A physics instructor performs a demonstration where a current-carrying wire is made to levitate. The wire...
A physics instructor performs a demonstration where a current-carrying wire is made to levitate. The wire has a mass per unit length of 0.590 g/cm and carries a current of 2.06 A horizontally to the south. Find the magnitude and direction of the minimum magnetic field needed to lift this wire vertically upward. (Enter the magnitude in T.)
Magnitude:?
Direction:?
Homework Answers
as we know
mg = Bi L
(m/L) g = B i
0.59* 10^-3 / 10^-2* 9.8 = B * 2.06
B = 0.2807 T
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field directed to the right
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Comment before rate in case any doubt, will reply for sure.. goodluck
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