# Voltage drop across ideal solenoidal inductor

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1. Jan 4, 2015

### genxium

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. Jan 4, 2015

### Staff: Mentor

Do you use t both for time and as parameter for the solenoid shape? That is confusing.

3. Jan 4, 2015

### genxium

Sorry for that :( I've updated the notations

4. Jan 4, 2015

### 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.

5. Jan 4, 2015

### genxium

@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?

6. Jan 4, 2015

### 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.