Correct usage of Ampère's law for calculating B-field outside parallel wires?

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Homework Help Overview

The problem involves calculating the magnetic field (B-field) outside two infinite parallel wires that carry equal but opposite currents. The focus is on the application of Ampère's law in this context.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants discuss the application of Ampère's law and the implications of enclosing both wires in an amperian loop. There is an exploration of the conditions under which the B-field can be considered constant along the loop.

Discussion Status

The discussion is examining the limitations of using Ampère's law for this scenario, particularly regarding the assumption of a constant B-field along the loop. Some participants express understanding of the need to reassess the formulas used in this context.

Contextual Notes

There is an emphasis on the distinction between the B-field inside and outside the wires, and the potential confusion arising from the opposing currents affecting the enclosed current in Ampère's law.

papercace
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Homework Statement


The problem is basically about tow infinite parallel wires separated by a distance ##d## with equally strong but opposite currents. You have to calculate the B-field outside the wires (not the field in between them).

Homework Equations


Ampères law:
##\oint \mathbf B \cdot d\mathbf l = \mu_0 I_{enc}##

B-field outside one infinite wire with current I:
##B=\frac{\mu_0 I}{2\pi s}##
where ##s## is the distance from the wire.

The Attempt at a Solution


Using the second formula on each wire and adding the resulting fields, we get the right answer, which obviously is bigger than zero. If we instead use Ampère's law, where we enclose both wires by an amperian circular loop, we get that the enclosed current is zero, since they run in opposite directions, which in turn makes the B-field equal to zero, which is obviously the wrong answer. In what way am I using Ampères law wrongly?
 
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papercace said:

Homework Statement


The problem is basically about tow infinite parallel wires separated by a distance ##d## with equally strong but opposite currents. You have to calculate the B-field outside the wires (not the field in between them).

Homework Equations


Ampères law:
##\oint \mathbf B \cdot d\mathbf l = \mu_0 I_{enc}##

B-field outside one infinite wire with current I:
##B=\frac{\mu_0 I}{2\pi s}##
where ##s## is the distance from the wire.

The Attempt at a Solution


Using the second formula on each wire and adding the resulting fields, we get the right answer, which obviously is bigger than zero. If we instead use Ampère's law, where we enclose both wires by an amperian circular loop, we get that the enclosed current is zero, since they run in opposite directions, which in turn makes the B-field equal to zero, which is obviously the wrong answer. In what way am I using Ampères law wrongly?
Yes, it is wrong. What does Ampère's law exactly state?
 
ehild said:
Yes, it is wrong. What does Ampère's law exactly state?
It states that the sum of the strength of the B-field in a tangential direction to the loop is proportional to the current enclosed by the loop.

If I may guess, we can only solve the integral analytically if the B-field in the tangential direction is assumed to be a constant over the entire loop, which is not the case with two parallel wires. Am I on going on the right track?
 
papercace said:
It states that the sum of the strength of the B-field in a tangential direction to the loop is proportional to the current enclosed by the loop.

If I may guess, we can only solve the integral analytically if the B-field in the tangential direction is assumed to be a constant over the entire loop, which is not the case with two parallel wires. Am I on going on the right track?
Yes. The tangential component of the B field is not constant along a loop enclosing both wires.
 
ehild said:
Yes. The tangential component of the B field is not constant along a loop enclosing both wires.
Now I understand. It's useful to take an extra hard look at the formulas when trying to use them. Thanks :)
 

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