What is the resultant force on a conducting plate in a magnetic field?

AI Thread Summary
The discussion focuses on the resultant force acting on a conducting plate within a magnetic field, particularly in the context of a lightning strike. It highlights that the free charges in the conductive plate will redistribute when subjected to a magnetic field, which influences the force experienced by the plate. The interaction between the magnetic field created by the lightning and the induced currents in the plate results in a downward pushing force. Participants suggest that understanding the forces between charged objects can help analyze the situation. The inquiry seeks to determine the magnitude of this force on the plate.
munozcim
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Dear all,

I've found in many books forces acting between 2 parallel wires, but not between a magnetic field and a surface. Let's now imagine a conducting plate which is within a magnetic field. Which is the resultant force acting on that panel?
Many thanks in advance!

Raúl
 
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Hi there,

Thinking about it, your question should boil down to a wire in an electric field. I assume that you plate is conductive. What happens to the "free" charges in a conductive plate? (come on this is simple).

From this result, you can apply a wire current within a magnetic field.

Cheers
 
If you are talking about two plates of evenly distributed current, I should think you can integrate the forces to find the answer.

It reminds me of the problem of the forces two equally and linearly uniformly charged rods exert on each other.
 
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Thanks fatra2 and espen180 for your quick reply. Actually I'm thinking about a lightning strike which is acting on a conductive plate. The current flows from the attachment point to the whole plate in a radial way in all directions. The plate can be assumed as isotropic.
The magnetic field created by the lightning strike acts on the conductive plate pushing it downwards. I would like to know the value of that pushing force.
cheers
 
But no matter what, once your lightning strikes the conductive plates, the extra free charges will redistribute on the edge of the plate. Therefore, applying current in a wire could be applied.

Cheers
 
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