Magnetic Flux

Transcript

0:00Introduction

Hello everyone! Let us start with a small puzzle. It is raining straight down, and you hold a ring under the rain. Held flat, like a plate, it catches lots of drops, right? Now stand the ring up on its edge. So how many drops pass through it now? Hold that thought for a minute. Magnetic flux works the same way, with field lines in place of raindrops. Every induced current in this unit comes from a change in this flux, and it comes up often in JEE problems. So, let's begin.

0:34Magnetic flux through a flat surface

So, what is magnetic flux? Write a tiny piece of area as a vector, d A with an arrow on top, pointing along the normal, the line perpendicular to the surface. Then d capital phi, the Greek letter phi, equals B vector dot d A vector, the dot product. Okay? Now take a flat surface of area capital A in a uniform field. The dot product gives capital phi equals B times A times cos theta, where theta, another Greek letter, is the angle between B and the normal. Right? Now let me ask you this. You slowly turn a loop in this field. When is the flux the largest, and when does it become zero? When the plane of the loop is perpendicular to the field, the flux is the largest, yes! Then the normal lies along B, so theta is zero degrees, the small raised circle meaning degrees. Cos zero is one, so phi equals B times A, like our ring held flat in the rain. Clear? And when the plane is parallel to the field, the normal is perpendicular to the field. So theta is ninety degrees, cos ninety is zero, and the flux is zero. That answers our puzzle: no rain passes through a ring on its edge. Now careful, students slip here: plane parallel means theta is ninety, not zero. Got it? The S I unit of flux is the weber, written W b. One weber equals one tesla metre squared, the small raised two meaning squared, which is also one volt second. With me? Now, a coil of N turns. The flux phi passes through every turn, so the total, called the flux linkage, is N times phi, also in weber.

2:16Closed surfaces and curved surfaces

Okay? Now, closed surfaces, like a balloon or a box. Field lines enter through some parts and leave through others. So what is the net magnetic flux through the whole closed surface? It is zero, exactly! The net magnetic flux through any closed surface is zero. Right? And the reason is simple: there are no magnetic monopoles, no lone north pole or south pole where field lines could start or end. So every field line that enters must also leave. Clear? Now a topper's trick from this. Put a bowl together with the flat surface across its rim, and you get a closed surface, with zero net flux. So the curved open surface carries the same flux as the flat surface with the same rim. Okay? So take a bowl of radius capital R, with its axis along the field. Forget the curve, and use the flat circle of area pi R squared. Phi curved, the small word curved naming the surface, equals pi R squared times B, the field. Got it?

3:22Non-uniform field: integrate

Now, what if the field is not uniform? Then we cannot just multiply. We add up the flux of tiny pieces: phi equals the integral of B vector dot d A vector, the long S sign being the integral. And where B lies along the normal, this is simply the integral of B, d A, with no dot product needed. Okay? The classic case is a current I in a long straight wire. Its field at distance r is B equals mu naught times I, over two pi r, where mu naught, the Greek mu with a small zero below, is the permeability of free space. Right? Now place a rectangle beside the wire, in the same plane as the wire, with small a, its length along the wire, and with its sides at distances d one and d two, the small one and two written below. In that plane, the field is perpendicular to the rectangle everywhere. Cut it into thin strips along the wire, so d A equals a times d r. With me? So the flux of each strip is B times a times d r, and B goes as one over r. That means we need the integral of d r over r. What does that give? The natural log, yes, exactly! From d one to d two, it gives the natural log of d two over d one, written l n on screen. So phi equals mu naught times I times a, over two pi, times the natural log of d two over d one. Clear?

4:43Example 1: flux through a loop at 30 degrees to the field

Let's try our first example. There is a circular loop of radius ten centimetres, in a field of zero point five tesla, placed so that its plane makes thirty degrees with the field. Find the flux through it. Pause the video and try it first. Done? Okay, let's check. The angle given is thirty degrees, so we put cos thirty into the formula, right? No, no, careful, that is the trap! Thirty degrees is the angle with the plane, but theta is measured from the normal. The normal is perpendicular to the plane, so theta is ninety minus thirty, which is sixty degrees. With me? Next, the area. Ten centimetres is zero point one metre, so the area is pi times zero point one squared, which is pi times ten to the power minus two square metre. Okay? Now the formula. Phi equals zero point five, times pi times ten to the power minus two, times cos sixty degrees, and cos sixty is one half. Right? Zero point five times one half is zero point two five. So phi is zero point two five pi times ten to the power minus two, which is approximately seven point eight five times ten to the power minus three weber, the wavy equals sign meaning approximately. Clear?

6:01Example 2: flux of a straight wire through a square loop

Now a harder one. There is a square loop of side ten centimetres, in the plane of a long straight wire carrying ten amperes. Its nearer side is parallel to the wire, and ten centimetres from it. Find the flux through the loop. Pause the video and try it first. Done? Okay, let's check. First, can we just use B times A times cos theta here? We cannot, because the wire's field is not uniform, so we use the integrated result. Here I is ten amperes, and a, the side along the wire, is zero point one metre. The near side is at ten centimetres, and the far side at ten plus ten, which is twenty centimetres. Okay? Mu naught is four pi times ten to the power minus seven tesla metre per ampere, the raised minus one meaning per. So mu naught over two pi is two times ten to the power minus seven, in the same unit. Right? Now substitute. Ten times zero point one is one, and twenty over ten is two, so phi is two times ten to the power minus seven, times l n two. The centimetres cancel inside the ratio, but the length a had to be in metres. Got it? The natural log of two is about zero point six nine three. So phi is approximately one point three nine times ten to the power minus seven weber. Non-uniform field, so we integrate, and the log does the work. With me?

7:28Quick revision: BA cos theta, closed surfaces, integrate

So, quick revision. First, phi equals B times A times cos theta, with theta from the normal, and flux linkage is N times phi, in weber. Second, the net magnetic flux through any closed surface is zero. Third, in a non-uniform field, integrate B dot d A over the surface. Now go and try the practice problems on this page. All the best!