# Two long parallel conductors flow in one direction current I = 6 A in each. The conductors were removed from each

Two long parallel conductors flow in one direction current I = 6 A in each. The conductors were removed from each other so that the distance between them became twice the original. What work per unit length of wires did the Ampere forces do?

Decision:
It is necessary to use this formula F = u0 * I1 * I2 / 2nR, since it shows the flow of the interaction force of currents flowing through two parallel conductors.
u0 is a constant value that is 4n * 10 ^ -7 H / A ^ 2.
We also need a nozzle, according to which we can find work to move one conductor with a current in a magnetic field that is created by another conductor.
A = integral R2 (top) R1 (bottom) of (u0 * I1 * I2 / 2nR) * dR.
From this we get A = (u0 * I1 * I2 / 2п) * ln R2 / R1.
It also turns out that I1 = I2 = I and R2 = 2R1.
We transform a little and get A = u0 * I ^ 2 / 2n * ln2.
Substitute all the values: A = 4n * 10 ^ -7 * 6 ^ 2 / 2n * ln2 = 50 * 10 ^ -7 J.
It should be taken into account that currents in conductors flow in one direction, which means that the conductors are attracted and the work is negative.

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