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Voltage drop across ideal solenoidal inductor

  1. Jan 4, 2015 #1
    First by ideal I mean zero resistance. I tried to verify Faraday's law in simple LCR circuit but ran into some conflicting results. Though the description below will be a little verbose, the configuration for this problem is extremely simple: assume that I put a solenoid ##\left\{\begin{array}{c} x(\tau) = r_0 \cdot cos (\omega \cdot \tau) \\ y(\tau) = r_0 \cdot sin (\omega \cdot \tau) \\ z(\tau) = v \cdot \tau \end{array}\right. ## in a 3D Cartesian coordinate. Given 2 points ##a, b## on this solenoid where ##a## corresponds to ##\tau=0## and ##b## corresponds to ##\tau=T##(large enough to contain more than 1 turn). All constants are positive if not otherwise specified.

    If a current ##I## flows in direction ##a \rightarrow b## then it yields a ##\textbf{B}## field to the ##z_+## direction. Now assume that ##\frac{\partial I}{\partial t} > 0##, according to Lenz's law, I shall expect the inductive ##\textbf{E}## field circulates clockwisely viewed in direction ##z_+ \rightarrow z_-##, i.e. ##\textbf{E}(\textbf{r}) = |E(\textbf{r})| \cdot sin\theta \cdot \hat{\textbf{x}} + (-|E(\textbf{r})| \cdot cos\theta) \cdot \hat{\textbf{y}}## where ##\theta## is the angle subtended by ##\textbf{r}## counter-clockwisely with respect to ##x-axis##.

    I suppose that the voltage DROP from a to b is positive in this case, i.e. ## \int_0^T \textbf{E} \cdot d \textbf{l} > 0##, however the path integral

    ## \int_0^T \textbf{E} \cdot d \textbf{l}##
    ##= \int_0^T -|E| r_0 \omega \cdot sin\theta \cdot sin(\omega \tau) - |E| r_0 \omega \cdot cos\theta \cdot cos(\omega \tau) \cdot d\tau ##
    ##= \int_0^T -|E| r_0 \omega \cdot [cos(\omega \tau) cos\theta + sin(\omega \tau) sin\theta] \cdot d\tau##
    ##= \int_0^T -|E| r_0 \omega \cdot cos(\omega \tau - \theta) \cdot d\tau##
    ##= -T \cdot |E| r_0 \omega < 0##

    where use has been made of ##\theta = \omega \cdot \tau## at every point on the path and ##d\textbf{l} = dx \cdot \hat{\textbf{x}} + dy \cdot \hat{\textbf{y}} + dz \cdot \hat{\textbf{z}}##. I checked the calculation for several times but still got the same result. This is confusing me badly, is anyone willing to help?
     
    Last edited: Jan 4, 2015
  2. jcsd
  3. Jan 4, 2015 #2

    mfb

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    2016 Award

    Staff: Mentor

    Do you use t both for time and as parameter for the solenoid shape? That is confusing.
     
  4. Jan 4, 2015 #3
    Sorry for that :( I've updated the notations
     
  5. Jan 4, 2015 #4

    Dale

    Staff: Mentor

    First, the path that you integrate over needs to be a closed path. This one is not.

    Second, the surface over which you calculate the flux must be bounded by the loop. For a solenoid that is a very complicated surface, I am not at all sure how you would calculate that.

    I would recommend simplifying a lot. Use a single flat loop of wire, and go from there.
     
  6. Jan 4, 2015 #5
    @mfb gave me satisfying answers about a flat loop in this thread. I agree that the surface over which the flux is calculated is complicated thus when reasoning the direction of the inductive ##E## field I used Lenz's Law instead of Faraday's Law. This is a method I learned in high school and I think it's an approximation regarding that each turn of the solenoid is approximately parallel to the ##XY## plane, i.e. ##v## is small in ##z(\tau) = v \cdot \tau##.

    If the math cannot be simplified for the solenoid, is there any other way to show that whether the voltage drops or increases from a to b along the solenoidal path?
     
  7. Jan 4, 2015 #6

    Dale

    Staff: Mentor

    If you have already done a flat loop then the easiest thing will be to consider a stack of several flat loops. Use the contour of one loop and the surface bounded by that loop.
     
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