Eddy-current braking system, find the new velocity

In summary, the problem involves a square loop being shot into a uniform magnetic field with certain given parameters. The goal is to find the velocity of the loop after 0.1 seconds, taking into account the loop's resistance and mass. Two possible approaches are discussed, one involving finding the current and using it to calculate the new velocity, and the other involving energy conservation.
  • #1
Parad0x88
74
0

Homework Statement


Consider the eddy-current braking. A square loop, with 10 cm side is shot with the velocity 10 m/s into the uniform magnetic field with magnitude 0.1 T. The field is perpendicular to the plane of the loop, and the loop starts entering magnetic field at t=0. The resistance of the loop is 1.00 Ohm and the mass is 1.0 g. Assume the loop is moving to the right along x-axis and that x(t=0)=0. Find the velocity of the loop 0.1 seconds later. Comment on assumptions and approximation, or venture into the realm of differential equations…


Homework Equations


A = 0.1m X 0.1 m = 0.01m2
V = 10m/s
B = 0.1T
R = 1Ω
m = 0.001kg
t = 0.1s
Flux = BA = 0.001Wb
ε = Flux/t = 0.01v
I = ε/R
l = 0.1m

The Attempt at a Solution


My first reflex was to find the current: I = .01v/1Ω = 0.01A

And then I wanted to find the new velocity with the formula above, problem is; IR/Bl = (0.01A X 1Ω)/(0.1T X .1m) = 10 m/s, so that doesn't work

And now I'm stumped, I can't really figure out this problem -_-
 
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  • #2
ε = Flux/t = 0.01v
Does your setup depend on the 0.1 seconds given in the problem statement? If not, why do you expect that the voltage depends on that time?
You have to use another value for the time here.

You could try energy conservation to calculate the velocity change.
 
  • #3
mfb said:
Does your setup depend on the 0.1 seconds given in the problem statement? If not, why do you expect that the voltage depends on that time?
You have to use another value for the time here.


If I follow you, the actualy EMF exists before we are breaking, that is why I shouldn't consider the .1 second, so it would be simply divided by 1s (since no other time restriction is given)?

Thus giving

0.001V, which would also give 0.001A, which would in turn give a new velocity of 0.1m/s

And I still believe it makes no sense, that's too big of a change for .1 second, and I am not even considering the mass in what I just did.

You could try energy conservation to calculate the velocity change.

If I take this approach, I could find Ek = 0.5mv2 = 0.05j

And then I assume I'd have to find the potential energy of the system once it has been slowed down after the 0.1 second (by using the 0.05j I found with law of conservation of energy), but again I am stumped and do not know how to go about finding that information.

Second method does make a lot more sense than the first one though.
 
  • #4
There is no relevant potential energy. You'll lose some energy due to the current flow in the curcuit.
 
  • #5
mfb said:
There is no relevant potential energy. You'll lose some energy due to the current flow in the curcuit.

If I lose energy to the current flow, how can I use energy conservation to calculate the velocity change like you previously mentioned?

Or can I simply go ahead right away and find the energy of the system at 0.1s, and then use that in the kinetic energy formula to find the velocity? (Don't know how to do that yet, but if I'm on the right track I'll work on that)
 
  • #6
kinetic energy before = kinetic energy afterwards + energy "lost" due to the resistance

"lost" in a technical sense here - the coil got warmer by a tiny amount.
 

FAQ: Eddy-current braking system, find the new velocity

1. What is an eddy-current braking system?

An eddy-current braking system is a type of electromagnetic braking mechanism used in vehicles to slow down or stop their motion. It works by using the principle of electromagnetic induction to create a resistive force that opposes the motion of the vehicle, thus slowing it down.

2. How does an eddy-current braking system work?

An eddy-current braking system works by using a series of electromagnets, which are powered by the vehicle's kinetic energy, to create a magnetic field around a metal disc or drum. As the metal disc or drum rotates, it creates eddy currents (small electric currents) in the metal, which in turn create a magnetic field that opposes the motion of the disc or drum. This resistance slows down the rotation of the disc or drum, ultimately stopping the vehicle.

3. What are the advantages of using an eddy-current braking system?

There are several advantages to using an eddy-current braking system. First, it is a relatively simple and lightweight system, making it ideal for use in vehicles. It also offers smooth and consistent braking, as there is no physical contact between the braking components. Additionally, eddy-current braking systems are highly efficient and require minimal maintenance.

4. What factors affect the new velocity in an eddy-current braking system?

The new velocity in an eddy-current braking system is affected by several factors, including the strength of the magnetic field created by the electromagnets, the speed and mass of the rotating disc or drum, and the amount of resistance generated by the eddy currents. Other factors, such as air resistance and the condition of the braking components, may also play a role.

5. How is the new velocity calculated in an eddy-current braking system?

To calculate the new velocity in an eddy-current braking system, the initial velocity of the vehicle, the strength of the magnetic field, and the amount of resistance generated by the eddy currents must be known. This information can then be used in equations that take into account the properties of the vehicle and the braking system to determine the new velocity. However, the exact calculation may vary depending on the specific design and setup of the eddy-current braking system.

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