Self-Inductance

Transcript

0:00Introduction

Hello everyone! Let's start with a small puzzle. Take a coil carrying a steady current, with no magnet and no other circuit near it. Now try to change that current quickly. Can anything there oppose you? Most students say no. But the coil itself fights back, and today we will see how. Self-inductance comes up often in JEE, so let's begin.

0:26Self-inductance: a coil's flux through itself

So, here is a coil of capital N turns, carrying a current, capital I, okay? Its magnetic field passes through its own turns. The double arrow means which gives, so we get a flux, capital phi, through each turn. Right? Now, if I double the current, what happens to this flux? It doubles too, yes, exactly! So the flux linkage, N times phi, is proportional to the current, and we write N times phi equals L times I, okay? The constant L is the self-inductance. Clear? Its unit is the henry, the capital H on screen. One henry is one weber per ampere, where W b is the weber, and A to the power minus one means per ampere. It is also one volt second per ampere, with V for volt and s for second, and one ohm second, where the Greek capital omega stands for ohm. Got it? Now here is a quick trap for you. If the current doubles, does the self-inductance double too? No, no, careful, it does not double! The flux linkage and the current both double, so their ratio, L, stays the same. L depends only on the geometry of the coil and the medium inside it, not on the current. With me?

1:39The back emf: the coil opposes the change

Now, back to our puzzle. What if the current changes? The flux linkage changes, so by Faraday's law an emf is induced in the coil itself. Curly E equals minus d of N phi, over d t, and since L is a constant, that is minus L times d I over d t. Okay? The minus sign is Lenz's law. This emf opposes the change in the current, so we call it a back emf. That is how the coil fights back. Right? Now suppose the current is falling. Does the back emf act against the current, or along it? Along it, yes, exactly! It opposes the change, not the current. When the current rises, curly E acts against it and slows the rise. When the current falls, curly E acts along it and tries to keep it going. Clear?

2:33Solenoid and toroid formulas

Now let us look at real coils. First, a long solenoid with capital N turns, length small l, and cross-section area capital A, okay? Small n, equal to N over l, is the number of turns per unit length. Right? Its self-inductance is mu nought times n squared times A times l, where mu nought, the Greek mu with a small zero below, is the permeability of free space. With n equal to N over l, this is mu nought times N squared times A, over l. Clear? Now tell me, if we double the number of turns, keeping the length fixed, what happens to the self-inductance? Many say it doubles, but it becomes four times! The arrow means changes to, and the sign before N squared means proportional to, so two squared gives four. Got it? With a core of relative permeability mu r, the Greek mu with a small r below, just multiply L by mu r. With me? And for a toroid of mean radius small r, L equals mu nought times N squared times A, over two pi r. Okay?

3:40Energy stored in an inductor

Now let us talk about energy. Building up a current against the back emf takes work, and the coil stores it. Capital U, the energy stored, equals one half times L times I squared. Right? This energy sits in the magnetic field. The energy per unit volume, small u, is B squared over two mu nought, where B is the magnetic field. Clear?

4:06An inductor in a circuit

Now, an inductor, L, in a circuit, with a current, small i, flowing from point A to point B, and increasing. Which end is at the higher potential? It is end A, yes, exactly! The back emf opposes the rise, so the potential drops from A to B, okay? V A minus V B, with the letters written small below, equals L times d i over d t, and the greater-than sign says it is positive. Right? If the coil also has a resistance, small r, just add i times r. Clear? And with no mutual coupling, inductors combine like resistors. In series, L is L one plus L two, the one and two written small. In parallel, one over L is one over L one, plus one over L two. Got it?

4:53Example 1: self-inductance of a 1000-turn solenoid

Now let us try our first example. Here is a solenoid with one thousand turns, a length of fifty centimetres, written c m, and a cross-section of four square centimetres, c m squared. Find its self-inductance. Pause the video and try it first. Done? Okay, let's check. The formula needs metres, so first we convert. Four square centimetres is how many square metres? It is four times ten to the power minus four, not ten to the power minus two! Squaring ten to the power minus two gives ten to the power minus four. So N is one thousand, and fifty centimetres is zero point five metres, small m for metre. Right? Now we substitute the values. Mu nought is four pi times ten to the power minus seven, N squared is ten to the power six, and capital A is four times ten to the power minus four, all over zero point five. With me? Four pi times four gives sixteen pi, and the powers of ten give ten to the power minus five. Dividing by zero point five doubles it, giving thirty two pi times ten to the power minus five henry. The wavy equals sign means approximately: about one point zero times ten to the power minus three henry, which is one point zero millihenry, m H on screen.

6:14Example 2: VA - VB for a rising and a falling current

Okay? Now let us try a harder one. Here is a branch from A to B, with three ohms and zero point five henry in series. The current from A to B is two amperes, increasing at four amperes per second, written A s to the power minus one. Find V A minus V B, then redo it for a current decreasing at the same rate. Pause the video and try it first. Done? Okay, let's check. Going from A to B along the current, the resistor drops i times capital R, and the inductor drops L times d i over d t. So V A minus V B is the sum of the two, okay? The current is increasing, so d i over d t is plus four. Three times two gives six, and zero point five times four gives two. Six plus two gives eight volts, capital V for volt. Right? Now the current is decreasing at the same rate. Is the answer still eight volts? No, careful, this is the trap! Now d i over d t is minus four, so the inductor term, zero point five times minus four, is minus two. Six minus two gives four volts. Just as we saw earlier, a falling current makes the back emf act along the current, to keep it going. Got it?

7:36Quick revision: N phi = LI, back emf, solenoid L, energy

So, let us do a quick revision. First, N phi equals L times I, and L depends only on geometry and medium, not on the current. Second, the back emf, minus L times d I over d t, opposes the change, and for a solenoid, L is mu nought times N squared times A, over l. Third, the energy is one half times L times I squared, and V A minus V B equals i times r, plus L times d i over d t. You have done really well today. Now go and try the practice problems on this page. All the best!