Magnetic Flux and Induced Voltage Interesing question

AI Thread Summary
The discussion revolves around calculating the speed of a cart at a specific point on a ramp, given various parameters such as mass, height, resistance, and magnetic field strength. The calculated initial speed at the bottom of the ramp is 2.8 m/s, but the expected speed at point 5w (1m along the flat surface) is 2 m/s, leading to confusion about the effects of the magnetic field. The participant attempts to apply kinematic equations but questions their applicability, suggesting that calculus may be necessary for a more accurate solution. The geometry of the system, including the slope and the flat surface where the magnetic field is applied, is also a point of discussion. The interaction between the magnetic field and the cart's motion is highlighted as a critical factor in understanding the speed discrepancy.
stonecoldgen
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Homework Statement


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



Homework Equations



\phi=BA
\epsilon=-N\Delta\phi/\Deltat


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?
 
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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.
 
Simon Bridge said:
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.

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...
 
This is why I asked about the geometry - isn't there a slope involved?
 
Simon Bridge said:
This is why I asked about the geometry - isn't there a slope involved?

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.
 
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?
 

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