Current - two parallel conductors

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SUMMARY

The discussion centers on calculating the greater current in two parallel wires, where one wire carries a current twice that of the other. Given the force of 7.0μN acting on a 2.0m length of one wire and the separation of 4.0mm, the relevant equation used is F(b) = (μ * I(1) * I(2) * l) / (2πa). The user initially attempted to solve for the current but recognized a potential oversight in their approach, indicating a need for clarification on the application of the formula.

PREREQUISITES
  • Understanding of electromagnetic force between parallel conductors
  • Familiarity with the formula F(b) = (μ * I(1) * I(2) * l) / (2πa)
  • Knowledge of units of measurement in electromagnetism (e.g., μN, mm)
  • Basic algebra for solving equations
NEXT STEPS
  • Review the principles of electromagnetic force and its applications
  • Study the derivation and implications of the formula for force between parallel wires
  • Explore the concept of current in electrical circuits and its measurement
  • Practice solving problems involving multiple currents and forces in parallel conductors
USEFUL FOR

Students studying electromagnetism, physics enthusiasts, and educators looking for problem-solving strategies related to parallel conductors and electromagnetic forces.

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Homework Statement


Two long parallel wires are separated by 4.0mm. The current in one of the wires is twice the other current. If the magnitude if the force on a 2.0m length of one of the wires is equal to 7.0μN, what is the greater of the two currents?

Homework Equations


F(b)=μ*I(1)*I(2)*l/2∏a

The Attempt at a Solution


I'm afraid I oversimplified my approach to this problem. I plugged the known values into the equation above and solved for I. I was then going to multiply that value by two. I am certain that is wrong, and that I am missing an important concept with regards to the problem. A shove in the right direction would be much appreciated!
 
Last edited:
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I think you are going correct. I just can't recall the formula for the force, so I will assume the one you posted is correct.
 

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