Magnetic Field in a Charging Capacitor

Transcript

0:00Introduction

Hello everyone! Let's start with a puzzle. We are charging a capacitor with round plates. No charge crosses the gap, and yet there is a magnetic field in it. Take one point inside the gap, at one third of the plate radius, and one point outside, at three times the radius. Where is the field stronger? Hold on to your answer. This setup comes up often in JEE, so let's begin.

0:27The displacement current in the gap

So, look at the left half of the figure. In the wire flows i c, the small c written below, the conduction current. The dashed ring round the wire is loop one. The plates carry plus q and minus q, and the red arrows are the electric field E, the little arrow on top meaning a vector. While the capacitor charges, this field grows. Okay? No charge jumps across the gap. But the red label says i d, the small d written below, the displacement current, equals epsilon naught, the Greek epsilon with a small zero, times d capital phi E by d t, the rate of change of electric flux, with the small E below for electric. It exists only while the flux changes, and in the gap it equals i, the charging current. Right? This displacement current is spread uniformly over the plate area, pi capital R squared, where capital R is the plate radius. Got it? Now draw a smaller circle in the gap, on the same axis, like the dashed loop two. Its radius, small r, is smaller than the plate radius. How much displacement current passes through it? Only its share of the area, yes, exactly! The less-than sign says small r is below capital R, and the circle gets pi r squared out of pi capital R squared. With me? The pi cancels, so the circle carries i times r squared over capital R squared. Everything else follows from this fraction. Clear?

2:01The Ampère–Maxwell law gives B

Now, the magnetic field. Its field lines are circles about the axis, the line through the centres of the plates. So on a circle of radius small r, B has the same size everywhere. Okay? So we apply the Ampere Maxwell law round that circle. B times two pi r, the length of the circle, equals mu naught, the Greek mu with a small zero, times the displacement current through it. In the gap there is no conduction current. Right? Inside, small r is less than or equal to capital R, shown by the less-than sign with a line below. Put in i times r squared over capital R squared, and divide by two pi r. The double arrow means which gives: B equals mu naught times i times r, over two pi capital R squared. Got it? Now go outside the plates. How much displacement current does a bigger circle enclose? All of it, the full i, that is correct! Here small r is greater than or equal to capital R, and the displacement current is spread only over the plates. So B equals mu naught times i, over two pi r, exactly the field of a long straight wire. Clear? And one more form, for inside the gap. B equals mu naught times epsilon naught times r, over two, times d E by d t, the rate at which the electric field grows. Use it when a question gives you d E by d t. With me?

3:31How B changes with the distance r

Now the right half of the figure: B up the vertical axis, and small r along the horizontal one. Tell me first: where between the plates is the field the strongest? Many say at the centre, on the axis. Be careful, that is the trap! On the axis small r is zero, so B is zero. Right? Inside, B is proportional to r; the curly sign means proportional to. So the graph rises in a straight line from zero, as the red label shows. Okay? B is greatest at the edge, where small r equals the plate radius. There it equals mu naught times i, over two pi capital R, marked by the dashed line. Got it? Outside, B is proportional to one over r, so it falls away, like the field of the wire. So the edge is the peak. With me? Now a trick that toppers use. Call the edge value B edge, the small word edge written below. Inside, B equals B edge times the fraction small r over capital R, nothing more. Clear? And outside, B is B edge times capital R over small r. Just compare the distance with the radius. Right?

4:49Example 1: B at 3 cm inside 6 cm plates with i = 0.5 A

Let's try our first example. Here is the question: a capacitor with circular plates of radius six centimetres is charged by a current of zero point five ampere. Find the field between the plates, three centimetres from the axis. Pause the video and try it first. Done? Okay, let's check. Many students write B equals mu naught times i, over two pi r, at once. Can we use that formula here? No, we cannot, and that is the trap! That wire formula holds only outside the plates. In metres, capital R is zero point zero six, i is zero point five ampere, and small r is zero point zero three. Small r is less than capital R, so the point is inside. Okay? Mu naught is four pi times ten to the power minus seven, so mu naught over two pi is two times ten to the power minus seven tesla metre per ampere; the raised minus one means per. Right? Substituting, we get two times ten to the power minus seven, times zero point five, times zero point zero three, over zero point zero six squared, the brackets holding the number squared. With me? Zero point five times zero point zero three is zero point zero one five, and zero point zero six squared is zero point zero zero three six. Dividing gives about four point one seven. Multiply that by two times ten to the power minus seven, and B comes to about eight point three times ten to the power minus seven tesla; the wavy equals sign means approximately. Well done if you got that! Got it?

6:33Example 2: B at R/3 inside vs 3R outside, ratio 1:1

Now a harder one: our puzzle from the start. Here is the question: a capacitor with plates of radius capital R is being charged. Find the ratio of the magnetic fields at small r equals capital R over three, and at small r equals three times capital R, from the axis. Pause the video and try it first. Done? Okay, let's check. The second point is nine times farther from the axis. So its field must be much weaker, right? No, no, wait, let's not guess; let's calculate! First, the edge value: B edge equals mu naught times i, over two pi capital R, okay? Inside, put small r equals capital R over three into the inside formula. B one, with a small one written below, is the edge value times one third; the dot means times. Right? Outside, put small r equals three capital R into the wire formula. B two is again the edge value times one third. With me? So B one to B two, the colon meaning to, is one to one. The fields are exactly equal; that is the surprise in our puzzle! With the toppers' trick it is instant: one third of the edge value on both sides. Clear?

7:55Quick revision: B inside, outside, zero on axis, max at edge

So, quick revision. First, in the gap the displacement current equals i, spread uniformly, and a circle of radius small r gets i times r squared over capital R squared. Second, inside, B equals mu naught times i times r, over two pi capital R squared; outside, mu naught times i over two pi r, like the wire. Third, B is zero on the axis and greatest at the edge. Now go and try the practice problems on this page. All the best!