length of horizontal wire = 1m ; current ( left to right ) =48A
second wire length =0.5m ; mass=0.5Kg ; height from 1st wire =0.30 m
(a) direction of current in second suspended wire = From Right to left ( because conductors carrying current in opposite directions repel each other)
(b) field at upper wire due to bottom wire
i = current in first wire r = distance of second wire from first
= permeability constant = 1.26 X 10^(-6) henry/m
B = 1.26 X 10^(-6) H/m X 48 A / 2 X 3.14 X 0.3 m = 32.1
(c) weight of the top wire that must be balanced = m g = 0.5 Kg X 9.8m/s^2 = 4.9 N
(d) Current in top wire to balance weight = ?
F = 4.9 N ( to balance wight of top wire) is required to be created at top wire due to bottom wire. This should be repulsive force
where l = length of 2nd wire at top
= current in wire at bottom , = current in top wire d= distance between the wires
F = 1.26 X 10^(-6) H/m X 0.5 m X 48 A X /(2) = 4.9N
= 2 X 3.14 X 0.3 m X 49 N /( 1.26 X 10^(-6) X ).5 X 48 A) = 3.0 X 10^(6) A
A horizontal wire of length 1.0 m carries a current of 48 A to the right....
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