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Planar circuit around a solenoid

  1. Jun 3, 2014 #1
    1. The problem statement, all variables and given/known data
    Grad student here, reviewing for my quals. It's been awhile since E&M and I think I need a refresher since this problem has me stumped.

    A planar circuit surrounds a solenoid and consists of two capacitors of capacitances [itex]C_1[/itex] and [itex]C_2[/itex] joined together by normal wires. The solenoid crosses the plane of the circuit in a patch of area A, and it produces a time-dependent magnetic field that is changing linearly with time: [itex]B(t) = B_0 + \frac{dB}{dt} t[/itex], the positive direction is coming up out of the paper. The field is uniform inside the solenoid and the magnetic field outside the solenoid is the be neglected.

    Before the field is applied the capacitors have zero charge. In equilibrium what are the charges [itex]Q_1[/itex] and [itex]Q_2[/itex] on the capacitors? Determine the signs.


    2. Relevant equations

    Q = CV
    Faraday's Law
    Lenz's Law


    3. The attempt at a solution

    I found the flux: [tex] \Phi = AB(t) = A(B_0 + \dot{B} t)[/tex]. I then used Faraday's law to find the induced voltage: [tex] \Delta V = - N \frac{\Delta \Phi_B}{\Delta t} = - \frac{\Delta BA}{\Delta t} = -A \frac{dB}{dt}[/tex].

    However, the capacitors are in series which means that they should have the same charge and different voltages, right? I feel like I'm combining two competing concepts here.

    As for the signs, I think that they should be opposite but I'm not sure, that's just intuition.
     
    Last edited: Jun 4, 2014
  2. jcsd
  3. Jun 4, 2014 #2
    Hi tourjete! Remember that Faraday's law gives you the voltage "all the way around the loop", not just across one of those capacitors.
     
  4. Jun 4, 2014 #3
    I'm assuming I'm supposed to use Kirchoff's Law's then? I did [itex]V_1 + V_2 -A \frac{dB}{dt} = 0[/itex], where [itex]V_1[/itex] and [itex]V_2[/itex] are the voltages across each capacitor. I also know that two capacitors in series have
    [tex]\frac{1}{C_{eq}} = \frac{V_1+V_2}{Q} = \frac{1}{C_1} + \frac{1}{C_2}[/tex]. Solvin Kirchoff's Law for [itex]V_1 + V_2[/itex], I get that [tex] Q_1 = Q_2 = \frac{A \dot{B} C_1 C_2}{C_1+C_2} [/tex]. Is this even remotely the correct approach? I'm still not 100% sure about the signs.
     
    Last edited: Jun 4, 2014
  5. Jun 4, 2014 #4
    I think that's basically correct. Normally Kirchoff's law is reserved for the situation where the voltage changes around the loop add up to zero - here they have to add up to [itex]A\dot{B}[/itex] (or [itex]-A\dot{B}[/itex] depending on which way you go around the loop).

    As for the "signs", I think the important thing is to draw a diagram of the circuit with + and - sides of the capacitors labelled correctly. You can figure that out from Lenz's law, or (equivalently) just from looking at the vectors in the appropriate Maxwell equation.
     
  6. Jun 4, 2014 #5
    okay, so it's impossible to know the signs of the charges without knowing whether [itex]\frac{dB}{dt}[/itex] is increasing or decreasing? (Thanks for the help, by the way, it was very useful)
     
  7. Jun 4, 2014 #6
    The important thing is which way the vector [itex]\frac{d \bf{B}}{dt}[/itex] is pointing.

    The relevant Maxwell equation is
    [tex]\nabla \times {\bf E} \; = \; - \frac{\partial \bf{B}}{\partial t}.[/tex]
    You can tell whether the charge in the circuit gets pushed around clockwise or counterclockwise by using the vector [itex]\frac{d \bf{B}}{dt}[/itex] together with some calc 3 hand rule for the curl :smile:
     
  8. Jun 6, 2014 #7

    rude man

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    YI agree with your post 3 expression for Q1 and Q2.

    Another way to determine capacitor voltage polarity is to think which way the current had to initially flow to get the capacitors charged up. You can use Lenz's law for this purpose. The initial B field generated by the transient current has to act so as to counter B-dot generated by the solenoid.
     
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