Part A: JEE Main Level

Transcript

0:00Introduction

Hello everyone! Okay, a small puzzle first. Two wires join the same two points: one straight, and one bent into a semicircle. Same current, same uniform magnetic field. The semicircle is longer, so it must feel a bigger force, right? Hold on to your answer. Today we solve quick JEE Main level problems on the magnetic effects of current, the kind that come up often, and we settle this puzzle very soon. So, let's begin.

0:30Idea 1: any shape, same end points

Idea number one is our puzzle. Here, a straight wire and a semicircular wire join the same two points, with the same current, in a uniform field perpendicular to both. Compare the forces on them. The rule for a straight wire is F equals I L cross B, with I the current, L the length of the wire and B the field. The arrows on top mean vectors, the vector L points along the current, and the cross means the cross product. In size, F equals ILB sine theta, with theta, the Greek letter, the angle between the wire and the field. Okay? Now, the semicircle is clearly longer. So more length means more force, right? No, no, wait, that is the trap! In a uniform field, the tiny pieces of any wire add up, as vectors, to the straight line from its start to its end. So a wire of any shape, from point P to point Q, feels F equals I times P Q, cross B, okay? The arrow over P Q means the straight line from P to Q, as a vector. Clear? Both our wires join the same points P and Q, with the current going from P to Q in each, so the forces are exactly equal, in size and in direction! Only the end points matter, not the length or the shape. So our puzzle is solved. With me?

1:54Idea 2: the radius of the circle

Now, idea number two. Here, a charged particle moves in a circle of radius two centimetres, written c m, in a field of zero point two tesla; the capital T stands for tesla. Find the radius if the field is raised to zero point four tesla, at the same speed. First, why does it move in a circle? With the velocity at right angles to the field, the force q v B is always at right angles to the velocity, so it only turns the particle, and q v B equals m v squared over r, with q the charge, v the speed, m the mass and r the radius. The double arrow means which gives: r equals m v over q B, okay? So tell me, the field doubles and nothing else changes. What happens to the radius? Yes, it gets smaller! B sits in the bottom, so with the same m, v and q, r is proportional to one over B; the sign like an open alpha means is proportional to. Right? So the field doubles and the radius halves: half of two centimetres is one centimetre. Stronger field, tighter circle. Got it?

3:06Idea 3: the period does not care about speed

Now, idea number three. Here, a proton and an electron move in the same field. Find the ratio of their periods. The period T, the time for one circle, is two pi r over the speed v, where pi is the Greek letter pi. Put in r equals m v over q B, and the v cancels! So T equals two pi m over q B, with no speed in it: fast or slow, one round takes the same time. Clear? So, the proton or the electron: which one takes longer to go round once? Yes, it is the proton, because it is much heavier. In the same field, T is proportional to m over mod q; the two vertical bars mean the modulus, the size of the charge without its sign. Plus e for the proton and minus e for the electron, so mod q is e for both, where e is the size of the electron's charge. Right? So the period ratio is just the mass ratio: T p over T e equals m p over m e, the small subscripts p and e meaning proton and electron. The wavy equals sign means approximately, and the ratio is about one thousand eight hundred and thirty six. With me?

4:22Example 1: field at centre of 31.4 cm wire bent into a loop

Alright, example one. Here, a wire thirty one point four centimetres long is bent into a single circular turn and carries two amperes; the capital A stands for ampere. Find the field at the centre. Pause the video and try it first. Done? Okay, let's check. We need the radius, so we put in thirty one point four centimetres, right? No, no, careful, that is the length of the wire, not the radius! Bent into one circle, the length becomes the circumference. So two pi R equals zero point three one four metres; change to metres first. With me? So R equals zero point three one four over two pi. With pi as three point one four, this gives zero point zero five metres, which is five centimetres. Okay? Now, at the centre of a loop, B equals mu naught I over two R, where mu naught, the Greek letter mu with a small zero, is the permeability of free space, four pi times ten to the power minus seven. So B equals four pi times ten to the power minus seven, times two, over, two times zero point zero five. Right? Two times zero point zero five is zero point one, so B equals eight pi times ten to the power minus six. Eight pi is about twenty five point one, so B is about two point five times ten to the power minus five tesla. Very good, that is our answer!

5:55Example 2: radius of alpha particle accelerated through 1 kV

Now a slightly harder one. Here, an alpha particle, the Greek letter alpha, is accelerated through one kilovolt, written k V, and enters a field of zero point five tesla at right angles. Find the radius. Pause the video and try it first. Done? Okay, let's check. The speed is not given, but the voltage is. From rest through a voltage V, it gains kinetic energy q V, so q V equals half m v squared. Put this v into m v over q B, and r equals one over B, times the square root of two m V over q. Okay? Now the key question. What is the charge q of an alpha particle? Yes, exactly, it is two e, twice the proton's charge! So q is three point two times ten to the power minus nineteen coulomb, with capital C for coulomb. Putting just e here is the common mistake, so the screen says not e. And the mass m is six point six eight times ten to the power minus twenty seven kilograms. Got it? Now the numbers: r equals one over zero point five, times the square root of, two times six point six eight times ten to the power minus twenty seven, times one thousand, over, three point two times ten to the power minus nineteen. The one thousand is one kilovolt in volts. With me? One over zero point five is two. Inside the root we get four point one seven five times ten to the power minus five, and its square root is about six point four six times ten to the power minus three. Right? Two times that gives about one point three times ten to the power minus two metres, which is one point three centimetres. Energy to speed, speed to radius, and the charge is two e. That's it, we are done!

7:52Quick revision: end points rule, r = mv/qB, alpha has 2e

So, quick revision. First, in a uniform field, any wire from P to Q feels the same force as a straight wire from P to Q would. Second, the radius is m v over q B, but the period, two pi m over q B, does not depend on speed. Third, at the centre of a loop, B is mu naught I over two R, and an alpha particle has charge two e. Now go and try the practice problems on this page. All the best!