Solve Current in Parallel Wires with Repulsion

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SUMMARY

The discussion focuses on calculating the current in two parallel wires that repel each other due to the same current flowing through them. The wires have a mass per unit length of 43 g/m and are supported by strings at an angle of 16°. The correct calculations reveal that the magnitude of the current is approximately 159.8 A, derived from the equation I^2 = [(2π)(d)(mg/L)] / [(4π)e-07] after determining the distance between the wires as 0.0167 m.

PREREQUISITES
  • Understanding of basic physics concepts such as forces and tension.
  • Familiarity with the equations of electromagnetism, specifically those related to current and magnetic fields.
  • Knowledge of trigonometry, particularly sine and cosine functions.
  • Ability to apply free body diagrams to analyze forces acting on objects.
NEXT STEPS
  • Review the derivation of the formula I^2 = [(2π)(d)(mg/L)] / [(4π)e-07] for parallel wires.
  • Learn about the principles of magnetic force between parallel currents.
  • Explore the application of free body diagrams in solving physics problems involving tension and angles.
  • Investigate the effects of varying current on the force between parallel wires.
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Students studying physics, particularly those focusing on electromagnetism and mechanics, as well as educators looking for practical examples of current interactions in parallel wires.

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Homework Statement


Two long parallel wires, each with a mass per unit length of 43 g/m, are supported in a horizontal plane by 6.0 cm long strings, as shown in Figure P19.64. Each wire carries the same current I, causing the wires to repel each other so that the angle between the supporting strings is 16°.
Determine the magnitude of each current.

Homework Equations


sin (theta) = opposite/hyp

I^2 = [(2pi)(d)(mg/L)] / [(4pi)e-07]


The Attempt at a Solution



First I found d (the distance between the wires). Which is 2[sin(8)](0.06m) = 0.0167m.
Then I substitute the given into this equation: I^2 = [(2pi)(d)(mg/L)] / [(4pi)e-07]. I = sq.root of [(2pi)(0.0167)(0.043)(9.8)] / [(4pi)e-07]. the answer I got is 132.64 A. What did I do wrong?

I tried using free body diagram.
T = mg/sin8
Tcos8 = [(I^2)(L)((4pi)e-7)] / [(2pi)d]
(mgcos8)/(sin8) = [(I^2)(L)((4pi)e-7)] / [(2pi)d]
(mg/L)(7.115) = [(I^2)((4pi)e-7)] / [(2pi)0.0167]
I = 159.8 A

Still incorrect.
 

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Ah shoot, I've made the most stupid mistake ever. I solved it.
 

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