Question

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

Final Answer

- $8.53\textrm{ N}, repulsive$
- This force is always repulsive so it doesn't create any risk for creating a short circuit. Also, it's small so there's no risk of breaking the wires.

### Solution video

# OpenStax College Physics, Chapter 22, Problem 50 (Problems & Exercises)

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Video Transcript

This is College Physics Answers with Shaun Dychko. There are two wires that bring current to a commuter train and one of them is called the hot wire because it has high voltage and the other one is neutral, which means it's basically ground but maybe not quite but it's really, you know, basically zero volts. So the current will go down the hot wire into the motor of the train and then back through the neutral wire. So the question is what force is there between these wires and we have to figure out the direction and also the magnitude of the force? So the wires carry 800 amps and so the current in the one wire has to be the same as the current in the other wire because they are part of the same circuit; there's, you know, a train here and then there's a generator over here. So yeah... we have the same current going in both wires so

*I 1*and*I 2*are the same— they are both 800 amps— the separation between the wires is 75.0 centimeters, which is 75.0 times 10 to the minus 2 meters, and the length of wire that we want to consider is 50.0 meters. So a 50.0 meter portion of the wire is the portion we consider to find the size of the force. So the force would be this permeability of free space times the two currents multiplied together times the length of the wire divided by 2*π*times the separation between the wires. So that's 4*π*times 10 to the minus 7 tesla meters per amp times 800 amps squared times 50.0 meters length divided by 2*π*times 75.0 times 10 to the minus 2 meters— separation between the wires— and that is 8.53 newtons. The next question is what is the direction? So I have drawn on here this in green, I have drawn the magnetic field due to current*I 2*, which is also green so I am having these colors correspond... you know, this magnetic field is green due to this current that I have labeled is green. And if you grab this bottom wire with your right hand with your thumb pointing left in the direction of current, your fingers will be pointing into the page at the position of this top wire and so that's how we know the magnetic field is going into the page there. And likewise if you grab the top wire with your right hand, thumb pointing to the right at this position of the bottom wire, your thumbs will be pointing into the page so likewise the magnetic field due to the top wire is going into the page here. So then we ask well what force would this green magnetic field be applying on this blue current? So your fingers will point in the direction of the magnetic field, your thumb points in the direction of the current to the right and your palm will be facing upwards so this is a force on wire one due to wire two. And likewise down here, thumb pointing to the left, fingers into the page, your palm is pushing down and so that's why this is a direction for force on wire two due to wire one; the force is repulsive, in other words. So since the force is always repulsive, it doesn't create any risk of creating a short circuit because if the force was attractive, it would bring the wires together and then maybe cause them to contact and have a short circuit but that's not gonna happen since the force is repulsive given that the currents are in opposite directions and also this force is a small magnitude force so it's not as though it's gonna push them right apart and break them or anything... it's a very small force of only 8.53 newtons.