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Question
A piece of straight wire has mass 20 g and length 1 m. It is to be levitated using a current of 1 A flowing through it and a perpendicular magnetic field B in a horizontal direction. What must be the magnitude B of the magnetic field?
Solution
Data: m = 20 g = 2 x 10-2 kg, `l` = 1 m, I = 1 A, g = 9.8 m/s2
To balance the wire, the magnitude of the upward magnetic force must be equal to the magnitude of the downward gravitational force.
∴ Fm = `"I"l`B = mg
Therefore, the magnitude of the magnetic field,
B = `"mg"/("I"l) = ((2 xx 10^-2)(9.8))/((1)(1))` = 0.196 T
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