Solenoids and Toroids

Transcript

0:00Introduction

Hello everyone! First, a small puzzle. Take a long solenoid, a coil wound like a spring, carrying a current. Now make it twice as long, with twice as many turns and the same current. Twice the turns, so twice the field inside, right? Hold on to your guess. Solenoids and toroids come up often in JEE magnetism, so let's begin.

0:26Long solenoid: Ampère's law does the work

So, look at the figure on the left: a long solenoid, cut along its length. The circles with a dot carry current out of the screen, and the circles with a cross carry it in. The blue arrows inside show B, the magnetic field, and small n is the number of turns per unit length, okay? Ampere's law holds for any closed loop, but it gives us B easily only when there is enough symmetry, and a long solenoid has it. Our loop is the dashed orange rectangle, with a side of length l inside, parallel to the axis, and the far side outside. Clear? Ampere's law adds up B along every side of this rectangle. So let me ask you this: which sides actually give anything? Only the inside side, yes! The short sides are perpendicular to the field inside, and outside, the field is almost zero. So B times l equals mu naught, times n times l in brackets, times I, the current, since n times l turns pass through the rectangle. Mu naught, the mu with a small zero, is the permeability of free space, okay? The l cancels on both sides, and B equals mu naught n I inside. Outside, B is almost zero; the wavy equals sign means approximately. Got it?

1:43What the field inside depends on

Look, mu naught n I has no radius and no position in it. So the field inside is uniform, the same everywhere, and independent of the radius. Okay? But what about the very end of a long solenoid, on its axis? Imagine a second, identical solenoid joined on there. How much field does each one give at the joint? Yes, each gives exactly half. Together they make one long solenoid, with mu naught n I at the joint, and by symmetry they share it equally. So at either end, B is half of mu naught n I, with me? Now, back to our puzzle. We doubled capital L, the length, and also capital N, the total number of turns. So twice the field, right? No, no, careful, the field does not change at all! What matters is small n, the turns per unit length, which is capital N over capital L, okay? Double both, and n stays the same, so B stays the same. Changing the radius does nothing either. Clear? Now one more useful fact. Wind wire of diameter d closely, turn touching turn, and n equals one over d, for each layer. Got it?

2:58A finite solenoid, on its axis between the ends

Now, for a shorter solenoid, at a point on its axis between its two ends, B equals mu naught n I over two, times cos theta one plus cos theta two, in brackets. Theta is the Greek letter, with small subscripts one and two, for the angles the two end rims subtend with the axis, seen from our point. Clear? And at the centre of a solenoid of length capital L and radius capital R, this becomes mu naught n I L, over the square root of L squared plus four R squared. Right?

3:27Toroid: a solenoid bent into a ring

Alright, now bend a long solenoid round into a ring. That is a toroid, on the right of the figure, with capital N turns on its core. Our Amperian loop is the dashed orange circle of radius small r, inside the core. Okay? By symmetry, B is the same all along this circle, and points along it. So how much current crosses the surface inside this circle? Yes, N times I, since each of the N turns passes through it once! So B times two pi r equals mu naught N I; the dot means times, and the double arrow means this gives. So B is mu naught N I over two pi r, in the core. Clear? Now, what is the field in the central hole, and outside the ring? The field is zero in both places. An Amperian circle in the hole, or one outside, encloses no net current. So the whole field stays in the core. Right? But careful, unlike a solenoid, this field depends on r; the symbol on screen means is proportional to. So B goes as one over r, and it is larger near the inner edge of the core. Got it?

4:36Putting a core inside

Now one last idea. Fill the solenoid with a core of relative permeability mu r, the mu with a small r, and the field becomes mu r times mu naught n I, okay? The magnetising field, capital H, equals n I, in amperes per metre; m to the power minus one means per metre. H does not depend on the core, and B equals mu H, with mu equal to mu r times mu naught. Clear?

5:01Example 1: field at centre and ends of a 50 cm solenoid

Alright, example one. Here is a solenoid fifty centimetres long, with five hundred turns, carrying two amperes; c m is centimetres, and capital A is amperes. Find the field near its centre and at its ends. Pause the video and try it first. Done? Okay, let's check. There are five hundred turns, so n is five hundred, right? Oh no, careful, n is turns per metre, not the total! Fifty centimetres is zero point five metres, small m for metres, so n is five hundred over zero point five, which gives one thousand per metre. With me? Near the centre, B equals mu naught n I: four pi times ten to the power minus seven, times one thousand, times two. That is eight pi times ten to the power minus four, about two point five one times ten to the power minus three tesla; capital T is tesla. Okay? At the ends, take half: four pi times ten to the power minus four, about one point two six times ten to the power minus three tesla. With n as five hundred, both answers would be half the right ones. Clear?

6:11Example 2: turns on a toroid from its field and mean radius

Now a slightly harder one. The field inside a toroid of mean radius ten centimetres is three times ten to the power minus three tesla, with one point five amperes. How many turns does it have? Pause the video and try it first. Done? Okay, let's check. Round the dashed circle, B times two pi r equals mu naught N I, so N equals B times two pi r, over mu naught I, got it? Now, before the numbers go in, what is two pi r here? First, ten centimetres is zero point one metres, so two pi r is zero point two pi metres. Keep centimetres by mistake, and you get a hundred times too many turns. Right? So N equals three times ten to the power minus three, times zero point two pi, over four pi times ten to the power minus seven, times one point five. With me? The pi on top cancels the pi below. The top gives six times ten to the power minus four, and the bottom, four times one point five, gives six times ten to the power minus seven. Divide, and the toroid has one thousand turns, as simple as that.

7:27Quick revision: solenoid, toroid and core formulas

So, quick revision. Long solenoid: mu naught n I inside, almost zero outside, half at the ends, and only n matters. Toroid: mu naught N I over two pi r in the core, zero in the hole and outside. And a core multiplies B by mu r, while H, which is n I, stays the same. Now go and try the practice problems on this page. All the best!