electromagnetismmagnetic fields
A copper pipe carries current I and the question asks if there is a magnetic field inside the pipe this is done by Ampere's law and equating the line integral to zero since the enclosed current in a loop is zero.
But this does not prove that the magnetic field can be perpendicular to the loop (i.e along the pipe). How can this be shown ??
The r in the Biot-Savart integral has to be taken from every current element to P, and for all the current elements that don't lie on that top line in your diagram (since you mentioned in a comment that the current is supposed to be present throughout the 3-dimensional wire), r should not be parallel to dl, the direction of current for that element, so in that case the cross-product will be nonzero.
It might be interesting to consider the case of a 1D wire though--according to the equation here the magnetic field from an infinite straight 1D current should go to infinity as the distance from the wire approaches zero, but for a point along the wire, it seems the cross-product in the Biot-Savart integral should be zero for every current element, since r and dl are always parallel. Probably the resolution to the 1D problem is that the Biot-Savart integrand also has an $ |r|^3$ in the denominator, so for the current element that passes through the point where we want to evaluate the field, the integrand becomes undefined, so either you can't use the Biot-Savart law there or the field itself is undefined at that point in classical electromagnetism.
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:
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 $x''=x'\cos θ$.)
Best Answer
No it doesn't prove that, but it does prove that any such field is nothing to do with the current. It could be the Earth's magnetic field or some other stray field, but it's not generated by the current.