Magnetic Force on Current-Carrying Wire - Online Tutor, Practice Problems & Exam Prep

Magnetic Force on Current-Carrying Wire

Hey guys. So in this video, we're going to talk about forces acting on current-carrying wires. Meaning if you have a wire and charge is moving through it, it has a current. And if that wire is inside of a magnetic field, it will feel a force. Let's check it out. Alright. So remember, charges can move in space and we've done this. If you have a little charge moving through space and it walks into a magnetic field, it will feel a magnetic force. Well, charges can also move inside of a wire and if you have charges inside of a wire, you have a current. You can think of a wire as simply a way to sort of restrict the path of charges so that they're not going in crazy directions. They have to move along the direction of the wire. But in a way, a wire is simply charges moving just the same. So remember, if charges are moving in space, they will be producing a new magnetic field. And we did this, the equation is μ₀qvsinθ/4πr2. Right? Just like how a charge moving freely through space will produce a new magnetic field away from itself, a charge moving in a wire, in other words, if you have a current-carrying wire, that is also going to produce a new magnetic field. Okay? And the equation for that is μ₀I/2πr. So it looks somewhat similar to this, but it's not exactly the same. We're not going to talk about this now, we're going to talk about this later but I want to make the point that they produce a charge, they produce a field away, the wire will produce a field away from itself. Remember, that charge does not only produce a field away from themselves, but they're also going to feel a force if they are in the presence of an existing magnetic field. And this is a really important theme of magnetism is that moving charges produce a field and if they are in an existing field, they will also feel a force. And just like how this happens for charges in space, this is also going to happen for charges moving in a wire. Okay? So if a charge is moving through space, it will feel a force given by qvBsinθ. Right? So for example, if you have a magnetic field this way and a charge moving this way, you can use your right hand rule and figure out that the force will be into the field. Okay. Similarly, if a charge is moving not through free space but it's moving inside of a wire, it is a current and that wire will feel a force as well. So, charge moving in an existing field feels a force and a wire in an existing field also feels a force because it's the same thing. Just and here it has to be moving and here it has to have a current. So it's a current-carrying wire, right? Just like here, current-carrying wire, and that force is going to be given by BILsinθ. K? And this is the big equation that we're going to talk about. I wanted to sort of do a big summary of all the different situations, but this is what we're going to talk about for the next few videos. Now most textbooks and professors write this equation in a different order. I like to write it this way just because there's so much crap to remember in this chapter. This spells Bill, which is easier to remember than some other combination. Okay? So ⇒F=BILsinθ. I has a direction which is the direction of the wire and B has a direction and theta is the angle between those two directions. The directions of the forces will be given always by the right-hand rule. Now remember, there's sort of this exception that if you have a negative charge if you have a negative charge that's moving freely, then you're going to use the left-hand rule. But in currents, because of sign convention, you are always going to use the right-hand rule every time. When you have a current, you don't have a negative current, right? So you're always going to use the right-hand rule no matter what. Alright. So the last thing we want to talk about here is that this force will cause the wire to bend slightly. Okay? So it'll cause the wire to bend slightly. So let's say here we have a wire here and the current is up. And here we have a wire between these two magnets and the current is moving down. Now you may remember that magnetic fields go from positive to negative on the outside. So positive to negative so it looks like this. But more importantly, we want to know what is the field around here. Well, it goes from north to south, from north to south. So it's going to look like this. So, as far as the wire is concerned, you have B field lines going this way or I guess this way. Okay? So what is the direction of the magnetic force that this wire will experience? Well, right-hand rule, right and guess what current is the direction of the motion so it's going to be your thumb and your B is going to be your fingers so here I have my B is going to go to the right and I want my current to go up. So this is already in the direction that I want, right? This matches the picture. So if I do this, notice that my palm is away from me, right? So please do this. It's away from you which means that it's into the page. So this force and the force on the wire will be into the page, everywhere the wire is being pushed into the page. Okay, we can't calculate this because it's with magnets but we're just talking about the direction. Here, you also have north to south so the magnetic field will look the same. And what do you think will be the direction of the force? Well hopefully you're thinking, well, if we're flipping the current, it must flip the force and that's correct. So the right-hand rule again, this is going to the right but now my currents down so the only way that I can keep these going to the right with my current going down is if I do this, right, and if I do this and you sort of bring it close to your face, right? This is weird but hopefully you see that your palm is facing you which means it's coming out of the page. So ⇒FB is coming out of the page. So ⇒FB is out of the page. So that's the direction of the force. Let's do an example. Here, there are two parts. So it says here a 2-meter-long wire, that's the length of the wire L equals 2 is passed through a constant magnetic field as shown. So there's a little piece of wire here and you might be wondering how's that current on a little piece of wire that's floating? Don't worry about that for now. I'll talk about that a little bit later. Just assume that this is possible, which it is. Okay? So there's a wire there. Somehow it's connected to a battery, but don't worry about it. So the piece of the wire there is 2 meters. It sits through a constant magnetic field. This magnetic field is going into the page by and you can tell by the little x's. And we want to know for part a, if the wire experiences a force of 3 newtons when it has a current of 4, what is the strength of the wire? What is the strength of the field? So what is B? So is there an equation that ties these guys together so that you can find B? And it's the equation we talked about earlier, F=BILsinθ. K. We're looking for B. B equals F/ILsinθ. Remember, the angle is the theta is the angle between the B and the I. I is either to the right or to the left. We actually don't know but it's either one of those. And the direction current is going this way or this way, right, this way or this way and the field is going into the page so they make 90 degrees with each other it's either this which is 90 degrees right here or it's this, right, which is also 90 degrees. So this angle here is 90 degrees. And by the way, generally speaking, if you don't see like a weird angle, it's probably 90, right? So let's plug these numbers in. F is 3, I is 4, L is 2. This is 3 over 8 and that is 0.375 Tesla. That's part a. Part b has to do with the direction. It says if the wire experiences a downward force, so if the wire experiences a downward force, what must the direction of the current be? So here's a wire and it's experiencing a force ⇒FB that's down. Right-hand rule, the force is down. So I want my palm to be down and I want my forefingers for the field to be going into the page, which into the page is away from you. Right? So I want my fingers like this and I want this force down. Notice what happens with my thumb which is it's going left and that's it. That tells me that the currents must be going left. So the direction of the current is left. That's all there is to this. Hopefully, it makes sense. Let's keep going.

Find Force on Current-Carrying Wire at an Angle

Hey guys. So in this example, we have a wire that sits on a magnetic field and it has a current. We want to know what the force on that wire is in 3 different scenarios. So let's see. The wire has a length of 2 meters, \( l = 2 \). The magnetic field strength is 3, \( b = 3 \), and it's directed in the negative y-axis; the magnetic field. We will determine the magnitude of the force, which in this case because it's in a wire, is going to be given by \( F = BIl \sin(\theta) \). By the way, we're also given that the current is 4, okay. It doesn't matter if it's a small 'i' or a big 'I', the current is 4 amps. That's a really ugly 4; four amps. Cool. So let's get to it.

The magnetic field, I like to draw the wire, and the current rather, is flowing in the negative y-axis, which means that the wire goes down this way, and then the current is going down in this direction here. So the equation is \( F = BIl \sin(\theta) \). And in these questions, the tricky part is the angle because I'm given \( b \), \( i \), and \( l \). So it's just plug and chug. Now the angle is going to be the angle between \( b \) and \( i \). They're both going down, so they're parallel to each other. So the angle is \( 0 \), and the \( \sin(0) \) is \( 0 \). The \( \sin(0) \), which means there is no force here because they're going parallel to each other. Remember, in currents, we use the right-hand rule always, and this is a reminder that you're supposed to have a \( 90^\circ \) angle between your \( b \) and your \( i \), right? A \( 90^\circ \) angle between your \( b \) and your \( i \). And here they're like that and that's not how it's supposed to be. In this situation, you have the maximum force. If you have an angle like this, you have less than maximum force, but you still get some force. And as you keep going this way, the force goes from max, max, at \( 90^\circ \) to smaller smaller smaller until you get here and you get \( 0 \). Okay? So if the angle is \( 0 \) because they're parallel or if the angle is \( 180^\circ \), which is antiparallel, opposite directions, there will be no force. Cool?

So no force on this one here. What about here? So \( B \) again is down, and the current is in the positive x-axis, which means the wire is horizontal, and the current is going in this direction. The angle between these two guys is \( 90^\circ \). So \( F_b \) is \( BIl \sin(90) \). \( \sin(90) \) is just \( 1 \). So really we only have \( BIl \). So \( 3 \times 4 \times 2 = 24 \) newtons. Very easy.

What about the direction of the force? Well, the right-hand rule because it's a wire. \( B \) is going down. \( I \) is going to the right. So you gotta do this. Okay? Away from me, \( B \) is down. So my palm, even though it's pointing at your face, you gotta do this yourself. Right? It's not my palm, it's your palm. Your palm is pointing away from you, which means it's going towards your page. So it's into the plane. Or into the page. Make sure you master your right-hand rule. Into the page is the direction, and the magnitude is 24.

Okay. So this one's a little bit more complicated because it's got an angle. And here we have \( B \) this way. And we have a wire in a direction that makes \( 53^\circ \) with the y-axis. So here is the positive y-axis up here. This makes \( 53^\circ \). Now this is a little tricky, because it's ambiguous. It's not totally clear whether it's \( 53^\circ \) with the y-axis this way or \( 53^\circ \) with the y-axis this way. But what you will see is that it actually doesn't matter when it comes to calculating this. Okay? So we're going to think of this as two possibilities that the current could be going this way or it could be going that way. And then we're going to calculate that. So the equation is \( F_B = BIl \sin(\theta) \). And the angle is the angle between the direction of the current and the direction of \( b \). So \( b \) is pointing down. So if you want, what you can do is you can draw \( b \) over here. Okay?

And what is the difference between these guys? So if you go counterclockwise here, this is \( 90^\circ \). And then this here is \( 37^\circ \). So \( 37 + 90 = 127^\circ \). \( 127^\circ \). Or you can go another, or you can go in this direction here, which would be negative, negative \( 127^\circ \). Or you can go all the way positive, and you can say that it's \( 90 + 90 + 53 = 233^\circ \). \( 180 + 53 = 233^\circ \). Did I get that right? Yeah. \( 233^\circ \) away from this. And the \( \sin \) of all these numbers will be the same. They may have different signs. One might be positive or negative. But here, we're just looking for the magnitude of this thing. So you can pick your choosing. I'm going to write that \( b = 3 \), \( I = 4 \), \( l = 2 \), \( \sin(127^\circ) \), positive. Again, whatever we get here, just think of this as an absolute value because we're just looking for the magnitude of this force. And if you do this, you get that the answer is 19.2 newtons. Okay?

So that's the force, and it would work whether you're going this way or this way. What about the direction? Right-hand rule, \( B \) is down. So let's do this. And I want this guy to be going in this direction, okay? I want this guy to be going this direction. So in the case here where I won, right? If you were going that way, you'd be going down, and you would have the \( I \) like this. Right? Which means my palm is pointing towards me, which means it's away from the page. So, for \( I_1 \), if it was going in that direction, and by the way, a question, a question on your test would tell you exactly which one, right? I just wanted to talk about both cases here. So in this case, you would be what did we say it was? We said it was, out of the plane. So out of the page. What about for \( I_2 \)? What if you were actually talking about this direction? Well, if you try to do \( B \) down and this \( I \) here, you can't put this thumb all the way over here, right? Like not without breaking it. And don't do that because now you're just messing yourself up, and you're breaking the right-hand rule anyway, right? So what can we do? Well, you have to do this, right? You have to do this, where now your fingers are still down following \( B \), and this guy is up like that, so this looks all kinds of weird. But my palm is away from me, which means it's going into into the page. So in this case, the direction of those two wires actually made a difference in terms of direction. It does make a difference for the direction, though not for the magnitude. Cool? Let's take a look at this one. Let's keep going.

A 5-m current-carrying wire (red line) is ran through a 4 T magnetic field (blue lines), as shown. The angle shown is 30°. What must the magnitude and direction of the current in the wire be when it feels a 3 N force directed into the page?

• A. 0.17 A @ 30° below the +x-axis
• B. 0.17 A @ 30° above the −x-axis
• C. 0.30 A @ 30° below the +x-axis
• D. 0.30 A @ 30° above the −x-axis
• E. 0.49 A @ 30° below the +x-axis
• F. 0.49 A @ 30° above the −x-axis
• G. 3.25 A @ 30° below the +x-axis
• H. 3.25 A @ 30° above the −x-axis