# Hat current is needed to provide a modest 40 kN force? plug and chug!

1. Nov 6, 2005

### mr_coffee

Consider the possibility of a new design for an electric train. The engine is driven by the force on a conducting axle due to the vertical component of Earth's magnetic field. Current is down one rail, through a conducting wheel, through the axle, through another conducting wheel, and then back to the source via the other rail.
a) What current is needed to provide a modest 40 kN force? Take the vertical component of Earth's field to be 10 µT and the length of the axle to be 3.0 m.

b) How much power would be lost for each ohm of resistance in the rails?

THe professor gave us the formula for this one and yet it didn't work, did I mess up somthing else?
He said for part a, use the following formula:
F = I*L*E-5;
I = F/(L*E-5);
I = 4000/(3*E-5);
I = 1.3E8

I don't know why its E-5 but i also tried E-6 and still didn't work...

and for part b he said to use
P = I^2;

But really I just need to find out why part A isn't working! Thanks!

2. Apr 23, 2006

### schrodinger07

F=iLB=i(3)(10E-6)
F=(30E-6)i
i=40000/(30E-6)=1.3333E9 Amps

This was the right answer for me...

and
P=i^2*R
i^2=(1.333E9)^2=1.77778e18 Watts

3. Apr 23, 2006

### Hootenanny

Staff Emeritus
$$40kN \neq 4000 N$$

~H

 Sorry didnt see schrodinger's post

Last edited: Apr 23, 2006