Field of Arcs, Circular Loops and Coils

Transcript

0:00Introduction

Hello everyone! First, a small puzzle. Take a current loop, with some magnetic field at its centre. Now walk along its axis, one radius away. How much of the field is left? Is it half? Hold on to your guess. Today we find the field of arcs, loops and coils, which comes up often in JEE magnetism. So, let's begin.

0:25Field of an arc at its centre

So, take an arc of radius R carrying a current I, and let us find the field at its centre. Break it into tiny elements d l; the arrow on top means a vector. Each one is R away from the centre, at ninety degrees to the radius; the little circle means degrees. Okay? Now, the Biot Savart law has the sine of the angle between the element and the line to the point. Here that angle is ninety degrees. So what is sine ninety? Yes, exactly, sine ninety is one! So d B equals mu naught over four pi, times I d l, over R squared. Mu naught, mu with a small zero, is the permeability of free space. By the right-hand rule, all these fields point the same way, so they simply add. Clear? Now, d l equals R times d phi, where phi, the Greek letter phi, is the angle of the arc at the centre; the double arrow means this gives. So B arc, with a small subscript arc, equals mu naught I phi, over four pi R, with phi in radians, never degrees. With me?

1:32Semicircle, full loop and coil

Now let us use it. What angle does a semicircle make at its centre, in radians? Yes, it is pi radians! Put phi equals pi, the pis cancel, and B equals mu naught I over four R, okay? For the full loop, phi is two pi, so B equals mu naught I over two R, at the centre. Remember this one well. Right? And a coil of N turns is N loops stacked together, so the fields add: mu naught N I over two R, at its centre. Clear?

2:06On the axis of a loop

Now move along the axis; look at the figure. The loop has centre O and radius R, the orange arrow I is the current, and P is on the axis; x, its distance from O, is marked below. The blue vector r runs from the element d l to P, and its length r is the square root of R squared plus x squared. And d l is perpendicular to r; the upside-down T means perpendicular. Got it? Each element gives a small field d B, the green arrow, tilted off the axis. The opposite element tilts its d B the other way. So what happens to the sideways parts? Yes, exactly, the perpendicular parts cancel in pairs, just as the figure says! Only the parts along the axis survive, and those all add. Clear? That axial part is d B times R over r; the dot means times. Here d B has the same Biot Savart form as before, with r in place of R, and every element gives the same amount, so when we add them up, all the d l's together make the whole length of the loop, two pi R, okay? Put it all together and simplify, and B equals mu naught I R squared, over two times R squared plus x squared, to the power three over two, as at the top right of the figure, with the centre value below it. For a coil of N turns, the field is N times bigger. With me?

3:33How fast the field falls along the axis

Now, back to our puzzle. Call the centre field B naught, the B with a small zero; it is mu naught N I over two R, okay? Walk one radius along the axis, so x equals R, and many students say half the field is left. Is that right? No, careful, it is less than half! R squared plus R squared is two R squared, and to the power three over two, that is two root two times R cubed. So the field is B naught over two root two, about zero point three five times B naught; the wavy equals sign means approximately. Got it? At x equals root three times R, R squared plus x squared is four R squared, and to the power three over two, that is eight R cubed. So the field drops to one eighth of B naught. Right?

4:24Topper's trick: arcs plus straight pieces

Now a topper's trick for wires made of arcs and straight pieces. Suppose a straight piece lies on a line passing right through the centre. What field does it give there? It gives zero field, absolutely nothing. Along that piece, d l is parallel to r; the two vertical bars mean parallel. So the angle is zero or one eighty degrees, and its sine is zero. Okay? So throw those pieces out, and add the arcs and the other straight pieces, each with its sign from the right-hand rule, into the plane or out of it. With me?

5:00Example 1: field at centre of a 4 cm, 2 A quarter-circle arc

Alright, example one. Here is a quarter-circle arc of radius four centimetres, c m, carrying two amperes, capital A for amperes. Find its field at the centre. Pause the video and try it first. Done? Okay, let's check. It is a quarter of a circle, so phi is ninety, right? Oh no, careful, ninety is in degrees, and we need radians. Quarter of two pi is pi over two. And four centimetres is zero point zero four metres, small m for metres. Got it? Now, B equals mu naught I phi over four pi R, and mu naught over four pi is just ten to the power minus seven. So B equals ten to the power minus seven, times two times pi over two, all over zero point zero four, okay? Two times pi over two is just pi. Then pi over zero point zero four, times ten to the power minus seven, gives about seven point eight five times ten to the power minus six tesla; capital T is tesla. Using ninety would give fifty seven times too much. Clear?

6:10Example 2: 3 cm, 10 A loop, field on axis 4 cm from centre

Now a slightly harder one. Take a loop of radius three centimetres carrying ten amperes. Find the field on its axis, four centimetres from the centre. Pause the video and try it first, and look for a famous triangle. Done? Okay, let's check. Here the radius is three centimetres and O P is four. So R squared plus x squared is nine plus sixteen, which is twenty five square centimetres, and r, the blue line, is five centimetres. Yes, the three, four, five triangle! Okay? Now careful, convert to metres first, because the value of mu naught is for metres. So B equals four pi times ten to the power minus seven, times ten, times zero point zero three squared, over two times zero point zero five cubed. See, it is r cubed downstairs. Right? Zero point zero three squared is nine times ten to the power minus four, and two times zero point zero five cubed is two point five times ten to the power minus four. Divide, and you get three point six. With me? Ten times three point six is thirty six, so B is four pi times ten to the power minus seven, times thirty six. That is about four point five two times ten to the power minus five tesla. Keep centimetres by mistake, and your answer is a hundred times too small. Clear?

7:37Quick revision: arc, loop centre and axis, field at x = R

So, quick revision. Arc at its centre: mu naught I phi over four pi R, phi in radians. Loop: mu naught N I over two R at the centre, and on the axis, mu naught NIR squared, over two times R squared plus x squared, to the power three over two. And one radius along the axis, only one over two root two of the centre field is left, not half. Now go and try the practice problems on this page. All the best!