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Is Biot-Savart inverse cube or inverse square law?

  1. Feb 13, 2015 #1
    I know we can represent it two different ways.
    First: [tex]\mathbf{B} = \frac{\mu_0}{4\pi}\int_C \frac{I d\mathbf{l} \times \mathbf{\hat r}}{|\mathbf{r}|^2}[/tex]

    If we open up unit vector, then it becomes:
    [tex]\mathbf{B} = \frac{\mu_0}{4\pi} \int_C \frac{I d\mathbf{l} \times \mathbf{r}}{|\mathbf{r}|^3}[/tex]

    I am trying to understand if there is a direct analogy between magnetic field and gravitational/electric field.

    I know that magnetic monopole doesn't exist so it is different from gravity and electricity so the formulas for magnetic field must be different. There mustn't be direct analogy, like it does between gravitational end electric field. But some textbooks say "biot-savart law is an inverse square law" and this makes me confused.

    Can you please tell me if biot-savart law must be taken as an inverse square or inverse cube law by conceptually?
    Can we say if there is a direct analogy between magnetic field and electric/gravitational field?

    Thanks...
     
  2. jcsd
  3. Feb 13, 2015 #2

    Philip Wood

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    Gold Member

    The first way of writing it is, imo, the clearest, and shows the law to be inverse square. [itex]\hat {\mathbf{r}}[/itex] is the symbol for a unit vector, a vector of unit length pointing in the direction of [itex]\mathbf{r}[/itex]. The second way of writing it carries an extra r on the bottom line to cancel the magnitude of the [itex]\mathbf{r}[/itex] on the top line, so as not to have to use the [itex]\hat {\mathbf{r}}[/itex] symbol.

    The main reason why the B-S equation is quite different from the inverse square law radial fields due to point charges or masses, is that a current element is not analogous to a point charge or a mass. The analogous magnetic quantity would be a monopole. You can write an equation for the magnetic field due to a monopole, and that does look just like the equations for fields due to point charges and masses. Most engineering and undergraduate Physics courses, though, start with the B-S law, rather than the law for monopoles, because current elements are considered more akin to everyday stuff than are hypothetical monopoles. There's more to it than this.
     
  4. Feb 13, 2015 #3

    jtbell

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    I would consider this to be analogous to $$\mathbf{E} = \frac{1}{4 \pi \epsilon_0} \int_V {\frac {\rho dV \mathbf{\hat r}}{|\mathbf{r}|^2}}$$ If we consider the electric field to be inverse-square with respect to a single infinitesimal element of charge, then surely we must consider the magnetic field to be inverse-square with respect to a single infinitesimal element of current.

    Of course, in practice we can have an isolated (nearly-)point charge that explicitly exhibits an inverse-square electric field, but we can't have an isolated "point current" that explicitly exhibits an inverse-square magnetic field.
     
    Last edited: Feb 13, 2015
  5. Feb 19, 2015 #4

    vanhees71

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    Well, be careful!

    Magneto static fields fall off faster with distance from the source (currents, including magnetization currents) than electrostatic fields. The reason is that there are no magnetic monopoles (at least up to now, nobody has found one, although there's no known principle explaning, why there shouldn't exist one). Thus the multipole expansion of the magnetic field starts with the dipole term, and thus the field falls with ##r^3## rather than ##r^2## as the Coulomb field of a charged object.
     
  6. Feb 24, 2015 #5
    A couple of decades ago, I worked out a derivation of Biot Savart's law based on Coulomb's law, and relativistic length contraction. This is analogous to how you can derive the force between two current carrying wires as being due to the different reference frames for the moving and stationary charges (in the wire) causing Lorentz contraction, resulting in a net charge, thus creating a Coulomb attraction.

    It has always amazed me that even the very slow average drift velocity of electrons in a wire (typically of the order of meters per hour), creates enough relativistic contraction to create the equivalence of a net charge capable of useful levels of force. This tells us how amazingly strong electric forces really are.
     
  7. Mar 8, 2015 #6

    vanhees71

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    Well, the drift velocity is very small (some millimeters per second), but the sheer number of conduction electrons is huge. Of course, the easiest way to derive the Biot-Savart law is by just integrating the magnetostatic equations
    $$\vec{\nabla} \times \vec{B}=\frac{1}{c} \vec{j}, \quad \vec{\nabla} \cdot \vec{B}=0,$$
    using the 2nd equation to introduce the vector potential,
    $$\vec{B}=\vec{\nabla} \times \vec{A}$$
    and using its gauge freedom to invoke the Coulomb-gauge condition,
    $$\vec{\nabla} \cdot \vec{A}=0.$$
    Then in Cartesian coordinates (!!!) you have
    $$\Delta \vec{A}=-\frac{1}{c} \vec{j}.$$
    From the analogous equation for the electrostatic scalar potential one knows the appropriate Green's function of the Laplace operator, leading to
    $$\vec{A}(\vec{x})=\frac{1}{4 \pi c} \int_{\mathbb{R}^3} \mathrm{d}^3 \vec{x}' \frac{\vec{j}(\vec{x}')}{|\vec{x}-\vec{x}'|}.$$
    Of course, all this holds if and only if the continuity equation (i.e., charge conservation) holds. From the Ampere law you immediately get it
    $$\vec{\nabla} \cdot \vec{j}=c \vec{\nabla} \cdot (\vec{\nabla} \times \vec{B})=0.$$
    Now, if you look for the solutions in terms of multipoles (spherical are simpler than Cartesian), you'll see that the expansion starts with ##l=1##, which is because there are no magnetic charges in the game, according to Gauss's law (the second equation above).
     
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