Cases when Ampere's circuital law fails to hold

[gumthakka]

Well, I guess that is pretty much my question.

 

[anuttarasammyak]

Ampere's law is written as
H=\frac{I}{2\pi r}
for infinite length current. Another form
\int_{\partial S} \mathbf{H}\cdot\mathbf{dl}=I
applies for any current, including closed loop current.

 

[kuruman]

Ampere's law (as extended by Maxwell)) is, in differential form,$$\vec\nabla\times \vec H=\vec J+\frac{\partial \vec D}{\partial t}.$$It always holds and the difference from case to case is in how you apply it. Specifically, if you want to transform the law into integral form, you will need a closed area $S$ with a contour $C$ bounding it. Then you can use Stokes' theorem to write the left hand side as $$\int_S (\vec\nabla \times \vec H)\cdot\hat n~ dA=\oint_C \vec H \cdot d\vec l.$$Whether you can do the line integral over the closed loop analytically depends on the symmetry of the situation; the validity of the law is independent of that.

 

[rude man]

Ampere's law always holds BUT if your wire is not of infinite length then there has to be a circuit to complete the current loop. Those extra conductors will contribute to the B field the same way the wire does that you're integrating around, spoiling the symmetry. But $\oint \bf H \cdot \bf dl = I$ always.

 

[Delta2]

The symmetry is broken when the wire is of finite length. Due to broken symmetry, the magnetic field will not be the same along the closed amperian loop on which we perform integration, so ampere's law integral can't be simplified to $B\cdot 2\pi r=\mu_0 I$. But it always hold that $\oint _{\partial S} \mathbf{B}\cdot \mathbf{dl}=\mu_0\left (I+ \epsilon_0\int\int_S \frac{\partial \mathbf{E}}{\partial t}\cdot\hat n dS\right )$ for any closed surface $S$ with boundary the loop $\partial S$.
It is just that $$\oint \mathbf{B}\cdot\mathbf{dl}\neq \mathbf{B}\cdot 2\pi r$$ when the (cylindrical) symmetry is broken.

 

[vanhees71]

Ampere's law always holds BUT if your wire is not of infinite length then there has to be a circuit to complete the current loop. Those extra conductors will contribute to the B field the same way the wire does that you're integrating around, spoiling the symmetry. But $\oint \bf H \cdot \bf dl = I$ always.

Of course this is not correct. The full Maxwell equation reads
$$\vec{\nabla} \times \vec{H} -\frac{1}{c} \partial_t \vec{D} = \frac{1}{c} \vec{j}_{\text{f}}.$$
There's the so-called "displacement current", i.e., the 2nd term on the left-hand side of the equation which invalidates the Ampere law, which does not have this very important term, which is what makes the Maxwell equations the complete laws governing all electromagnetism rather than Ampere's predecessor theory.

 

[kuruman]

The symmetry is broken when the wire is of finite length. Due to broken symmetry, the magnetic field will not be the same along the closed amperian loop on which we perform integration, so ampere's law integral can't be simplified to $B\cdot 2\pi r=\mu_0 I$. But it always hold that $\oint _{\partial S} \mathbf{B}\cdot \mathbf{dl}=\mu_0\left (I+ \epsilon_0\int\int_S \frac{\partial \mathbf{E}}{\partial t}\cdot\hat n dS\right )$ for any closed surface $S$ with boundary the loop $\partial S$.
It is just that $$\oint \mathbf{B}\cdot\mathbf{dl}\neq \mathbf{B}\cdot 2\pi r$$ when the (cylindrical) symmetry is broken.

We should not forget that the "unbroken" symmetry in the infinite wire is only an approximation motivated by one's desire to apply Ampere's law and get the answer $B=\dfrac{\mu_0I}{2\pi r}.~$ A current requires a closed loop, else there will be violation of charge conservation. For a loop where $a$ is a linear parameter determining the size of the loop, the approximation is valid close to the wire where $r<<a.$