The Ampere's Circuital law states

$$\oint B\cdot d\ell~=~ \mu_0I$$

We can use it to derive the magnetic field of an infinitely long current carrying wire easily. My question is, why does the wire need to be infinitely long? I know it has something to do with $B$ being constant and tangential to the loop at every point for easy evaluation of the integral, but I can't find an explanation to my question.


We use the idealized case of an infinitely long current to be able to justify (by symmetry) that the strength of the field will only depend on the radial coordinate $r$, so that it can be taken out of the integral, since we are only integrating over the angle which parametrizes a circle around the wire:

$$ \oint B\cdot d\ell = B \int_0^{2\pi}r\ d\theta = 2\pi r B = \mu_0 I\implies B=\frac{\mu_0I}{2\pi r}$$

If the wire is not infinitely long, you can move towards the end of it, where it is obvious that the $B$-field should not just depend on the radial coordinate, so our simple calculation fails. In practice, one can very often use this ideal case as a good approximation for the field close to the wire - so long as the distance from the wire is much smaller than the length of the wire the effect is pretty much that of an infinite wire.

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  • $\begingroup$ "where it is obvious that the B-field should not just depend on the radial coordinate, so our simple calculation fails": I think if you apply Biot-Savart law you can see toward the end too it is still tangential to the circular loop. And that is where I think OP's question is. I think the probable answer is: we can't take only finite straight current carrying wire without violating continuity equation. $\endgroup$ – user22180 Aug 12 '14 at 12:21
  • $\begingroup$ @user22180 For this system, the natural coordinates are cylindrical. Take the wire to be along $x=y=0\implies r=0$. Then, the field of an infinite wire has translational symmetry in the $z-$direction, but a finite wire does not. This means that, for a finite wire, $B=B(r,z)$, so our simple formula cannot be right. It is clear that the field right next to the middle of a finite wire cannot be the same as the field next to and above the wire. $\endgroup$ – Danu Aug 12 '14 at 12:26
  • $\begingroup$ see the magnetic field is still tangential. And I agree with you that in the case of finite wire the magnitude of the $\vec{B}$ depends on z also. But it doesn't depend on $\phi$ which is important in this case of line integral. As it doesn't depend on $\phi$ you can still take the magnetic field out of the integral and you have B(r,z)=$\frac{\mu_0 I}{2\pi r}$. But this is inconsistent as left hand side shows dependence on Z (so our intuition is) but right hand side shows no dependence on Z. I think this is what OP's question is. $\endgroup$ – user22180 Aug 12 '14 at 14:33
  • $\begingroup$ I think you are getting this inconsistency as you are breaking continuity equation by taking only a finite current carrying wire alone. $\endgroup$ – user22180 Aug 12 '14 at 14:34
  • $\begingroup$ @user22180 While I understand your point ('where does the current come from?!'), I was merely pointing out that the independence of $\phi$ is irrelevant. For a finite wire, $B=\frac{\mu_0I}{2\pi r}$ is simply wrong, since the field above the wire will get weaker and weaker which is NOT reflected in this formula. Mathematically, we must have $B\to 0$ as $|z|\to \infty$ for any finite wire. So, the simple solution is really wrong, and this has nothing to do with the continuity equation. $\endgroup$ – Danu Aug 12 '14 at 14:43

First of all, I would suggest you to read the comments I have made in the Danu's answer to check whether I have understood your question or not.

See, $\oint B\cdot d\ell~=~ \mu_0I$ has been derived only on the basis of $\vec{\nabla}\times \vec{B}=\mu_0 \vec{J}$. But actually the Maxwell equation is $\vec{\nabla}\times \vec{B}=\mu_0 (\vec{J}+\epsilon_0 \frac{\partial \vec{E}}{\partial t})$ which is consistent with the continuity eqution.

Now in the case of infinite wire $\vec{\nabla}\times \vec{B}=\mu_0 \vec{J}$ is sufficient as at the region of interest $\vec{\nabla} \cdot \vec{J}=0$.

But in the case of finite wire alone there is accumulation of charge in the finite wire so that $\vec{\nabla} \cdot \vec{J}+ \frac{\partial \rho}{\partial t}=0$. So relevant Maxwell equation is $\vec{\nabla}\times \vec{B}=\mu_0 (\vec{J}+\epsilon_0 \frac{\partial \vec{E}}{\partial t})$ . See the Ampere's law in this case is given by this, not by the $\oint B\cdot d\ell~=~ \mu_0I$. So you are applying wrong formula and that is why you are getting wrong answer.

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