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.