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Induced current with two solenoids

  1. Apr 30, 2013 #1
    1. The problem statement, all variables and given/known data

    so theres two solenoids on the same axis of symmetry, and the smaller one inside the larger one has a resistance of 11 ohms, 1000 turns, and radius .02 meters. the outer solenoid has n 18000 turns per meter, and had a current that is increasing at a constant rate with dI/dt=.7 Amps/sec. The inner solenoid is connected to an ammeter.

    3. The attempt at a solution

    how would you find the induced current in the inner solenoid? i have an equation i tried using:
    Iind=[(μ0)(n(turns per meter))(Imax)(A)]/[(R)(Δt)]
    made μ0=4pi(10-7), A=pi(0.022) R=11, and since dI/dt=0.7 A/s i just made change in time equal 1 sec and Imax equal 0.7 Amps. One thing I think that may be messed up is the value for n, in the problem were given how many exact turns it has which is N, and no length of the solenoid(s). But the equation i have uses n which is turns per meter. So i just put for n (18000).

    Got 1.91(10-6) amps, but the correct answer is 1.81(10-3) amps...tried a few other things and im just kind of lost. Can somebody help me understand this?
     
    Last edited: Apr 30, 2013
  2. jcsd
  3. May 1, 2013 #2
    Thanks for the help guys -_-
     
  4. May 1, 2013 #3
    From Ampere's Law you can derive that the magnetic field B in a solenoid is:

    [itex]B_{solenoid} = μ_{0} n I[/itex]

    Then, you can use Faraday's Law of Induction to find the electromotive force in the second solenoid:

    [itex]ε = - \frac{\partial \Phi_{B}}{\partial t} = - μ_{0} n \frac{dI}{dt} A N[/itex],

    where A is the area of the second solenoid and N is the number of loops(this is actually an integral but since B is approximately constant, then the flux is just the magnetic field times the area. Since the solenoid has many loops, the area is multiplied by N since you're calculating the flux through each loop.

    This gives you [itex]ε[/itex], from which you can calculate the induced current since [itex]ε = I_{induced} R[/itex].

    This should give you enough info to solve the problem.

    Also, the negative sign indicates the induced current is in the opposite direction, but for your case you can just take the absolute value.
     
  5. May 1, 2013 #4

    rude man

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    1. Compute the flux inside the larger coil as a function of applied current.
    2. Use Farady's law to determine the emf generated in the smaller coil.
    3. Use i = emf/R for the current of the smaller coil. (An ammeter shorts the smaller coil).

    Don't start with an equation with delta t in it. You're using equations without understanding their origin. What's important is to understand what's going on, not pulling equations you "have" to hopefully solve a given problem.
     
  6. May 1, 2013 #5
    unfortunately i have a teacher that just makes us watch him derive equations and then doesnt explain anything or how it worked so i get a little lost. but i got the answer now, thanks
     
  7. May 1, 2013 #6
    seriously, "rude_man" is right. I dont get why some people on this site have to act so condescending about physics. @rude_man, i am new to this subject...i tried really hard to understand it myself but was still having trouble, so i came here. i dont need people like you giving me attitude like that. it happens way too much on this site. Thanks LoadedAnvils, you didnt care to criticize me and simply explained how the problem works and how to find the answer. now i understand it. much appreciated
     
  8. May 1, 2013 #7
    No problem. Physics is a very hard subject (I had a hard time with it even though I liked the subject), and sometimes the only way to understand something is to be walked through it step by step.
     
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