Ampere's Law and Conductor: Finding ∫B.dl for a Semicircle in the x-z Plane

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SUMMARY

The forum discussion centers on applying Ampere's Law to a semicircular conductor carrying current 'I' in the x-y plane. The integral ∫B.dl for a circular path in the x-z plane was initially calculated incorrectly as zero due to the absence of current through the loop. Participants clarified that to apply Ampere's Law correctly, one must consider the semicircular path as part of a complete circuit, including a straight line from B to A. The final expression for the integral is established as (1-√3/2)μ0I, demonstrating the importance of superposition and the correct interpretation of magnetic fields from both the semicircular and straight conductors.

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  • Ampere's Law
  • Biot-Savart Law
  • Magnetic field concepts
  • Vector calculus for line integrals
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  • #31
TSny said:
Yes, everything looks good. I was misinterpreting how you were using the word "equivalent". I thought you were implying that the semicircle current (alone) and the D loop produce the same result for the integral of Bnet around the circular path.

What if you replaced the semicircular current from A to B with a current of a different shape, such as shown in the attachment?

Are the points A and B along the axis of the integration circle?
 
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  • #32
If A and B are along the axis I would say the OP's a is correct but not b or c.

(I might as well benefit from this too!) :biggrin:
 
  • #33
rude man said:
If A and B are along the axis I would say the OP's a is correct but not b or c.

I wish TSny is a bit lenient in his marking :wink:. I somehow have the feeling that getting one answer correct should be enough to pass the test :-p.
 
  • #34
rude man said:
Are the points A and B along the axis of the integration circle?

Yes, I should have made that clear. A and B are still in the same location relative to the path of integration.
 
  • #35
Tanya Sharma said:
I wish TSny is a bit lenient in his marking :wink:. I somehow have the feeling that getting one answer correct should be enough to pass the test :-p.

Well, I'm retired. So, I refuse to do any more grading! :-p
But, if you were to use Tanya's suggested grading scheme, then you both passed. Tanya gets A+ on this one. (That is, her answers agree with mine, anyway.)
 
  • #36
Wow ! I can't believe I topped the class :-p

Have to say the Prof. is really cool :biggrin:.
 
  • #37
TSny said:
Well, I'm retired. So, I refuse to do any more grading! :-p
But, if you were to use Tanya's suggested grading scheme, then you both passed. Tanya gets A+ on this one. (That is, her answers agree with mine, anyway.)

Tanya's answers for b and c seem to have the wrong sign, otherwise I agree.

A downward current gives a positive B circulation as I understand it.
 
  • #38
Tanya Sharma said:
Wow ! I can't believe I topped the class :-p

Have to say the Prof. is really cool :biggrin:.

Congrats! And yes, we owe Dr. T. one - or actually several by now!
 

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