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Inductive charging: Emf induced in two coil loops

  1. Oct 22, 2016 #1
    1. The problem statement, all variables and given/known data
    So I'm trying to figure out this problem:

    The base of our charging station is composed of a coil with N1 turns and area A1 connected to a 120 VAC, 60 Hz source. The device has a smaller coil with N2 turns and area A2, which when attached will fit completely inside the charging coil.

    A) What is the EMF of the base coil?

    B) what is the EMF of the device coil?

    C) how many turns do you need In the second coil to get a 6 vac RMS source?

    2. Relevant equations

    B_loop=N*mu*I/(2*R)
    EMF=d(flux)/dt

    3. The attempt at a solution

    A) What is the EMF of the base coil?

    This should be 0, right? No changing flux is actually going through this coil.

    B) what is the EMF of the device coil?

    B1 from the outer loop is : N1mu*I/(2*R1), so the RMS EMF through this loop is d/dt(B1*A2)=N1*mu*A2*omega*I_peak/(2*sqrt(2)*R1).

    C) how many turns do you need In the second coil to get a 6 vac RMS source?

    ?

    My questions are:

    did I do A) right? Is the EMF 0 in the base coil?

    And as for part B), the problem states that the resistance of the loop is negligible, and my final answer is in terms of I. Unfortunately the problem gave me the voltage (120V) of the ac power source, so how do I get the current?

    And on Part C), I don't see how to factor in the number of coils on the second loop.. did I do part B) correct?

    Thanks so much in advance!
     
  2. jcsd
  3. Oct 22, 2016 #2

    Henryk

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    Gold Member

    This is not correct. The base coils is connected to an AC source. There will be an AC current, which will produce the time-varying (AC) magnetic field and induce the EMF in the base coil. Assuming the resistance of the base coil is negligible (perfect inductor), the induced EMF is equal to the source voltage, that is 120 V AC.
    You don't really need the current. but you have to make a second assumption: the field is uniform within the base coil.
    In that case, the induced EMF in the base coil is given by: $$ EMF_{BASE} = N_1*A_1* \frac {dB}{dt} $$
    I hope I gave you enough hints to answer B and C
    Henryk
     
  4. Oct 24, 2016 #3
    Hi HenryK! Thanks for your quick response! So in response to your second hint, I agree that the EMF is proportional to dB/dt. But doesn't dB/dt = d/dt(N1mu*I(t)/(2*R1))? This definitely depends on I, so how do I get a value for this?

    Thanks again for your help!
     
  5. Oct 25, 2016 #4

    Henryk

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    Gold Member

    Hi,

    Yes, your formula is correct, but as I said, you don't need to know the value of the current.
    I gave you a hint, the EMF induced in the base coil is equal to the applied voltage (true for a perfect inductor) and it is given by
    $$ EMF_{BASE} = N_1 * A_1 * \frac{dB}{dt} $$
    The EMF of the device coil is equal to
    $$ EMF_{DEVICE} = N_2 * A_2* \frac{dB}{dt} $$
    Assuming the uniform magnetic field inside, it is the same for both coils. Therefore
    $$ \frac {EMF_{BASE}} {N_1 * A_1} = \frac {EMF_{DEVICE}} {N_2 * A_2} $$

    It's now easy to get your answers.
     
  6. Oct 26, 2016 #5
    I see exactly what you mean now! I think I understand how it all works now. Thank you so much for your help HenryK, I really appreciate it!
     
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