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Ch 27: Magnetic Field and Magnetic Forces

Chapter 27, Problem 27

A straight, 2.5-m wire carries a typical household current of 1.5 A (in one direction) at a location where the earth's magnetic field is 0.55 gauss from south to north. Find the magnitude and direction of the force that our planet's magnetic field exerts on this wire if it is oriented so that the current in it is running (a) from west to east, (b) vertically upward, (c) from north to south. (d) Is the magnetic force ever large enough to cause significant effects under normal household conditions?

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Hey, everyone. So this problem is working with magnetic forces. Let's see what they're giving us. And what they're asking us, we have a one m long conductor, it's transporting a given current and it's brought between hem holds coils. We have a magnetic field that is uniform directed along the positive y axis and has a given magnitude. We were asked to calculate the force exerted by the coil if the current is flowing in a number of different directions. So there's three parts, four parts of this problem, three parts where the current is in a different direction. And then the fourth part is asking if the magnetic force is large enough to deform the rod. So the first thing we can do is recall that the magnetic force on a current carrying wire is given by the force vector equals I the current times the length across the magnetic field. And so the current is going to stay the same for parts 12 and three, that's 1.5 amps. The magnetic field is going to stay the same as well. That's 80 micro tesla, I'm going to rewrite that as eight times 10 to the minus five Tesla. And we were told that in the positive Y direction, so that's going to be in the J direction or the J component. And then our lengths are what's changing for each of these problems. So L one we are told is along the positive X direction. So the magnitude isn't gonna change, that's still going to be one m. But for one, we are in the positive X direction. So that's I 42, we are in the positive Z direction that's K. And then for three, we are in the negative Y direction. Some negative one shame. And so all we have to do now is solve this equation three times to get our answers. So let's take L 1/1. So we have Equals 1.5 Amps times one m. I cross eight times 10 to the minus five Tesla in the positive J direction. Solve that out and we get 1.2 times 10 to the minus four newtons in the positive K direction. So if we look at our potential answers, we, if we're very confident, we can just choose B because that is the only one that has the correct answer for part one. But let's keep working through each of them, make sure that we're on the right track. So this was for one, 42, we have again, the same current, our L two is one m in the Z direction positive Z direction. And then the same magnetic field magnitude and direction eight times 10 to the -5 was We solved that out and we get negative 1.2 times 10 to the -4 newtons in the eye direction. And so we look up b and yes, our answer for number two also aligns. So I think we're on the right track for part three, we have again, the same current and then our length, our current, the direction of our current is traveling in the negative J direction negative Y. So when we cross that with our magnetic field, We are left with zero. So the whole term goes to zero and of course is zero. And that also aligns with choice B. So we're looking good. And then the last part of the question is if the magnetic force would deform the rod. And so the answer to that is no, we're dealing with very, very small forces here 1.2 times 10 to the minus four newtons. And that is not large enough to deform the rod. So that is the answer to this problem. That's all we have for you on this one. We'll see you in the next video.