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, a 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 dead across the tracks acting as if a resistor R is connected as a load. Find The current as a function of time. What is the current after the generator attains the steady-state condition?

Solution

For this Problem we will represent the large circle as a pair of straight parallel wires and the cart as a single wire. This is illustrated below

FIGURE

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

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

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

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

. . . . .

In conclusion it can be seen that a penguin driven, polar bear killing generator would be a viable option for alternative energy in Canada.