Induced Current in Aluminum Ring Around Solenoid

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SUMMARY

The discussion focuses on calculating the induced current in an aluminum ring surrounding a solenoid. The solenoid has 1000 turns per meter and a current increase rate of 270 A/s. Using the formula for magnetic field strength (B = μ * n * I) and Faraday's Law, participants explore the relationship between changing magnetic flux and induced current. The correct approach involves calculating the change in magnetic flux and applying it to determine the induced current accurately.

PREREQUISITES
  • Understanding of Faraday's Law of Electromagnetic Induction
  • Familiarity with magnetic field calculations (B = μ * n * I)
  • Knowledge of area calculations for circular shapes
  • Basic concepts of electrical resistance and current
NEXT STEPS
  • Study the application of Faraday's Law in different electromagnetic scenarios
  • Learn about the relationship between magnetic flux and induced electromotive force (EMF)
  • Explore the effects of resistance on induced current in conductive materials
  • Investigate the principles of solenoids and their magnetic fields in detail
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Students and professionals in physics, electrical engineering, and anyone interested in electromagnetic theory and applications.

Punchlinegirl
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An aluminum ring of radius 5 cm and resistance 0.003 ohms is placed around the center of a long air-core solenoid with 1000 turns per meter and a smaller radius of 3 cm. If the current in the solenoid is increasing at a constant rate of 270 A/s, what is the induced current in the ring?

B= [tex]\mu*n*I[/tex]
change in B/change in time = [tex]\mu[/tex] n *change in current/change in time
= [tex]4 \pi e-7)(1000)(270)[/tex] = .339
then change in flux/change in time= A*change in B/change in t
A= [tex]\pi*r^2[/tex]
So A= (.03)^2 *3.14
then multiply that by .339 to get 9.58 e -4.
Then I divided this by 3 e -4 to get the current and found that it was 3.19..
which wasn't right.. can someone please help me? Thanks in advance!
 
Last edited:
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I think the question wants you to use Faraday's law. Incidently, that's how I would go about it.

-Hoot
 
So if I use Faraday's Law, would I do
change in flux= B*cos (theta)A
Where B= [tex]\mu[/tex] I* n
and would A be the big area minus the small? [tex]\pi[/tex] (.05^2)-(.03^2)?
 

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