To see why I think the law fails, please check the following situation.

The law states that $d\mathbf B=k\dfrac{id\mathbf l \times \mathbf r}{r^3}$.

Consider a current element through the centre of a circular loop. We want to calculate the magnetic field at a point on the boundary of the loop. In a closed loop, the Maxwell-Ampere Law states implies that $$c^2\times(\text{the circulation of B around it})=\frac{(\text{current through it})}{\epsilon_0}+(\text{rate of change of electric flux through it})\>.$$

In the situation described, the rate of change of flux can be calculated by noticing that a charge of magnitude idt flows from one side of the loop to the other. Since the charge is always extremely close to the centre, the flux through the loop before the charge crosses can be calculated by imagining a hemisphere (as in the figure ,drawn on the side opposite to the charge). As the flux through the hemisphere is $idt/2\epsilon_0$(due to the symmetry of a sphere) and the total flux through the surface comprising the loop and the hemisphere is zero, the flux through the loop is $idt/2\epsilon_0$.

When the charge crosses over, the flux is the same but with opposite sign.Hence,the total change in electric flux is $-idt/2\epsilon_0$. This implies that the time-derivative of the flux is $-i/2\epsilon_0$ which cancels the other term in the equation, as $i$ is the current flowing through the loop, making the circulation of $B$ zero. Then, from symmetry, we can conclude that $B$ is zero at any point on the boundary of the loop, which contradicts Biot-Savart’s Law. However this problem does not arise if the current element does not cross the loop.

I do not know if I have made a mistake. Please explain it if I have. Otherwise an explanation of this failure is welcome.

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    $\begingroup$ Can you include argument (or whatever it is) in your post? $\endgroup$ – garyp Nov 1 '16 at 18:20
  • $\begingroup$ Welcome on Physics SE :) Please always make your posts self-contained - so post your full paradox/problem/failure here. $\endgroup$ – Sanya Nov 1 '16 at 18:22
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    $\begingroup$ Some hints: 1) Use mathjax, 2) Use proper English punctuation: put a period at the end of each sentence, followed by a space, and put a space after each comma, but never before a comma. $\endgroup$ – DanielSank Nov 4 '16 at 18:34

It seems that you have a misunderstanding of the flux as used in Ampere's Law. First I want to restate Ampere's law in the integral form, $$ \oint_C \mathbf B\cdot d\mathbf l = \mu_0 \iint_S \mathbf J\cdot d\mathbf S \>. $$ The integral on the right is done over a surface $S$ that has the closed loop $C$ as its boundary. Here $S$ is not a closed surface as the one used in Gauss's law. There is no unique choice for this surface. In fact, one may chose a hemisphere with $C$ as its boundary, or one may simply choose a disk. For any current element that passes through the loop, it only crosses such a surface once, so you do not need to calculate the flux twice as you would for a closed surface. Therefore the total flux as represented by the integral on the right is nonzero.


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