Magnetic Flux and Induced Voltage Interesing question

stonecoldgen

1. The problem statement, all variables and given/known data
assuming that friction is negligible and tthat electromagnetic forces are constant, (when they occur) calculate the speed of the cart at point 5w

this is part IV of a question, so in the other parts and in the given, i found out that:

mass of the cart=0.5kg
number of loops=150
hieght of the rampo=0.4m
height of the wire=0.1
width=0.2m
Resistance: 2.5 ohms
Magnetic field=1.2T



2. Relevant equations

[itex]\phi[/itex]=BA
[itex]\epsilon[/itex]=-N[itex]\Delta\phi[/itex]/[itex]\Delta[/itex]t


3. The attempt at a solution
The velocity of the cart at the bottom of the ramp is 2.8m/s
the Voltage of the cart as it enters the field is 50V
the current induced is 20A
The force of teh cart as it enters the field is 2.4N, making its acceleration 4.8m/s/s


Normal kinematics don't seem to work well, any tip?

Simon Bridge

What does "5w" refer to?
What is the geometry of the system?

You have not actually shown us how you attempted the problem - only the numbers you got as a result.

stonecoldgen

5w refers to 5(.2)=1m

what i did was the following:

The acceleration is 4.8m/s/s, this is sort of like a railgun.

so

V₂²=v₁²+2ad

and i plugged in the humbers

V₂²=(2.8)²+2(4.8)(1)

and it turns out that supposedly v₂=4.18

when it's supposed to be 2



so now this makes me think why this isn't exactly like a railgun (the velocity decrased)

any ideas?




for some reason I think this can only be solved with calculus, just a feeling, a weird feeling though...

Simon Bridge

This is why I asked about the geometry - isn't there a slope involved?

stonecoldgen

There is a slope and then comes the magnetic field. The magnetic field lies on a flat surface. I already calculated the speed the cart gets when it reaches the flat part.

Simon Bridge

OK - so if your initial speed is 2.8m/s, and the final speed is supposed to be 2m/s, then what is the effect of the magnetic field on the cart?