How Do Faraday's Laws Relate to Magnetic Energy Dissipation in a Cylinder?

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Rider4
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Homework Statement


A thin cylinder of radius r, thickness s, and length L, is made of metal with a resistivity of ρ. The cylinder is resting in a uniform magnetic field B0, with the field being in the same direction as the axis of the cylinder. The cylinder is then removed from the field. Find the current I0 that would be flowing initially, then find the rate at which energy would be dissipated by the resistance of the cylinder to the current flowing around it, and finally, find how long it would take at that rate for the magnetic energy in the cylinder to be dissipated.


Homework Equations





The Attempt at a Solution


I used B0=(μ0I0)/(2pi(r))
Then I solved for I0=((B0)(2pi(r))/(μ0)
I'm not too sure of the equations that I need to solve parts 2 and 3 of the problem. Any help is greatly appreciated.
 
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Rider4 said:

Homework Statement


A thin cylinder of radius r, thickness s, and length L, is made of metal with a resistivity of ρ. The cylinder is resting in a uniform magnetic field B0, with the field being in the same direction as the axis of the cylinder. The cylinder is then removed from the field. Find the current I0 that would be flowing initially, then find the rate at which energy would be dissipated by the resistance of the cylinder to the current flowing around it, and finally, find how long it would take at that rate for the magnetic energy in the cylinder to be dissipated.


Homework Equations





The Attempt at a Solution


I used B0=(μ0I0)/(2pi(r))
Then I solved for I0=((B0)(2pi(r))/(μ0)
I'm not too sure of the equations that I need to solve parts 2 and 3 of the problem. Any help is greatly appreciated.

You're trying to use an equation that relates the B field due to a current i (aka Ampere's law). Here the B field is not generated by the current you're looking for. The current you're looking for is the result of the external B field and the motion imparted to the cylinder as it's being removed from the B field.

Think Faraday instead.