Mutual Inductance

Transcript

0:00Introduction

Hello everyone! Let's start with a puzzle. Picture a tiny loop at the centre of a big loop, both in the same plane, with a changing current in the tiny loop. Its field spreads over the big loop in a messy way. So is there a two-line way to find the emf induced in the big loop? By the end of this video, you will do it yourself. Mutual inductance comes up often in JEE, so let's begin.

0:27Mutual inductance: one coil's flux through another

So, take two coils, with a current, capital I one, in coil one. The small one written below the I is a subscript, the coil number. The double arrow means which gives: its field gives a flux, capital phi two, the Greek letter phi, through each turn of coil two. Right? Now tell me this. If I double the current in coil one, what happens to the flux through coil two? It doubles too, yes, exactly! So the flux linkage of coil two, with its N two turns, is proportional to I one, and N two phi two equals M times I one. Here M is the mutual inductance, in henry, the capital H on screen. Clear? Now here is a surprising fact. Put the current in coil two instead, and find the flux linkage of coil one. You get exactly the same M! This will solve our puzzle. Got it? Now let the current I one change. By Faraday's law, the emf induced in coil two, curly E two, equals minus M times d I one over d t. With me? The minus sign is Lenz's law. The induced current in coil two flows so that its own flux opposes the change. So when I one rises, the field of coil two's induced current acts against coil one's field, and when I one falls, it acts along it. Okay?

1:48Two standard formulas for M

Now, let's find M for real coils. First, a coil of N two turns, wound on a long solenoid with small n one turns per metre and area capital A, okay? Put the current in the solenoid, find the flux through the coil, and M equals mu nought times n one times N two times capital A, as you see. Mu nought, the Greek mu with a small zero below, is the permeability of free space. Right? Now here is a trap. Two coaxial solenoids, one inside the other. Whose area goes into this formula? The inner one, yes, exactly! The flux that links both coils passes only through the inner solenoid's cross-section. Clear? Second set-up: a small coil, radius small r and N two turns, at the centre of a large coil, radius capital R and N one turns. Coplanar means they lie in the same plane. With me? Here M equals mu nought times N one times N two times pi r squared, over two R, okay? And pi r squared is just the small coil's area. Got it?

2:54Coupling coefficient and coils in series

Next, M equals k times the square root of L one times L two, the two self-inductances, where k is the coupling coefficient. The signs say k lies between zero and one, both included. Okay? Now join the coupled coils in series, with their fluxes aiding each other. Is the total just L one plus L two plus M? No, no, careful, the extra term is two M! Coil one links coil two through M, and coil two links coil one through the same M, so M comes in twice. So L equals L one plus L two plus two M, okay? And when the fluxes oppose, the sign flips, so L equals L one plus L two minus two M, with me?

3:38Topper's trick: put the current where the field is easy

Now here is a trick that toppers use. Since M is the same both ways, you choose which coil carries the current. So put it where the field is easy to find: the long solenoid, the large loop, or the straight wire. Okay? Then find the flux through the other coil, and divide, so M equals N two phi two, over I one, and that's it!

4:00Example 1: M for a 100-turn coil on a long solenoid

Now our first example, with a coil of one hundred turns, wound on a long solenoid of two thousand turns per metre, with a cross-section of five square centimetres, written c m squared. Find the mutual inductance. Pause the video and try it first. Done? Okay, let's check. Five square centimetres is how many square metres? It is five times ten to the power minus four, because we square the ten to the power minus two! And n one is two thousand per metre, written m to the power minus one, while N two is one hundred. Right? Now substitute into M equals mu nought times n one times N two times capital A, okay? Mu nought is four pi times ten to the power minus seven, times two thousand, times one hundred, times five times ten to the power minus four. With me? Here is a neat shortcut: two thousand times one hundred, times five times ten to the power minus four, gives exactly one hundred. Nice, right? So M is four pi times ten to the power minus five henry. The wavy equals sign means approximately, so about one point two six times ten to the power minus four henry, or zero point one two six millihenry, the m H on screen. Okay?

5:17Example 2: peak emf in a large loop around a small loop

Now let's solve our puzzle from the start. Here is a small loop of area one square centimetre, at the centre of a large circular loop of radius twenty centimetres, coplanar with it. The current in it is small i equals two sine of one hundred pi times t, in amperes, written capital A on screen. Find the peak emf in the large loop. Pause the video and try it first. Done? Okay, let's check. Can you directly find the flux of the small loop's field through the large loop? That is really hard, right? But M is the same both ways, so use our trick, and put the current in the large loop, where the field at the centre is easy. Clear? Our coplanar formula, with one turn each, and capital A, the small loop's area, for pi r squared, gives M equals mu nought times A, over two R, okay? Four pi times ten to the power minus seven, times ten to the power minus four, over zero point four, which is two R in metres, gives three point one four times ten to the power minus ten henry. Right? Now, the emf depends on d i over d t, and our current is a sine function. So what is the greatest value of d i over d t? When we differentiate, d i over d t equals two hundred pi times cos of one hundred pi times t. Cos is at most one, so the peak is omega times I nought. Here omega, the Greek letter, is one hundred pi, and I nought, with a small zero below, is the peak current of two amperes. With me? So the peak emf, curly E nought, is M times omega times I nought. Three point one four times ten to the power minus ten, times one hundred pi, times two, gives about two point zero times ten to the power minus seven volts, the capital V on screen. And our messy puzzle is done in just two lines!

7:12Quick revision: M both ways, solenoid M, k, coils in series

So, quick revision. First, N two phi two equals M times I one, curly E two is minus M times d I one over d t, and M is the same both ways. Second, for a solenoid, M is mu nought times n one times N two times capital A, and M is k times the square root of L one times L two, with k from zero to one. Third, coils in series give L one plus L two, plus or minus two M, and put the current where the field is easy. Now try the practice problems on this page. All the best!