Magnetic field due to infinite plane

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The discussion focuses on calculating the magnetic induction vector B for an infinite plane with a uniform current density i. Participants suggest using Ampere's circuital law to derive the solution, proposing a rectangular Amperian loop that effectively captures the current through the loop. The magnetic field H is determined to be parallel to the plane and perpendicular to the current direction, with its magnitude expressed as H = i/2. The direction of the magnetic field is confirmed to be downward towards the plane, with opposite directions on either side. The conversation concludes with an acknowledgment of the correct interpretation of the magnetic field's orientation.
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Homework Statement


Find the magnitude and direction of the magnetic induction vector ##\textbf{B}## of an infinite plane carrying a current of linear density ##\textbf{i}##; the vector ##\textbf{i}## is same at all points of the plane.

Homework Equations


The Attempt at a Solution


I can do this with integration but I have seen the same question solved before. The solution was done using Ampere's circuital law. I seem to have forgotten the way it was solved. How can I do this using Ampere's circuital law. I don't even have the slightest idea about what shape should be the amperian loop.

If we divide the plane into thin infinite current carrying wire and consider wires at equal distances from the symmetrical axis, it can be deduced from symmetry that the direction of magnetic field is downwards towards the plane.

Thanks!
 

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The following may be something like what you had in mind, so forgive any redundant observations, but:

Could you not run a rectangular Amperian loop around a section of plane of length L and depth D, D << L. The depth dimension penetrates the sheet on both sides. So then the current thru this loop is i*L and ampere's law gives 2L*H = i*L. The H field is parallel to the sheet but perpendicular to i everywher which is what I think you also concluded, with H = i/2, i in amp/m. The H field points in opposite directions on opposite sides of the plane.
 
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Sorry for the late reply.

rude man said:
The following may be something like what you had in mind, so forgive any redundant observations, but:

Could you not run a rectangular Amperian loop around a section of plane of length L and depth D, D << L. The depth dimension penetrates the sheet on both sides. So then the current thru this loop is i*L and ampere's law gives 2L*H = i*L. The H field is parallel to the sheet but perpendicular to i everywher which is what I think you also concluded, with H = i/2, i in amp/m. The H field points in opposite directions on opposite sides of the plane.

Thanks rude man, that worked! :smile:

I did not conclude that the direction of field is perpendicular to ##i## but instead I said that it is perpendicular to the plane of infinite sheet which is completely wrong. :redface:

Thank you!
 
Thread 'Correct statement about size of wire to produce larger extension'
The answer is (B) but I don't really understand why. Based on formula of Young Modulus: $$x=\frac{FL}{AE}$$ The second wire made of the same material so it means they have same Young Modulus. Larger extension means larger value of ##x## so to get larger value of ##x## we can increase ##F## and ##L## and decrease ##A## I am not sure whether there is change in ##F## for first and second wire so I will just assume ##F## does not change. It leaves (B) and (C) as possible options so why is (C)...

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