Force on a Current-Carrying Conductor

Transcript

0:00Introduction

Hello everyone! Okay, a small puzzle before we start. Take two wires with the same two ends. One is straight. The other is bent into a long, wild zigzag, many times longer. Both carry the same current, in the same uniform magnetic field. Now tell me, which wire feels the bigger force? The long zigzag one, obviously, right? More wire, more force. Today we will find out if that is really true. The force on a current-carrying wire comes up often in JEE problems. So, let's begin.

0:36From moving charges to a wire

So, a current is just a stream of moving charges, right? And each moving charge q, with velocity v, feels the Lorentz force, q v cross B, where B is the magnetic field. The small arrows on top mark vectors, and the cross sign is the cross product. Okay? Now think. Millions of charges move inside the wire, and every one feels this force. So what force does the whole wire feel? The sum of all these little forces, exactly! For a straight wire of length L, carrying a current I in a uniform field, F vector equals I times L vector, cross B vector. And L vector points along the current. Clear? Its size is F equals I times L times B, times sine theta. Theta, the Greek letter, is the angle between the wire and the field. With me? So let me ask you this: at what angle is the force the largest, and when is it zero? At ninety degrees, and the small circle on screen means degrees, sine theta is one, so the force is the largest, F equals I times L times B, okay? And when the wire runs along the field, theta is zero, sine zero is zero, so there is no force at all. The same holds at one hundred eighty degrees, with the current exactly opposite to the field. Got it? And for a tiny piece of the wire, the small d l, the force is d F equals I times d l, cross B, okay? The small d just means a tiny bit. We need this form when the wire is not straight. Clear?

2:09Any shape, same two ends, same force

Now, back to our puzzle. Take a uniform field, and a wire of any shape at all, going from a to b. Zigzag or straight, which one feels more force? Neither of them, they feel exactly the same force! F vector equals I times a b vector, cross B, where the arrow over a b means the straight arrow from a to b. The reason is simple: add all the little d l pieces head to tail, and they give the arrow from a to b. The field is the same everywhere, so it comes out common. So all that extra zigzag length does not matter at all. Right? Now, what if the wire comes back to where it started, a closed loop? What is the arrow from a to b now? It is zero, yes! The end point is the same as the start point, so a b vector is the zero vector, the zero with an arrow on top. So a closed loop in a uniform field feels no net force. Okay? So a current loop in a uniform field just sits there, still, right? Oh no, careful, this is the trap, because no net force only means the loop does not move off as a whole. It may still feel a torque, a turning effect, and turn. That torque comes later in this chapter. Got it? Now, all this was for a uniform field. In a non-uniform field, a loop can feel a net force, for example a loop near a long straight wire, or in the diverging field of a magnet. Clear? And one last fact. Any current-carrying loop in a field tends to change its shape so that it encloses as much magnetic flux as possible. So a flexible loop, like a loose ring of wire, becomes a circle. With me?

3:53Example 1: 50 cm, 4 A wire in 0.2 T at 90 and 30 degrees

Okay, time for the examples. Read this one with me. Take a straight wire, fifty centimetres long, carrying four amperes, perpendicular to a field of zero point two tesla. Find the force on it, and the force if the wire makes thirty degrees with the field. On screen, c m is centimetres, capital A is amperes, and capital T is tesla. Pause the video and try it first. Done? Okay, let's check. F equals I times L times B, so four times fifty times zero point two, which gives forty newtons, right? Oh no, careful, that is the classic trap, because L must be in metres. Fifty centimetres is zero point five metre. And the wire is perpendicular to the field, so theta is ninety degrees. Okay? So F equals I times L times B: four times zero point five times zero point two, which gives zero point four newton. The capital N on screen stands for newton. Clear? Now the wire makes thirty degrees with the field. Do we multiply by sine thirty or by cos thirty? It is sine thirty, because theta is the angle between the wire and the field. Sine thirty is zero point five, so zero point four times zero point five gives zero point two newton. So the force is exactly halved. Got it?

5:17Example 2: any-shape wire from origin to (3, 4, 0) m, find F

Now a harder one. This time, a wire of arbitrary shape runs from the origin to the point three metres, four metres, zero, and carries two amperes. The field, B vector, is zero point one k hat tesla, where k hat, the k with a little cap, is the unit vector along the z axis. Find the force on it. Pause the video and try it first. Done? Okay, let's check. The shape of the wire is not given at all. So how can we find the force? Remember our puzzle: in a uniform field, the shape does not matter! Just use the straight arrow from the origin to the end point. So a b vector equals three i hat plus four j hat, in metres, where i hat and j hat, again with little caps, are the unit vectors along x and y. Okay? So F vector equals I times a b vector, cross B: two, times three i hat plus four j hat in brackets, cross zero point one k hat. Right? Now the key step. i hat cross k hat, is it plus j hat, or minus j hat? It is minus j hat, because i, k, j goes the wrong way round the cycle i, j, k. And j hat cross k hat is plus i hat, the right way round. Clear? Two times zero point one is zero point two. Three i hat cross k hat gives minus three j hat, and four j hat cross k hat gives plus four i hat. So F vector is zero point two, times four i hat minus three j hat, which is zero point eight i hat, minus zero point six j hat, newton. Got it? Now let us find the size of the force. The two vertical bars mean magnitude, and the tick-shaped sign is the square root. Zero point eight squared is zero point six four, and zero point six squared is zero point three six. They add up to one, and the root of one is one, so the force is exactly one newton. See, just like our zigzag puzzle, we never needed the shape at all! With me?

7:23Quick revision: ILB sin theta, any shape, closed loop F = 0

So, quick revision. The force on a wire is I times L vector cross B, of size ILB sine theta, with L in metres. In a uniform field, a wire of any shape feels the same force as the straight wire from a to b. And a closed loop in a uniform field feels no net force, but it may still feel a torque. Now go and try the practice problems on this page. You can do them, all the best!