- #1
genxium
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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?
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?
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