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Current of a coil surrounding a solenoid

  1. Apr 13, 2007 #1
    A 40-turn, 4.0-cm-diameter coil with resistance R=.40 ohms surrounds a 3.0-cm-diameter solenoid. The solenoid is 20 cm long and has 200 turns. The 60 Hz current through the solenoid is I = I_0 * sin(2pi*f*t).

    What is I_0 if the maximum current in the coil is 0.20 A?


    Basically, there is a coil with N=40 turns, radius = .02 m, resistance = .40 ohms, and I_max = .20 A. That coil surrounds a solenoid with N = 200 turns, radius = .015 m, and length = .2 m. The current through the solenoid is given by I = I_0 * sin(2*pi*60*t). The problem asks for I_0.

    I'm pretty sure it has something to do with V = -L*dI/dt.

    I tried L = mu_0*(40*200)^2*.015^2*pi/.2
    and dI/dt = 2*60*pi*I_0*cos(2*60*pi*t) = 2*60*pi*I_0 (we want initial current so t -> 0)
    and V = IR = .2*.4

    I am doing something wrong. There are no examples in the text like this, and I am pretty stuck. Any help would be appreciated. Thanks :)
     
  2. jcsd
  3. Apr 13, 2007 #2

    Päällikkö

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    Varying current in the solenoid gives rise to a magnetic field, which in turn, by Faraday's law, induces an emf in the coil. The coil being conductive, current starts to flow. So basically, what you're dealing with is a transformer.

    Assuming I understood the problem correctly, you should start by calculating the magnetic field (caused by the solenoid) inside the coil.

    V = -L*dI/dt is a bit unrelated as what you're interested in is the mutual inductance M: V2 = M dI1/dt, which is determined from the geometry of the problem, as described above.
     
  4. Apr 13, 2007 #3
    Wow, I was way off.

    If 1=sol. and 2=coil:

    To find the flux of the outer coil, would this be correct: mu0*N1*I1/length * pi*r2^2 * N2 ? and then what do I do with it?

    And is it ok to say M = sqrt(L1*L2) ?

    And what about dI/dt ?

    Maybe I'm just tired, but I'm still pretty confused.
     
  5. Apr 13, 2007 #4

    Päällikkö

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    Well, the solenoid's the smaller one in diameter, so the flux caused by it all goes through the coil, ie.
    [tex]\varphi_2 = N_2 \varphi_1 = N_2 B_1 A_1 = N_2 \mu_0 I_1 \left(\frac{N_1}{l}\right) \pi r_1^2[/tex]
    so almost the same as what you got. Do notice that I1 is a function of time.

    Now use Faraday's law.

    No. Using that requires some assumptions about the geometry (if I remember right the coils must be tightly wound [ie. have the same area] around iron [none of the magnetic flux may "escape"]).


    Would you happen to have the correct answer somewhere?
     
    Last edited: Apr 13, 2007
  6. Apr 13, 2007 #5
    That's actually what I meant for the flux. I accidentally put r2 instead of r1.

    I don't have the answer, but I can submit it online and know instantly if I'm correct or not. I've already used 3 out of 8 attempts.
     
  7. Apr 14, 2008 #6
    Hmmm... Let's revive this one.

    I got as far as this:

    Flux through coil = (Ncoil*Mu0*Nsolen*I(t)*Pi*r^2)/L

    I can then input my function for I into that equation and plug it into Farraday's law:

    ((Ncoil*Mu0*Nsolen*I(0)*sin(2Pift)*Pi*r^2)/L)/t

    But without a time, I'm not sure where that gets me.

    I know I need to get the emf value so I can solve for I ( I = emf/R), but I'm not quite getting this. Any help?
     
  8. Apr 15, 2008 #7
    Anyone out there care to help? I can't seem to make any progress on this.
     
  9. Apr 15, 2008 #8
    I'm pretty sure that this problem has to do with the R-L time constant. For clarification though what does I_0 represent (in words please)?

    Chris
     
  10. Apr 16, 2008 #9
    I am not the original poster but I_0 refers to "I not", the current at t=0. They are trying to show 0 as a subscript.
     
  11. Apr 16, 2008 #10

    alphysicist

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    Hi vertabatt,

    Here I believe I_0 refers to the amplitude of the current in the solenoid; at t=0 this current is zero since it has a sine function.

    You had already found in your earlier post that

    [tex]
    \Phi(t)= \frac{N_{\rm coil} \mu_0 N_{\rm sol} I(t) \pi r^2}{L}
    [/tex]

    You do have a time dependent function inside [itex]I(t)[/itex]; finding the time-dependent emf using Faraday's law involves taking the derivative of the flux with respect to time.

    (In your comment about not having a time I think you were considering the average induced emf which involves the change in flux divided by the time period, but here we need the instantaneous induced emf.)
     
  12. Apr 16, 2008 #11
    Okay, so I took the derivative and got:

    120pi*cos(120pi*t)((N(coil)mu(0)N(solenoid)I(0)pi*r^2)/L)


    Do I need to solve for t by knowing that cos(120pit) is a max when 120pi*t = 0?

    Even then it would seem that I'd be left with two unknowns: the emf and I(0)...


    (sorry about the formatting, I can't get a grasp on how to do the fancy stuff)
     
  13. Apr 16, 2008 #12

    alphysicist

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    The expression you got gives the emf induced in the coil. Now we want to use that to relate if to the other information they gave which was the maximum induced current in the coil; in a previous post you had already stated how to get the induced current (I=E/R).

    When you get the expression for induced current as a function of time you can then use the fact that the cosine function has a maximum value of 1 to solve for I_0.
     
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