1. Not finding help here? Sign up for a free 30min 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!

Ampere's law problem

  1. Apr 23, 2014 #1
    1. The problem statement, all variables and given/known data
    A conductor carrying current ##I## is in the form of a semicircle AB of radius ##R## and lying in xy-plane with it’s centre at origin as shown in the figure. Find the magnitude of ##\oint \vec{B}\cdot \vec{dl}## for the circle ##3x^2+3z^2=R^2## in the xz plane due to curve AB.

    (Ans: ##(2-\sqrt{3})\frac{\mu_0 I}{2}##)

    2. Relevant equations



    3. The attempt at a solution
    From Ampere's law:
    $$\oint \vec{B}\cdot \vec{dl}=\mu_0 I_{enclosed}$$
    There is no current passing through the loop ##3x^2+3z^2=R^2##, so for this loop, RHS for ampere's law is zero hence answer should be zero but it isn't. :confused:

    Any help is appreciated. Thanks!
     

    Attached Files:

    • bdl.png
      bdl.png
      File size:
      3.6 KB
      Views:
      55
    Last edited: Apr 23, 2014
  2. jcsd
  3. Apr 23, 2014 #2
    You're right that Ienclosed = 0. But there is a magnetic field generated at each point of the circle caused by the current in the wire.

    What then is the sum of the length x this magnetic field along the circle?
     
  4. Apr 23, 2014 #3
    Maxwell's displacement current
     
  5. Apr 23, 2014 #4

    TSny

    User Avatar
    Homework Helper
    Gold Member

    Ampere's law holds only for a complete circuit of current.

    Try to think of a nice way to complete the circuit so that you can use Ampere's law and the superposition principle to get the answer without too much work.
     
  6. Apr 24, 2014 #5
    I can't think of any approach. :(

    Working backwards from the answer, the RHS of ampere's law must be ##\mu_0(1-\sqrt{3}/2)I##. This would mean that an "effective current" of ##I(1-\sqrt{3}/2)## passes from the circle. But I don't see how to obtain this effective current. :confused:

    This is going to be silly. Finding magnetic field with the help of long current carrying wire is a standard textbook exercise. Do we assume that the circuit is complete in that case?
     
  7. Apr 24, 2014 #6
    Yes, every application of Ampere's law assumes that either the circuit is complete (the circuit completion may be at infinite, have no effect, and be omitted) or Maxwell's displacement current has been included in which case the circuit doesn't have to be complete.
     
  8. Apr 24, 2014 #7
    Thanks dauto!

    Can you please give me a few hints to start with?
     
  9. Apr 24, 2014 #8
    Have you ever heard of Maxwell's displacement current before? That was my hint...
     
  10. Apr 24, 2014 #9
    I have only heard of it but never applied it to problem solving. There isn't much about it in my book.

    The displacement current is given by:
    $$I_d=\epsilon_0\frac{d\phi_E}{dt}$$
    where ##\phi_E## is the electric flux. How do I find ##\phi_E##?
     
  11. Apr 24, 2014 #10
    The problem with the incomplete circuit is that the charges seem to appear at one end of the circuit and disappear at the other end. That's not possible. Clearly if you start with no charges at the ends of the circuit, overtime a positive charge will build up at point B while a negative charge builds up at A. Those charges create an electric flux through the circle in the xz-plane that builds over time. That building flux means you have a Maxwell displacement current going through the circle explaining how the line integral of the magnetic field does not vanish even though there is no physical current going through the loop. Ampere's law only works if Maxwell's displacement current is included along with the actual physical current.
     
  12. Apr 24, 2014 #11
    If I understand correctly, I can assume a point charge +q at B and -q at A and find the flux through the loop. Once I have the flux, I can differentiate it wrt to time to get the displacement current, right?
     
  13. Apr 24, 2014 #12
    Right. the point charges at A and B are changing overtime because the electric current is transferring charges from A to B.
     
  14. Apr 24, 2014 #13
    Thanks dauto! :)

    I found
    $$\phi_E=\frac{q}{\epsilon_0}\left(1-\frac{\sqrt{3}}{2}\right)$$
    Differentiating wrt to time gives me the correct displacement current.

    Btw, do you have some idea about the TSny's approach? I still cannot think how to use the superposition principle here.
     
  15. Apr 24, 2014 #14
    I think he want's you to close the circuit by connecting A to B with a straight line right through the center of the loop. That off course adds an extra physical current going through the loop so that's a different problem. But that new problem can be though of as a superposition of two separate circuits. In one circuit the current flows from A to B along the round about - That's your original problem. In the other circuit the current flows back right through the center. One other possibility is that instead of connecting through the center, you might actually connect A to -∞ along the y-axis while connecting B to +∞ along the y-axis. You'll have to ask him to figure which one he had in mind.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Ampere's law problem
  1. Ampere Law Problem (Replies: 4)

  2. Amperes Law Problem (Replies: 1)

  3. Ampere's Law Problem (Replies: 1)

Loading...