The Bar Magnet as a Magnetic Dipole

Transcript

0:00Introduction

Hello everyone! Okay, a small puzzle before we start. Take a bar magnet, north at one end and south at the other, and snap it exactly in the middle. Do you now hold one piece that is only north, and another that is only south? Most people say yes, right? Today we will find out what really happens. Questions on the bar magnet come up often in JEE, so let's begin.

0:27A bar magnet is a small solenoid

So, here is the big idea. The bar magnet behaves exactly like a short solenoid, the coil we met earlier. Outside, its field lines leave the north end, the capital N, and return at the south end, the capital S, okay? Now let me ask you this. What happens to these lines inside the magnet? Which way do they run? They run from south to north, yes, exactly! So every magnetic field line is a closed loop, with no start and no end. Right? We describe the magnet by its magnetic moment, m, with the little arrow on top that marks a vector. It points from the south end to the north end. Clear? In the pole picture, q sub m, the q with a small m below it, is the pole strength, and two l is the magnetic length, the distance between the poles. Then m equals q sub m times two l, where the dot means times. Okay?

1:24Field of a short magnet

Let r be the distance from the centre of the magnet. For a short magnet, r is much bigger than the magnet itself. Take a point on its axis, the line through both poles. There, the field B equals mu naught over four pi, times two m over r cubed, and it points along m. Mu naught, the Greek mu with a small zero, is the permeability of free space. Got it? Now move sideways, to the equatorial line, through the centre and perpendicular to the axis. At the same r, is the field bigger or smaller here? Yes, it is smaller, exactly half! Here B equals mu naught over four pi, times m over r cubed, with no two, so axial to equatorial is two to one. And careful, here the field points opposite to m. With me?

2:16When the magnet is not short

But what if the magnet is not short? Let its magnetic length be two l, with r measured from the centre. On the axis, B equals mu naught over four pi, times two times m times r, over r squared minus l squared, whole squared. Right? On the equatorial line, B equals mu naught over four pi, times m, over r squared plus l squared, to the power three over two. Minus on the axis, plus on the equator. Got it?

2:45In a uniform field

Now put the magnet in a uniform magnetic field, B, that is the same everywhere. It feels a torque, tau, the Greek letter tau with the little arrow on top, equal to m cross B, where the cross means the cross product. This torque turns the magnet. Clear? Its potential energy, U, equals minus m dot B, where the dot is the dot product. So the energy is lowest when m points along the field. With me? So, there is a torque. Is there also a net force, pulling the whole magnet along? The answer is no. In a uniform field the net force is zero. So the magnet turns, but it does not move off as a whole.

3:28Cutting and bending

Okay? Now, back to our puzzle. Cut a bar magnet in half, straight across. Do you get a piece with only a north pole? The answer is no, never, because field lines are closed loops, so each piece is a complete magnet with its own two poles. Cut across into n pieces, and each keeps the same pole strength, q sub m. But its length is the original length over n, so its moment is m over n. With me? Now cut along the length instead, into n thin strips. What happens to the pole strength? Now the pole strength is shared, so each strip has q sub m over n. The length stays the same, so the moment is again m over n. Same moment both ways, but the pole strength drops only in the second cut, a common trap. Right? And if you bend a thin magnet into an arc, the pole strength stays, but the poles come closer. So the new moment, m prime, with the little dash, is q sub m times the chord, the straight distance between the ends. Clear? For a semicircle, the chord is the diameter, and the new moment is two m over pi, smaller than m, as it should be. Got it?

4:42Example 1: short magnet field on axis and equatorial line

Okay, time for the examples. Here, a short magnet has moment zero point five ampere metre squared, written capital A m squared. Find the field twenty centimetres away, first on its axis, then on its equatorial line. Pause the video and try it first. Done? Okay, let's check. Mu naught over four pi is ten to the power minus seven, in tesla metre per ampere, and capital T stands for tesla. And r is twenty centimetres, which is zero point two metre. Okay? Now, r cubed. What is zero point two cubed? It is zero point zero zero eight metre cubed. Many students write zero point zero eight here, and then the answer comes out ten times too small. So be careful with the zeros, right? Part a, the axial point: ten to the power minus seven, times two times zero point five, over zero point zero zero eight. Two times zero point five is one, and one over zero point zero zero eight gives one hundred and twenty five. So B is one point two five times ten to the power minus five tesla. Clear? Part b, the equatorial point. Just halve it, using the two to one ratio. So B is six point two five times ten to the power minus six tesla. Got it?

6:07Example 2: 10 cm magnet axial field, exact vs short dipole

Now a harder one. Take a magnet ten centimetres long, so l is five centimetres, with moment two ampere metre squared. Find the axial field fifteen centimetres from its centre, exactly, and with the short-dipole formula. Pause the video and try it first. Done? Okay, let's check. Is this magnet really short, compared with fifteen centimetres? It is not short at all, so the exact answer needs the formula with both r and l. In metres, r is zero point one five, and l is zero point zero five. Okay? On the axis, ten to the power minus seven, times two times m times r, that is two times two times zero point one five, over r squared minus l squared, whole squared, which is zero point zero two two five minus zero point zero zero two five, squared. With me? The top gives zero point six, and the bracket gives zero point zero two, squared. Working it out, B is one point five times ten to the power minus four tesla. Clear? Now the short-dipole formula: ten to the power minus seven, times two m, which is four, over zero point one five cubed. The wavy equals sign means about, so B is about one point one nine times ten to the power minus four tesla. Right? See, the short-dipole answer is clearly smaller. The formula is poor here, because r is only three times l, so keep it for magnets that really are short. Got it?

7:43Quick revision: moment S to N, 2:1 ratio, cutting a magnet

So, quick revision. The bar magnet is like a short solenoid, its moment points from south to north, and m equals q sub m times two l. For a short magnet, the axial field is mu naught over four pi, times two m over r cubed, and the equatorial field is half of it, opposite to m. Cut across, the pole strength stays; cut along, it is shared; the moment is m over n either way. Now go and try the practice problems on this page. All the best!