No.
F= weight bottom wire=mg m=.01kg/meter
Your UoI1*I2/2pi*d is force per length, so set that equal to the weight of the bottom wire per length.
Two parallel wires carry currents I1=100A and I2 (they run in the same direction). The top wire is held in position; the bottom wire is prevented from moving sideways but can slide up and down without friction. if the wires have a mass of 10g/meter of length, calculate I2 such that the lower wire levitates with a position of 4cm below the top wire.
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would i use..
F=Uo*I1*I2*l/2pi*d=ma
and if so would a=0 and so Uo*I1*I2/2pi*d=0
..im really unsure with these concepts any guidance would be greatly appreciated!
1 answer
1 answer