Thick Wires, Cylinders and Current Sheets

Transcript

0:00Introduction

Hello everyone! First, here is a small puzzle. Take a thick copper wire, with a steady current spread evenly through it. Where is its magnetic field the strongest? Right on the axis, in the very middle? Many students say yes. Hold that thought. Thick wires, tubes and current sheets come up often in JEE problems. So, let's begin.

0:24Solid wire: inside and outside

Here is a long solid wire of radius capital R, carrying a current capital I, spread uniformly. By symmetry, B has the same size all round a circle centred on the axis, and it points along that circle. So take a circle of radius small r inside the wire, small r less than capital R, right? Now tell me, does this circle enclose the full current, or only a part of it? Only a part, that is correct. The current goes by area, pi small r squared over pi capital R squared. So I enclosed, the I with the small label enc, equals capital I times small r squared, over capital R squared. Clear? Now Ampere's law. It holds for any closed loop, but it gives B easily only when there is enough symmetry, and here we have it. It says: the integral of B dot d l round a closed loop, which the small circle on the integral sign shows; the arrows mean vectors. B is constant along our loop, so this is B times two pi r, equal to mu naught times I enclosed. Mu naught, the Greek letter mu with a small zero, is the permeability of free space. Okay? The double arrow gives the result: inside, B equals mu naught times I times small r, over two pi capital R squared. So inside, the field grows in proportion to r. With me? Outside, small r is greater than capital R and the whole current is enclosed, so B equals mu naught I over two pi r, the thin-wire result. Got it? Now, back to our puzzle. The field rises inside and falls outside. So where is it the strongest? At the surface, yes, exactly! There B reaches its largest value, mu naught I over two pi capital R, and on the axis, where small r is zero, the field is zero. So the middle is the weakest point, not the strongest! Clear?

2:19Hollow tubes and the coaxial cable

Now let us hollow it out: a thin-walled tube carrying a current capital I, okay? Draw a circle inside the tube. How much current does it enclose? None at all, very good! All the current flows in the wall. So inside a thin tube the field is zero, and outside it is mu naught I over two pi r again. Right? Next, a thick tube, inner radius small a, outer radius small b, with the current spread uniformly over the metal. For r between a and b, only the metal from a out to r is enclosed, so B equals mu naught I over two pi r, times r squared minus a squared, over b squared minus a squared. Okay? And the coaxial cable: an inner wire carries a current one way, and an outer tube carries an equal current back. Now, a circle between them encloses the inner current, so there is a field there. But a circle outside both encloses the two currents, which cancel, so the field outside is zero.

3:17Non-uniform current and current sheets

Clear? Now suppose the current is not uniform: J of r, the current density, changes with r. Cut the wire into thin rings of radius r prime, the r with a little dash, and width d r prime. Then I enclosed is the integral from zero to r of J of r prime, times two pi r prime, d r prime. Okay? Then B equals mu naught I enclosed, over two pi r. Find the enclosed current, then divide. Got it? Now a different shape: an infinite flat sheet, with current capital K per unit width. Its field is mu naught K over two. Right? So tell me, as you move away from the sheet, does its field get weaker, like a wire's? No, it does not, surprising but true! The field has the same size on each side, in opposite directions, and is independent of distance. With me?

4:11Topper's trick: cavities by superposition

Now a trick that toppers use. Suppose the conductor has a cavity along its length, so the symmetry is broken. Treat it as a full conductor with a uniform current density J, the arrow meaning a vector, plus a cylinder filling the cavity with minus J, okay? In the cavity the two cancel, leaving exactly a hole. Clear? Add the fields of the two solid cylinders. Inside a cylindrical cavity whose axis is displaced by the vector d from the conductor's axis, B equals mu naught times J times d, over two. Got it? And the beautiful part: this field is uniform, the same everywhere in the cavity, and perpendicular to d. Remember it, right?

4:54Example 1: field inside, at and outside a 2 mm radius wire

Time for example one. Here is a long wire of radius two millimetres, m m on screen, carrying ten amperes, capital A, uniformly. Find the field at its surface, and at one millimetre and four millimetres from the axis. Pause the video and try it first. Done? Okay, let's check. Mu naught over two pi is two times ten to the power minus seven in S I units, and capital R, two millimetres, is zero point zero zero two metres, small m on screen. Clear? At the surface, B is two times ten to the power minus seven, times ten, over zero point zero zero two. That is ten to the power minus three tesla, capital T on screen. Right? Now one millimetre from the axis. Put r equal to one millimetre in mu naught I over two pi r, and we get two times ten to the power minus three tesla, don't we? Oh no, careful, that formula is only for outside! One millimetre is inside, where B is proportional to r; the sign like a stretched alpha means proportional to. Half the radius gives half the surface field, five times ten to the power minus four tesla. Got it? Four millimetres is outside, where B is proportional to one over r. Twice the radius gives half the surface field, again five times ten to the power minus four tesla. One point inside, one outside, and the same field, very nice!

6:24Example 2: field inside a thick tube at 1.5 cm

Now example two, a little harder. Here is a thick tube, inner radius one centimetre and outer radius two centimetres, c m on screen, carrying ten amperes spread uniformly. Find the field one point five centimetres from the axis. Pause the video and try it first. Done? Okay, let's check. The point is inside the metal, so we need the enclosed current. By area, it is ten times one point five squared, over two squared, isn't it? No, no, wait, the hole in the middle carries no current! So I enclosed is ten times one point five squared minus one squared, over two squared minus one squared. That is ten times one point two five over three, which gives twenty five over six amperes. Right? Now use B equals mu naught I enclosed over two pi r, with r as zero point zero one five metres. B is two times ten to the power minus seven, times twenty five over six, all over zero point zero one five, okay? That gives about five point six times ten to the power minus five tesla; the wavy equals sign means approximately. And that is the answer, perfect!

7:40Quick revision: wire, tubes, current sheet and cavity

So, quick revision. Solid wire: B is proportional to r inside, to one over r outside, and largest at the surface. Tubes, cables and non-uniform currents: find the enclosed current, then B equals mu naught I enclosed, over two pi r. An infinite sheet gives mu naught K over two at any distance, and a cavity has a uniform field, mu naught times J times d, over two. Now go and try the practice problems on this page!