Mixed Problems: Set 1 (Induction and Motional EMF)

Transcript

0:00Introduction

Hello everyone! Here is a small puzzle. Take a metal rod on two smooth rails, in a magnetic field, with the rails joined by a resistance. You push the rod once, and let go. No friction, so it slides forever, right? Keep your answer in mind; we settle it in example two. Mixed problems like these, on Faraday's law and motional emf, come up often in JEE Main. So, let's begin.

0:28Idea 1: Faraday's law

First, Faraday's law. The curly E is the induced emf, capital N the number of turns, and capital phi, the Greek letter, the magnetic flux, so d phi over d t is how fast the flux changes. The minus sign is Lenz's law: the induced current opposes the change in flux. Okay? Now our loop: its flux is four times t squared, minus three times t, in milliweber, the m W b on screen. At t equal to one second, the small s, the flux is four minus three, one milliweber. So the emf is one millivolt, right? No, no, wait, that is exactly the trap! The emf depends on how fast the flux changes, not on the flux itself. So differentiate: d phi over d t is eight times t, minus three, in milliweber per second; the s to the power minus one on screen means per second. With me? Put t equal to one: eight minus three gives five. One milliweber per second is one millivolt, so the emf is five millivolts, the m V on screen. The vertical bars mean modulus, the size; the sign only gives the direction. Here d phi over d t is positive, so the flux is growing, and by Lenz's law the induced current flows the way that opposes this growth. Clear?

1:53Idea 2: the charge does not care about time

Next idea. The flux through each turn of a one hundred turn coil changes by five times ten to the power minus four weber, the W b on screen, and the resistance is twenty ohms, the Greek capital omega. We want the charge, but no time is given! Can we still find it? Yes, we can! The current I is the emf over the resistance R, and the emf is N times d phi over d t. Add up current times time, and the d t cancels: q equals N times the change in flux, over R, okay? Capital delta means change in, and the double arrow means gives. Change the flux faster, and the current is bigger, but for less time, so the charge stays the same. The emf and the current do depend on the time, though. Got it? One hundred times five times ten to the power minus four is five times ten to the power minus two. Over twenty, that is two point five times ten to the power minus three coulombs, capital C, or two point five millicoulombs, the m C on screen.

3:01Idea 3: motional emf

Right? Now idea three, motional emf. Take a straight rod of length small l, moving with velocity v, in a magnetic field B, okay? If all three are mutually perpendicular, the emf is B times l times v; if not, use only the components that are mutually perpendicular. Clear? Now the aircraft: its wingspan is forty metres, the small m, and it flies horizontally at nine hundred kilometres per hour, k m h to the power minus one, where the vertical field is five times ten to the power minus five tesla, the capital T on screen. The wings and the velocity are both horizontal, so which field direction is perpendicular to both? Vertical, yes, exactly! So only the vertical field counts; a horizontal field gives no emf between the wing tips. With me? One more trap: do not put in nine hundred as it is. Nine hundred times one thousand, over three thousand six hundred, gives two hundred and fifty metres per second. Clear? Now B l v: five times ten to the power minus five, times forty, times two hundred and fifty. Forty times two hundred and fifty is ten thousand, so the emf is zero point five volt, the capital V on screen, which is half a volt. Okay?

4:24Example 1: rod on rails, induced current and retarding force

Alright, example one. Here a rod slides at ten metres per second on rails half a metre apart, in a field of zero point two tesla, with the rails joined by four ohms. Find the current, and the force needed to keep the speed constant. Pause the video and try it first. Done? Okay, let's check. In the figure, the crosses show the field B, with its vector arrow, into the page; R is the resistance, the rails are small l apart, and the rod moves right with speed v. All three are mutually perpendicular, so B l v is zero point two times zero point five times ten, which is one volt. Clear? The current is the emf over R: one over four, zero point two five ampere, the capital A on screen. Which way? As the rod moves right, the flux into the page through the loop grows, so the induced current makes a field out of the page to oppose it. That is anticlockwise, as the orange arrows show, with I going up the rod. With me? Now, why is any force needed to keep the speed constant? Because the rod carries a current in a field, it feels a magnetic force, B times I times l. By Lenz's law this force opposes the motion, so we must push forward with an equal force. Okay? Zero point two, times zero point two five, times zero point five, gives zero point zero two five newton, the capital N here, which is twenty five millinewtons. Got it?

6:00Example 2: how far a rod slides on rails before it stops

Now, back to our puzzle, with numbers. Here a rod of mass zero point two kilogram, the k g on screen, lies on rails half a metre apart, in a field of one tesla, with one ohm of resistance. Given five metres per second, how far does it go before stopping? Pause the video and try it first. Done? Okay, let's check. After the push, the only force along the rails is the retarding force at the bottom of the figure: since I equals B l v over R, the force B I l becomes B squared l squared v, over R, against v. Right? So the rod slows down. And as it slows, what happens to this force? It gets smaller too, since it is proportional to v, so the speed dies away exponentially: v equals v naught, times e to the power minus t over tau. Here v naught, with the small zero, is the starting speed, and tau, the Greek letter, is the time constant, m R over B squared l squared, okay? Over all time, the total distance is finite: x equals v naught times tau, which is m times v naught times R, over B squared l squared. Clear? On top, zero point two times five times one is one; below, one squared times zero point five squared is zero point two five. One over zero point two five gives four metres; forget to square the half metre, and you get two, which is wrong. So the puzzle is settled: no friction, yet the rod does not slide forever. Its speed dies away to zero, and in all it covers only four metres, its kinetic energy turned into heat. With me?

7:47Quick revision: Faraday, charge flown, Blv and rod on rails

So, quick revision. Faraday's law: the emf is minus N d phi over d t, and the charge is N times the change in flux over R, whatever the time. Motional emf is B l v, with field, rod and velocity mutually perpendicular, in S I units. On rails, B squared l squared v over R opposes the motion, so with no push, the rod slides a total distance of v naught times tau. Now go and try the practice problems on this page. All the best!