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Based on this question

Ampere's law and Biot-Savart law gives different terms for magnetic field in middle of a current running in a loop

I made the following square Amperian loop: enter image description here enter image description here enter image description here

where I have colored the section of the line integral that would cancel each other, as well as side 4 that canceled itself.

This is bogus though isn't it? There is NOTHING that stops me from using a rectangle, such that the line integral of side 1 I want left over, doesn't have a 2r, but rather.... whatever the length of the rectangle is.

Can anyone point out WHY one cannot use Ampere's law in this problem? Is this figure I have misleading, in that there is no straight B field line in the middle? Just circular loop B's to the left and right?

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    $\begingroup$ It is just a theoretical exercise I wanted my students to do... and the fact that I'm struggling with it is alarming haha. I am ok with the answer "it isn't symmetric enough, we need Biot Savart" but I want to know explicitly why my method here is bogus. $\endgroup$
    – Lopey Tall
    Mar 21, 2023 at 23:15

3 Answers 3

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The issue is with your cancellations that were a bit cavalier. The circulation is oriented, so the integration of side 2 and side 3 do not cancel but rather add up. Similarly, the integration along side 4 does not cancel.

Indeed, taking $z$ the axis of the loop in the plane $z=0$, you have from symmetry: $$ B_x(x,y,z)=- B_x(x,y,-z) \\ B_y(x,y,z)=- B_y(x,y,-z) \\ B_z(x,y,z)= B_z(x,y,-z) $$ The line elements have opposing orientations on side 2 and 3, as well as the projected magnetic field of the line element. Thus both contributions are equal and do not cancel.

The same is true for side 4. The line element keeps the same orientation. From the previous symmetry, the projection of the field on the line element also has the same orientation for opposing $z$ so again there is no cancellation.

This loop therefore cannot be used to calculate the field. The only useful application I can think of is by extending all the sides to infinity. Contributions 2, 3 and 4 vanish due to the fast decay of the field. You can therefore check that the integral of the field along the axis (given by Biot-Savart) coincides with Ampere’s law.

Hope this helps.

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  • $\begingroup$ It was indeed helpful! Thank you! $\endgroup$
    – Lopey Tall
    Mar 22, 2023 at 3:39
  • $\begingroup$ The integral over side 4 isn't zero. The line element keeps the same orientation, and so does the field component along the line element (downward). $\endgroup$
    – Puk
    Mar 22, 2023 at 15:15
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    $\begingroup$ Yes messed up with the signs. Thanks! $\endgroup$
    – LPZ
    Mar 22, 2023 at 16:31
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I think using the Maxwell's equations in their differential form is better in this case:

$\vec\triangledown \times \vec H = \vec J$.If you want to find the magnetic field $\vec H$ you first find the cross product $\vec\triangledown \times \vec H$ and then compare by parts your result with $\vec J$

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You're perfectly entitled to apply Ampère's law around your rectangle. It tells you that $$\int_{\text{side 1}} \vec B.d\vec l+\int_{\text{side 2}} \vec B.d\vec l+\int_{\text{side 3}} \vec B.d\vec l+\int_{\text{side 4}} \vec B.d\vec l=\mu_0 I.$$ (The integrals are evaluated cyclically around the Ampèrian loop. Those over sides 2 and 3 are equal.)

But $\vec B$ varies in magnitude along side 1, and in both magnitude and direction along sides 2, 3, 4, so you can't deduce the value of $\vec B$ at any particular point from the Law.

[One thing you can do is to check that the field at points along the axis of the loop, as calculated from the B-S rule, is consistent with Ampère's law. Take for your Ampèrian loop the current-loop's axis running from from a huge negative displacement to a huge positive displacement and returning via the other 3 sides of a huge rectangle, along which the field is negligible because they are so far from the current loop. The integral of the field all along the axis (and therefore all round the rectangle) is, you'll find, $\mu_0 I$.]

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