OpenStax College Physics, Chapter 22, Problem 52 (Problems & Exercises)

This is College Physics Answers with Shaun Dychko. A 2.50 meter segment of wire is carrying 1000 amps and it experiences a force of 4.00 newtons and it's a repulsive force we are told from this other wire which is carrying some unknown current I 2 that we want to figure out and the separation between these two wires is 5.00 centimeters, which is 5.00 times 10 to the minus 2 meters and... yeah! So to figure out the direction of this current, we will put our thumb in the direction of the current I 1 and our palm in the direction of the force that it experiences, which is repulsive and that's up the page here, and that makes our fingers point into the page and so that's how we know that the magnetic field is pointing into the page because that's the magnetic field direction needed to cause a force upwards on this wire with the current going to the right. And to have a magnetic field point in that direction, we grab this bottom wire with our fingers pointing into the page in order to indicate that direction at the position of this top wire and our thumb ends up pointing to the left and so that's how we know that's the direction for this current I 2. So the next question is what's the magnitude of the current in I 2? Well, the force between the wires is the length of wire times the permeability of free space times the two currents multiplied together divided by 2π times the separation between the wires. So we can solve this for I 2 by multiplying both sides by 2πr over lμ naughtI 1 and then I 2 is 2πrF over lμ naughtI 1. So that's 2π times 5.00 times 10 to the minus 2 meters times 4.00 newtons divided by 2.50 meters times 4π times 10 to the minus 7 tesla meters per amp times a 1000 amps and that is 400 amps in the opposite direction to whatever I 1 was.