Mixed Problems: Set 3 (Forces, Torques and Magnetism)

Transcript

0:00Introduction

Hello everyone! Okay, a small puzzle first. Take a wire bent into the shape of an L, with arms of three and four centimetres, in a uniform magnetic field perpendicular to its plane. That is seven centimetres of wire, so the force is the same as on a straight wire seven centimetres long, right? Keep your answer; we will check it in the second example. Today is Set three of our mixed JEE Main problems: forces, torques and magnetism, topics that come up often. So, let's begin.

0:33Forces on current-carrying wires

So, first the tool kit for forces. Take a straight wire of length capital L, carrying current I, in a field B; it feels the force I L cross B, where the arrows on top mark vectors and the cross sign is the cross product. Its size is ILB sine theta, where theta, the Greek letter, is the angle between the wire and the field. Okay? Now a beautiful result. In a uniform field, a wire of any shape feels the same force as a straight wire joining its two ends. Hold that thought for our L-shaped wire. Right? So for a closed loop, the two ends meet, the joining line has zero length, and the net force is zero, though a torque may remain. Clear? Next, two long parallel wires a distance d apart, with currents I one and I two; the small one and two are subscripts. The force per unit length, F over small l, is mu naught I one I two, over two pi d. Mu naught, the Greek mu with a small zero, is the permeability of free space, and pi is the circle constant. Got it? Now, let both currents flow the same way. Like charges repel, so like currents should repel too, right? No, no, wait, it is just the opposite! Parallel currents attract, and antiparallel currents repel. Remember this trap. With me?

1:56Torque on a coil

Now torques. In the figure, the dark line is a coil seen edge-on, in a uniform field B pointing right. The dot in a circle means current out of the page, and the cross in a circle, current into it. The two red forces, each I times small l times B for a side of length small l, are equal and opposite but not on one line. They are b sine theta apart, where small b is the width of the coil, so together they turn the coil. Okay? The coil's magnetic moment, mu vector, the Greek letter mu, is N times I times A vector, for N turns of area A each, with current I flowing. It points along the normal, the green arrow; n hat, with the little cap, is the unit normal. Its unit is ampere metre squared. Right? The torque, tau, this Greek letter, is mu vector cross B vector, of size mu B sine theta. Beside the figure, the same torque is written NIAB sine theta, with the energy U, which we skip today. Clear? Now, the angle theta. Is it measured from the plane of the coil, or from the normal? From the normal, exactly! In the figure, theta sits between the green arrow mu and the field, not between the dark line and the field. Our first example tests this. With me? And the work to turn the coil from theta one to theta two is W equals mu B, times cos theta one, minus cos theta two. Got it?

3:27Magnetic materials

Now, magnetism in materials. Chi, the Greek letter that looks like an x, is the susceptibility, and mu r, mu with a small r below it, is the relative permeability. They are linked by mu r equals one plus chi. Okay? Diamagnetic materials have chi small and negative, so mu r is less than one; the pointed sign means less than. Right? Paramagnetic materials have chi small and positive, and they obey Curie's law: chi equals capital C over capital T, where capital C is a constant and capital T is the temperature in kelvin. Clear? And ferromagnetic materials have chi large and positive. Got it? Now a quick one. If you double the temperature of a paramagnetic salt, what happens to its chi? It halves, very good! Chi goes as one over T, so doubling the temperature halves chi. With me?

4:22Example 1: moment and torque of a 100-turn coil in 0.4 T

Now, Example one, an easy one. Read it with me: a coil of one hundred turns, of area five square centimetres, carries two amperes. Find its magnetic moment, and the torque in zero point four tesla with its plane parallel to the field. On screen, c m squared is square centimetres, capital A is ampere, and capital T is tesla. Pause the video and try it first. Done? Okay, let's check. First the area, capital A, which here means area, not ampere. One square centimetre is ten to the power minus four square metres, so five square centimetres is five times ten to the power minus four square metres; here the cross sign simply means times. Right? Now multiply N, I and A to get mu. One hundred times two times five is one thousand, and one thousand times ten to the power minus four gives zero point one ampere metre squared. Clear? Now the torque. The plane of the coil is parallel to the field, so the angle is zero, and the torque is zero, right? Oh no, careful, remember the figure! Theta is measured from the normal, and with the plane along the field, the normal is at ninety degrees; the small circle means degrees. So sine theta is one, the largest torque, not zero. With me? So tau equals mu B, which is zero point one times zero point four, giving zero point zero four newton metre, the N m on screen. That's it, both parts done. Got it?

5:57Example 2: force on an L-shaped wire with 3 cm and 4 cm arms

Now Example two, our puzzle from the start. Read it with me: a wire bent into an L, with perpendicular arms of three centimetres and four centimetres, carries two amperes in a uniform field of zero point five tesla, perpendicular to its plane. Find the force on it. Pause the video and try it first. Done? Okay, let's check. The wire is three plus four, seven centimetres long, so we put seven centimetres into ILB, right? No, no, wait, the field is uniform! So the bent wire feels the same force as a straight wire joining its two ends. Okay? The two arms are perpendicular, so the joining line is the hypotenuse: the square root of three squared plus four squared; the tick-shaped sign is the square root. That is five centimetres, or zero point zero five metre. Right? The upside-down T sign means perpendicular: the field is perpendicular to the plane, so also to this line, and theta is ninety degrees. So F equals ILB: two times zero point zero five times zero point five. Clear? That gives zero point zero five newton, the capital N on screen. See, puzzle solved: seven centimetres of wire, but the force of a five centimetre wire! With me? Here is a quick check that toppers like. The two arms feel zero point zero three newton and zero point zero four newton, at right angles to each other. Add them as vectors: the square root of zero point zero three squared plus zero point zero four squared is zero point zero five, the same answer again. Got it?

7:36Quick revision: wire forces, coil torque, magnetic materials

So, quick revision. The force on a wire is ILB sine theta; in a uniform field, join the ends; and parallel currents attract. The moment is NIA, and the torque is mu B sine theta, with theta from the normal. And mu r equals one plus chi, with chi equal to C over T for a paramagnet. Now go to the practice problems on this page, give each one or two minutes, and begin. All the best!