Asked by Raven
A thin copper rod 1m long has a mass of 50g. What should be the direction of magnetic field and the maximum current in the rod that would allow it to levitate above the ground in a magnetic field of 0.1T.
Answers
Answered by
Shegs
F = IBL
F = mg
mg = IBL
I = mg / BL
( (0.05 kg)(9.8 m/s^2) / (0.1 T)(1 m) )
= 4.9 A
F = mg
mg = IBL
I = mg / BL
( (0.05 kg)(9.8 m/s^2) / (0.1 T)(1 m) )
= 4.9 A
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