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22. 10 Magnetic Force between Two ParallelConductors(a) The hot and neutral wires supplying DC power to a light-rail commuter train carry 800 A and are separated by 75.0 $\mathrm{cm} .$ What is the magnitude and direction of the force between 50.0 $\mathrm{m}$ of these wires? (b) Discuss the practical consequences of this force, if any.

Part $(a) :$ The Force between the two wires is 8.53 $\mathrm{N}$Part $(\mathrm{b}) :$ As the force is repulsive, this will prevent the two wires from touching each other and creating a short circuit.

Physics 102 Electricity and Magnetism

Chapter 22

Magnetism

Magnetic Field and Magnetic Forces

Sources of Magnetic field

Electromagnetic Induction

Inductance

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University of Michigan - Ann Arbor

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the force per unit link between two parallel conductors would be equaling you, not the magnetic permeability in a vacuum or the readings of the permeability of free space times the current through the first conductor times the current through the second conductor divided by two pi r Again, this is only for two parallel conductors. And so we can say that we're going to rewrite the rewrite, this formula to find f This would be mu not ice of one ice of two times the length divided by again two pi R and so we can solve. This would be equaling four pi times 10 to the negative seventh Tesla's meters per amp. Multiplied bye 800 amps. Both of these conductors have 800 amps, so we can say 800 amps squared, multiplied by 50 meters, divided by two pi times the radius of 0.750 meters and the force is equaling 8.53 Newtons. And here because the currents are in the opposite direction, this would be repulsive. So this would be repulsive force. This would be our answer for part a four part B. The question is essentially asking us what would be the practical consequences in this case, Um, currents are in the opposite direction. Force is repulsive and nature. And so, uh, no risk of the wires touching and shorting the circuit. That is the end of the solution. Thank you for

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