Retarding force of eddy currents in a disc

In summary, the braking force generated by an eddy current is proportional to the rotational speed of the disc.
  • #1
fysiikka111
41
0

Homework Statement


How to calculate braking force generated by eddy currents. If there is a disc of radius r with conductivity K, with a magnet located at a distance r-d from the center of the disc with a magnetic field B, what is the retarding force of the magnetic field created by the eddy currents for a given rotational velocity v of the disc.

Homework Equations


Faraday's law of induction
Ampere-Maxwell law

The Attempt at a Solution


Haven't been able to find information that shows how to explicitly solve for the retarding force produced by an eddy current.
Thanks
 
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  • #2
Well, think of what happens when there is no current in the disc, and it starts spinning. The moving electrons spinning with the disk see a changing magnetic field, and the corresponding Lorentz force causes them to drift in the disk with some current density [tex] \vec{J} [/tex].

These electrons drifting within the disk interact with the magnetic field and ultimately produce a force opposing the disk's motion. It's best to assume steady state, and not worry about 'closure' of the Eddy currents. Just suppose that the current density is constant.

Once you know the current density, which will vary radially due to the outer portion of the disk moving faster than the inner portion, you can take the cross product [tex] \vec{J}\times\vec{B} [/tex] to get the force per unit volume in the disk as a function of radius. It should be easy to see where to go from there.
 
  • #3
quarky2001 said:
Well, think of what happens when there is no current in the disc, and it starts spinning. The moving electrons spinning with the disk see a changing magnetic field, and the corresponding Lorentz force causes them to drift in the disk with some current density [tex] \vec{J} [/tex].

These electrons drifting within the disk interact with the magnetic field and ultimately produce a force opposing the disk's motion. It's best to assume steady state, and not worry about 'closure' of the Eddy currents. Just suppose that the current density is constant.

Once you know the current density, which will vary radially due to the outer portion of the disk moving faster than the inner portion, you can take the cross product [tex] \vec{J}\times\vec{B} [/tex] to get the force per unit volume in the disk as a function of radius. It should be easy to see where to go from there.

Thanks for the reply!
So, the magnetic force and the electric force of the charges will be equal and opposite at steady state. From the Lorentz equation at equilibrium, or steady state
[tex]q\mathbf{E}=q(\mathbf{v}\times\mathbf{B})\longrightarrow\frac{q\mathbf{J}}{\sigma}=q(\mathbf{v}\times\mathbf{B})\longrightarrow\mathbf{J}=\sigma(\mathbf{v}\times\mathbf{B})[/tex]
where [tex]\sigma[/tex] is conductivity.
Where did you get the force equals cross product of J and B equation from?
Thanks
 
  • #4
I think the the source of [tex] \vec{f} = \vec{J}\times\vec{B} [/tex] can be seen if you just look at the standard formula of [tex] \vec{F}=I\vec{l}\times\vec{B} [/tex] and think about how you would adapt it to deal with a volume element rather than a full "chunk" of conductor.

If you have a copy of Griffith's Electrodynamics (3rd edition), there's a good explanation of the J x B formula on page 212.
 
  • #5
Thanks. On page 298 of Griffiths' Electrodynamics he says that eddy currents are difficult to calculate. Would a reasonable approximation be that the retarding force is proportional to velocity, like a viscous damper?
 
  • #6
I think the the source of
f⃗ =J⃗ ×B⃗
can be seen if you just look at the standard formula of
F⃗ = Il⃗ ×B⃗
and think about how you would adapt it to deal with a volume element rather than a full "chunk" of conductor.

If you have a copy of Griffith's Electrodynamics (3rd edition), there's a good explanation of the J x B formula on page 212.

EDIT:

Would a further approximation be accurate (F = the braking force)?

F = σ*v*B2 , where v is the rotating speed of the disc (average).

OR, should I move back to the Farraday Law for integrating solutions?

Thanks for any tips!
 
Last edited:
  • #7
The solution would need to be integrated.

Using an average rotational velocity would yield an approximation of the right order of magnitude, but it wouldn't be very accurate - that's because when you think about the disc's area, there is a larger area of the disk moving at a faster speed (the edges) and a smaller area moving at a slower speed (near the axis). Taking an average speed (presumably at half the disc's radius) would underestimate the total retarding force.
 
  • #8
The solution would need to be integrated.

Using an average rotational velocity would yield an approximation of the right order of magnitude, but it wouldn't be very accurate - that's because when you think about the disc's area, there is a larger area of the disk moving at a faster speed (the edges) and a smaller area moving at a slower speed (near the axis). Taking an average speed (presumably at half the disc's radius) would underestimate the total retarding force.

Thanks for tip! I solved it through dependence on r , and integrated it at the end.

I understand that this will be a rough approximation, because of the different speeds of the disc. I'll look for some constant that I can add to the equation to make it more accurate and will work with the average speed.
 

1. What is the retarding force of eddy currents in a disc?

The retarding force of eddy currents in a disc is a force that opposes the motion of a conductive disc when it moves through a magnetic field. This force is caused by the interaction between the magnetic field and the induced eddy currents, which create their own magnetic field that opposes the original field.

2. How is the retarding force of eddy currents in a disc calculated?

The retarding force of eddy currents in a disc can be calculated using the formula F = B^2 x ω x t^2 x σ x A, where B is the magnetic field strength, ω is the angular velocity of the disc, t is the thickness of the disc, σ is the conductivity of the disc material, and A is the area of the disc.

3. What factors affect the strength of the retarding force of eddy currents in a disc?

The strength of the retarding force of eddy currents in a disc is affected by several factors, including the strength of the magnetic field, the conductivity and thickness of the disc material, and the disc's angular velocity. Additionally, the presence of any non-uniformities in the magnetic field or the disc material can also affect the strength of the force.

4. How does the retarding force of eddy currents in a disc affect the disc's motion?

The retarding force of eddy currents in a disc acts in the direction opposite to the disc's motion, causing it to slow down. This force can also cause the disc to experience a torque, which can result in a change in the disc's rotational speed or direction.

5. Can the retarding force of eddy currents in a disc be utilized in any practical applications?

Yes, the retarding force of eddy currents in a disc can be utilized in various practical applications, such as in eddy current brakes for trains and roller coasters, magnetic dampers for vehicles, and magnetic bearings for machines. It can also be used in non-contact braking systems for industrial equipment and in speed governors for turbines.

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