jubba
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jubba said:anybody know why
Nope. The eddy current would need to flow around the enclosed surface with the changing flux. Break that outside path, and there is no total eddy current generating an opposing flux and force.jubba said:i know that the open loop won't have as high a current as the closed loop of copper but should it still have some eddy currents. Afterall it is still a piece of metal experiencing a change in flux through it.
Thats why i thought it should be
p,r,q