Magnetic Field and the Biot–Savart Law

Transcript

0:00Introduction

Hello everyone! Here is a small puzzle. Put a charge on your table and keep it still. It makes an electric field, and nothing more, right? Now give it a push, and a magnetic field appears. But this field does not point along the motion, or away from the charge. So where does it point? Hold that question; the Biot Savart law answers it today. Every field in this chapter comes from this law, and it comes up often in JEE problems. So, let's begin.

0:30Moving charges make a magnetic field

So, a charge at rest produces only an electric field, E with a little arrow on top, the arrow marking a vector. Okay? But a moving charge also produces a magnetic field, capital B with an arrow. And a current is just moving charges, so every current makes one too. Right? The S I unit is the tesla, capital T, and one tesla is one newton per ampere per metre, the raised minus ones meaning per. The older unit, the gauss, capital G, is ten to the power minus four tesla. Got it? So, is one tesla a big field, or a small one? It is big, very big! The Earth's field is only about zero point three to zero point six gauss, while a lab magnet gives about one tesla, which is ten thousand gauss. With me?

1:23The Biot–Savart law

Now, the main law. Cut a wire into tiny pieces. Each piece is a current element, I d l, where d l with an arrow is a tiny length along the current. Okay? The Biot Savart law gives the tiny field d B of this element at a point, with the vector r running from the element to the point. d B vector equals mu naught over four pi, times I d l cross r, over r cubed. Mu naught is the Greek mu with a small zero below, and the cross is the cross product. Right? Now tell me, what is the magnitude of a cross product? It is the product of the two sizes, times the sine of the angle between them, exactly! So one r cancels, and d B equals mu naught over four pi, times I d l sine theta, over r squared, with theta, the Greek letter, the angle between d l and r. Clear? And mu naught, the permeability of free space, is four pi times ten to the power minus seven tesla metre per ampere. So mu naught over four pi is just ten to the power minus seven. Got it?

2:30What the law tells us

Now, d B is proportional to one over r squared, that curly sign meaning proportional to, like Coulomb's law. So does d B also point along r, like the electric field of a charge? No, no, careful, it does not! Because of the cross product, d B is perpendicular to both d l and r. And that answers our opening puzzle: the field points sideways. With me? Now take a point straight ahead, on the element's own line. What is the field there? It is zero, yes, exactly! There theta is zero or one hundred and eighty degrees, the small raised circle meaning degrees, and the sine of both is zero. Okay? And sine theta is largest at ninety degrees, so, at a given distance r, d B is largest when the point is straight out to the side of the element.

3:24A single moving charge

For a point charge q moving with velocity v, just replace I d l by q v. So B vector equals mu naught over four pi, times q v cross r, over r cubed. Clear? Now, mu naught times epsilon naught, the Greek epsilon with a small zero, equals one over c squared, where c is the speed of light. Okay? So for a slowly moving charge, B vector is v cross E, over c squared. That makes it weaker than E over c by a factor v over c, and for a slow charge, v is tiny next to c. Got it?

4:00Direction: the right-hand rule

Now the direction, with a topper's shortcut: the right-hand rule. Grip a straight wire with your right hand, thumb along the current, and your curled fingers show the field. Okay? For a loop, the roles swap: fingers along the current, and the thumb gives the field at the centre. Right? On paper, the dot in a circle means out of the page, and the cross in a circle means into the page. Clear?

4:26Example 1: field at (0.3, 0.4, 0) m of a 1 cm, 10 A element

Now, Example one. One centimetre of wire at the origin carries ten amperes along plus x. Find its field at the point zero point three metre, zero point four metre, zero. Pause the video and try it first. Done? Okay, let's check. d l is zero point zero one metre, and r is the square root of zero point three squared plus zero point four squared, which is zero point five metre. Now, is theta ninety degrees here? No, it is not, and this is the trap! Theta is measured from d l, along the x axis, and the point is zero point four metre from that axis. So sine theta is zero point four over zero point five, which is zero point eight. Clear? Ten times zero point zero one times zero point eight is zero point zero eight. Dividing by zero point two five, which is r squared, gives zero point three two. Times ten to the power minus seven, d B is three point two times ten to the power minus eight tesla. Got it? Finally, the direction, from d l cross r. d l is along i hat, the unit vector along x, the little cap on top being called a hat. And r has a part along j hat, along y, and i hat cross j hat is k hat, so the field is along plus z. With me?

5:51Example 2: vector dB at r = 3i + 4k cm, using j cross i = -k

Now Example two, a step harder. An element I d l, two amperes times one millimetre along j hat, sits at the origin. Find d B at r vector equals three i hat plus four k hat, in centimetres. Pause the video and try it first. Done? Okay, let's check. For d l cross r, we first need j hat cross i hat. Is that plus k hat, or minus k hat? It is minus k hat, very good, because j then i goes against the order i, j, k. So j hat cross three i hat is minus three k hat, and j hat cross four k hat is plus four i hat. Okay? The size of r is the square root of nine plus sixteen, which is five centimetres. One millimetre is ten to the power minus three metre, and one centimetre, ten to the power minus two. And careful, in the vector form we divide by r cubed, not r squared. Right? On top we get two times ten to the power minus five, and below, one point two five times ten to the power minus four. Dividing them gives zero point one six, and times ten to the power minus seven, that is one point six times ten to the power minus eight. So d B is one point six times ten to the power minus eight, times four i hat minus three k hat, which is six point four i hat minus four point eight k hat, times ten to the power minus eight tesla. Clear? And the magnitude, between the two vertical bars. Four i hat minus three k hat has size five, so it is one point six times ten to the power minus eight, times five, which gives eight times ten to the power minus eight tesla. Got it?

7:43Quick revision: dB formula, direction, zero on the line

So, quick revision. First, d B equals mu naught over four pi, times I d l sine theta, over r squared, and mu naught over four pi is ten to the power minus seven. Second, d B is perpendicular to both d l and r, and zero on the element's line. Third, the direction comes from d l cross r, or the right-hand rule. Now go and try the practice problems on this page. All the best!