A coil of 25 turns of wire is suspended by a thread

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SUMMARY

The discussion centers on calculating the current flowing through a coil of 25 turns of wire suspended by a thread, which experiences a reduction in tension of 0.04 N when current flows. The correct calculation involves using the formula F = IBL, where B is the magnetic field strength, and L is the length of the wire in the coil. The correct current value is determined to be 0.16 amps, factoring in the total number of turns in the coil. Participants emphasize the importance of correctly applying the number of turns in the calculations to arrive at the accurate solution.

PREREQUISITES
  • Understanding of electromagnetic force (F = IBL)
  • Knowledge of magnetic field strength (B = nμI)
  • Basic algebra for solving equations
  • Familiarity with coil configurations and their effects on current
NEXT STEPS
  • Study the principles of electromagnetism and the Lorentz force law
  • Learn about the effects of multiple turns in coils on magnetic fields
  • Explore the relationship between current, magnetic fields, and force in coils
  • Practice problems involving coils and magnetic forces to reinforce understanding
USEFUL FOR

Students studying physics, particularly those focusing on electromagnetism, as well as educators seeking to clarify concepts related to coils and current calculations.

Astraithious
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Homework Statement



A coil of 25 turns of wire is suspended by a thread. When a current flows through the coil, the tension in the thread is reduced by 4.0 ´10-2N.
http://members.shaw.ca/barry-barclay/Self-Tests/test11/Magnet20.gif

Homework Equations


F=IBL
B=nμI[/B]

The Attempt at a Solution


I have been driving myself nuts with this question haha but here is my attempt.

The force against gravity has to come from the current so I tried

0.04 = I (0.065)(0.15)
I = 4.1a
, which is incorrect

then I tried combining the two

to get
I=√F/Nμ

but that left me with 35.68

So I am at a loss as to how to go about this , any help would be great! or any formulas I am missing. I have not done calc yet this is all supposed to be simple algebra however this process for this particular question eludes me and it is the last of the set of 190 , I want it done! haha

Correct answer is .16 amps I know it has to go counterclockwise for the force to be pushing up
 
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Astraithious said:
1
The force against gravity has to come from the current so I tried

0.04 = I (0.065)(0.15)
I = 4.1a
, which is incorrect

Did you take into account that there are 25 wraps of wire?
 
Haha I did however I did 0.04 = I 0.065x25x0.15, not 0.04 = I 0.065x(25x0.15)

Thank you so much lol, i knew it was simple but it was killing my soul with frustration
 

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