A long horizontal wire carries a current of = 52 . A second wire, made of 1.00--diameter copper wire and parallel to the first, is kept in suspension magnetically 5.0 below.What is the magnitude of the current in the lower wire?
The lower wire is kept in suspension magnetically means that the downward force of gravity is equal to the upward magnetic force. Its NOT an easy problem so we need to solve it in small steps. Let's find the two forces acting on the lower wire: (*) Downward force of gravity = mg = (volume) x (density) x (gravitational acceleration) = (pi r^2 L) x(8920)x(9.8) = 0.06896L We assume the radius of the wire is 0.5 mm or 0.0005m, copper density is 8960 N/m^3, gravitational acceleration = 9.8 m/s^2 (**) Upward magnetic force = mu(zero) x i(one) x i(two) x L / (2 pi d) = 2.08x10^(-4) L i(two) mu(zero) is a constant = 4 pi 10^(-7), i(one) is the current in the first wire = 52A, i(two) is the unknown current in the second wire, d is the distance between the wires = 5 mm or 0.005m The forces are equal, therefore 0.06896L = 2.08x10^(-4) L i(two) We divide both sides by L (the unknown length of the lower wire) and now we can solve for the current i(two) ANSWER: i(two) = 331.5 A
There is 1V across ampere meter. To find the current.... Current = 1V / meter internal resistance
