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 $\overrightarrow{B}$ 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?

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

$\overrightarrow{F}=\underset{c}{\int }I\overrightarrow{d}l\times \overrightarrow{B}⟹{\overrightarrow{F}}_2=\underset{0}{\overset{L}{\int }}I\overrightarrow{d}l\times \overrightarrow{B}⟹{\overrightarrow{F}}_2=-I(t)BL\hat{i}$

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

And $e_m(t)=\underset{o}{\overset{L}{\int }}(\overrightarrow{v}\times \overrightarrow{B})\overrightarrow{d}l⟹e_m(t)=-vLB$

$I(t)=\frac{vLB}{R}{\overrightarrow{F}}_2=-I(t)BL\hat{i}⟹{\overrightarrow{F}}_2=-\frac{vLB}{R}BL\hat{i}⟹{\overrightarrow{F}}_2=-\frac{v{L}^2{B}^2}{R}\hat{i}$

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

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

${ℒ}\{\dot{v}+v\left(\frac{L^2{B}^2}{Rm}\right)-\frac{F_1}{m}\}⟹sV(s)+\frac{L^2{B}^2}{Rm}V(s)-\frac{F_1}{ms}$

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

$sV(s)+\psi V(s)-\frac{F_1}{ms}=0$

$V(s)(s+\psi )=\frac{F_1}{ms}⟹V(s)=\frac{F_1}{ms(s+\psi )}$

Using partial fraction expansion

$\frac{F_1}{ms(s+\psi )}⟹\frac{F_1/m\psi }{s}-\frac{F_1/m\psi }{s+\psi }$

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

$V(t)={ℒ}\{\frac{F_1/m\psi }{s}-\frac{F_1/m\psi }{s+\psi }\}⟹V(t)=\frac{F_0}{m\psi }(u(t)-e^{-\psi t}u(t))⟹V(t)=\frac{F_0}{m\psi }(1-e^{-\psi t})$

$V(t)=\frac{F_0}{m\frac{L^2{B}^2}{Rm}}(1-e^{-\frac{L^2{B}^2}{Rm}t})⟹V(t)=\frac{F_{0}R}{L^2{B}^2}(1-e^{-\frac{L^2{B}^2}{Rm}t})$

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

So we can substitute in V(t) to get

$e_m(t)=\frac{F_{0}R}{LB}(1-e^{-\frac{L^2{B}^2}{Rm}t})$

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

$I(t)=\frac{\frac{F_{0}R}{LB}(1-e^{-\frac{L^2{B}^2}{Rm}t})}{R}⟹I(t)=\frac{F_0}{LB}(1-e^{-\frac{L^2{B}^2}{Rm}t})$

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

$\underset{t\to \mathrm{\infty }}{\lim }I(t)=\frac{F_0}{LB}(1-e^{-\mathrm{\infty }})⟹I(\mathrm{\infty })=\frac{F_0}{LB}$

So the Steady-State Current = $\frac{F_0}{LB}$

$Inconclusionitcanbeseenthatapenguindriven,polarbearkillinggenerator$

$wouldbeaviableoptionforalternativeenergyinCanada.$

Review Comments and Status

Kirk Betz Read and approved 1-26-10

Will Griffith Approved 1-27-10