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Why am I wrong? If the length of the path is ##2\pi s## and we 'shake' it, the perimeter does not change (which is indeed true).No. As has already been explained in this thread.
Why am I wrong? If the length of the path is ##2\pi s## and we 'shake' it, the perimeter does not change (which is indeed true).No. As has already been explained in this thread.
Because the perimeter has absolutely nothing to do with it. You have already seen that the perimeter has absolutely no influence on the integral for the result of the integral for the circle and you seem to argue that it will have something to do with the final result for a shape different from the circle, yet it should give you the same as the circle. You are simply wrong and the reason as has been pointed out repeatedly is that it is only for the circle that you get the integral ##\oint d\ell## at all.Why am I wrong? If the length of the path is ##2\pi s## and we 'shake' it, the perimeter does not change (which is indeed true).
So if I am not mistaken, if the path were to be different from a circle, the Ampère's Law'd perfectly hold:Because the perimeter has absolutely nothing to do with it. You have already seen that the perimeter has absolutely no influence on the integral for the result of the integral for the circle and you seem to argue that it will have something to do with the final result for a shape different from the circle, yet it should give you the same as the circle. You are simply wrong and the reason as has been pointed out repeatedly is that it is only for the circle that you get the integral ##\oint d\ell## at all.
Gauss's law is just as impressive except that it involves a surface integral instead of a line integral.WoW, this is impressive...
You can test this theoretically by yourself. Consider the usual circle of radius ##R## with a current-carrying current ##I## at its center. Now consider a square Amperian loop of side ##R\sqrt{2}~##inscribed in the circle. Calculate ##\oint \vec B\cdot d\vec l## around the square loop and show that it is equal to ##\mu_0~I##. Hint: Exploit the symmetry of the problem. You only need to find ##\int \vec B\cdot d\vec l## along one side of the square and multiply the result by 4.I'd like to test this, do you know about any famous example of non-circular Amperian Loop?
See #10 and #11 ...You can test this theoretically by yourself. Consider the usual circle of radius ##R## with a current-carrying current ##I## at its center. Now consider a square Amperian loop of side ##R\sqrt{2}~##inscribed in the circle. Calculate ##\oint \vec B\cdot d\vec l## around the square loop and show that it is equal to ##\mu_0~I##. Hint: Exploit the symmetry of the problem. You only need to find ##\int \vec B\cdot d\vec l## along one side of the square and multiply the result by 4.
Oops, I replied without rereading all the posts to check for duplicates.See #10 and #11 ...
Let's focus on getting the magnetic field due to one of the four edges, which is a straight wire.You can test this theoretically by yourself. Consider the usual circle of radius ##R## with a current-carrying current ##I## at its center. Now consider a square Amperian loop of side ##R\sqrt{2}~##inscribed in the circle. Calculate ##\oint \vec B\cdot d\vec l## around the square loop and show that it is equal to ##\mu_0~I##. Hint: Exploit the symmetry of the problem. You only need to find ##\int \vec B\cdot d\vec l## along one side of the square and multiply the result by 4.
Sorry, I did not think about it.See #10 and #11 ...
You are missing that you you are not looking for the "magnetic field due to the entire square". You need to do the line integral ##\oint \vec B \cdot d\vec l## around the square knowing that ##\vec B = \frac{\mu_0~I}{2\pi r}\hat {\theta}##. Note that along the side of the square ##\vec B## varies both in magnitude and direction.But I should get μI\mu I. I must be missing something...
But isn't ##B = \frac{\mu I}{4\pi R'}(sin \theta_2 - sin \theta_1)##? If I am not mistaken, ##\vec B = \frac{\mu_0~I}{2\pi r}\hat {\theta}## is the magnetic field due to an infinite straight wire (which is not the case here; we're dealing with a finite straight wire).You are missing that you you are not looking for the "magnetic field due to the entire square". You need to do the line integral ##\oint \vec B \cdot d\vec l## around the square knowing that ##\vec B = \frac{\mu_0~I}{2\pi r}\hat {\theta}##. Note that along the side of the square ##\vec B## varies both in magnitude and direction.
No it isn't.By symmetry, the magnitude of ##B## is constant around the Amperian square loop.
You are not right. Look at the drawing below that shows the situation. The magnetic field has constant magnitude on the dashed circle but not on the square because the distance to the center of the square varies as you move along the side of the square. You need to find an expression for vector ##\vec B## at an arbitrary point on the side shown, find ##d\vec l## (for this side it is ##dx~\hat x##), find the dot product ##\vec B \cdot d\vec l## (note that ##\vec B## and ##d \vec l## are not parallel) and add all such dot products continuously (i.e. integrate) from one corner of the square to the other. Then and only then you multiply the result by 4. Note: The wire carrying current into the screen and the circle of radius ##R## enclosing the square are not shown for clarity.Am I right? If not, please let me know and I'll keep trying.
Thank you, enlightening explanation and drawing (may you tell what app did you use to make it up? :) )You are not right. Look at the drawing below that shows the situation. The magnetic field has constant magnitude on the dashed circle but not on the square because the distance to the center of the square varies as you move along the side of the square. You need to find an expression for vector ##\vec B## at an arbitrary point on the side shown, find ##d\vec l## (for this side it is ##dx~\hat x##), find the dot product ##\vec B \cdot d\vec l## (note that ##\vec B## and ##d \vec l## are not parallel) and add all such dot products continuously (i.e. integrate) from one corner of the square to the other. Then and only then you multiply the result by 4. Note: The wire carrying current into the screen and the circle of radius ##R## enclosing the square are not shown for clarity.
View attachment 240643
My bad, the magnetic field isn't constant; I'm unsure whether the following is correct:You need to find an expression for vector ##\vec B## at an arbitrary point on the side shown
I used PowerPoint and saved the figure as a .PNG file.Thank you, enlightening explanation and drawing (may you tell what app did you use to make it up? :) )
You're doing fine. Now you have to make sure that there is only one variable under the integral sign. For the time being you have three, ##\alpha##, ##x## and ##r##. Do the trig.Once here I feel like the idea is good; it is just about figuring out how rr depends on xx so that I can solve the integral.
Am I going right?
Do the trig.
OK Let's see now.Forget the upper triangle. It's not needed and I am not sure what ##z## is all about and why it is there. You need to calculate the ##\vec B \cdot d\vec l## along the horizontal side of the bottom right triangle. Label the angles of the bottom triangle, they have to be related to ##\alpha## and ##\beta##. Also, don't worry about the law of cosines; you have a right triangle so the Pythagorean theorem works just as well to relate the square of one side to the squares of the other two.