Ampere's law -- application with two conducting loops

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SUMMARY

The discussion focuses on applying Ampere's Law to two conducting loops, specifically regarding the assignment of current directions. In Loop 1, the user determines that the current i1 is negative and i2 is positive based on the right-hand rule, resulting in a magnetic field calculation of -2.5 μT·m. In Loop 2, confusion arises over the current directions at integration points, with the user questioning whether opposing currents cancel each other out. The recommendation is to visualize the system with directional arrows for clarity.

PREREQUISITES
  • Understanding of Ampere's Law and its mathematical formulation.
  • Familiarity with the right-hand rule for determining current direction.
  • Basic knowledge of magnetic fields and their calculations.
  • Ability to interpret and draw circuit diagrams with current directions.
NEXT STEPS
  • Study the application of Ampere's Law in complex circuit configurations.
  • Learn how to visualize magnetic fields using vector diagrams.
  • Explore the concept of current cancellation in parallel circuits.
  • Investigate advanced topics in electromagnetism, such as Biot-Savart Law.
USEFUL FOR

Physics students, electrical engineers, and anyone studying electromagnetism who seeks to deepen their understanding of current direction and magnetic field interactions in conducting loops.

Physicslearner500039
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Homework Statement
Figure 29-67 shows two closed paths wrapped around two conducting loops carrying currents
i j = 5.0 A and i2 = 3.0 A. What is the value of the integral ~ B· d'S for (a) path 1 and (b) path2?
Relevant Equations
No specific equations
I am trying to get more confidence on the direction of current using Amperes law, the problem statement is
Amp3.JPG

Loop1:
My first task was to assign the direction of current. If I wrap around the my right hand fingers in the direction of integration the thumb is pointing up hence Positive Y direction current is +Ve and Negative Y direction current is -Ve. Hence i1 is -ve and i2 is +Ve.
∫B.ds = (-5+3)*μ = -2*4*π*10^-7 = -2.5u T.m.
Is everything ok?

Loop2:
The real confusion happens here while assigning the current direction I have shown the integration points
Amp3_2.JPG

Right hand rules for integration points
A: Up current is positive.
B: Up current is negative.
C: Up current is positive.
So which direction should i consider positive?
For side 1 the current is one direction and for side 2 it is opposite direction. Do they always cancel? Please advise.
 
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I suggest that you draw a picture of the system as seen from above, including the currents going in/out of the paper and the loop with directional arrows.
 
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