Example: Metal Cart

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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?

Emec cart polarBear2.png

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>


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. Emec cart topview.png

Emec cart endview.png

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


 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(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

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