the circuit to be:

    B = μoI/(2π) * (1/z + 1/(d-z))

    pointing out of the page. The 1/z term is the contribution from the bottom rail, and the 1/(d-z) term is the contribution from the top rail. My issue is when I integrate this expression over the area of the circuit to find the flux, I end up with a ln(0) term.

    I am not sure how to approach this problem now. I've tried an Amperian loop as well but I was having a difficult time finding a shape that I could integrate over.
  2. jcsd
  3. Apr 7, 2015 #2


    User Avatar
    Gold Member

    Hmmm... I see your problem. Honestly, what I would be tempted to do is this: assume the rails carry current through the center of the rail (as opposed to the surface) and then integrate from the surface (r) to the surface of the other rail (r+d) and then bam, ln(0) turns into ln(r). I'm gonna think about it some more to see if there's a better way (I'm sure there is) so don't take that and run just yet.
    Biot Svart law seems ideal to me, unless someone can provide a reason why it isn't.
  4. Apr 7, 2015 #3


    User Avatar
    Gold Member

    I'm not thinking of anything else at the moment. One other thing you need to do, though, is make sure you account for the other 2 sides of the circuit. You have top and bottom, but what about left and right?
  5. Apr 7, 2015 #4
    Thanks for the reply. I too came to the conclusion of integrating from the surface of the bottom of the rail to the surface of the top rail. This gets rid of the problem of including an infinite field in the integral. Using this approach I end up with a flux of:

    Φ= μo*I*x/(2π) * ln (d/r -1)

    This seems okay. As far as the other 2 sides of the circuit go, the problem statement says we can ignore the contributions from the sides of the circuit, so we have a free pass there.

    The formula for the magnetic field from the infinite wire comes from the Biot Savart law so it is somewhat used in the problem. I don't see how we could apply it directly to this system without separating the wires and finding the same formula used above, but that doesn't mean there isn't a way! Do you see anything we could do with it? Thanks again.
  6. Apr 7, 2015 #5


    User Avatar
    Gold Member

    That's actually a particular application of the Biot-Savart law. The original is ##\int_a^b \frac{\mu_0 I \vec{\ell}\times d\vec{r}}{4\pi r^2}## I was just mentioning it to let you know you were on the right track.
    ##\frac{\mu_0 I}{2\pi r}## is a solution of the Biot-Savart law for the field of an infinite wire at a distance r.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted