In both Griffiths and Jackson, Ampere's Law (or conversely the curl of the magnetic field) is derived by applying Stokes' Theorem to the surface integral of the current density J. The argument relies on the fact that $$I_{enc} = \int_{S} J \cdot da$$ but I am struggling to see how this could be true in general for any current density and any surface bounded by $\partial{S}$.

Of course this will be true for a straight wire passing through the center of an Amperian loop, when the vector J is parallel to the normal vector of the simplest surface enclosed by the loop, but what if the wire is angled slightly with respect to the plane? Using the same flat surface, the current enclosed will now be

$$\int_{S} I\hspace{.12cm}\delta(x)\begin{bmatrix} x \\ y \\ z \\ \end{bmatrix} \cdot \hat{n} \hspace{.12cm} da$$ which is clearly not the same value. So does Ampere's Law only account for the magnetic field produced by vectors parallel to the normal?

And my second question is this. How does Stokes' Theorem hold in the second case, where the curl is only non-zero along a single line? Wouldn't a different surface extending into a third dimension with a normal that is parallel to the current at the intersection give a different value for the surface integral, since the actual area of the surface doesn't matter? Obviously there is a serious flaw in my mathematical understanding of the material, but it isn't obvious to me.



1 Answer 1


Your first equation, which is a mere definition for the current, applies to regular vector fields $\mathbf J$. It is not straightforward to apply it to Dirac delta-functions. If you consider a finite-size wire, when it is angled wrt $S$ the intersection is larger by a factor $1/\cos θ$ which compensates exactly the $\cos θ$ factor arising from the dot product.

Now for your second question. Let's start by writing the current density correctly for an infinitesimally thin wire placed along the $z$ axis (it is defined by equations $x=0, y=0$, hence the two delta-functions): $$\mathbf j=I δ(x)δ(y)\hat z.$$ Suppose the surface of integration $S$ is a rectangle in a plane $(x',y)$ that's tilted by an angle $θ$ with respect to the $(x,y)$ plane. Viewed from the side:

Side view, in the (x,z) plane

The normal $\mathbf n$ (along a $z'$ axis angled by $θ$ wrt $z$) has coordinates $(\sin θ,0,\cos θ)$ in the $(x,y,z)$ system. So $$\iint_S \mathbf j·\mathbf n\,\mathrm dS=\iint_S Iδ(x)δ(y)\cos θ\,\mathrm dS.$$

Then we have to express the delta-functions in $x'y$ coordinates. $y$ is unchanged. Equation $x=0$ becomes $x'\cos θ+z'\sin θ=0$, so $δ(x)$ is replaced by $δ(x'\cos θ+z'\sin θ)$. On $S$, $z'$ is always zero, so finally we have to calculate $$\iint_S Iδ(x'\cos θ)δ(y)\cos θ\,\mathrm dS.$$ Now, you should know $δ(k x')=\frac 1{|k|}δ(x')$, hence $$\iint_S Iδ(x'\cos θ)δ(y)\cos θ\,\mathrm dS=\iint_S Iδ(x')δ(y)\,\mathrm dS=I.$$ (Alternatively, express $\mathrm dS=\mathrm dx'\,\mathrm dy$ and substitute variable $x''=x'\cos θ$.)

  • $\begingroup$ Thanks. This is very helpful. I don't have much formal education in delta functions - could you point me in the right direction in this case? As for my second question, it really is just a restatement of the first. If you have a single, infinitesimally small wire, how can the value of the integral over any surface bounded by $\partial{S}$ be the same? Can't the normal vector at the point of intersection literally point in any direction, depending on the surface? How can $$\int_{S}J\cdot n dS$$ be constant for any S if J is some delta function and n can point in any direction? $\endgroup$
    – JAustin
    Commented Apr 29, 2016 at 23:43
  • $\begingroup$ In other words, if the curl of B, $\mu J$, is zero-valued everywhere except along a single infinitesimally small (straight) wire where the magnitude and direction of J is fixed, how can the surface integral over any surface bounded by $\partial{S}$ with any possible normal be the same? If J is fixed, and n can be any vector, how does Stokes' Theorem hold? Thanks! $\endgroup$
    – JAustin
    Commented Apr 29, 2016 at 23:53

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.