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Anderson Video - Magnetic Force Between Parallel Wires

Professor Anderson
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And what we need to know is something that was introduced yesterday, but let's say it again. Currents, themselves, produce B fields. Okay? B fields affect currents, but currents also produce B fields. So, if I think about a wire and it's carrying current I, there is a B field around that wire. And the B field around the wire looks like this. The dashed line indicates going behind the wire. The solid indicates coming out in front of it. And, we can figure out the direction of B based on the right-hand rule. Okay? So if you take your thumb and put it in the direction of I, then your fingers wrap around in the direction of B. Okay? So put your -- look at the computer monitor -- put your thumb in the direction of I, wrap your fingers around, that should be the direction of B. Okay? Now, if that thing is making a circle, then the B field has to have the same strength at any radius from the line. And, we know what that strength is. We talked about it yesterday. The strength of B is this: mu naught I over 2 Pi r. The direction is determined from the right-hand rule, but it's a slightly different right-hand rule. It's a slightly modified right-hand rule, which is the following: Put your thumb in the direction of the current, and your fingers curl around in the direction of B. Okay? So my thumb is going up in the the direction of the current; my fingers are wrapping around in the the direction of B. Okay? We call this the right-hand rule for a reason. It works with your right hand; it doesn't work with your left hand. Okay? If you do your left hand, you will get the exact opposite of what you're intended. So, you have to use your right hand to do it. I'm, of course, using my left hand in reality but then when we flip it, on the monitor it looks like my right hand. Okay? And so, I'm finding this challenging over here to do this. Alright. What is this mu naught? Mu naught is the permeability of free space. Just like with the electric field, we had the permittivity of free space. Now, we have mu naught which is the permeability. And it's 4 pi times 10 to the minus 7. And, the funky units on it are tesla meter per amp. All right. So, that's the strength of the B field at any radius, at any distance from a current-carrying wire. All right. So maybe that'll help us because it's telling us we have a long current-carrying wire, and then there is some force per unit length on the second parallel wire. All right. so let's hang on to some of this. We're gonna say B equals mu naught I over 2 pi R And now, let's add a second wire and see if we can make some sense of it. I've got a B here that is going around as we indicated in that picture. Now we're going to take a second wire and run current through it. Now the question is, are those wires attracted together, or are they pushed apart? And let's think about this B field for a second at the point of this wire. Okay? This B field it's not just at that radius. It's at any radius you want to pick. And so, at this wire this B field is in fact doing what? It is going into the page. Okay? Going into the page. If it's going into the page, then there must be a force on this current carrying wire. And, the force has to be I L B sine-theta. And I determined that from the right-hand rule. So, I hold up my right hand and I put my finger straight in the direction of I. And then, the B field is going into the screen, [inaudible] which way? Is it towards the other wire or is it away? It's towards. There is a force pulling it towards the other wire. Okay? That's exactly what they said in the problem, right? They said an upward current exerts an attractive force on the second wire. Okay? And now, we need to figure out what the current is in the second wire. And what they tell us in the problem is the following: they give us the force per unit length. Force per unit length is: We can use my numbers -- 8.8 times 10 to the minus four Newtons per meter. They also tell us that the distance between the wires which is our R is 5 centimeters. So, 5 centimeters equals 0.05 meters And we need to figure out what this current is in this wire. Okay, what's the magnitude of the current in the second wire? So, to be clear, we better call these two different things. Let's call this one I1 and this one I2. And that means this down here is I1, and this B field here is B1. What is the B field due to the first wire at this location of the second wire? Okay. So, let's see if we can put this stuff together. What we know is force is ILB sine theta. That's the magnitude of the force. All right. Let's divide F over L -- equals I B sine theta. And now, we can start being specific about what currents and what magnetic fields we're talking about. If we're talking about the force on this second wire, then we're worrying about the current I2 in that wire. But, it feels a force due to the B field from the other one, so that should be B1. And, we know that sine of theta, let's see, theta is going to be 90 degrees, right? I is up. B is into the screen and so theta is going to be 90 degrees. And so now we can plug in for B1. B1 was mu naught I1 divided by 2 Pi r. We have sine of 90 degrees. But, that's just 1 and so we get: mu naught over 2 pi times I1 I2 all over 2 pi r, whoops. I already wrote the 2pi, just over r. Okay? And now, we can solve this equation for I2. Okay? That's what we're looking for. So let's move it over here and we'll solve it for I2. Alright. So we have: F over L equals mu naught over 2 pi, I1 I2, all over r. And we want to solve this for I2. So let's see. I got a multiply by 2 Pi r. I have to divide by mu naught. And, I have to divide by I1. And that should be my I2. And so now, we can calculate I2. I2 is F over L, which we're given. R we're given. I1 we're given. F over L was this: 8.8 times 10 to the minus 4. We have a 2 pi. We have R which is 0.05. And then, we have mu naught which we know is 4 pi times 10 to the minus 7 in SI units. And then, I1 for our problem is 22 amps. All right? So, somebody punch that into your calculator and tell me what you get. Let's see if we can approximate it. It's kind of fun to approximate these things. So, what do we got we got? We got the Pi's, those are going to drop out. That's nice. We got a 2 that drops out with that, if I give that a 2 right there. And so, we're gonna get 8.8. 0.05 is 1 over 20. Then, we have a 10 to the minus 4 and we're going to divide the whole thing by 2 times 22 which is 44. And then, we're going to end up with another 10 to the 7 up on top. So let's see. What do we get? 8.8 over 20. That is really close to 1/2. If I divide a half by 44, what do we get? That's 1 over 88, which is pretty close to 0.01 -- a little bit more. how about 0.01, eh let's take a guess, 2. And then, we've got a 10 to the minus 4 and a 10 to the 7 which we get a 10 to the 3. So, I'm gonna say that I2 is approximately 1, 2, 3. 12 amps. That's my guess. Anybody get a real number? Nobody got an answer yet? >> (student speaking) I got 10. >> 10? Okay. That's pretty close to 12. Anybody else get 10? 10? Okay. All right. So, our guess was a little off, but that's fine Let's try 10. 10 units are amps. Should we click submit? Let's click it. Correct. All right. I got one point on my homework, excellent. Okay. The part B of this is: What is the direction of the current in the second wire? If you remember our picture, we had current in the first wire. And, at the current in the second wire was up and it was an attractive force So, we're gonna say upward for the second part, part B. Submit. It's also correct. Okay? Good problem. All right. Why don't we wrap it for today and I will see you guys tomorrow online. Cheers!