Magnetic Materials

Transcript

0:00Introduction

Hello everyone! Okay, a small puzzle first. Bring a strong magnet near an iron nail, and the nail is pulled in hard. Near aluminium, only a very weak pull. But bismuth, or even water, is pushed away, very weakly! Same magnet, three different behaviours. So what is going on? By the end of this video, you will know. This topic comes up often in JEE, so let's begin.

0:28Magnetisation, intensity and susceptibility

So, in a magnetic field, a material becomes magnetised: its atoms, tiny current loops, get a net moment. The moment per unit volume is the magnetisation, capital M, the little arrow on top marking a vector. Its unit is ampere per metre, written A m to the power minus one. Okay? Next, the magnetic intensity, H, also with the vector arrow, fixed by the free currents in the wires. In a solenoid, H equals n times I, with n the turns per unit length and I the current. Right? Now, the field B, also a vector with its arrow, equals mu naught, times H plus M in brackets. Mu naught, the Greek mu with a small zero, is the permeability of free space. And M equals chi times H, where chi, the Greek letter like a curly x, is the susceptibility. Now put this into the first formula. What do you get for B? Yes, B equals mu naught times H, times one plus chi. One plus chi is the relative permeability, mu sub r, the mu with a small r below. So B equals mu naught times mu sub r times H, or mu times H, mu being the permeability. Clear?

1:48Diamagnetic: negative susceptibility, Meissner effect

Now the three kinds of material. First, diamagnetic: chi is small and negative, so mu sub r is less than one, the less-than sign. Examples are bismuth, copper, water, and sodium chloride, N a C l. With me? Their atoms have no permanent moment; the field changes the orbital motion of the electrons and induces moments opposite to it. So in a non-uniform field, which way does a diamagnet move, towards the strong field, or away from it? Yes, exactly, it is pushed from the stronger field to the weaker field, and that solves our puzzle: bismuth and water are diamagnetic! And for a diamagnet, chi does not depend on the temperature, the capital T on screen, okay? Now, a superconductor pushes the field out completely, so B inside is zero. Using our formula, B equals mu naught, times one plus chi, times H, what must chi be? Chi must be minus one, so that one plus chi is zero. So a superconductor is a perfect diamagnet, with chi equal to minus one and B equal to zero inside. This is the Meissner effect. Got it?

3:01Paramagnetic and ferromagnetic

Next, paramagnetic materials. Chi is small but positive, so mu sub r is slightly greater than one, the greater-than sign. Right? Now the atoms have permanent moments. The field tries to line them up, but thermal agitation, the random jiggling from heat, fights it, so only partly. So, if you heat a paramagnet, does its susceptibility go up or down? It goes down, yes, since more heat means less alignment. That is Curie's law: chi equals capital C over T, with C a constant and T the temperature in kelvin. Examples are aluminium, sodium and oxygen, so the aluminium in our puzzle is paramagnetic, which is why the pull was weak. Clear? And third, ferromagnetic: chi is large and positive, from ten to the power two up to ten to the power five. The moments line up on their own in small regions called domains; an applied field grows the domains along it and turns the others. With me? Above the Curie temperature, T sub C, the domain alignment is destroyed, and it turns paramagnetic, with chi equal to C over, T minus T sub C, okay? Careful, the bottom is T minus T sub C, not just T, a common slip. Examples are iron, cobalt, nickel and gadolinium, and there is our iron nail, pulled in hard. Right?

4:26Hysteresis: soft iron versus steel

Now this graph for a ferromagnet has B on the vertical axis and H on the horizontal axis. The dashed green curve is the first magnetisation, rising to saturation at the top right. Now bring H back to zero, along the arrow. Does B return to zero too? No, it does not, some B is left behind. This is the retentivity, B sub r, the B with a small r, at the top red dot on the vertical axis. Okay? To bring B down to zero, we need a reverse field. Its size is the coercivity, H sub c, at the red dot marked minus H sub c on the left. Clear? Go once round the loop, and energy is lost as heat. The area of the loop is the energy lost per unit volume, per cycle. Right? So for a transformer core, which goes round this loop again and again, we want a thin loop: soft iron, with low retentivity and low coercivity. Got it? For a permanent magnet, we want the opposite: steel or alnico, with high retentivity and high coercivity, so they keep their magnetism. With me? To demagnetise a magnet of coercivity H sub c in a solenoid with n turns per metre, we need n times I equal to H sub c. So the current I is H sub c over n. Okay?

5:48Example 1: mu r and mu for a susceptibility of 599

Alright, time for the examples. Here, a material has susceptibility chi equal to five hundred and ninety nine. Find mu sub r and mu. Pause the video and try it first. Done? Okay, let's check. Mu sub r is one plus chi, one plus five hundred and ninety nine, which gives six hundred. It is a pure ratio, with no unit. Right? Now mu is mu sub r times mu naught, six hundred times four pi times ten to the power minus seven. That is about seven point five times ten to the power minus four tesla metre per ampere, written T m A to the power minus one; the wavy equals sign means about. Clear?

6:34Example 2: H, B and M in a solenoid core of mu r 1000

Now a harder one. Here, a long solenoid with five hundred turns per metre carries zero point five ampere, the capital A on screen, and its core has mu sub r equal to one thousand. Find H, B and M in the core. Pause the video and try it first. Done? Okay, let's check. H depends only on the free current, so it is n times I, five hundred times zero point five, which gives two hundred and fifty ampere per metre. Right? Now B equals mu sub r times mu naught times H, so one thousand, times four pi times ten to the power minus seven, times two hundred and fifty. That is zero point one pi, about zero point three one tesla, the capital T on screen. Clear? Now, M equals chi times H, and chi is also one thousand, right? No, no, careful, that is exactly the trap here. Mu sub r is one plus chi, so chi equals mu sub r minus one, and that is nine hundred and ninety nine. For a paramagnet, with mu sub r close to one, this slip changes everything. Got it? So M equals nine hundred and ninety nine times two hundred and fifty, about two point five times ten to the power five ampere per metre, nearly a thousand times H, okay?

8:00Quick revision: mu r = 1 + chi, dia, para, ferro, hysteresis

So, quick revision. B equals mu naught, times H plus M, with M equal to chi times H, and mu sub r equal to one plus chi. Chi is small and negative for diamagnets, small and positive for paramagnets, where chi equals C over T, and large for ferromagnets. And from hysteresis, soft iron for cores, steel and alnico for permanent magnets. You now know all three kinds of material, well done! Now try the practice problems on this page. All the best!