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Connect a Battery to a Solenoid - Current through a rectangular coil

  1. Apr 2, 2012 #1
    1. The problem statement, all variables and given/known data

    Connect a battery to a solenoid
    A cylindrical solenoid 40 cm long with a radius of 8 mm has 250 tightly-wound turns of wire uniformly distributed along its length (see the figure). Around the middle of the solenoid is a two-turn rectangular loop 3 cm by 2 cm made of resistive wire having a resistance of 105 ohms. One microsecond after connecting the loose wire to the battery to form a series circuit with the battery and a 20 resistor, what is the magnitude of the current in the rectangular loop and its direction (clockwise or counter-clockwise in the diagram)? (The battery has an emf of 9 V.)

    23-086-solenoid2.jpg

    2. Relevant equations

    emf = IR
    For a solenoid,
    N = Number of turns
    I = Current
    L = Solenoid Length
    B = μ0*N*I/L
    n = perpendicular unit vector
    [itex]\Phi[/itex] = ∫B*nDA
    emf = d[itex]\Phi[/itex]/dt

    3. The attempt at a solution
    I began by finding the current in the solenoid.

    9 V/20 [itex]\Omega[/itex] = .45

    Then, I used this current to solve for the magnetic field the solenoid produces.

    μ0*250*.45/.4 = 3.53*10^-4

    Once I found this, I used it to find the magnetic flux.

    [itex]\Phi[/itex] = 3.53*10^-4*∏*.008^2 = 7.106*10^-8

    Then I found d[itex]\Phi[/itex]/dt, which is just the flux over the change in time since starting flux was 0.

    7.106*10^-8/(10^-6) = 7.106 * 10^-2

    This was the emf, so this *2 is the emf of the rectangular coil (since the rectangular coil has two turns).

    So emf(coil) = .142122

    And the current through the rectangular coil equals this emf divided by its resistance.

    .142122/105 = .13535

    But this wasn't correct.
     
  2. jcsd
  3. Apr 2, 2012 #2

    tiny-tim

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    Welcome to PF!

    Hi LordessCass! Welcome to PF! :smile:
    But isn't this an inductor, in an RL ciruit?
     
  4. Apr 2, 2012 #3
    Hi there! Thanks for the welcome! :)

    So does that mean that I'd need to use the formula:

    I = emf(battery)/R (1-e^(-R/L*t))

    instead? Where R is the resistance of the resistor and L is the inductance proportionality constant?
     
  5. Apr 2, 2012 #4

    tiny-tim

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  6. Apr 2, 2012 #5
    Thanks! I tested that out, though, and it doesn't look like that's my only problem because I still got the answer wrong. Is there anything else I'm doing incorrectly?
     
  7. Apr 2, 2012 #6

    gneill

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    What expression are you using for the magnetic flux on the outside of the solenoid? :wink:
     
  8. Apr 2, 2012 #7
    Hmm. I was using B*A, but there wouldn't be an A of the outside of a solenoid. I tried a slightly different tact where I found the current using the formula I listed above, and then used:

    emf(inducted) = μ0*N^2/d*∏R^2*dI/dt

    And then divided the result by the resistance of the rectangular coil, but that didn't produce the right answer either. Is the answer dependent on the number of turns in the rectangular coil, maybe?

    EDIT: Oh, I realized perhaps I should use the area of the rectangular coil, or maybe that area minus the area enclosed in the solenoid, as my A for the magnetic flux. Is that on the right track?
     
    Last edited: Apr 2, 2012
  9. Apr 3, 2012 #8
    I really can't figure this out. I've tried using the area of the rectangle as my area for the flux instead of inside the solenoid, I've tried using the current equation given for an RL circuit, and other associated things. I even looked up some problems I thought were comparable and solved this one the same way, but I'm not coming up with the right answer. Here's a different-numbers version of the same problem which I do have the answer to (but I have no idea how they're getting it, since my methods aren't working):

    A cylindrical solenoid 39 cm long with a radius of 7 mm has 350 tightly-wound turns of wire uniformly distributed along its length (see the figure). Around the middle of the solenoid is a two-turn rectangular loop 3 cm by 2 cm made of resistive wire having a resistance of 150 ohms. One microsecond after connecting the loose wire to the battery to form a series circuit with the battery and a 20 resistor, what is the magnitude of the current in the rectangular loop and its direction (clockwise or counter-clockwise in the diagram)? (The battery has an emf of 9 V.)

    The answer to this one ended up being 0.000247 A.
     
  10. Apr 3, 2012 #9

    gneill

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    Let's do this one with the known answer step-by-step then, and maybe we can spot where things are going awry.

    To begin with, what are your calculations for the inductance of the solenoid and the time constant for the current?
     
  11. Apr 3, 2012 #10
    Okay! So first I'll find the current going through the solenoid so I can find its inductance. I'll use the formula for an RL circuit current:

    I = emf(battery)/R*(1-e^(-R/L)t)

    L = μ0N^2/d*∏R^2

    So I'll find L first because I'll need it for I:
    L = 4∏*10^-7*(350)^2/.39*∏(.007)^2 = 6.07613 * 10^-5

    So then to plug in to find the current I get:

    I = 9/20*(1-e^(-20/(6.07613*10^-5)*10^-6) = .12621 A

    With this current, I plug into the induced emf formula of

    emf(ind) = L * dI/dt

    dI/dt is just the current I found since it was 0 before, so I get:

    6.07613 * 10^-5 * .12621 = 7.6688 * 10^-6 V as my inductance of the solenoid.
     
  12. Apr 3, 2012 #11

    gneill

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    Ah. Here's a problem: You want the instantaneous rate of change of the current, not the average over the time. How can you find dI/dt at t=1μs?
     
  13. Apr 3, 2012 #12
    Ah, so it's what I got for my emf, but also divided by the change in time, which is 10^-6 seconds?

    So for my emf instead of 7.6688 * 10^-6 V, I'll have 7.6688 V?

    I don't think that was my only problem, but if that was one of them, I'm glad to know.
     
  14. Apr 3, 2012 #13

    gneill

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    Well, you shouldn't need the emf across the coil at all; You already determined the inductance of the coil when you wrote: L = 4∏*10^-7*(350)^2/.39*∏(.007)^2 = 6.07613 * 10^-5.
    L is the inductance of the coil.

    You also wrote a formula for the coil current with respect to time: I = emf(battery)/R*(1-e^(-R/L)t). Why not find dI/dt from that?
     
  15. Apr 3, 2012 #14
    Oh, so dI/dt is:

    dI/dt = emf(battery)/R(R/L * e^((-R/L)t)) = emf(battery)/L * e^((-R/L)t)

    So dI/dt is

    9/(6.07613*10^-5) * e^((-20/(6.07613*10^-5))*10^-6) = 106577.23 A/s

    So now that I have dI/dt and I have L, how do I use that to find the current if it doesn't involve the emf?
     
  16. Apr 3, 2012 #15

    gneill

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    That's excellent.
    That's easy: you don't need the current :smile: You need the rate of change of the current in order to find the rate of change of the magnetic field.

    How might you use your rate of change of current in order to find the rate of change of B?
     
  17. Apr 3, 2012 #16
    Oh, so could I use the equation for B, but differentiate it so it's the equation for dB/dt, which becomes:

    dB/dt = μ0/(4∏)*N*dI/dt /d

    Which, if I plugged everything in, would give me:

    dB/dt = 10 ^ -7 * 350 * 106577.23 /.39 = 9.5646 T/s
     
  18. Apr 3, 2012 #17

    gneill

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    Why the ##4\pi##? Isn't B = μ0*N*I/L?
     
  19. Apr 3, 2012 #18
    Ah, right. So I would have to multiply what I got before by 4∏ for it to be accurate, which would give me:

    120.192 = dB/dt

    So I'll go out on a limb here and state that then I can find the emf (which is just d/dt(magnetic flux)) by multiplying this rate of change by the area encased by the solenoid, so

    120.192*∏*.007^2 = .18502

    And then divide this emf by the resistance, then multiply by 2 because there are 2 coils in the rectangle, so

    .18502/150 = 1.2335 * 10^-4 *2 = 2.467 * 10^-4 or .0002467 = current of the rectangular coil. Is that accurate?
     
    Last edited: Apr 3, 2012
  20. Apr 3, 2012 #19

    gneill

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    Correct
    You seem to have misplaced the decimal point. The digits are correct.
    Correct procedure.
     
  21. Apr 3, 2012 #20
    Sweet! I just tried that out on another one too and that worked. Thanks so much!
     
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