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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.

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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
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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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