# Example: Metal Cart

## Problem

A DC generator is built using a metal cart with metallic wheels that travel around a set of perfectly conducting rails in a large circle. The rails are L m apart and there is a uniform magnetic $\vec B$ field normal to the plane. The cart has a penguin,with mass m, and is driven by a rocket engine having a constant thrust $F_1$. A wet polar bear, having stumbled out of a shack where he recently had a bad experience with a battery, lays dying across the tracks acting as a load resistance R over the rails. Find The current as a function of time. What is the current after the generator attains the steady-state condition?

Problem loosely based on 2.6 from Electric Machinery and Transformers, 3rd ed <ref>Guru and Huseyin, Electric Machinery and Transformers, 3rd ed. (New York: Oxford University Press, 2001)</ref>

## Solution

For this Problem the large circle will be represented by a pair of parallel wires and the cart as a single wire. This is illustrated below in the top and end view figures.

We have two forces, $F_1$ being the force from the rocket engine and $F_2$ being the force caused by the current in the conductor and the Magnetic Field. The resulting Force $F_t$ is simply the sum of $F_1$ and $F_2$

$F_2$ can be found using Ampere's Law

$\vec F=\int\limits_{c} I ~\vec dl\times \vec B~~~~~\Longrightarrow~~~~~ \vec F_2=\int\limits_{0}^{L} I ~\vec dl\times \vec B ~~~~~\Longrightarrow~~~~~ \vec F_2=- I(t) ~B~L ~~ \hat i$

We can also say that $I(t)=\frac{-e_m(t)}{R}$

And $~~ {e_m(t)}= \int\limits_{o}^{L} (\vec v \times \vec B)~\vec dl ~~~~~\Longrightarrow~~~~~ {e_m(t)}=-v~L~B$

$I(t)=\frac{v~L~B}{R} ~~~~~~~~~~~~~~~~~~~~~~\vec F_2=- I(t) ~B~L ~~ \hat i~~~~~\Longrightarrow~~~~~\vec F_2=- \frac{v~L~B}{R} ~B~L ~~ \hat i ~~~~~\Longrightarrow~~~~~\vec F_2=- \frac{v~L^2~B^2}{R} ~~ \hat i$

$\dot v=\frac{F_1}{m} ~\hat i~+\frac {F_2}{m}~ \hat i ~~~~~\Longrightarrow~~~~~ \dot v= ~\frac{F_1}{m} ~~\hat i~- \frac{v~L^2~B^2}{R m}~~\hat i ~~~~~\Longrightarrow~~~~~ \dot v~ +~ v \left(\frac{L^2~B^2}{R~m}\right) ~-~\frac{F_1}{m}~=~0$

Now we have a lovely differential equation to work with! To attempt to find the current we will take the Laplace transform.

$\mathcal{L}\left\{ \dot v ~+~v\left(\frac{L^2~B^2}{R~m}\right)-\frac{F_1}{m} \right\} ~~~~~\Longrightarrow~~~~~ s~V(s)~+~\frac{L^2B^2}{R~m} V(s) ~-~\frac{F_1}{m~s}$

Lets title and substitute in the variable $~~~~\psi=\frac{L^2~B^2}{R ~m}~~$ to simplify things

$s~V(s)~+~\psi~V(s)~-~\frac{F_1}{m~s}~=~0$

$V(s)~(s~+~\psi)~=~\frac{F_1}{m~s} ~~~~~\Longrightarrow~~~~~ V(s)~=~\frac {F_1}{m~s~(s~+~\psi)}$

Using partial fraction expansion

$\frac {F_1}{m~s~(s~+~\psi)}~~~~~~\Longrightarrow~~~~~~\frac{F_1/m\psi}{s}~-~\frac{F_1/m\psi}{s~+~\psi}$

$V(s)=\frac{F_1/m\psi}{s}~-~\frac{F_1/m\psi}{s~+~\psi}$

$V(t)= \mathcal{L}\left\{ \frac{F_1/m\psi}{s}~-~\frac{F_1/m\psi}{s~+~\psi} \right\} ~~~~~\Longrightarrow~~~~~ V(t)=\frac {F_0}{m~\psi}\left(u(t)~-~e^{-\psi~t} ~u(t)\right) ~~~~~\Longrightarrow~~~~~ V(t)= \frac {F_0}{m~\psi}\left(1~-~e^{-\psi~t}\right)$

$V(t)=\frac {F_0}{m~ \frac{L^2~B^2}{R ~m}}\left(1~-~e^{-\frac{L^2~B^2}{R ~m}~t}\right) ~~~~~\Longrightarrow~~~~~ V(t)=\frac {F_0~R}{L^2~B^2}\left(1~-~e^{-\frac{L^2~B^2}{R ~m}~t}\right)$

We know that $~~ e_m(t)~=~V(t)LB$

So we can substitute in V(t) to get

$e_m(t)= \frac {F_0~R}{L~B}\left(1~-~e^{-\frac{L^2~B^2}{R ~m}~t}\right)$

And we know that $~~I(t)~=~\frac{e_m(t)}{R}$

$I(t)=\frac{\frac {F_0~R}{L~B}\left(1~-~e^{-\frac{L^2~B^2}{R ~m}~t}\right)}{R}~~~~~\Longrightarrow~~~~~ I(t)~=~\frac {F_0}{L~B}\left(1~-~e^{-\frac{L^2~B^2}{R ~m}~t}\right)$

To find the steady-state current we simply look at the limit of I(t) as $t \rightarrow \infty$

$\lim_{t\rightarrow \infty} I(t)=\frac{F_0}{L~B} \left( 1- e^{-\infty}\right) ~~~~~\Longrightarrow~~~~~ I(\infty)=\frac{F_0}{L~B}$

So the Steady-State Current = $\frac{F_0}{L~B}$

$In~conclusion~it ~can ~be ~seen ~that ~a ~penguin ~driven, ~polar ~bear ~killing ~generator$

$~would ~be ~a ~viable ~option ~for ~alternative ~energy ~in ~Canada.$

<references />

## Reviewed by

Kirk Betz Read and approved 1-26-10

Will Griffith Approved 1-27-10