Work done by permanent magnet

  • Thread starter korneld
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  • #1
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Main Question or Discussion Point

Hi,

I have two questions about permanent magnets:


1. How do you calculate the work done by a permanent magnet on, say, an iron plate?

2. I am aware that the force exerted on an object by a magnet depends on the surface area. Is it also affected by the thickness of the object to a certain extent? What is the minimum thickness after which thickness is not an issue?


Thanks.
 

Answers and Replies

  • #2
The work is force times distance (when linear) or better, the integral of force to the distance it acted upon.

Thickness of what, the metal plate of the magnet.
 
  • #3
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Thanks for that.

But the problem is that the force increases as the distance between the objects decreases. I thought there might be a formula that takes the increasing flux into account.

Second, I meant to ask about the thickness of the plate.
 
  • #4
russ_watters
Mentor
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It is an inverse-square relationship between distance and force.

Btw, your OP asks about work. You do understand that work is force times distance moved, right? Static forces do no work.
 
  • #5
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Yes, no movement means no work done.

What I am looking to find out is if a ferromagnetic plate is "sucked in" by a magnet, how much kinetic energy is gained. Also, what the minimum thickness of this plate would have to be to take advantage the full force of this magnet.
 
  • #6
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Use conservation of energy, viz. the work done by the magnet is equal to vertical component of its displacement multiplied by the objects' weight.
 
  • #7
Meir Achuz
Science Advisor
Homework Helper
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Yes, no movement means no work done.

What I am looking to find out is if a ferromagnetic plate is "sucked in" by a magnet, how much kinetic energy is gained. Also, what the minimum thickness of this plate would have to be to take advantage the full force of this magnet.
Unless the plate starts out very close to the magnet, you have to do a complicated integral of F(x)dx. The force ~1/7 (like Van der Waals) at large distance and becomes indep of x when x<<R (for a magnet with end radius R).
You can decide on the miimum thickness of the plate by solving the following
electrostatics problem: Consider two parallel identical uniformly charged
disks of radius R, a distance L apart. When the distance x above one disk is large enough so that you can neglect the charge on the other plate is the same as when the plate is thick enough. The plate thickness L will depend on x, R and what you mean by "full force"
 
  • #8
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Unless the plate starts out very close to the magnet, you have to do a complicated integral of F(x)dx. The force ~1/7 (like Van der Waals) at large distance and becomes indep of x when x<<R (for a magnet with end radius R).
You can decide on the miimum thickness of the plate by solving the following
electrostatics problem: Consider two parallel identical uniformly charged
disks of radius R, a distance L apart. When the distance x above one disk is large enough so that you can neglect the charge on the other plate is the same as when the plate is thick enough. The plate thickness L will depend on x, R and what you mean by "full force"

Thanks for the info.

'... and what you mean by "full force"': I've read on one permanent magnet manufacturer's website (which now I can't seem to find) that the material to be attracted by the magnet has to to have a certain thickness, but beyond that point thickness is irrelevant. I am assuming that below it, the magnetic field will have a lesser effect on the material.
 

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