Mixed Problems: Set 1 (Fields of Currents)

Transcript

0:00Introduction

Hello everyone! Here is a small puzzle. Two flat coils share the same centre. The small coil has ten turns, the big coil has twenty, and the same current flows in both, in opposite senses. The big coil has twice the turns, so at the centre its field must win, right? Keep that answer in mind. This is our first set of mixed problems, on the fields of wires, arcs and loops, which come up often in JEE Main. So, let's begin.

0:30Idea 1: a long straight wire

Let us warm up: find the field four centimetres from a long wire carrying eight amperes; c m is centimetre, and capital A is ampere. Mu naught, the Greek letter mu with a small zero below, is the permeability of free space, four pi times ten to the power minus seven tesla metre per ampere, written T m A to the power minus one, where the cross means times. So mu naught over four pi is ten to the power minus seven, and mu naught over two pi is two times ten to the power minus seven, okay? For a long straight wire, B equals mu naught I over two pi d, where capital I is the current, and small d the perpendicular distance. For a semi-infinite wire, with the point P level with its end, on the perpendicular through that end, the field is exactly half, mu naught I over four pi d. We need this in example two. Right? So, for our wire, what do we put in for d: four, or zero point zero four? Zero point zero four, that is correct, because the formula works in metres, the small m on screen. So B equals two times ten to the power minus seven, times eight, over zero point zero four. Clear? Eight over zero point zero four is two hundred, and two hundred times two times ten to the power minus seven gives four times ten to the power minus five tesla. Put in four instead, and the answer is a hundred times too small. Always change to metres first. Got it?

1:53Idea 2: arcs, loops and coils

Now, idea number two: an arc of radius capital R, turning through an angle phi, the Greek letter phi, in radians. Every piece of the arc is at the same distance from the centre and gives a field the same way, so they simply add: B equals mu naught I phi, over four pi R, okay? Now tell me, for a semicircle, what is phi in radians? It is pi, half of a full turn, very good! Put phi equal to pi, and the two pis cancel, so B equals mu naught I over four R, as you see. Clear? And a full loop, two pi, gives mu naught I over two R, right? For capital N turns, each turn adds the same field, so a coil gives mu naught N I over two R, as shown. With me?

2:38Back to the puzzle: two coils

So, back to our puzzle, now with numbers. The small coil has ten turns of radius ten centimetres, the big coil has twenty turns of radius twenty centimetres, and they are concentric and coplanar, which means the same centre and the same plane. For direction, use the right-hand rule: curl your fingers along the current, and the thumb gives the field at the centre. With opposite senses, the two thumbs point opposite ways. Okay? So, before you calculate, which coil gives the bigger field at the centre? Let us not guess; let us calculate. Each coil gives mu naught N I over two R, and the fields subtract, so take mu naught I over two outside the bracket. With the radii in metres: ten over zero point one, minus twenty over zero point two. With me? Ten over zero point one is one hundred, and twenty over zero point two is also one hundred. So neither coil wins: the field at the centre is exactly zero! Right? The big coil has twice the turns, but also twice the radius. The field depends on N over R, not on N alone; that is where the puzzle tricked us. Got it?

3:48Example 1: zero field between parallel 3 A and 6 A wires

Alright, example one. Two long parallel wires, six centimetres apart, carry three amperes and six amperes in the same direction. Where is the field zero? Pause the video and try it first. Done? Okay, let's check. First, where can the field be zero at all: between the wires, or outside them? Between them, yes, exactly! Grip each wire with your right hand, thumb along the current: between the wires, the two fields are opposite, so they can cancel, while outside they point the same way and only add. Okay? Now, is the zero point in the middle, at three centimetres? No, no, wait, the currents are not equal. Let the point be x centimetres from the three ampere wire, so six minus x from the other. Set the two fields, mu naught I over two pi d, equal: mu naught over two pi cancels, and three over x equals six over six minus x. The units cancel, so centimetres are fine. With me? Cross-multiply: three times, six minus x in brackets, equals six x. Open the bracket and collect the x terms: eighteen equals nine x, so x equals two, where the double arrow means gives. Clear? So the field is zero between the wires, two centimetres from the three ampere wire. Not the middle: it sits nearer the smaller current, which must be closer to give the same field. Got it?

5:18Example 2: field at centre of a 2 cm, 4 A hairpin wire

Now a harder one. Take a hairpin wire: two long parallel wires, joined by a semicircle of radius two centimetres, carrying four amperes. Find the field at the centre of the semicircle. Pause the video and try it first. Done? Okay, let's check. There are three pieces: two straight wires and the semicircle. Many students write mu naught I over two pi R for each straight wire, as if it were a long wire. Is that correct? No, no, careful, each straight part starts at the end of the semicircle and runs one way only. So it is semi-infinite, and the centre sits level with its end, on the perpendicular through that end, at distance d equal to R, the radius. From idea one, that gives half, mu naught I over four pi R, okay? The semicircle gives mu naught I over four R, or, multiplying top and bottom by pi, mu naught I pi over four pi R, the same form as the wires. Right? But do these three fields add, or do some cancel? They all add, and here is why. The current comes in along one wire, goes round the semicircle, and leaves along the other, so it turns round the centre the same way all along. By the right-hand rule, all three fields point the same way. With me? Adding gives mu naught I times pi plus two, over four pi R, okay? With mu naught over four pi as ten to the power minus seven, and R as zero point zero two metre, that is ten to the power minus seven, times four times pi plus two, over zero point zero two. Clear? Pi plus two is about five point one four, and four over zero point zero two is two hundred. Two hundred times five point one four, times ten to the power minus seven, gives about one point zero three times ten to the power minus four tesla, where the wavy equals sign means approximately. Very good, well done!

7:12Quick revision: wire, semi-infinite wire, arc, coil, N/R

So, quick revision. Long wire: mu naught I over two pi d, with d in metres; a semi-infinite wire, seen from a point level with its end, on the perpendicular through that end, gives half. Arc: mu naught I phi over four pi R, phi in radians; a coil gives mu naught N I over two R, so what counts is N over R, not N alone. Fields cancel only where they are opposite, so add the pieces with the right-hand rule. Now go and try the practice problems on this page. All the best!