Torque on a Current Loop; Magnetic Moment

Transcript

0:00Introduction

Hello everyone! Okay, a small puzzle first. Earlier in this chapter, we saw that a closed current loop in a uniform magnetic field feels no net force. So such a loop should just sit there, perfectly still, right? Hmm, but the coil of an electric motor sits in a field like this, and it spins. So how can a loop with zero net force turn? This comes up often in JEE problems, so let's begin.

0:29Why the loop turns: a couple

So, look at the figure: a rectangular coil seen edge-on, in a uniform field B pointing right; the little arrows over the letters mark vectors. The dot in a circle means current out of the page, and the cross in a circle, current into it. These two long sides, perpendicular to the page, each of length l, carry current I, so each feels a force I times l times B, one up, one down, equal and opposite. Okay? Equal and opposite forces cancel, so nothing happens, right? Look at the figure. Do the two forces act along the same line? They do not, and that is the whole secret! Equal and opposite forces on different lines form a couple, and a couple turns things. That is how the motor coil spins. The arm of the couple, the perpendicular distance between them, is b sine theta. Here b is the short side, and theta, the Greek letter, is the angle between the field and the normal to the coil, the green arrow, which we will meet in a minute as the magnetic moment. Clear? So the torque, tau, this Greek letter, is force times arm: I times l times B, times b sine theta. And l times b is the area, capital A, so tau equals I times A times B, times sine theta. With me? And here is the nice part: this is true for a plane loop of any shape at all. Got it?

1:49Magnetic moment: the whole loop as one vector

Now, for N turns, each of area A, carrying current I, the magnetic dipole moment, mu vector, Greek letter mu, equals N times I times A vector. Its unit is ampere metre squared, the same as joule per tesla; T to the power minus one means per tesla. Okay? The area vector is normal to the loop. But a normal can point two ways, up or down. Which way do we take? We take it by the right-hand rule, yes! Curl your right-hand fingers along the current, and the thumb gives the direction of the area vector, and of mu vector. Right? So tau vector equals mu vector cross B vector, the cross sign being the cross product, and its size is mu times B times sine theta. Clear? Now a quick trap. Suppose the plane of the coil lies along the field. Then the torque is zero, right? Oh no, careful, theta is measured from the normal, not from the plane! The normal is at ninety degrees to the field, the small circle meaning degrees, so sine theta is one. The double arrow means therefore, and the torque is mu times B, the largest. Got it?

2:58Energy: stable and unstable positions

Now let us talk about energy. The potential energy is U equals minus mu vector dot B vector, the dot being the dot product, so U equals minus mu times B times cos theta. With me? At theta equals zero, and at one hundred and eighty degrees, sine theta is zero, so the torque is zero in both, and both are equilibrium positions. But which one is stable? Theta equals zero, with mu along the field, exactly! There U is minus mu B, the lowest energy, and the torque pulls the loop back after a small push. Okay? At one hundred and eighty degrees, mu points against the field, and U is plus mu B, the highest. The smallest push flips it over, so this one is unstable. Right? And the work to turn the loop from theta one to theta two, the small one and two being subscripts, is W equals mu B, times cos theta one, minus cos theta two. For a half turn, the bracket is one minus minus one, which is two, so going from zero to one hundred and eighty degrees, as the arrow shows, needs two mu B of work. Clear?

4:02Example 1: mu and torque, 50 turns of 4 cm2 at 30 degrees

Okay, the first example. We have a coil of fifty turns, each of area four square centimetres, carrying two amperes. Find its magnetic moment, and the torque in a zero point five tesla field, when its normal makes thirty degrees with the field. On screen, c m squared is square centimetres, capital A is amperes, capital T is tesla. Pause the video and try it first. Done? Okay, let's check. One centimetre is ten to the power minus two metre, so four square centimetres is four times ten to the minus two square metre, right? Oh no, careful, the centimetre is squared, so the ten to the minus two gets squared too. That gives ten to the minus four, so the area is four times ten to the minus four square metre. With me? Now multiply N, I and A to get mu. Fifty times two is one hundred, and one hundred times four is four hundred. Four hundred times ten to the minus four gives zero point zero four ampere metre squared. Okay? For the torque, thirty degrees is the angle between the normal and the field, exactly our theta, so we use sine thirty, which is zero point five. Zero point zero four times zero point five times zero point five gives zero point zero one newton metre; that is the N m on screen. Got it?

5:25Example 2: torque and energy of a loop in B = 0.3i + 0.4k T

Now a harder one, in vectors. We have a square loop of side ten centimetres in the x y plane, carrying two amperes, anticlockwise as seen from plus z. The field is zero point three i hat, plus zero point four k hat, tesla; i hat and k hat, with little caps, are unit vectors along x and z. Find the torque and the potential energy. Pause the video and try it first. Done? Okay, let's check. First, which way does mu vector point? Along plus k hat, very good! Curl your right-hand fingers anticlockwise, as seen from plus z, and the thumb points up the z axis. Ten centimetres is zero point one metre, so the area is zero point one squared, which is zero point zero one square metre. Right? With just one turn, mu vector is two times zero point zero one, which gives zero point zero two k hat, ampere metre squared. Clear? Now the torque is mu vector cross B: zero point zero two k hat, cross the bracket, zero point three i hat plus zero point four k hat. Tell me, what is k hat cross k hat? It is the zero vector, yes, since any vector crossed with itself is zero. And k hat cross i hat is plus j hat, round the cycle i, j, k. So zero point zero two times zero point three gives zero point zero zero six, and tau vector is six times ten to the minus three j hat, newton metre. With me? And the energy: U equals minus mu vector dot B vector. Mu has only a k hat part, so only the k hat part of the field counts: minus zero point zero two times zero point four, which gives minus eight times ten to the minus three joule. So i hat gave the torque, and k hat the energy.

7:16Quick revision: mu = NIA, tau = mu cross B, U = -mu dot B

Okay? So, quick revision. The magnetic moment is N times I times A vector, by the right-hand rule, and the torque is mu vector cross B vector. Theta is measured from the normal, not from the plane. And U equals minus mu dot B: stable at zero, unstable at one hundred and eighty degrees, with two mu B of work between them. Now go and try the practice problems on this page. You can do them, all the best!