Field of a Straight Current

Transcript

0:00Introduction

Hello everyone! Let us start with a small puzzle. Take a straight wire carrying a current, and make it infinitely long. Every tiny piece of it adds a little magnetic field at a nearby point. Infinitely many pieces, all adding up, so the field must be infinite, right? Hold that thought for a while. This result comes up often in JEE, inside loops and bent wires. So, let's begin.

0:27Setting up: one wire, one point

So, here is the set-up. The thick line, from end A up to end B, is our wire, and the orange arrow shows the current, capital I, flowing up. We want the field at the point P, to the right of the wire. The dashed perpendicular from P meets the wire at N, and its length, small d, is our distance. The ends make angles at P with this perpendicular: the Greek letter alpha towards end A, and beta towards end B, okay? Now take one tiny element, the orange piece, I times d l. The blue arrow, r, runs from the element to the point P, and the little arrows on top of d l and r mark them as vectors. Beside the figure is the Biot-Savart law from the last lesson; mu naught, the Greek letter mu with a small zero, is the permeability of free space, and theta, the Greek letter theta, is the angle between d l and r. Measure everything with one angle, the Greek letter phi, taken from the dashed line P N, and a few lines of trigonometry give d B equals mu naught I over four pi d, times cos phi, d phi. With me? Now let me ask you this. Every piece gives its own little field, so do they all point the same way, or do we need vector addition? They all point the same way, that is the beauty of it. Each d B is perpendicular to the plane containing the wire and the point P, which in the figure means into the page: the circle with a cross, next to B with an arrow on top. So the contributions simply add. Got it?

2:04Add them up: the finite-wire formula

Now we add. Phi runs from minus alpha at end A, to plus beta at end B, and the integral of cos phi is sine phi. So B equals mu naught I over four pi d, times sine alpha plus sine beta. This is our formula, printed beside the figure too. Okay? Now a very important question. These angles, alpha and beta: are they measured from the wire, or from the perpendicular P N, the dashed line? From the perpendicular P N, yes, exactly! Phi was measured from P N, so alpha and beta are too; in the figure, both open out from the dashed line. Right? Now, here is the trap. Some books write cos theta one minus cos theta two, with small subscripts one and two, and those angles are measured from the wire itself. Both forms are correct, but mixing them is the commonest error in this topic. Clear?

3:01Special cases from one formula

Now, back to our puzzle. Make the wire infinitely long in both directions. What do alpha and beta become? They become ninety degrees each, that is correct! Sine ninety is one, so B equals mu naught I over two pi d. And there is our puzzle solved: the far pieces are seen at angles close to ninety degrees, where cos phi is almost zero, and phi can never go past ninety degrees, so the sum stays finite. The field lines are circles around the wire. With me? Next, a semi-infinite wire, with P on the perpendicular through its end. Alpha is zero and beta is ninety degrees, so B is mu naught I over four pi d, exactly half of the infinite wire. That makes sense, right? If P lies on the line of the wire, or on its extension, then B is zero: d l and r lie along the same line, so sine theta is zero. Okay? And if both ends lie on the same side of N, the angles overlap, so use sine beta minus sine alpha, with alpha for the nearer end and beta for the farther end.

4:13Topper's trick: loops made of straight sides

Clear? Now a trick that toppers use. Any loop of straight sides is just finite wires, end to end. For a regular polygon with n sides, the sides all add at the centre, and B equals n times mu naught I over two pi capital R, times tan of pi over n, where capital R is the distance from the centre to a corner. Okay? For a square of side small a, each side is small a over two away, with alpha and beta both forty five degrees. So B equals two root two times mu naught I, over pi times a; the tick-shaped sign is the square root, so that is root two. Right? Now let n tend to infinity; the arrow means tends to, and the sideways eight is infinity. Then B tends to mu naught I over two capital R, the field at the centre of a circular loop. So a circle is a polygon with very many sides. Beautiful, isn't it?

5:07Example 1: 30 cm, 6 A wire, field 20 cm from its midpoint

Time for example one. Here is a straight wire, thirty centimetres long, carrying six amperes; c m is centimetres and capital A is amperes. Find the field at a point twenty centimetres from its midpoint, on the perpendicular bisector. Pause the video and try it first. Done? Okay, let's check. Here d is twenty centimetres, or zero point two metres, small m on screen. Each half of the wire is fifteen centimetres, so each end is the square root of fifteen squared plus twenty squared away, which gives twenty five centimetres. That is a three four five triangle! Okay? Measured from the perpendicular, sine alpha is fifteen over twenty five, which is zero point six. By symmetry, sine beta is the same. Clear? Now plug in, using mu naught over four pi equal to ten to the power minus seven. So B equals ten to the power minus seven times six, over zero point two, times zero point six plus zero point six. With me? That is three times ten to the power minus six, times one point two, so B equals three point six times ten to the power minus six tesla, capital T on screen. Very good!

6:27Example 2: same wire, field 20 cm from one end, alpha = 0

Now example two, a little harder. Same wire, thirty centimetres and six amperes. But now the point is twenty centimetres from one end, on the perpendicular to the wire through that end. Pause the video and try it first. Done? Okay, let's check. The line from our point to the near end makes ninety degrees with the wire. So alpha is ninety degrees, right? Oh no, careful, this is exactly the trap we talked about! Alpha is measured from the perpendicular, and here the line to the near end is the perpendicular itself. So alpha is zero, and the whole thirty centimetres of wire lies on one side. Got it? For the far end, sine beta is thirty, over the square root of thirty squared plus twenty squared. That works out to zero point eight three two. Okay? Here d is again twenty centimetres, zero point two metres. So B equals ten to the power minus seven times six, over zero point two, times zero point eight three two. That is three times ten to the power minus six, times zero point eight three two, which gives two point five times ten to the power minus six tesla. Opposite the end, the field is smaller than the three point six times ten to the power minus six tesla of example one. Perfect!

7:54Quick revision: finite, infinite, semi-infinite wire, loops

So, quick revision. For a finite wire, B equals mu naught I over four pi d, times sine alpha plus sine beta, with the angles from the perpendicular. An infinite wire gives mu naught I over two pi d, a semi-infinite one, at a point opposite its end, half of that, and a point on the line of the wire gets zero. And for loops of straight sides, add the fields of the sides, which all point the same way. Now go and try the practice problems on this page. You can do them. All the best!