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Solenoid with an aluminum ring around it! (induced emf, induced current, mag. field?)

  1. Jun 8, 2012 #1
    1. The problem statement, all variables and given/known data

    31-p-009.gif

    An aluminum ring of radius r1 = 5.00 cm and a resistance of 2.65 x 10^-4 Ω is placed around one end of a long air-core solenoid with 970 turns per meter and radius r2 = 3.00 cm as shown in the figure. Assume the axial component of the field produced by the solenoid is one-half as strong over the area of the end of the solenoid as at the center of the solenoid. Also assume the solenoid produces negligible field outside its cross-sectional area. The current in the solenoid is increasing at a rate of 270 A/s.


    2. Relevant equations


    ε = d(magnetic flux) / dt


    magnetic flux = ∫(B)(dA)


    I = ε / R


    magnetic field in a solenoid:

    B = μ0(turns / length)(current)


    Ampere's law? :

    ∫B ds = μ0I


    (+ some other formula / combination i'm probably missing)


    3. The attempt at a solution
    I think I mostly need help just finding the induced emf (ε). I know you're to somehow find the magnetic flux of what I believe is the ring.


    Afterwards you find the magnetic flux and you can divide it by time (rather the time interval in which the magnetic flux changes) giving you the induced emf?

    But it's complicated because there's changing current instead, which is in the solenoid, and I am not sure what to do about it, or how it relates to the change in flux over time.


    At the end you divide the induced emf by the resistance of the ring giving you the induced current.
     
    Last edited: Jun 8, 2012
  2. jcsd
  3. Jun 8, 2012 #2

    tiny-tim

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    Hi Color_of_Cyan! :smile:
    I don't see the difficulty :confused:

    B = μ0(turns / length)(current),

    so dB/dt = μ0(turns / length)d(current)/dt​
     
  4. Jun 9, 2012 #3
    Re: solenoid with an aluminum ring around it! (induced emf, induced current, mag. fie

    Sorry if I sound dumb:


    If I solve for flux how do I account for the magnetic field that you now show is changing? (I'm really not too good with calculus either just so you know)


    I'm thinking since flux = ∫B dA

    Then flux = ∫[ (dB/dt)(dA) ] ? (Not sure what to do about this)

    On the subject of flux of the RING, the area is still πr2 right, even though it's a ring?

    I guess the flux is changing as well since the magnetic field is changing.


    I'm thinking something along the lines that

    flux = μ0(970 turns / meter)[ π(0.05m)2]

    then emf is just = μ0(970 turns / meter)[ π(0.05m)2](270A / time change in s)

    ?
     
  5. Jun 9, 2012 #4

    tiny-tim

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    flux = ∫B dA

    so d(flux)/dt = ∫[ (dB/dt)(dA) ]
    what matters is the flux through the ring

    but the question tells you not to bother to calculate it properly …

    it tells you to assume that the flux through the ring (radius r1) is simply the flux through radius r2, and also that the B field there is half the B field at the centre of the solenoid
     
  6. Jun 9, 2012 #5
    Re: solenoid with an aluminum ring around it! (induced emf, induced current, mag. fie

    Yes I meant flux through the ring, my bad.
    I can't see where you got that the flux through the ring is the same as the flux through r2 though


    So, the flux would be the same using the area of the solenoid end though? Also multiply by 1/2 ?

    So would it be

    flux = ∫B dA

    emf = d(flux)/dt

    emf = (1/2)(μ0)(970 turns / meter)(270 A / s)[∏(0.03m)2]

    ?
     
  7. Jun 10, 2012 #6

    tiny-tim

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    from …
    since the ring is perpendicular to the axis, and since the field is obviously cylindrically symmetrical, that means only the axial component is relevant to the flux through the ring

    and it specifically tells you to ignore any field outside r2

    so the flux through the ring (r1) is the same as the flux through r2 :wink:
    looks good! :smile:
     
  8. Jun 10, 2012 #7
    Re: solenoid with an aluminum ring around it! (induced emf, induced current, mag. fie

    Thanks.
     
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