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Ampere's Law (enclosed current?)

  1. Nov 3, 2009 #1
    Hi, I was hoping someone could clarify Ampere's Law for me.
    The equation says that the only current that contributes to the magnetic field
    is current enclosed by the loop you select. But say you had a straight wire running through
    a solenoid, each having a current. If you selected a circular loop with a radius that is less
    than the solenoid radius but encloses the wire, wouldn't the magnetic field on that circle be
    changed by the solenoid's magnetic field created by its current, despite the solenoid's current not being enclosed?

    I am sort of confused and my prof. just kind of brushed me off saying: it cancels out.

    Could someone clarify either his statement or my misunderstanding? Any help is appreciated.

    Actually now that I am typing it this is something I didn't quite understand regarding Gauss's Law either, which also has only "enclosed" charge contributing to the Electric field even when there are outside charges. I feel like both explanations should be similar...?
  2. jcsd
  3. Nov 3, 2009 #2


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    Good question. Keep thinking like that.

    The key here has to do with the direction of the B-fields involved. The direction of the B-field from the straight wire circulates around the wire according to the right-hand rule, and is basically circulating around the straight wire in the same way as the solenoid coils are oriented.

    The B-field from the solenoid coils runs down the length of the solenoid, parallel to the straight wire. Do you see how these two fields are basically orthogonal, and independent?
    Last edited: Nov 4, 2009
  4. Nov 4, 2009 #3
    Thanks for the response. Um. I see how the B-fields will be orthogonal but I
    don't see how that necessarily implies their being independent.
    I understand that the field generated by the solenoid is not dependent on the field generated by the inner wire (and vice versa), but I don't see why the total current in the Ampere's Law Equation does not include the outer current since the magnetic field seems to be the vector sum of the inner and outer current-induced magnetic fields.
    To add a bit of context to this question I had a homework problem regarding
    the magnetic field at a certain radius less than the radius of the solenoid and it asked
    for the radius where the angle of the magnetic field was a certain value.
    I got the correct answer by considering the magnetic fields of both, but that answer seemed to contradict what I would get if I simply used Ampere's Law.
    Last edited: Nov 4, 2009
  5. Nov 4, 2009 #4


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    Look at the integral form of Ampere's law:

    [tex]\oint_C \mathbf{B} \cdot \mathrm{d}\boldsymbol{\ell} = \iint_S (\mu_0 \mathbf{J}+ \mu_0 \epsilon_0 \frac{\partial }{\partial t}\mathbf{E}) \cdot \mathrm{d} \mathbf{A}[/tex]

    If you choose a surface that has the inner wire's current in the same direction as its normal, then the currents in the solenoid will be parallel to the surface, orthogonal to the normal. So the solenoid's currents will not be included. On the right-hand side, we likewise see that the boundary of the surface will only be have line elements parallel to the magnetic field from the inner wire since the solenoid's magnetic field is in the same direction as the inner wire's current.

    So when you draw out your surfaces, you will see that the currents and fields produced by the inner wire are completely independent of the currents and fields from the solenoid for Ampere' Law. Of course, a true solenoid has wires that are inclining and a non-uniform field inside the solenoid due to the finite length. Both of these features will remove the strict independence since a very small factor of current and field will now intersect the surface and boundary chosen for the inner wire.
  6. Nov 6, 2009 #5
    Ah, sorry for the late reply. I have to admit that I have never seen the right side of that equation. We just learned that

    B dl = u I (enclosed)

    I suppose your's is the more rigorous formulation since I'm only in a freshman E&M course.
    But after talking to my TA again and the previous post I think I understand it better.
    I can't say I precisely understand that your form of the equation since I have never done double integrals and have only grazed partial derivatives, but I think you are presenting the same basic idea as berkeman.
    Okay. Well, this is interesting and I'll definitely look at it more when I don't have midterms :)
    Thank you both.
  7. Nov 6, 2009 #6


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    Staff: Mentor

    You will do well. Keep asking the right questions.
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