Shadowing Practice: LET THERE BE... Voltage? | Maxwell's Equation #2 Explained for Beginners - Learn English Speaking with Video

What's up you lot, Parth here and today I'm going to be talking about one of Maxwell's four equations of electromagnetism.

Now, I've already made a video about one of them in the past and you guys seem to enjoy it, so check it out up here if you haven't seen it already.

Although don't worry, you don't need to have seen it to understand what I'm going to be talking about in this one.

But still go and check it out if you haven't already, it's actually quite a good video if I do say so myself.

Anyway, let's get into it.

Well, no, hang on, I can't do that yet.

I have to say first of all, Thank you so much for 7,000 subscribers.

That is ridiculous.

We are well on our way to 10k, which is, you know, pretty cool.

So thank you so much if you've subscribed already.

And if you haven't, then please do consider it, depending on whether or not you like this video.

So without further ado, let's get into the video so you can figure out whether or not you like it.

So if you don't know, Maxwell's equations are a series of four equations compiled by James Clerk Maxwell that describe everything to do with electricity and magnetism,

collectively known as electromagnetism.

Now, in the previous video I made about Maxwell's equations, which I linked up there earlier, I discussed one of the four Maxwell equations, and that one dealt with the fact that there is no such thing as a magnetic monopole.

And as a consequence of this, we discussed the idea that magnetic field lines must not have a start or an end point.

They must be closed loops.

Now that sounds relatively complicated, especially if you don't know what any of that means, but I tried to make it as easy to understand for somebody who's never seen this kind of stuff before.

So that's what I'm going to be trying to do in this video as well.

Now, the reason that I brought up the previous video is because in that one I used differential notation.

I discussed what these little upside down triangles mean.

However, some of you in the comments said

that you would like to see the next Maxwell equation discussed in the integral form, which looks a little bit something like this for the previous one that I discussed.

And that actually really helps me out because the differential notation

and the integral notation are just two different ways of writing exactly the same thing.

They're just two ways of looking at the same thing from different angles.

And the reason that your request helps me out is because, well, the equation that we're going to be looking at today is actually much easier to discuss in integral notation.

So thank you for that.

Okay, so the equation that we're looking at today is this one.

This is what it looks like, and hopefully by the end of the video, we'll be able to understand exactly what all of the symbols in this equation mean.

So let's start with this S-shaped symbol, which we can see on both sides of the equation.

It's known as the integral symbol.

Some of you will have seen this symbol already from your studies of calculus, but if you haven't already seen the symbol, then let's quickly go over what it means.

To do this, let's first imagine that we're dealing with a mathematical function.

Let's say we're dealing with y is equal to x, one of the simplest functions that we can think of.

This function is such that whatever the x value at a particular point along that line, the y value is exactly the same as the x value.

And this is true along the entire length of the line.

So that's why the function is known as y is equal to x, because on that line, at every point, the y value is exactly the same as the x value.

Now let's imagine that for some reason we want to try

and find the area underneath this line and above the horizontal or x axis.

And we want to do this between x is equal to 0 and a specific value of x.

Let's say x is equal to 10.

So now what we've drawn is a shaded triangle.

So we can work out this shaded triangle's area by recalling how to calculate the area of a triangle.

We can remember that the area of a triangle is found by multiplying half by the base by the height.

Now we know that the base of the triangle is 10 units long

because we're calculating the area of the triangle from x is

equal to 0 all the way to x is equal to 10.

However, we don't immediately know the height of the triangle, but we kind of do.

Because we know that the function we were dealing with is y is equal to x.

Therefore if we pick any point on the line, and in this case we'll pick the most convenient one, then we know that the x value is equal to the y value.

This means that if we pick the point where x is equal to 10, then y must also be equal to 10.

Therefore we can see that the height of the triangle is 10 units.

And in doing so we can calculate the area of the triangle as half multiplied by the base, which is 10, multiplied by the height, which is also 10.

And that ends up being 50 units squared.

Now this is easy enough, right?

But this kind of stuff only works with relatively simple functions, such as y is equal to x.

What do we do in a situation where we've got a more complicated function?

Let's say y is equal to x squared.

Now this curve, y is equal to x squared, is defined in such a way that whatever the x value is at a certain point on the curve, the y value is that x value squared.

That's why it's known as y is equal to x squared.

For example, if we take the point x is equal to 2, then on the curve, y is equal to 4.

And if we take the value x is equal to 5, then on the curve y is equal to 25.

So coming back to what we were trying to do earlier, let's now try and find the area underneath the curve above the horizontal axis

and between x is equal to 0 and x is equal to 10.

Now we don't have a triangle anymore, so it's not quite as simple as it was before.

In fact, we've got this whole curvy bit, right?

That's what's preventing us from calculating just the area of a triangle.

However, we could calculate the area of this triangle

that I've drawn in right now and just say that this is an estimate of the actual area.

Admittedly, it's a pretty poor estimate and it's an overestimate of the actual area

because we're also saying that the area under the function includes this shaded bit here, when in reality it doesn't.

So we've overestimated the area.

Now, a way to improve our estimate is to not say

that the area is a triangle but rather divide up the area under the curve into trapezia.

Then what we can do is to find out the area of each trapezium

and add it up all together to find the area underneath the curve

or at least an estimate of the area underneath the curve.

Now the reason

that this is still an estimate of the area albeit a better one than before is

because we have straight lines at the top of each trapezium rather than the curve

which is what we'd have if we had exactly the area that we're trying to find. But anyway, so the way to find the area of each trapezium is

to multiply the base length of the trapezium by the height exactly down the middle of the trapezium.

And we can do this for each different trapezium

and then add up all the areas to find a better estimate of the area underneath the curve.

Now, another thing that we're going to do is to call the base of each trapezium dx.

This is just a notation thing, but it'll become clear why we do this in a minute.

So don't worry about why it's called dx.

It's just that's what we're going to call that length.

And so essentially what we're doing is dividing the area

that we're trying to find into trapezia such

that the base length of each trapezium is the same as the base length of every other trapezium.

Each trapezium has a base length of dx.

Then we can work out the middle height of let's say the first trapezium.

So we can say

that the center of the base of the trapezium has a x coordinate x1

and then we can follow it up to the curve

and we can work out what the y value is

because we know that y on the curve is equal to x squared.

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So the area of the first trapezium is the base dx multiplied by the middle height which is x1 squared.

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The area of the second trapezium is dx multiplied by x2 squared.

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The area of the third trapezium is dx multiplied by x3 squared and so on and so forth.

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And to find an estimate for the total area underneath the curve we add up all of these areas.

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So we get dx multiplied by x1 squared plus dx multiplied by x2 squared plus so on and so forth.

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And then we can factorize.

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We factorize up dx

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and what we've got in brackets is x1 squared plus x2 squared plus x3 squared

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and so on and so Then what we can do is improve our estimate of the area even further.

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What we can do is to divide up the area into much thinner trapezia.

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We'll still say that each trapezium has a base of dx, but this time we've got a lot more of them.

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Yes, there are still problems because we're still using straight lines to approximate curves, but these problems are much smaller than before.

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So increasing the number of trapezia is reducing the error in our estimate.

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Our estimate is getting better and better the more trapezia we have.

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And we can keep doing this until we have large numbers of trapezia trapezia representing the area underneath the curve.

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Then the total area under the curve becomes the sum

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of the base dx multiplied by x1 squared plus x2 squared plus x3 squared plus however many midpoint x's

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that we have squared.

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And a short way to write this is as the sum of xn squared, where n takes the numbers 1, 2, 3, 4, 5, however many that we have,

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and this weird e looking symbol which is actually the Greek letter sigma represents a sum over all of these terms.

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Now don't worry too much about this bit, the next bit is what's important.

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Basically what we can do is to then say

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that our area is now divided up into infinitely many trapezia and each trapezium is infinitesimally thin.

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In other words it's as thin as can be, we can make it as thin as we want it, and

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if we were to zoom into the curve the base of the trapezium would still be thinner than we can ever imagine.

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So to reiterate, we've divided the area of the curve into infinitely many trapezia.

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And if we divide it up into infinitely many trapezia, then now what we have is no longer an estimate for the area of the curve, we have the exact area of the curve.

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Because remember, as we increase the number of trapezia representing the area of the curve, our estimates got better and better and better.

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And if we make dx so small that we have infinitely many trapezia, then we now have the exact area under the curve.

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In this situation, the squiggly e-sign, the sigma, becomes an integral sign.

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And that's where the integral sign comes from.

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It represents an adding up of lots of little bits and pieces to give us the overall whole thing.

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And the reason I said it so abstractly was because it doesn't necessarily have to represent the area underneath the curve.

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The idea is that you add up lots of tiny elements to give us a whole thing.

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Now of course when we're doing this properly and rigorously the symbols are slightly different.

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We use delta x to represent the non-infinitesimally thick trapezia bases, but that's not really relevant to the video.

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So let's move on.

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Like I said, the integral sign is used to add up lots of little

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tiny elements to give us the total whatever it is that we're trying to find.

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And I described it very visually using the area underneath the curve, but this can be used for many different things, including the equation that we were looking at earlier.

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And when we apply this train of thought to our equation, we see integral signs on both sides of the equation.

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This means that on both sides, we're adding up lots of little things to find something bigger.

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But what are these little things on each side?

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Well, to understand what we've got inside these integral signs on each side of the equation, we need to remember the meanings of B and E.

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If you've seen my previous video on Maxwell's equations, you'll recall that E represents the electric field that we're trying to study, and B represents the magnetic field that we're trying to study.

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Both E and B, the electric field and magnetic field, are vector fields.

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That's why they have arrows on the top of them to signify that they're vector fields.

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Now, when we say vector fields, what we actually mean is

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that at every point in the region of space being studied where the E field

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and the B field Exist each of these fields has a vector value In other words, there's a vector or an arrow

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that can be used to represent the value of this field E field

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or B field at every point in space in the case

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of the electric field the vector at each point Represents the force

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that would be felt by a positive test charge placed at

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that point In other words if we were to take a positive charge

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and place it at

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that particular point at the base of the arrow the force experienced by the charge would be towards the right

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and it would be a fairly large force.

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Whereas if we were to place it here instead, then it would experience a small force to the left.

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The B field or the magnetic field is similar, but it represents the force experienced by a north pole of a magnet placed in the magnetic field.

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If the explanation was a bit too complicated, then definitely check out the previous video.

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I went into a lot more detail and did it a little bit more slowly as well.

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However, let's carry on.

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So now we can see that we're integrating something to do with the B field

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and the E field on either side of the equation.

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So let's clarify what we mean by dl and ds.

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If we think back to our explanation about integral signs, we can see that they look a little bit like the dx's, which is what we use to label the bases of the trapezia.

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In this case, we're not talking about using trapezia to work out the area of anything anymore, but we are using dl and ds, exactly like we use dx.

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They are the little elements of something over which we'll be integrating.

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Now that sounds a little bit vague, but before we look at dl and ds specifically, we need to realize that both dl and ds are also vectors.

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They have arrows over the top of them.

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And the dot that's been pointed out on either side of the equation doesn't just represent a normal multiplication.

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It represents a sort of vector multiplication known as the scalar product or dot product.

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And the reason it's known as a scalar product even though we're dealing with vectors is

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because when you multiply these two vectors in a special way you get a scalar value.

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And it preserves the relationship between the two vectors that we multiply together.

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Let's not go into too much detail about that here.

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But then on the left hand side we've got the electric field multiplied by this dl vector.

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And then we integrate this whole expression.

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So what do we mean by dl?

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Well dl is a very small vector that lies on the perimeter of a surface that we shall call s.

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Now the surface that we call s is going to become very important relatively soon so let's keep an eye on it.

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But in the meantime let's say that this surface s is a circle

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and on its perimeter we are talking about little elements dl that go all the way around the surface s.

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In other words dl is representing a little length of the perimeter of the surface.

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And so what we're doing here on the left hand side is to multiply the electric field by the dl vector,

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which tells us how much the electric field is pointing in the direction of the dl vector at any particular point.

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So let's say that at this point the electric field is pointing in the direction of the dl vector.

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Well, in this case the electric field and the dl vector are going to multiply and because they're aligned, this means that the contribution of the electric field is going to be large at this point.

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So we're looking at the contribution of the electric field at every single dl along the perimeter of our circular surface.

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This means we're adding up the behavior of the electric field along the entire length of the circle.

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And the reason that we're talking about this circle in the first place is

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because that circle will become very important pretty soon.

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Now let's quickly move on to ds.

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Now ds is a similar kind of vector, but this ds represents an area.

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Specifically, every single ds is a vector that is perpendicular to a little area element inside the circle that we're considering.

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In other words, if we take a very tiny area, which we'll call dA for dArea, then ds is the vector that's perpendicular to that that represents this area.

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And the reason that we use a vector to represent an area is

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because that's one way of simplifying the description of an area.

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Instead of thinking about an area as a two-dimensional surface, we can now just think of it as one vector.

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So what we're doing here is to multiply this area vector by the magnetic field now.

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In other words, what we're looking at is the contribution of the magnetic field within every single little area element.

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And then once we add it all up, we add up all the contributions of the magnetic field to the little area elements within the circle that we're looking at.

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And therefore, we're looking at the total contribution of the magnetic field within the circle.

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Now, there's a complicated reason for why we're looking at the electric field along the perimeter of the circle, and why we're instead looking at the magnetic field going through the circle.

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And hence why we're using DL when talking about the electric field and DS when talking about the magnetic field.

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But the idea is, on the left hand side we've got the contribution of the electric field along the perimeter of the circle, and on the right we've got the contribution of the magnetic field through the circle.

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And it's important, by the way, that all of the DLs that add up are around the perimeter of the surface

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that we're considering on the right hand side.

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In other words, when we add up all the little DAs or DSs, depending on how you want to think about it, the area that we find will have the perimeter that we get when we add up all of the DLs together.

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Now, there's a little bit of an elephant in the room.

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We haven't talked about the d divided by dt thing on the right hand side of the equation.

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Well, all this means is that we're finding the rate of change of whatever is inside the integral.

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Think about it this way.

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You know when you find the velocity of an object, we're trying to find the distance it travels divided by the time taken for that object to travel that distance.

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In other words, the velocity of an object is delta x, that's the change in distance, divided by delta t, that's the time interval over which that change in distance occurs.

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In this case, we're doing exactly the same thing, but a little bit more rigorously and a little bit more calculus-ly.

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If that's a word, which it's not.

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Basically the d in the numerator is the same thing as the delta in the delta x part of the velocity equation, and the dt is the same thing as the delta t.

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In other words, we're finding the rate of change of this whole integral, or how fast that integral changes over time.

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And the negative sign basically just flips the sign, it's just the negative value of this rate of change.

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And by the way, the phrase rate of change simply means how quickly or slowly something is changing.

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So on the right hand side we're finding the negative of how quickly the integral is changing.

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Where the integral is the contribution of the magnetic field within the area of the circle that we're thinking about.

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Now that all sounds very complicated but this is where it gets simpler.

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What if I was to tell you that the perimeter of the circle that we were considering is a loop of wire.

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And that wire can carry a current.

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What if I also tell you

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that the left hand side of the equation gives us the EMF or voltage generated in that coil of wire.

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This is because the voltage in a circuit is very closely related to the electric field in that circuit.

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After all, electric fields are created by electric charges, and a voltage is just a measure of how electric charges want to move.

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Kind of like how gravitational fields are a measure of how objects with mass want to move in that gravitational field, right?

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So now we're thinking about a coil of wire, and on the left-hand side we have the EMF generated in that coil of wire.

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And we're talking about magnetic fields going through that wire, and more specifically, the rate of change of the total contribution of the magnetic field within that wire.

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This might be sounding a little bit familiar to some of you right now.

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This equation is a very thorough and very glorified way of encoding the effects of electromagnetic induction.

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That's the process where a changing magnetic field through a conducting coil causes an EMF to be generated within that coil.

237

Some of you might even have done an experiment where you take a bar magnet

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and you move it in and out of a conducting piece of wire.

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And if that wire forms a closed loop, then we can see that there's a current passing through the wire.

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That current flows because we generate a voltage or EMF across the coils

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when we move the bar magnet in or out of the coil.

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In other words, a changing magnetic field through the area of the coil causes an EMF across the coil.

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So this equation, once again, is essentially talking about electromagnetic induction, whilst also accounting for all of the effects that could possibly come about because of it.

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And the way it does this is by being very mathematically rigorous.

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We've used integrals and vectors and all this kind of neat but complicated stuff

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to describe something which at high school we would have just learned as you move a magnet in

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and out of a coil and we see an EMF generator.

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And this stuff is actually quite tricky to understand, so if there's something that I haven't fully explained properly, then let me know in the comments down below as well and I'll try and clarify.

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And as always, if I've got something wrong, let me know as well.

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But yeah, so because this Maxwell equation is describing the effect of electromagnetic induction, this particular equation is known as the Maxwell-Faraday equation,

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because Faraday's law of electromagnetic induction, if you don't know what that is, then definitely check out some other YouTube videos about it.

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It's actually a really cool effect and if you don't know about it already then I'd highly recommend learning about it.

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In high school we basically learned that the faster that the magnetic flux, that's the amount of magnetic field passing through a certain area, changes, the larger the EMF that's generated.

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In other words, the quicker we move our bar magnet in and out of the coil, the larger the EMF that's generated.

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And it's important to note that it's a change in magnetic flux through the area of the coil that makes the difference, not just putting a magnetic field through the coil.

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That in itself is not enough.

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That magnetic field must change and there are different ways to do this.

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You could of course move the magnet in and out of the coil

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but you could also change the area of the coil or rotate the coil.

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And whatever we choose to do, the right hand side of the equation is accounting for the fact that the magnetic field through the coil is changing.

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That's why we've got the d by dt term because it's calculating the rate of change of that magnetic field.

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It's also accounting for the vector nature of the magnetic field

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because the amount of magnetic field passing through an area that's perpendicular to it

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is going to be different to the magnetic field passing through

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that same area if the area is slightly rotated relative to the magnetic field.

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And of course, there's a lot more intricate details that I haven't gone into, but we don't need to know about those.

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The fact is that we now have a better understanding, hopefully, of this Maxwell equation, or rather, the Maxwell-Faraday equation.

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And with all of that being said, guys, I hope you enjoyed this video.

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Once again, if there's something you don't understand, let me know in the comments below.

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If you've got any other misconceptions about physics, then let me know in the comments as well.

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And if I can clarify anything or if I've got anything wrong, tell me down below also.

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I know I've been away a little bit, I've only been uploading once every three weeks or so, but you know work life is busy

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and I've been playing a lot of badminton

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and doing a bit of music on the side as well and sort of enjoying life a little bit.

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So I'm gonna try and come back to YouTube because I miss doing it, but I hope you guys are okay with having a video every two

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or three weeks in the meantime and hopefully I'll be producing more videos

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because like I said I genuinely really miss making videos, but again that depends on work commitments and anyway I'm I'm going to stop rambling.

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Thank you so much for watching and I'll see you next time.

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Bye.

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Woordenschat en spreektips bij deze les

Deze video bevat 279 zinnen en 4695 woorden om na te spreken. Het gesproken deel duurt 19:36. De spreker praat snel, ongeveer 239 woorden per minuut, dus reken op verbonden en ingeslikte klanken. 82% van de woorden hoort bij de 3.000 meest gebruikte Engelse woorden; de rest kun je beter vooraf bekijken.

Belangrijke woorden in deze video

15 woorden uit de video die het leren waard zijn, met uitspraak en betekenis:

Woord	Uitspraak	Betekenis
magnetic bijvoeglijk naamwoord	/mæɡˈnɛtɪk/	magnetisch
equation zelfstandig naamwoord	/ɪˈkweɪ.ʒən/	vergelijking
vector zelfstandig naamwoord	/ˈvɛktɚ/	vector
curve zelfstandig naamwoord	/kɚv/	curve, kromming
coil zelfstandig naamwoord	/kɔɪl/	spiraal, veer
integral bijvoeglijk naamwoord	/ˈɪntəɡɹəl/	integraal
triangle zelfstandig naamwoord	/ˈtɹaɪˌæŋɡəl/	driehoek
estimate zelfstandig naamwoord	/ˈɛs.tɪ.mət/	schatting
wire zelfstandig naamwoord	/ˈwaɪɚ/	draad
contribution zelfstandig naamwoord	/ˌkɑntɹɪˈbjuʃən/	bijdrage, contributie
divide werkwoord	/dɪˈvaɪd/	verdelen
perimeter zelfstandig naamwoord	/pɪˈɹɪm.ɪ.tə/	omtrek, perimeter
symbol zelfstandig naamwoord	/ˈsɪmbəl/	symbool
complicated bijvoeglijk naamwoord	/ˈkɑm.plɪˌkeɪ.tɪd/	ingewikkeld, gecompliceerd
delta zelfstandig naamwoord	/ˈdɛltə/	delta

Phrasal verbs die je zult horen

Woord	Uitspraak	Betekenis
work out werkwoord	/ˌwɝk ˈaʊt/	uitrekenen
check out werkwoord		onderzoeken
bring up werkwoord		opvoeden
end up werkwoord		belanden, terechtkomen
figure out werkwoord		uitvogelen, er achter komen
find out werkwoord		ontdekken, onderzoeken

Uitspraak om op te letten

De spreker gebruikt 109 samentrekkingen en verkorte vormen, zoals we're, we've, don't. Spreek ze kort uit, zoals je ze hoort.

De “th”-klanken: underneath /ˌʌndɚˈniθ/, thin /ˈθɪn/, thick /θɪk/, mathematical /ˌmæθ(.ə)ˈmæt.ɪ.kəl/, thorough /ˈθʌɹoʊ/
De klanken “sh” en “zh”: equation /ɪˈkweɪ.ʒən/, contribution /ˌkɑntɹɪˈbjuʃən/, induction /ɪnˈdʌkʃən/, shade /ʃeɪd/, explanation /ˌɛkspləˈneɪʃən/
Lange woorden — let op de klemtoon: contribution /ˌkɑntɹɪˈbjuʃən/, perimeter /pɪˈɹɪm.ɪ.tə/, complicated /ˈkɑm.plɪˌkeɪ.tɪd/, electromagnetic /ɪˌlɛk.tɹoʊ.mæɡˈnɛ.tɪk/, infinitely /ˈɪnfɪnɪtli/

Zo oefen je met deze video

Luister de hele video één keer zonder te spreken en noteer de woorden die je niet kent.
Begin op 0,75× snelheid, spreek zin voor zin na en ga terug naar normale snelheid zodra het makkelijk gaat.
Neem jezelf op en vergelijk met het origineel; let daarbij op woorden als magnetic, equation, vector.

Grammatica in deze video

De structuren die de spreker het meest gebruikt, met de exacte woorden uit de video:

Structuur	In de video
Present perfect continuous have/has been + -ing — iets dat eerder begon en nog bezig is	I've only been uploading · I've been playing
Present perfect have/has + voltooid deelwoord — iets uit het verleden dat nu nog telt	I've already made · you've subscribed · have seen
Lijdende vorm be + voltooid deelwoord — het gaat om wat er gebeurt, niet om wie het doet	must be closed · is known · is found
Betrekkelijke bijzinnen who / which + zin — extra informatie over een persoon of ding	base, which is · height which is · symbol which is

Wat is de Shadowing-techniek?

Shadowing is een wetenschappelijk onderbouwde taalleermethode die oorspronkelijk is ontwikkeld voor professionele tolkentraining en gepopulariseerd door polyglot Dr. Alexander Arguelles. De methode is eenvoudig maar krachtig: je luistert naar native Engelse audio en herhaalt het onmiddellijk hardop — als een schaduw die de spreker volgt met slechts 1–2 seconden vertraging. In tegenstelling tot passief luisteren of grammaticadrills, dwingt shadowing je hersenen en mondspieren om echte spraakpatronen tegelijkertijd te verwerken en te reproduceren. Onderzoek toont aan dat het de uitspraaknauwkeurigheid, intonatie, ritme, verbonden spraak, luisterbegrip en spreekvaardigheid aanzienlijk verbetert — waardoor het een van de meest effectieve methoden is voor IELTS Speaking-voorbereiding en echte Engelse communicatie.

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