Ampère's Circuital Law

Transcript

0:00Introduction

Hello everyone! Okay, a small puzzle first. Draw a closed loop around a wire carrying a current. Now put a second wire just outside the loop, carrying ten amperes. Its magnetic field reaches every point of your loop, right? And yet, in today's law, that wire counts for nothing. How can that be? Today we learn Ampere's circuital law, a much shorter route to magnetic fields that comes up often in JEE problems. So, let's begin.

0:31Ampère's circuital law

So, the Biot Savart law always works, but it needs a long integral every time, right? When the current has high symmetry, there is a much shorter route. Here is Ampere's circuital law: the integral of B dot d l round a closed path equals mu naught times I enclosed. B, with a small arrow on top, is the magnetic field vector, and d l, with an arrow, is a tiny step along the path; the dot is the dot product. Mu naught, the Greek letter mu with a small zero, is the permeability of free space. Okay? The integral sign with a small circle on it means we go once round a closed path. We choose this path ourselves, and we call it an Amperian loop. With me? And I, with the small subscript e n c, short for enclosed, is the enclosed current: the net current crossing any surface bounded by the loop. We read it as I enclosed.

1:23Which current counts as positive?

Clear? Now, I enclosed is a net current, so signs matter. Choose the direction in which you go round the loop, and curl the fingers of your right hand along it. Got it? Your thumb now points through the loop. Suppose one wire carries current along your thumb, and another carries current the opposite way. Do the two currents simply add? No, these two currents subtract! The current along the thumb counts as positive, the plus sign, and the one against it as negative, the minus sign. I enclosed is the net current, with these signs. Right?

1:59Using the law: choose a smart loop

So how do we use this law? The whole skill is choosing the loop. Pick one on which B is constant and parallel to d l. Okay? Or, on some parts of the loop, B is perpendicular to d l, or zero. Those parts give nothing at all. With me? Then the integral is simply B times the length; the cross sign just means times. Clear? Let us try the simplest case in the course: a long straight wire carrying a current I, okay? Which loop would you choose? Yes, exactly, a circle of radius r, centred on the wire! By symmetry, B is the same everywhere on it and tangent to it. So B times two pi r equals mu naught I; this dot just means times. Got it? So B equals mu naught I over two pi r, the same field the Biot Savart law gave, but in one line. That is the power of symmetry. Right?

2:54What the integral sees, and what it ignores

Now, a current outside the loop does change B on the loop, but its contributions to the integral cancel as we go round. So it changes B, not the integral of B dot d l. Remember this for our puzzle. Got it? Next, if the loop winds k times round a current I, each winding gives mu naught I, so k windings give k times mu naught I, okay? And one condition. The law holds for steady currents. When charge is piling up, it needs an extra term, the displacement current, from the unit on electromagnetic waves. With me?

3:27Gauss's law for magnetism

Now, its partner, Gauss's law for magnetism. Here the circle on the integral means a closed surface, and d A, with an arrow on top, is a tiny piece of it. The magnetic flux, the integral of B dot d A, is zero for every closed surface. Got it? There are no magnetic monopoles; field lines always form closed loops. So take a closed surface round just one pole of a long bar magnet. What is the net flux through it? It is zero, even here, because every line that enters also leaves. Non-zero would need a lone pole, and that does not exist. Right?

4:05Example 1: net enclosed current with an outside 10 A wire

Alright, example one, our puzzle. Here a closed loop encloses two wires: five amperes in the positive sense, and two amperes in the negative sense; capital A is for amperes. Then a third wire, carrying ten amperes, lies outside. Find the integral of B dot d l. Pause the video and try it first. Done? Okay, let's check. Many students add everything: five plus two plus ten, so seventeen mu naught. Is that right? No, no, careful, that is a double trap! The two ampere current is in the negative sense, so we subtract it, and the ten ampere wire is not enclosed at all. So I enclosed is five minus two, which is three amperes. Okay? So the integral is three mu naught. Mu naught is four pi times ten to the power minus seven, so this is three point seven seven times ten to the power minus six tesla metre; capital T for tesla, small m for metre. Got it? Now a follow-up from the course. The ten ampere wire is not enclosed, so is the magnetic field zero on the loop? No, it is not zero. The outside wire changes B at every point of the loop, but its contributions cancel as we go round, so only the line integral is unaffected. That is how the ten ampere wire counts for nothing, and that solves our puzzle. Clear?

5:34Example 2: B dot dl along a quarter circle and a chord

Now a harder one: a long wire carries a current I along the z axis. Find the integral of B dot d l along a quarter circle of radius r centred on the wire, and along a straight line from the point r, zero, zero to the point zero, r, zero, with x, y and z coordinates in brackets. These paths are open, so the integral has no circle. Pause the video and try it first. Done? Okay, let's check. On the quarter circle, B is mu naught I over two pi r, tangent and constant, so the integral is B times the length. So how long is a quarter circle of radius r? Yes, one quarter of two pi r, which is pi r over two. So mu naught I over two pi r, times pi r over two; the pi and the r cancel, and we get mu naught I over four. Right? Now the straight line. Here the distance from the wire keeps changing, so B changes in size and direction. Integrating directly is painful. Is there a shortcut? Yes, use Ampere's law itself! Go out along the arc and come back along the straight line: that is a closed loop. The wire at the centre lies outside the region between them, so I enclosed is zero, and so is the integral round this loop. With me? So coming back along the line cancels the arc exactly, and going forward along the line gives the same, mu naught I over four. Got it? Here is the course's rule behind it: for a path that does not go round the wire, the integral depends only on the angle swept round the wire, here ninety degrees, the little circle meaning degrees. One full turn gives mu naught I, so a quarter turn gives mu naught I over four.

7:22Quick revision: Ampere's law, smart loops, no monopoles

Clear? So, quick revision. The integral of B dot d l round a closed loop is mu naught times the enclosed current, with signs, for steady currents. To use it, choose a loop where B is constant and parallel to d l, or perpendicular. And no monopoles, so the flux of B through every closed surface is zero. Now go and try the practice problems on this page. All the best!