Induced Electric Field and Eddy Currents

Transcript

0:00Introduction

Hello everyone! Here is a small puzzle. Imagine a magnetic field inside a long cylinder, and a charge at rest outside it, where there is no magnetic field. So the magnetic force on it is zero, right? Now we increase the field inside. Does the charge feel any force now? Hold on to your guess; you will calculate the answer in this video. This idea is useful in JEE problems, so let's begin.

0:29A changing magnetic field makes an electric field

So, look at the figure. The grey crosses show a magnetic field, B, into the page, only inside the shaded circle of radius small a, shown by the black arrow, and it is increasing. There is no wire here. So tell me, is anything induced at all? Yes, an electric field is induced! Faraday's law says the closed integral of E dot d l equals minus d phi over d t. The circle on the integral sign means once around a loop, the arrows mean vectors, the dot is the dot product, and phi is the magnetic flux. Okay? And it needs no conductor at all; it exists even in empty space. Clear? Now, the orange circles labelled E with an arrow are its field lines, and the figure says they go anticlockwise. Can you see why, using Lenz's law? Let us check it. The minus sign is Lenz's law: the flux into the page is increasing, so the field circulates the way a current would flow to make its own field out of the page. By the right-hand rule, that is anticlockwise. Got it?

1:34Uniform field in a cylinder: how strong is E?

Now, for a uniform field filling a cylinder like this, how strong is E? Take a circle of radius r around the axis. By symmetry, E is the same all along it, so the left side is E times two pi r, equal to the size of d phi over d t; the vertical bars mean modulus. Inside the field, what is the flux through this circle? It is pi times r squared, times B, yes! So for r less than or equal to small a, the sign with a line under it, E equals r over two, times d B over d t. On the graph, E rises in a straight line; the curly sign means proportional to. Okay? Now take a circle outside, with r greater than small a. The flux is pi times r squared, times B again, right? No, no, careful, that is the trap! The field fills only the circle of radius small a, so the flux stays B times pi times small a squared. So for r greater than or equal to small a, E equals small a squared over two times r, times d B over d t, falling as one over r on the graph. At r equal to small a, both formulas give the peak value marked on the graph, small a over two, times d B over d t. With me? And here is the surprise. Outside the region, B is zero, yet E is not zero; the slashed equals sign means not equal to. So a charge out there can feel a force. Clear?

3:01Closed field lines: no potential

Now, a question that catches many students. In the electrostatic field, the work done around any closed loop is zero. So take a charge q once around a closed loop in our induced field. Is the work zero again? No, it is not zero, and that is the big difference! It is q times curly E, the emf, which is q times the modulus of d phi over d t. Got it? So its lines are closed loops, the field is non-conservative, and no potential can be defined for it. And this work is what drives a current around a closed conducting loop.

3:39Eddy currents

Okay? Now, a solid piece of metal. When the flux through it changes, the induced emf drives currents circulating inside the conductor. These are eddy currents. With me? So why does a copper plate swinging between the poles of a magnet stop so quickly? Because of eddy currents, exactly! They heat the conductor and oppose the motion, so the plate's kinetic energy turns into heat. Now, if we cut slots in the plate, more damping or less? There is much less damping, because the slots break up the big current loops, so the eddy currents become much smaller. Clear? Eddy currents are useful in induction furnaces, which use their heat, and in electromagnetic brakes. They also give the damping in galvanometers and speedometers. Okay? But in transformer cores they are harmful, so the core is made of thin insulated laminations, which cut up the current paths. Got it?

4:36Example 1: induced E at 2 cm and 10 cm, region radius 5 cm

Now, our first example. Here, a uniform field fills a cylindrical region of radius five centimetres, and increases at zero point one tesla per second, the capital T, small s, to the power minus one. Find the induced electric field at two and at ten centimetres from the axis. Pause the video and try it first. Are you done? Okay, let's check. Small a is five centimetres, and two is less than five, that sign on screen, so this point is inside. So E equals r over two, times d B over d t. Okay? Zero point zero two over two, times zero point one, gives one point zero times ten to the power minus three volts per metre, the capital V, small m, to the power minus one. Now ten centimetres, with the same formula, right? Oh no, careful, ten is more than five, the sign for greater than, so this point is outside! So we need E equals small a squared over two times r, times d B over d t. Clear? Small a squared is twenty five times ten to the power minus four, and two times r is zero point two. So E is one point two five times ten to the power minus three volts per metre, even bigger than at two centimetres, though B is zero out here! So the charge in our puzzle does feel a force. Okay?

6:01Example 2: emf across a 16 cm chord rod in a changing B

Now a harder one. In a region of radius ten centimetres, the field grows at zero point five tesla per second. Then, a rod sixteen centimetres long is placed as a chord, six centimetres from the centre. Find the emf between its ends. Pause the video and try it first. Are you done? Okay, let's check. Our law needs a closed loop, but the rod is just a straight piece. So where is the loop? We simply make one ourselves! Join both ends of the rod to the centre, and the rod becomes one side of a closed triangle. Do the two radial lines add any emf? No, they add nothing! The induced field is tangential, along circles around the centre, so it is perpendicular to every radial line, and gives no emf along them. Clear? So the whole emf of the triangle sits in the rod, and it equals the area times d B over d t. The area is half times base times height, with base sixteen centimetres, and height the radius, ten centimetres, right? No, no, wait, the height is the distance from the centre to the rod, six centimetres. So curly E is a half, times zero point one six, times zero point zero six, times zero point five. That is two point four times ten to the power minus three volts, or two point four millivolts, the m V on screen. If you got this, very well done!

7:31Quick revision: E inside and outside, eddy currents

So, quick revision. First, a changing magnetic field makes a non-conservative electric field with closed lines, even in empty space. Second, E is r over two, times d B over d t, inside, and small a squared over two times r, times d B over d t, outside. Third, eddy currents heat and oppose motion, and laminations reduce them. Now try the four practice problems on this page. All the best!