Part A: JEE Main Level

Transcript

0:00Introduction

Hello everyone! Okay, a small puzzle first. Pull a coil out of a magnetic field, and a current flows round its circuit. Now do it twice as fast. Twice the emf, so twice the charge flows, right? Hold on to that; our first example settles it. This is part A of our induction and AC problem bank, at JEE Main level. So, let's begin.

0:27Tool 1: Faraday's law and Lenz's law

So, tool number one. Here, the flux through each turn of a one thousand turn coil falls from three times ten to the power minus five weber, written W b, to one times ten to the power minus five weber, in zero point zero one second, the small s. Find the average emf. Faraday's law says the curly E, the emf, equals minus N, times delta capital phi over delta t, with N the number of turns, capital phi, the Greek letter, for the flux through one turn, and capital delta, the triangle, for the change. The minus sign is Lenz's law. Okay? Now, which flux goes in? The starting flux, three times ten to the power minus five, right? No, no, careful, that is the trap! The emf comes from the change in flux. The flux goes from three to one, so delta capital phi is minus two times ten to the power minus five; the two vertical bars mean its size, three minus one, times ten to the power minus five, which gives two times ten to the power minus five weber per turn. With me? So the size of the emf is one thousand times two times ten to the power minus five, over zero point zero one, which gives two volts, V for volt, right? And the direction? The flux is falling. Does the induced field point against the original field, or along it? Along it, yes, exactly! By Lenz's law, the induced current opposes the change in flux, not the flux itself. The flux is falling, so the induced current makes its own field along the original field, holding the flux up. Clear?

2:02Tool 2: a rod sliding on rails

Now, tool number two. Here, a rod slides at ten metres per second, written m s to the power minus one, on rails zero point two metres apart, in a field of zero point five tesla, capital T, with circuit resistance one ohm, the Greek capital omega. Find the power dissipated. In the figure, the crosses show the field B, with a small arrow on top because it is a vector, pointing into the page; the rod moves at speed v, l is the gap between the rails, and R is the resistance. Okay? The emf is B l v, but only when B, l and v are mutually perpendicular, as here; otherwise, use the perpendicular components. Got it? So the emf is zero point five times zero point two times ten, which gives one volt. Right? The power P is the emf squared over R, so one squared over one gives one watt, capital W for watt. The red arrows show the induced current, which is B l v over R, okay? Which way does it flow? The rod moves right, so the area and the flux into the page grow, and by Lenz's law the current makes its own field out of the page, so it flows anticlockwise, up the rod. That current brings the retarding force written below, so your push turns into this heat. With me?

3:11Tool 3: mains voltage, rms and peak

Now, tool number three. Here, the mains voltage is two hundred and thirty volts. Find its peak value. So, is two hundred and thirty volts the highest it ever reaches? No, it is not, and that is the trap! AC values are quoted as rms, root mean square, so V rms, the V with a small r m s below, is two hundred and thirty volts. Okay? For a sinusoidal voltage, and only for a sinusoid, V rms equals V naught over root two, where V naught, the V with a small zero, is the peak value. Clear? So V naught is two hundred and thirty times root two, and the wavy equals sign means approximately: about three hundred and twenty five volts. Right?

3:54Example 1: charge through a coil pulled out of a field

Alright, example one, and it settles our puzzle. Here, a coil of two hundred turns, area five square centimetres, written c m squared, and circuit resistance fifteen ohm, is removed from a field of zero point three tesla normal to it. Find the charge that flows. Pause the video and try it first. Done? Okay, let's check. Notice, no time is given at all. So are we stuck? No, we are not. Charge q is current times time, I times delta t, and the current is the emf over R, okay? Put in Faraday's law and the delta t cancels: q equals N times the change in flux, over R, with no time in it. Twice as fast means twice the current for half the time, so the same charge. That solves our puzzle. Got it? Normal means the field is at right angles to the coil's plane, so the flux through one turn is B times A inside and zero outside: the change is B times A, right? Now careful: five square centimetres is five times ten to the power minus four square metres, because a centimetre is ten to the power minus two metres, and we square that. With me? So q equals NBA over R, which is two hundred times zero point three times five times ten to the power minus four, over fifteen. Clear? The top is zero point zero three, and zero point zero three over fifteen gives two times ten to the power minus three coulomb, capital C, which is two millicoulomb, written m C, our final answer.

5:22Example 2: series LCR, current, power factor and power

Now a slightly harder one. Here, a series circuit has R equals twenty four ohm, X L equals fifty ohm and X C equals eighteen ohm, the reactances of the inductor and the capacitor, the X with a small L or C below. It is connected to two hundred volts AC. Find the current, the power factor and the power. Pause the video and try it first. Done? Okay, let's check. In the figure, on the left are the voltage phasors: V R lies along the current I, V L points up, V C points down, and the slanted arrow is the supply voltage V, the sum of all three. On the right, the impedance triangle has R along the bottom, X L minus X C up the side, and Z, the impedance, on the slant, at angle phi, the Greek letter. So Z is the square root of R squared plus, X L minus X C, squared. Okay? Now tell me, why not simply add all three, twenty four plus fifty plus eighteen, to get ninety two ohm? No, no, wait, that is the trap! V L and V C are opposite, so the reactances subtract: fifty minus eighteen gives thirty two ohm, and R adds at right angles. Right? So Z is the square root of twenty four squared plus thirty two squared, which is root one thousand six hundred, so forty ohm. Got it? The current is V over Z, so two hundred over forty gives five amperes, capital A, the rms current, since AC values are quoted as rms. With me? The power factor, cos phi, equals R over Z, so twenty four over forty gives zero point six. And X L is bigger than X C, so the circuit is inductive, and the current lags the voltage by phi. Clear? Finally, the average power is V rms times I rms times cos phi, so two hundred times five times zero point six gives six hundred watts, and that's it, all done!

7:14Quick revision: emf vs charge, Blv, rms to peak, impedance

So, quick revision. First, the size of the average emf is N times the change in flux over the time, but the charge, N times the change in flux over R, does not depend on the speed. Second, a rod on rails gives B l v, with B, l and v mutually perpendicular. Third, the peak is root two times the rms for sinusoids, and Z is the square root of R squared plus, X L minus X C, squared. You handled some real JEE Main questions today, so well done. Now try the practice problems on this page. All the best!