JEE Advanced, AIEEE and Earlier JEE Main Questions: Waves

Transcript

0:00Introduction

Hello everyone! Here is a small puzzle. Take two constants of empty space, mu naught and epsilon naught. Divide the first by the second, and take the square root. The answer comes out in ohms, the unit of resistance! How can that be? Today we solve past JEE questions on electromagnetic waves, from JEE Advanced, A I triple E and the earlier JEE Main papers. Questions like these come up often, so let's begin.

0:32Transverse waves and their directions

In the figure, the wave travels along the x axis; the electric field, E with a small arrow for a vector, swings along the y axis, and the magnetic field, B, along the z axis, in phase. Now, A I triple E two thousand and two asked: which property shows that these waves are transverse? Polarisation, yes, exactly! Only transverse waves can be polarised, and these waves can be, so they are transverse. Okay? Next, A I triple E twenty twelve. X with an arrow is the direction of polarisation. The electric field sets the direction of polarisation, so X is parallel to E; the two upright lines mean parallel. Right? Now, travel is perpendicular to both fields, so it lies along a cross product, the small cross on screen. But in which order: E cross B, or B cross E? It is E cross B, and the order matters! So k with an arrow, the direction of travel, is parallel to E cross B, as on screen. E, B and the direction of travel, in that order, form a right handed set. Swap them, and B cross E points exactly backwards, a common slip.

1:47Entering a medium: what changes?

Clear? Now the speed, from IITJEE nineteen ninety two. In vacuum, c equals one over the square root of mu naught epsilon naught; the tick shaped sign is the square root. Mu naught, the Greek mu with a small zero, is the permeability of free space, and epsilon naught, the curly e with a small zero, the permittivity. In a medium of permeability mu and permittivity epsilon, the speed, small v, is one over the square root of mu epsilon. With me? So tell me, what is the refractive index, n, of this medium? It is c over v, yes, exactly! So n equals the square root of, mu epsilon over mu naught epsilon naught. Got it? Next, A I triple E two thousand and four: a wave goes from vacuum into a dielectric of relative permittivity four, written epsilon r, with a small r below. Taking it as non magnetic, as the screen notes, n is the square root of epsilon r, the square root of four, which is two. Okay? Its frequency is three point zero megahertz, the M H z on screen, and inside it moves at half the speed. So what gets halved: the frequency, or the wavelength? Many students halve the frequency, but careful, it does not change at all! The source fixes it, so it stays three point zero megahertz. Right? So the wavelength changes instead. Lambda, the Greek letter for wavelength, equals lambda naught, the wavelength in vacuum, over n. With n equal to two, it is halved. Clear?

3:21Dimensions from two relations

Now dimensions, from JEE Advanced twenty eighteen. How are the dimensions of E and B related, in terms of length and time? In an electromagnetic wave, E equals c times B, and c is a length over a time. The square brackets mean dimensions, L is length, T is time, and the double arrow means which gives. So the dimensions of E are those of B, times L, times T to the power minus one. Okay? The same paper asked about mu naught and epsilon naught. Use mu naught epsilon naught equals one over c squared, so mu naught has the dimensions of epsilon naught to the power minus one, times L to the power minus two, times T squared. With me? Now a topper's trick, our puzzle itself. The square root of mu naught over epsilon naught is about three hundred and seventy seven ohms; the wavy equals sign means approximately, and the capital omega means ohms. It has the dimensions of a resistance: V over I, voltage over current. Got it?

4:24Example 1: peak E field from a peak B of 20 nT

Now Example one, the easier one, from JEE Main twenty thirteen. The peak magnetic field of a travelling electromagnetic wave is twenty nanotesla, the n T on screen; find the peak electric field. Pause the video and try it first. Done? Okay, let's check. For peak values, E naught equals c times B naught; the small zero means the peak value. Right? So we just put B naught equals twenty in the formula, correct? No, no, wait, that is the trap! Nano means ten to the power minus nine, so B naught is twenty times ten to the power minus nine tesla, the capital T on screen. Got it? Now we multiply: three times ten to the power eight, times twenty times ten to the power minus nine. Three times twenty is sixty, and the powers of ten leave ten to the power minus one, so sixty times ten to the power minus one gives six. So E naught is six volts per metre, the V m to the power minus one on screen. Easy, isn't it?

5:28Example 2: dimension check with mu0, epsilon0, c, V and I

Now Example two, the harder one, from JEE Advanced twenty fifteen. Capital V is the potential difference, capital I the current, and c the speed of light. Which of the four relations on screen are dimensionally correct? Use the topper's trick, and pause the video to try it first. Done? Okay, let's check. By our trick, mu naught over epsilon naught has the dimensions of a resistance squared, capital R squared, which is V squared over I squared. Right? The first relation, rearranged, says mu naught over epsilon naught equals V squared over I squared. It matches, so the first is correct. Clear? Now the second: epsilon naught times I equals mu naught times V, with the same two constants. So it should be fine too, right? Oh no, careful, it is wrong! For that, epsilon naught over mu naught would have to be V over I, a resistance. But it is the flip of R squared, I squared over V squared. Got it? Now the third, which has c in it. Since c is one over the square root of mu naught epsilon naught, epsilon naught times c is the square root of epsilon naught over mu naught, which is one over R, so epsilon naught times c times V is V over R, a current, and the third is correct! With me? The fourth works the same way: mu naught times c is the square root of mu naught over epsilon naught, which is R itself. So mu naught times c times I is R times I, a voltage, which cannot equal epsilon naught times V; the crossed equals sign means not equal to. The fourth is wrong. Clear? So the answer is the first and the third. And our puzzle is solved: the square root of mu naught over epsilon naught is in ohms because its dimensions are a voltage over a current, just like any resistance. Perfect!

7:19Quick revision: polarisation, n = c/v, E0 = cB0, 377 ohms

So, quick revision. Polarisation proves the waves are transverse; it is along E, and travel is along E cross B, in that order. In a medium, n is the square root of mu epsilon over mu naught epsilon naught; the frequency stays fixed, the wavelength becomes lambda naught over n. For dimensions: E naught equals c times B naught, mu naught epsilon naught is one over c squared, and the square root of mu naught over epsilon naught is about three hundred and seventy seven ohms. Next, try the recent JEE Main questions. All the best!