Overview

Transcript

0:00Introduction

Hello everyone! Okay, one small puzzle before we start. You know that two long parallel wires, with currents in the same direction, attract each other. Now take two beams of electrons, moving side by side, in the same direction. Each beam is a current, and both currents point the same way. So the two beams should attract each other, right? Hold on to your guess for a few minutes. Today, we will first see how to use this unit's problem bank, and then settle this puzzle with a concept check from it. Questions like this come up often in JEE, so let's begin.

0:37What is in this chapter

So, first, let us see what this chapter is. It is the problem bank for the whole unit on magnetic effects of current and magnetism: mixed solved problems, in three parts. Okay? Part A has twelve solved problems, at JEE Main level. Part B has twelve more solved problems, at JEE Advanced level, and they combine several ideas of the unit. With me? And Part C has eleven concept checks, on the statements most often tested as true or false, or as assertion and reason. After each part, there are practice problems on the page for you: four after Part A, and three each after the other two parts. Got it?

1:19Part A: a 30-minute test

Now, when should you attempt Part A of this chapter? On the very first day of the unit, just to have a look? No, not on the first day. The course says to attempt it only after Chapters one to six, okay? That is because its problems are mixed: the fields of loops, wires and solenoids, charged particles moving in fields, force and torque on currents, and magnetism, all together. In the exam, no question tells you which chapter it comes from. Right? So finish every chapter first, and then attempt Part A as a real thirty minute test: solve each problem yourself, and only then check the solution. Clear? Now, quick maths. Twelve problems, and thirty minutes on the clock. So how much time can you give each problem? Thirty over twelve gives two point five, so you have two and a half minutes for each problem. So keep moving. With me?

2:16Part C: two concept checks

Now, a taste of Part C, with two checks from the course. The first one: can a static magnetic field, one that does not change with time, change the kinetic energy of a charged particle? The answer is no. The magnetic force is always perpendicular to the velocity, so it does no work. It turns the direction of motion, but the speed, and so the kinetic energy, stays the same. Okay? Now the second check, and it is our puzzle from the start. Two electron beams move side by side, in the same direction. Do they attract or repel? They repel, so if you said attract, you fell for the trap. As currents, the beams are parallel, so yes, there is a magnetic attraction. But electrons are like charges, and they also repel each other electrically. The magnetic attraction is weaker than this electric repulsion by the factor v squared over c squared, where v is the speed of the electrons and c is the speed of light. Since v is always less than c, the repulsion wins. Got it?

3:21Example 1: field at centre of 31.4 cm wire bent into a loop

Now, Example one, and it is from Part A, okay? Here, a wire thirty one point four centimetres long, the c m on screen, is bent into a single circular turn, and it carries two ampere; the capital A means ampere. Find the magnetic field at the centre. Pause the video, and try it first. Done? Okay, let's check. The whole wire becomes the circle, so its length is the circumference, two pi R, where capital R is the radius. And change to metres first: thirty one point four centimetres is zero point three one four metre, and the small m means metre. Right? So, R equals zero point three one four over two pi. With pi as three point one four, that gives zero point zero five metre, or five centimetres. Clear? Now, which formula? On screen you see mu naught, the Greek letter mu with a small zero below it: the permeability of free space. Capital I is the current. So, is the field mu naught I over two pi R, or mu naught I over two R? For the centre of a circular loop, B, the magnetic field, is mu naught I over two R, yes! The other one, the one with the pi, is the field of a long straight wire, not of a loop. Do not mix these two up. With me? Now put in the numbers. On top, mu naught, four pi times ten to the power minus seven, times the current, two. The bottom is two times zero point zero five, which is zero point one. That gives eight pi times ten to the power minus six tesla, and the capital T means tesla. Okay? Eight pi is about twenty five point one, so B is about two point five times ten to the power minus five tesla; the wavy equals sign means approximately equal. That fits easily in two and a half minutes. Got it?

5:20Example 2: max electron speed to miss the solenoid winding

Now Example two, from Part B, a harder one that mixes two chapters. An electron is emitted from the axis of a long solenoid. The solenoid has radius capital R, small n turns per metre, and current capital I, okay? The electron's speed is v, perpendicular to the axis. Find the largest v for which it does not strike the winding. Pause the video, and try it first. Done? Okay, let's check. Inside a long solenoid, the field B is mu naught n I, uniform, and along the axis. Right? The electron moves perpendicular to this field; the upside down T on screen means perpendicular. So it moves in a circle of radius r equals m v over e B, where m is the electron's mass and e is the size of its charge. And remember Part C: the magnetic force does no work, so the speed stays v all the way round. Clear? Now, many students write: the radius r must be less than capital R, and that is it. The angled sign on screen means less than. Is that enough? No, no, wait, that is the trap! The electron starts on the axis, so its circle passes through the axis. The farthest point of the circle from the axis is a full diameter away, two r, and not just r. With me? So, to miss the winding, we need two times m v over e B to be less than capital R, the radius of the solenoid. Okay? Now we just solve this for v. The largest speed, v max, the v with the small word max below it, equals e B R over two m. Put in B equals mu naught n I, and v max equals e mu naught n I R over two m. With just r less than R, your answer would be twice too large. Got it?

7:16Quick revision: loop-centre field, no work, 2r less than R

So, quick revision. Part A is a thirty minute test after Chapters one to six; Part B is at JEE Advanced level; and Part C has the concept checks. At the centre of a loop, the field is mu naught I over two R, not the straight wire's mu naught I over two pi r; and a magnetic force does no work. And a charge that starts on the axis of a solenoid reaches two r from it, so we need two r to be less than capital R, the solenoid's radius. Now set a thirty minute clock, open Part A, and begin. All the best!