How Does Ampere's Law Treat External Currents in Magnetic Field Calculations?

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Homework Help Overview

The discussion revolves around the application of Ampere's Law in calculating magnetic fields, particularly focusing on how external currents are treated in different scenarios. Participants are examining the implications of the law when applied to various configurations of current-carrying wires.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants explore the definition of the integration curve in Ampere's Law and its impact on the calculated magnetic field. Questions arise regarding the treatment of external currents and whether they influence the results when only certain wires are enclosed by the integration path.

Discussion Status

Some participants have offered clarifications regarding the application of Ampere's Law, noting that the magnetic field is determined by the net current enclosed by the integration path. There is an acknowledgment of the confusion surrounding the influence of external currents, and a productive dialogue is ongoing about the nature of the magnetic field in relation to all currents present.

Contextual Notes

There appears to be some uncertainty regarding the interpretation of Ampere's Law and the role of external currents in magnetic field calculations. Participants are questioning their understanding of the law's application in different scenarios, which may reflect imposed homework constraints or assumptions about the nature of magnetic fields.

cupid.callin
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In ampere circuital law, ∫B.dl = μo inet
i includes current "only passing through the loop "

also B is the net mag field at any point "all to any current anywhere"

now look at the pic.

In A, it is very easy to find field using ampere law ... its μi/2πr

Also if in B is you apply the law, B is again μi/2πr ... how is this possible
its like the external current doesn't make any difference!

Plz explain me this !
 

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Ampere's law relies on what you define to be your curve of integration. Ampere's law states that the closed-loop integral Bdl over a given curve is proportional to the net current ENCLOSED by said loop.

In A you only have 1 wire enclosed by your curve of integration (red circle). In B, yes there are 2 wires, but only one is enclosed by your curve of integration, thus the same answer as in A.
 
Clever-Name said:
Ampere's law relies on what you define to be your curve of integration. Ampere's law states that the closed-loop integral Bdl over a given curve is proportional to the net current ENCLOSED by said loop.

In A you only have 1 wire enclosed by your curve of integration (red circle). In B, yes there are 2 wires, but only one is enclosed by your curve of integration, thus the same answer as in A.

So you are saying that i studied it wrong that in Bdl B is not due to all the currents existing in space?
 
cupid.callin said:
So you are saying that i studied it wrong that in Bdl B is not due to all the currents existing in space?
No, no, you are correct. B is due to the combined effect from all the currents.

cupid.callin said:
Also if in B is you apply the law, B is again μi/2πr ... how is this possible
its like the external current doesn't make any difference!
The integral \oint \vec{B} \cdot \vec{dl} gives the same result, but you cannot pull |\vec{B}| out like that because it is not constant in this case, unlike in the first!
 
Oh yes! you are right!

How can i ignore that thing !

Dumb of me !

Thanks a lot Fightfish !
 

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