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Relationship between Bit-Savart and Ampere laws

  1. Jan 23, 2015 #1
    I've tried to relate Biot-Savart's Law to Ampere's and I've found a contradiction, which I guess is due to a naive use of Ampere's.

    If ## \int \vec B · d\vec l = \mu_0 I_{enc} ## is applied to a circle of radius R around a current element ##Id\vec l## we have ## B·2\pi R = \mu_{0} I_{enc} ##, which gives ## B = \frac{\mu_0 I_{enc}}{2\pi R} ##, different from Biot-Savart. I'm guessins we can't use AMpere in this situation, but I can't put my finger on the exact reason.
     
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  3. Jan 23, 2015 #2

    Dale

    Staff: Mentor

    An isolated current element is impossible, it does not conserve charge. Instead of a current element you need to use a loop. Try an infinitely long straight wire with a return path at infinity.
     
  4. Jan 23, 2015 #3

    jtbell

    User Avatar

    Staff: Mentor

    One problem here is with the concept of Ienc which is often stated in first-year textbooks as "the current through the loop" (circle in this case). What this means precisely is: "the current that passes through any surface whose boundary is the loop." The problem is that with a finite current element, you can construct some surfaces that the current "pierces", and some that that the current does not "pierce" (i.e. the surface is curved in such a way as to avoid the current element entirely). So "current through the loop" is ambiguous for a finite current element. It depends on which surface you use. If you use a complete current loop instead of a finite element, you avoid the ambiguity.

    As DaleSpam also noted, an isolated current element (or any finite-length current-carrying wire) does not conserve charge all by itself. You can make it conserve charge by attaching a charged object to each end of the wire. As the current flows, one charge becomes more negative and the other becomes more positive. As these charges change, the electric field that they produce also changes. This changing electric field is associated with a magnetic field (in addition to the magnetic field associated with the current), according to the term that Maxwell added to Ampere's Law in order to make his equations "complete." This term compensates for the different possible values of "current through the loop".
    $$\oint {\vec B \cdot d \vec l} = \mu_0 \int {\vec J \cdot d \vec a} + \mu_0 \epsilon_0 \frac {d}{dt} \int {\vec E \cdot d \vec a} \\
    \oint {\vec B \cdot d \vec l} = \mu_0 i_\textrm{enc} + \mu_0 \epsilon_0 \frac {d}{dt} \int {\vec E \cdot d \vec a}$$
     
    Last edited: Jan 23, 2015
  5. Jan 23, 2015 #4
    Wow, that was enlightening. Many thanks you both!
     
  6. Jan 24, 2015 #5

    vanhees71

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    Science Advisor
    2016 Award

    Caveat concerning #3: This analysis is only valid for a surface on the right-hand side at rest. Otherwise there's an additional line integral missing. You find the correct laws always starting from the local Maxwell equations which are the fundamental equations anyway. Here you start from the Ampere-Maxwell law (written in SI units, sigh):
    $$\vec{\nabla} \times \vec{B}=\mu_0 \vec{J} + \frac{1}{c^2} \partial_t \vec{E}.$$
    Now using Stokes's integral theorem this gives the correct general form of the corresponding integral equation,
    $$\int_{\partial A} \mathrm{d} \vec{r} \cdot \vec{B}=\mu_0 \vec{J} + \frac{1}{c^2} \int_A \mathrm{d}^2 \vec{a} \cdot \partial_t \vec{E}.$$
    Now to take the time derivative outside of the integral, one must note that
    $$\frac{\mathrm{d}}{\mathrm{d} t} \int_A \mathrm{d}^2 \vec{a} \cdot \vec{E}=\int_{A} \mathrm{d}^2 \vec{a} \cdot \left [\partial_t \vec{E}+\vec{v} (\vec{\nabla} \cdot \vec{E}) \right ] - \int_{\partial A} \mathrm{d} \vec{r} \cdot (\vec{v} \times \vec{E}).$$
    Here ##\vec{v}## is the velocity field of the moving surface. If the surface is at rest, there's of course no problem, and you can simply put the time derivative out of the integral!
     
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