Conduction and Displacement Current in a Material

Transcript

0:00Introduction

Hello everyone! Here is a small puzzle to start with. The fan in your room runs on alternating current, so the electric field inside its copper wire keeps changing all the time. And we know that a changing electric field gives a displacement current. So tell me, why does nobody ever add a displacement current inside a wire? Is it zero, or is it just very, very small? By the end of this video, you will put an actual number on it. This idea is useful for JEE questions on displacement current, so let's begin.

0:35Two current densities in one material

So, take any material at all. It has a conductivity, sigma, the Greek letter that looks like a small o with a tail. And it has a permittivity, epsilon, the Greek letter that looks like a curly e. Okay? Now we put an electric field in it: E equals E naught sine omega t. E naught, the E with a small zero below it, is the amplitude, the peak value of the field. And omega, the Greek letter that looks like a curly w, is the angular frequency. Now tell me, how many kinds of current flow in this material? There are two kinds, yes, exactly! First, the charges in the material actually move, and that is the conduction current. J with a small c below it is the conduction current density, the current per unit area. J c equals sigma times E, which is sigma E naught sine omega t. Right? And what about the second kind? Look, this field is not steady; it keeps changing with time. So what does a changing electric field give us? It gives a displacement current, that is correct! J with a small d below it is the displacement current density. It equals epsilon times the rate of change of E with time; the curly d symbols on screen mean a partial derivative, simply how fast E changes with time. Differentiate sine omega t, and you get omega times cos omega t. So J d equals epsilon omega E naught cos omega t. With me? Now look at the two answers side by side. One goes as sine omega t, the other as cos omega t. So what is the phase difference between them? It is ninety degrees, absolutely! The small circle on screen means degrees. So when the conduction current is at its peak, the displacement current is zero, and the other way round. Keep this in mind; it can come up as a quick one-line question too.

2:40Which current wins?

Clear? Now the big question: which of these two currents is bigger? Let us compare the amplitudes, the peak values. J c max over J d max, where the small max below each J means its maximum value, equals sigma E naught, over epsilon omega E naught. E naught cancels, so the ratio of amplitudes is just sigma over epsilon omega. Okay? So then, at every single instant, J c over J d is also sigma over epsilon omega, right? Oh no, careful, that is the trap here! Sigma over epsilon omega is the ratio of the peak values only. At any instant, we divide sine omega t by cos omega t, and that gives tan omega t. So J c over J d equals sigma over epsilon omega, times tan omega t. Got it? Now, a good conductor like copper has a huge sigma. At ordinary frequencies, which current wins there? The conduction current wins, by a very long way! Sigma over epsilon omega is enormous, so in a good conductor at ordinary frequencies, the displacement current is negligible. Right? And a good insulator is exactly the opposite. Its sigma is tiny, so the conduction current is tiny too, and the displacement current dominates. So two extremes, and one ratio tells them apart. Clear?

4:03Example 1: copper at 50 Hz, ratio of the two amplitudes

Now our first example, and it settles the puzzle from the start. For copper, sigma is five point eight times ten to the power seven siemens per metre. The capital S is the siemens, and m to the power minus one means per metre. Epsilon is approximately epsilon naught; the wavy equals sign means approximately, and epsilon naught, the epsilon with a small zero, is the permittivity of free space. Find the ratio of the amplitudes of the conduction and displacement current densities at fifty hertz, the H z on screen. Pause the video and try it first. Done? Okay, let's check. We need sigma over epsilon naught omega. The frequency is fifty hertz, so can we put fifty straight in for omega? No, no, that is the classic slip! Omega is the angular frequency, and it equals two pi f, where f is the frequency in hertz. Two pi times fifty is one hundred pi, in radians per second. Okay? Now we substitute. Epsilon naught is eight point eight five times ten to the power minus twelve. So the ratio is five point eight times ten to the power seven, over eight point eight five times ten to the power minus twelve, times one hundred pi. Right? The bottom, eight point eight five times ten to the power minus twelve, times one hundred pi, is about two point seven eight times ten to the power minus nine. Dividing, the ratio comes to about two point one times ten to the power sixteen. See, the peak conduction current is more than ten to the power sixteen times the peak displacement current! So in the copper wire of your fan, the displacement current is not zero, but it is completely negligible, and that solves our puzzle.

5:55Example 2: sea water, frequency where amplitudes are equal

Clear? Now a harder one, where we work backwards. Sea water has sigma equal to four siemens per metre, and epsilon equal to eighty times epsilon naught. At what frequency are the amplitudes of the two current densities equal? Pause the video and try it first. Done? Okay, let's check. First of all, what do equal amplitudes mean, in symbols? The two peak values are equal, yes, exactly! Sigma E naught equals epsilon omega E naught. E naught cancels, and the double arrow means which gives sigma equals epsilon omega. In other words, the ratio sigma over epsilon omega is exactly one. Okay? So sigma over epsilon is our frequency, and we are done, right? No, no, wait, sigma over epsilon gives omega, not f! Omega equals two pi f, so f equals sigma over two pi epsilon. Forget the two pi, and your answer is more than six times too big. Got it? Now we substitute. The frequency is four, over two pi times eighty times eight point eight five times ten to the power minus twelve. That bottom works out to about four point four five times ten to the power minus nine. Right? So four, over four point four five times ten to the power minus nine, gives about nine times ten to the power eight hertz. That is about nine hundred megahertz; in M H z on screen, the capital M means mega, a million. See how the same ratio, sigma over epsilon omega, answered both questions? Clear?

7:40Quick revision: 90 degree phase, sigma over epsilon omega

So, quick revision. First, J c equals sigma E, and J d equals epsilon times the rate of change of E, and they differ in phase by ninety degrees. Second, the ratio of the amplitudes is sigma over epsilon omega, and at an instant you multiply it by tan omega t. Third, in a good conductor the displacement current is negligible, and in a good insulator it dominates. Now go and try the practice problems on this page. All the best!