Sal derives the formula F=ILB to determine the force on a current carrying wire. Created by Sal Khan.
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- what if in f=BIL,length of the conductor is infinite .(6 votes)
- Ah, good question
If you look at diagrams of this situation you will see that the conductor may be infinite in length but the only part of the conductor that we are interested in is the length L which is only equal to the length of the conductor which is INSIDE the magnetic field.
- will there be deflection in the wire due to the force applied on it.(10 votes)
- Yes. In fact there are electric motors that use this deflection to turn.
- I'm confused. Wouldn't the force point downward because current is the movement of elections... So we would have a negative current move right magnetic field going in to page and force down. OR in cases like this do we just always assume that the direction of current is positive charges moving in the currents direction?(4 votes)
- Current is defined as the direction positive charges are moving. So if you have electrons moving in some direction, the current will be moving in the opposite direction. The direction of the current in all formulas is always the direction of positive charge or opposite direction of negative charge.(5 votes)
- What exactly denotes time as a scalar quantity? How do you look at an equation and know exactly what is a vector quantity and what is scalar, other than memory?(2 votes)
- Scalar quantity is the quantity which has only magnitude but no direction. Vectors are those which has both magnitude as well as direction.(5 votes)
- If the formula (ilb) applies, then a particle never experiences a force because its length is zero! if its a formula for a conductor of finite length, then how could we derive a formula for a conductor carrying current from a formula that applies for a particle?(3 votes)
- F = IL x B right?
But I = q/t
and L = v*t where v is the velocity of the particle
Substitute those in and you will get
F = qv x B for a particle(2 votes)
- I am a little confused about the right hand rule as well. On video "magnetism 3", Sal drew a hand with the palm facing in while on this video the hand's palm was facing out. If Sal had drawn a hand with the palm facing in for this video, the direction would be different. How do I know which way to face my hand palm when applying the right hand rule?(2 votes)
- When applying the right hand rule to a F = q(v x B) equation, start by pointing your fingers in the direction of the particle's path (v), and have your palm face the direction of the magnetic field (B). This orientation should give you the proper direction of the Force when you give the thumbs-up sign.
Does this help?(4 votes)
- If you were to move a wire through an electric field, would it produce a greater force than a stationary wire?(3 votes)
- It should, because even if we are dealing with current here there'll be opposing forces relative to the velocity, I don't know it we can somehow relate it to the electric flux or something.(1 vote)
- what is the direction of drift velocity in current carryin wire?(2 votes)
- So, there's a 10N force pushing/pulling that 2m piece of wire 90 Degrees from both the direction of the 5a current and the magnetic field vector. Why doesn't the wire just fly off in that direction, after all, 10N would accelerate 10 kg 1m/s/s ?(2 votes)
- Could someone demonstrate how the units in F=ILB make sense?(1 vote)
Let's explore the repercussions of this equation some more. So what was the equation? It was that the force of a magnetic field on a moving charged particle is equal to the charge-- that's not what I wanted to do-- is equal to the charge of the particle-- and that's just a scalar quantity-- times the velocity-- the cross product of the velocity of the particle-- with the magnetic field. Now, isn't the velocity vector just the same thing as the distance vector divided by time? So the velocity vector is equal to-- let's call the distance that the electron travels, l. Distance divided by time. So we could rewrite that, that the force vector is equal to the charge times-- and I'm doing this on purpose-- 1 over time, right? Times the distance vector taken-- you take the cross product with the magnetic field. All I did is I rewrote velocity as per time times distance, or distance per time. And this is a scalar quantity, at least for our purposes, time only has a magnitude. Maybe we could call it change in time. But it doesn't have a direction. We're not going at an angle in time. So we could take the scalar quantity out. It doesn't affect this vector cross product. So what we get left with is, force is equal to charge per time times-- and this is just a regular times, because this is just a number, it's not a vector-- times the cross product of the distance vector and the magnetic field. And what is charge per time? Coulombs per second? Well, that's just current. So we get that force is equal to current times the distance that the current is flowing along, taken-- and you take the cross product of that with the magnetic field. And sometimes this is written as a capital L because it's a vector and all that, but we started with a lower case l, so we'll stay with the lower case l. So let's see if we can apply this formula, which is really the same thing as this. We just took the division by time and took it out of velocity so we get distance. And we took it and we divided the coulombs, or we took the charge divided by that. So we took charge divided by time, or charge per unit of time, you get current. So this is really just another way of writing this. It's not even a new formula. You could almost prove it to yourself if you ever forget it. But let's see if we can use this to figure out the effect that a magnetic field has on a current carrying wire. So let me-- actually, I probably want to put this up at the top, just so that I have space to draw a current carrying wire. So let me rewrite it in green. So you're familiar with the formula in all colors. So now our new derivation is that the force of a magnetic field on a current carrying wire is equal to the current in the wire-- and that's just a scalar quantity, although it could be positive or negative depending on the direction. Well, current is always a positive number, but if this current is going in the opposite direction as our distance vector, then it might be negative. But I wouldn't worry about that for now. Let's just assume this is a current in the direction of the distance vector. So it's a scalar quantity current times our distance vector l, or maybe the length of the conductor. You take the cross product of l with the magnetic field vector. So let's see if we can apply that. Let's say that we have a wire. Actually, let's do the magnetic field first. I've been doing a lot of magnetic fields that pop out of the screen. Let's do a magnetic field that goes into the screen. And those are even easier to draw. They're just x's. Now, why is it an x? Because you're looking at the rear end of an arrow. That's why it's an x. And that's why a circle with a dot means a field or a vector coming out of the window. Because if an arrow was shot at you, all you would see is the tip of the arrow with maybe a little circle around it. But anyway, this shows us a vector going into the screen. So this is our magnetic field. That is B. I don't know, let's assign some value. Let's say that the magnitude of B is equal to 1 tesla. And let's say I have a wire going through that magnetic field. Let's say the wire is going along or it's in the plane of your computer monitor. Let me just draw a wire going through the magnetic field. And my question to you-- let me tell you a little bit of information about this wire. Let's say the wire is carrying a current. So I is going in that direction. And it is carrying a current of-- I'm just making up numbers-- 5 amperes, or 5 coulombs per second. My question to you is, what is the net force of this magnetic field on a section of this wire? And let's make this section of the wire, I don't know, let's say it's a 2 meter section of wire. So obviously the more wire you have, the more charged moving particles you'll have. So the larger a section you have, the more of a force you'll have on that longer piece of wire. So we have to pick our length. So we want to know, what is the force of the magnetic field on this section of wire? From here to here. So let's just go to this formula. The force is equal to the current. So that's 5 amperes. And remember, just from what we learned about electricity, the current is the direction that notional positive charges would travel in, and suits us fine. Because when we did the first equation, we cared about the direction a positive charge would go in. And if it was an electron or a negative charge, we would put a negative sign there. So that works fine. But if you ever have to visualize things as they maybe are in reality, but when you talk about electrons it's hard to say that they really are reality, because they're almost more an idea than an object. But it's always good to remember that when the current is flowing in this direction, that would be true. Because if they were positive charges moving, but we know it's a negative charge moving in the opposite direction. Or you can think of it as, maybe, holes. Well, I don't want to get into that. But anyway, the current-- you could visualize it if you like as positive charges going in this direction. So the current is going this direction. So you could view this distance vector that we care about. Its magnitude is 2 meters. Because that's the length of wire in question. And its direction is the direction of the current. So let me-- this is l. Sometimes I get a little carried away on tangents. So that is l. It's 2 meters in that direction. I is 5 amperes. And we already figured out that the magnetic field is 1 tesla. So what's this going to be equal to? So the force is going to be equal to-- we're using all SI units, so we don't have to convert anything-- 5 amperes times 2 meters in that direction. I won't specify right now, let's just say that's a magnitude of l. Actually, let me write it. Well, 2 meters times the magnetic field, 1 tesla. And so when you take a cross product of something, this is just a reminder. l cross B. That's equal to the magnitude of l times the magnitude of B times the sine of the angle between them times some unit directional vector that we figure out with the right hand rule. So we already did the magnitude of the distance vector. That was 2 meters. We did the magnitude of the magnetic field. And what's the sine of the angle between them? Well, if the magnetic field is going into the screen, if it's going straight into the screen, you could imagine a bunch of arrows shooting into the screen. Those are the vectors. While our distance vector, or this l is in the screen, they actually are perpendicular, in 3 dimensions. So this angle is 90 degrees. So this actually just becomes 1. So in terms of the magnitude, we're done. The l cross B magnitude is 2 times 1 tesla. And then we multiply that times the current. And then we actually have the magnitude of the force. The magnitude of this force is going to be equal to 5 amperes times 2 meters times 1 tesla. Which is equal to 10 newtons. And then the only question left is, what is the direction of the force that the magnetic field is exerting? And this is where we break out the right hand rule. And it's no different. You could just imagine one of the positive particles moving in that direction, and just use the right hand rule. So let's take our hand out. And if we-- let me draw a hand. A right hand. So this is my right hand. If I have my thumb sticking out like that. So the l is going to be my index finger. The first thing in the cross product. And then the B is the magnetic field. That's going into the screen. So you can't see it. All you can take my word for it is that my middle finger is pointed downwards into the screen and then my other fingers are just doing something else. And there you have it. Your thumb is actually the direction of the force. Your index finger is the direction of-- we'll say l for these purposes. And then the magnetic field is going into it, so you can't see my middle finger but it's pointing downwards. I could draw a little x there, to show it's going downwards. And then the force is what my thumb is doing. So the force on this wire, or at least on that section of wire, is going to be perpendicular to the direction of the current. And that direction is going to be a 10 newton force. Anyway, I've run out of time.