Force Between Parallel Currents

Transcript

0:00Introduction

Hello everyone! Let us start with a small puzzle. Two long wires run one above the other, and nothing touches the upper wire. Can it float in the air, held up only by the currents? And should the two currents flow the same way, or opposite ways? Keep that question in mind. Forces between currents come up often in JEE, so let's begin.

0:25One wire makes a field, the other feels a force

So, take two long parallel wires, a distance d apart, with currents I one and I two; the small one and two below just number the wires. Wire one makes a field at wire two, B one, equal to mu naught I one over two pi d. Mu naught, the Greek letter mu with a small zero, is the permeability of free space, and pi is the circle constant. This field is perpendicular to wire two. Okay? Now, from the last lesson, the force on a wire is ILB sine theta, where theta is the angle between the wire and the field. So what force does each metre of wire two feel? It is just I two times B one, exactly! The angle is ninety degrees, and sine ninety is one. Capital F over small l is the force per unit length. Right? Put in B one, and F over l equals mu naught I one I two, over two pi d. I one and I two sit together, so each wire feels the same force per metre. Write this down.

1:28Same direction: attract or repel?

Clear? Now the direction, with both currents flowing the same way. Like charges repel, so like currents should repel too, right? Do the wires attract or repel? Let us check with I two, l cross B one. The small arrows mark vectors, and the cross sign between them is the cross product. Take the direction of B one from the right-hand thumb rule, and the force on wire two points towards wire one. So parallel currents attract! With me? Now reverse one current, and the force flips. Antiparallel currents, flowing in opposite directions, repel each other. Got it? So for charges, like ones repel, and for currents, like ones attract, the exact opposite. Mixing these up is a classic trap, so don't fall for it. Okay?

2:18The force that defined the ampere

Now a number to keep handy. Mu naught is four pi times ten to the power minus seven, tesla metre per ampere, written T m A to the power minus one; between numbers, the cross sign means times. So mu naught over two pi is two times ten to the power minus seven. Right? Two long parallel wires, one metre apart, carry one ampere each. What force does each metre feel? It is two times ten to the power minus seven newton per metre, written N m to the power minus one. In fact, this very force was used to define the ampere. Clear?

2:55A rod perpendicular to a long wire

Now a different shape. Take a rod with current I two, in the plane of the long wire and perpendicular to it, stretching from distance a to distance b. Okay? But the field of the long wire gets weaker as you move away from it. So can one value of B work for the whole rod? No, it cannot, so cut the rod into tiny pieces. Take a piece of length d r at distance r. The field there is perpendicular to it, so its force, d F, is mu naught I one I two over two pi r, times d r. Right? Now add the pieces from a to b with an integral, the tall S-shaped sign. The integral of d r over r is the natural log, the l n on screen. So F equals mu naught I one I two over two pi, times the natural log of b over a. Got it?

3:47A rectangular loop near a long wire

Now a rectangular loop with current I two. Its two sides of length l are parallel to the long wire, at distances d and d plus b; here b is the width of the loop. Okay? Say the near side's current runs the same way as the wire's; then the far side's current runs the opposite way. One side is pulled in, the other pushed out. Which force wins? The near side wins, exactly, because it is closer! With its current parallel to the wire's, it is attracted, and the far side is repelled more weakly. With me? So the net force is mu naught I one I two l, over two pi, times the big bracket: one over d, minus one over the sum d plus b. That is the pull on the near side, minus the push on the far side. Right? The two short sides feel equal and opposite forces, so they cancel. And the net force points towards the wire when the near side's current is parallel to the wire's current. Clear?

4:46Example 1: 20 A and 10 A wires 10 cm apart, force per metre

Now two examples from the course. Two long parallel wires, ten centimetres apart, carry twenty amperes and ten amperes in the same direction. Find the force per metre between them. Pause the video and try it first. Done? Okay, let's check. First, change to metres: ten centimetres is zero point one metre. Put in ten, and your answer is a hundred times too small. Right? Now use mu naught over two pi equals two times ten to the power minus seven. F over l equals two times ten to the power minus seven, times twenty, times ten, over zero point one. Okay? Two times twenty times ten is four hundred, so the top is four times ten to the power minus five. Dividing by zero point one gives four times ten to the power minus four newton per metre. Do these two wires attract or repel? They attract, because the two currents flow in the same direction.

5:49Example 2: current to float a 5 g/m wire 1 cm above 50 A

Got it? Now, remember the floating wire from the start? There is a long horizontal wire carrying fifty amperes. Another parallel wire, of mass five grams per metre, is to float one centimetre above it. What current must it carry, and in which direction? Pause the video and try it first. Done? Okay, let's check. Many students jump to the same direction straight away. But first, which way must the force on the upper wire point? It must point upward, to balance the weight, exactly! An upward push from the wire below is a repulsion, so the currents must be opposite. The same direction would pull it down. Okay? Now the weight of one metre. Five grams is five times ten to the power minus three kilogram, times nine point eight, which is g, the acceleration due to gravity. Right? Balance the two forces per metre: two times ten to the power minus seven, times fifty, times capital I, the current we want, over zero point zero one, equals five times ten to the power minus three, times nine point eight. With me? On the left, the numbers give ten to the power minus three. On the right, four point nine times ten to the power minus two. The double arrow on screen means "which gives", and capital I comes out as forty nine amperes, opposite to the lower current. So our puzzle is solved. Got it? And one more check that toppers make. Nudge the floating wire down a little: the repulsion grows and pushes it back up. So this balance is stable. Clear?

7:28Quick revision: mu0 I1 I2 over 2 pi d, like currents attract

So, quick revision. Between two long parallel wires, the force per metre is mu naught I one I two over two pi d, with mu naught over two pi equal to two times ten to the power minus seven. Parallel currents attract, and antiparallel currents repel. For a rod, you get the natural log of b over a, and for a loop, the near side wins. Now go and try the practice problems on this page. All the best!