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Homework Help: Induced current

  1. Apr 10, 2006 #1
    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: Apr 10, 2006
  2. jcsd
  3. Apr 10, 2006 #2

    Hootenanny

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    I think the question wants you to use Faraday's law. Incidently, thats how I would go about it.

    -Hoot
     
  4. Apr 10, 2006 #3
    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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