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Homework Help: EM , magnetic field at centre of a carrying circular loop

  1. Apr 1, 2014 #1
    The question is to find the magnetic field at the centre of a current carrying circular loop of radius R, where the current = I

    Okay so I'm trying to do this by both Amp's Law and Biot Savarts Law, and I can't get my answers to agree.

    First method - Biot Savarts Law:

    B=[itex]\frac{I\mu_{0}}{4\pi}[/itex][itex]\int[/itex][itex]\frac{dl X \hat{n}}{n^{2}}[/itex]


    dl[itex]X[/itex][itex]\hat{n}[/itex]=dl (as |n|=R is always perpendicular to a given line element dl)

    => B=[itex]\frac{I\mu_{0}}{4\pi}[/itex][itex]\int[/itex][itex]\frac{dl X \hat{n}}{n^{2}}[/itex]=[itex]\frac{I\mu_{0}}{4R^{2}\pi}[/itex][itex]\int[/itex]dl=[itex]\frac{I\mu_{0}}{4R^{2}\pi}[/itex]2R[itex]\pi[/itex]=[itex]\frac{\mu_{0}I}{2R}[/itex]

    Second method - Ampere's Law:


    So B[itex]\oint[/itex]dl=[itex]\mu_{0}[/itex]I


    => B=[itex]\frac{\mu_{0}I}{2πR}[/itex]

    Thanks in advance.
  2. jcsd
  3. Apr 2, 2014 #2
    In your Ampere's Law attempt, what is the loop you are using to determine the line integral? How did you arrive at [itex]2\pi R[/itex] in the third line?
  4. Apr 3, 2014 #3
    An amperian loop running over the circular loop of wire
  5. Apr 3, 2014 #4
    Ampère's law only applies in magnetostatics. That is, [itex]\vec{B}\cdot d\vec{l}[/itex] must be constant for all [itex]d\vec{l}[/itex] in the loop. Is that the case for your ampèrian loop?
  6. Apr 3, 2014 #5
    Oh okay, because [itex]I[/itex] constantly changes direction so does the direction of the magnetic field, although its magnitude is equal, and so B can not be taken outside the integral in line 3 of original post. Am I correct in thinking it is not possible to find any amperian loop were [itex]B[/itex].[itex]dl[/itex] is constant?
  7. Apr 3, 2014 #6
    Correct. Ampere's Law can only be applied when B•dl is constant along the entire loop. It can be used for situations such as an infinitely long wire or solenoid because symmetry ensures that B is constant. There is no such symmetry in a single loop, so that's why Ampere's law isn't useful in this case.
  8. Apr 3, 2014 #7


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    Homework Helper

    yeah. also, you can derive the Biot-Savart law from Ampere's law, if you want some practice at integration.
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