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Faraday's Law, Induction Help.

  1. Oct 30, 2008 #1
    1. The problem statement, all variables and given/known data

    A coil of 15 turns and radius 10cm surrounds a long solenoid of radius 2cm and 1x10^3 turns/m. The current in the solenoid changes as I = (5A)sin(120t). Find the induced emf in the 15-turn coil as a function of time.


    2. Relevant equations

    emf = -Nd[flux]/dt

    [tex]\Phi = \oint[B*dA][/tex]


    3. The attempt at a solution

    Edit: I figured out what I was doing wrong. However, I am still puzzled on one aspect of this problem

    When I figured out what I did wrong I came up with:

    [tex]B = \munI[/tex]
    [tex]Emf = \frac{d}{dt}(NBAcos\theta) = \frac{d}{dt}(NBA)[/tex] cos(theta) = 1 in this case
    [tex]Therefore:
    Emf = NA\mun(I)[/tex] ...and we have I so it's a simple derivative from there

    What I don't get is why the .1m radius doesn't really come into play. Unless I missed something or am not thinking here at all, it's not necessary for this problem, other than to tell us the coils are outside the solenoid.
     
    Last edited: Oct 30, 2008
  2. jcsd
  3. Oct 30, 2008 #2
  4. Oct 30, 2008 #3
    What area is the one you call A? The area of the solenoid i suppose.
    When you calculate the flux through the coil you should use (in general) the area of the coil.
    But... for a "long solenoid", the field on the outside of the solenoid is very weak and is usually neglected in elementary problems.
    The field of the solenoid B=I*N/L (no area) is inside the solenoid - does not depend upon the area of the solenoid.
     
  5. Oct 30, 2008 #4
    The A I am using is the area of the solenoid...at least the cross sectional area.

    Also, let me correct a mistake in that last equation:

    [tex]
    Therefore:
    Emf = NA(\mu)n\frac{d}{dt}(I)
    [/tex]
     
  6. Oct 30, 2008 #5
    Then you should see why.
    When you calculate the flux through something, you should use the area of the "something".
    Here you use the area (cross-section) of the solenoid instead because the field outside the solenoid is neglected (and you don't have a formula for the field outside anyway, in elementary textbooks)
     
  7. Oct 30, 2008 #6
    Ok, so it doesn't matter that the coil of 15 turns is .08 cm away from the solenoid, we only need the solenoid radius for the flux...hmm, I must have missed that in class.
     
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