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Biot-Savart law states that the static magnetic field created by a constant current density $\mathbf{j}$ is

$$\mathbf{B}(\mathbf{r}) = \frac{\mu_0}{4\pi} \iiint\frac{\mathbf{j}\times (\mathbf{r}' - \mathbf{r})}{|\mathbf{r}' - \mathbf{r}|^3}\textrm{d}\tau$$

and can be derived using Maxwell's law for the magnetic field, which state in this case that

$$\begin{cases} \nabla\cdot\mathbf{B} &= 0 \\ \nabla\times\mathbf{B} &= \mu_0 \mathbf{j} \end{cases}$$

However, one could add a uniform field to $\mathbf{B}$, and the resulting magnetic field would still verify these two previous laws, so in particular, Biot-Savart law could be

$$\mathbf{B}(\mathbf{r}) = \frac{\mu_0}{4\pi} \iiint\frac{\mathbf{j}\times (\mathbf{r}' - \mathbf{r})}{|\mathbf{r}' - \mathbf{r}|^3}\textrm{d}\tau + \mathbf{B}_0$$

where $\mathbf{B}_0$ is a uniform constant field. Then, why in general is this uniform field null ? For a finite current density (ie. a real one), it can be deduced by energy cosideration: such a density can be obtained by inputing a finite amount of energy, and if the magnetic field uniform and non-null at infinite distances, energy would be infinite. Yet, what about an infinite distribution ? I would say that, generally speaking, one can't say that $\mathbf{B}_0=0$ using only Maxwell's laws, then could you think of a device where $\mathbf{B}_0\neq 0$ ?

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  • $\begingroup$ along with the div and curl you also need the "radiation" conditions at infinity that your $B_0$ does not satisfy. $\endgroup$
    – hyportnex
    Jul 23, 2017 at 20:45
  • $\begingroup$ What are these conditions ? Are they in Maxwell's theory of electromagnetism ? $\endgroup$
    – Spirine
    Jul 23, 2017 at 21:06
  • $\begingroup$ read this related physics.stackexchange.com/questions/346877/… $\endgroup$
    – hyportnex
    Jul 23, 2017 at 21:16
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    $\begingroup$ Not really; that $B_0=0$ is not arbitrary, a differential equation needs boundary conditions and those two together, equation and BC, give you the solution. The right boundary conditions are no more arbitrary than the differential equations themselves. As @ZeroTheHero correctly points out an everywhere constant field has infinite energy, and such wrong solutions of the differential equation must be excluded by any physically sensible set of boundary conditions. $\endgroup$
    – hyportnex
    Jul 23, 2017 at 21:31
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    $\begingroup$ Why is infinite energy an issue for, for example, infinite devices? How would you deal with an infinite plane with uniform and constant current ? Would you say that this situation is not feasible, so physics can't work here? $\endgroup$
    – Spirine
    Jul 23, 2017 at 21:45

1 Answer 1

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As a generale rule, differential equations and integral equations are not equivalent. They can be derived one from another only if suitable boundary conditions hold, so that the Stoke's theorem (or whatever other trick you are using to turn the former ones into the latter ones) can be applied.

From the practical point of view one must almost always require that fields vanish at infinity (or some sort of similar physical requirements on the physical significance of the electric and magnetic fluxes): since $\mathbf{B}(\mathbf{r})$ does the only way this might still hold is that $\mathbf{B}_0$ be null everywhere.

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  • $\begingroup$ The purpose of my question was to know more about the almost of your second paragraph, and to justify why $\mathbf{B}$ vanishes at infinity for every - in a sense that remains to be specified - current density. $\endgroup$
    – Spirine
    Jul 24, 2017 at 15:21
  • $\begingroup$ Well, first of all the boundary terms after the integration in Stoke's theorem must vanish on the contour of the domain: this means $\pm \infty$, and thus the additional constant must be zero at infinity too, but since it is a constant then its value is zero everywhere else as well. $\endgroup$
    – gented
    Jul 24, 2017 at 15:43
  • $\begingroup$ I don't understand why you can say what you're assuming, could you prove it please ? $\endgroup$
    – Spirine
    Jul 24, 2017 at 15:46
  • $\begingroup$ What's there to be proven? I don't understand. $\endgroup$
    – gented
    Jul 24, 2017 at 15:49
  • $\begingroup$ You said "the boundary terms (...) must vanish", why ? $\endgroup$
    – Spirine
    Jul 24, 2017 at 15:50

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