maxwell’s equations: นี่คือโพสต์ที่เกี่ยวข้องกับหัวข้อนี้

20–1

Waves in free space; plane waves

In Chapter 18 we had reached the point where we had the

Maxwell equations in complete form. All there is to know about the

classical theory of the electric and magnetic fields can be found in the

four equations:

\begin{equation}

\begin{alignedat}{6}

&\text{I.}&\;\;

&\FLPdiv{\FLPE}&&\;=\frac{\rho}{\epsO}\quad&\quad

&\text{II.}&\;\;

&\FLPcurl{\FLPE}&&\;=-\ddp{\FLPB}{t}\\[1ex]
&\text{III.}&\;\;

&\FLPdiv{\FLPB}&&\;=0\quad&\quad

&\text{IV.}&\;\;

c^2&\FLPcurl{\FLPB}&&\;=\frac{\FLPj}{\epsO}+\ddp{\FLPE}{t}

\end{alignedat}

\label{Eq:II:20:1}

\end{equation}

\begin{equation}

\begin{alignedat}{3}

&\text{I.}&&\FLPdiv{\FLPE}&\;=&\;\frac{\rho}{\epsO}\\[1ex]
&\text{II.}&&\FLPcurl{\FLPE}&\;=&\;-\ddp{\FLPB}{t}\\[1ex]
&\text{III.}&&\FLPdiv{\FLPB}&\;=&\;0\\[1ex]
&\text{IV.}&\quad c^2&\FLPcurl{\FLPB}&\;=&\;\frac{\FLPj}{\epsO}+\ddp{\FLPE}{t}

\end{alignedat}

\label{Eq:II:20:1}

\end{equation}

When we put all these equations together, a remarkable new phenomenon

occurs: fields generated by moving charges can leave the sources and

travel alone through space. We considered a special example in which

an infinite current sheet is suddenly turned on. After the current has

been on for the time $t$, there are uniform electric and magnetic

fields extending out the distance $ct$ from the source. Suppose that

the current sheet lies in the $yz$-plane with a surface current

density $J$ going toward positive $y$. The electric field will have

only a $y$-component, and the magnetic field, only a

$z$-component. The field components are given by

\begin{equation}

\label{Eq:II:20:2}

E_y=cB_z=-\frac{J}{2\epsO c},

\end{equation}

for positive values of $x$ less than $ct$. For larger $x$ the fields

are zero. There are, of course, similar fields extending the same

distance from the current sheet in the negative $x$-direction. In

Fig. 20–1 we show a graph of the magnitude of the

fields as a function of $x$ at the instant $t$. As time goes on, the

“wavefront” at $ct$ moves outward in $x$ at the constant

velocity $c$.

Now consider the following sequence of events. We turn on a current of

unit strength for a while, then suddenly increase the current strength

to three units, and hold it constant at this value. What do the fields

look like then? We can see what the fields will look like in the

following way. First, we imagine a current of unit strength that is

turned on at $t=0$ and left constant forever. The fields for

positive $x$ are then given by the graph in part (a) of Fig. 20–2.

Next, we ask what would happen if we turn on a steady current of two

units at the time $t_1$.

The fields in this case will be twice as high as before, but will

extend out in $x$ only the distance $c(t-t_1)$, as shown in part (b)

of the figure. When we add these two solutions, using the principle of

superposition, we find that the sum of the two sources is a current of

one unit for the time from zero to $t_1$ and a current of three units

for times greater than $t_1$. At the time $t$ the fields will vary

with $x$ as shown in part (c) of Fig. 20–2.

Now let’s take a more complicated problem. Consider a current which is

turned on to one unit for a while, then turned up to three units, and

later turned off to zero. What are the fields for such a current? We

can find the solution in the same way—by adding the solutions of

three separate problems. First, we find the fields for a step current

of unit strength. (We have solved that problem already.) Next, we

find the fields produced by a step current of two units. Finally, we

solve for the fields of a step current of minus three

units. When we add the three solutions, we will have a current which

is one unit strong from $t=0$ to some later time, say $t_1$, then

three units strong until a still later time $t_2$, and then turned

off—that is, to zero. A graph of the current as a function of time

is shown in Fig. 20–3(a). When we add the three

solutions for the electric field, we find that its variation with $x$,

at a given instant $t$, is as shown in Fig. 20–3(b). The

field is an exact representation of the current. The field distribution

in space is a nice graph of the current variation with time—only drawn

backwards. As time goes on the whole picture moves outward at the

speed $c$, so there is a little blob of field, travelling toward positive $x$,

which contains a completely detailed memory of the history of all the

current variations. If we were to stand miles away, we could tell from

the variation of the electric or magnetic field exactly how the current

had varied at the source.

You will also notice that long after all activity at the source has

completely stopped and all charges and currents are zero, the block of

field continues to travel through space. We have a distribution of

electric and magnetic fields that exist independently of any charges

or currents. That is the new effect that comes from the complete set

of Maxwell’s equations. If we want, we can give a complete

mathematical representation of the analysis we have just done by

writing that the electric field at a given place and a given time is

proportional to the current at the source, only not at the same

time, but at the earlier time $t-x/c$. We can write

\begin{equation}

\label{Eq:II:20:3}

E_y(t)=-\frac{J(t-x/c)}{2\epsO c}.

\end{equation}

We have, believe it or not, already derived this same equation from

another point of view in Vol. I, when we were dealing with the

theory of the index of refraction. Then, we had to figure out what

fields were produced by a thin layer of oscillating dipoles in a sheet

of dielectric material with the dipoles set in motion by the electric

field of an incoming electromagnetic wave. Our problem was to

calculate the combined fields of the original wave and the waves

radiated by the oscillating dipoles. How could we have calculated the

fields generated by moving charges when we didn’t have Maxwell’s

equations? At that time we took as our starting point (without any

derivation) a formula for the radiation fields produced at large

distances from an accelerating point charge. If you will look in

Chapter 31 of Vol. I, you will see that

Eq. (31.9) there is just the same as the

Eq. (20.3) that we have just written down. Although our

earlier derivation was correct only at large distances from the source,

we see now that the same result continues to be correct even right up to

the source.

We want now to look in a general way at the behavior of electric and

magnetic fields in empty space far away from the sources, i.e., from

the currents and charges. Very near the sources—near enough so that

during the delay in transmission, the source has not had time to

change much—the fields are very much the same as we have found in

what we called the electrostatic or magnetostatic cases. If we go out

to distances large enough so that the delays become important,

however, the nature of the fields can be radically different from the

solutions we have found. In a sense, the fields begin to take on a

character of their own when they have gone a long way from all the

sources. So we can begin by discussing the behavior of the fields in a

region where there are no currents or charges.

Suppose we ask: What kind of fields can there be in regions where

$\rho$ and $\FLPj$ are both zero? In Chapter 18 we saw

that the physics of Maxwell’s equations could also be expressed in

terms of differential equations for the scalar and vector potentials:

\begin{align}

\label{Eq:II:20:4}

\nabla^2\phi-

\frac{1}{c^2}\,\frac{\partial^2\phi}{\partial t^2}&=

-\frac{\rho}{\epsO},\\[1ex]
\label{Eq:II:20:5}

\nabla^2\FLPA-

\frac{1}{c^2}\,\frac{\partial^2\FLPA}{\partial t^2}&=

-\frac{\FLPj}{\epsO c^2}.

\end{align}

If $\rho$ and $\FLPj$ are zero, these equations take on the simpler

form

\begin{align}

\label{Eq:II:20:6}

\nabla^2\phi-

\frac{1}{c^2}\,\frac{\partial^2\phi}{\partial t^2}&=0,\\[1ex]
\label{Eq:II:20:7}

\nabla^2\FLPA-

\frac{1}{c^2}\,\frac{\partial^2\FLPA}{\partial t^2}&=\FLPzero.

\end{align}

Thus in free space the scalar potential $\phi$ and each component of

the vector potential $\FLPA$ all satisfy the same mathematical

equation. Suppose we let $\psi$ (psi) stand for any one of the four

quantities $\phi$, $A_x$, $A_y$, $A_z$; then we want to investigate

the general solutions of the following equation:

\begin{equation}

\label{Eq:II:20:8}

\nabla^2\psi-

\frac{1}{c^2}\,\frac{\partial^2\psi}{\partial t^2}=0.

\end{equation}

This equation is called the three-dimensional wave

equation—three-dimensional, because the function $\psi$ may depend

in general on $x$, $y$, and $z$, and we need to worry about variations

in all three coordinates. This is made clear if we write out

explicitly the three terms of the Laplacian operator:

\begin{equation}

\label{Eq:II:20:9}

\frac{\partial^2\psi}{\partial x^2}+

\frac{\partial^2\psi}{\partial y^2}+

\frac{\partial^2\psi}{\partial z^2}-

\frac{1}{c^2}\,\frac{\partial^2\psi}{\partial t^2}=0.

\end{equation}

In free space, the electric fields $\FLPE$ and $\FLPB$ also satisfy

the wave equation. For example, since $\FLPB=\FLPcurl{\FLPA}$, we can

get a differential equation for $\FLPB$ by taking the curl of

Eq. (20.7). Since the Laplacian is a scalar operator, the

order of the Laplacian and curl operations can be interchanged:

\begin{equation*}

\FLPcurl{(\nabla^2\FLPA)}=\nabla^2(\FLPcurl{\FLPA})=\nabla^2\FLPB.

\end{equation*}

Similarly, the order of the operations curl and $\ddpl{}{t}$ can be

interchanged:

\begin{equation*}

\FLPcurl{\frac{1}{c^2}\,\frac{\partial^2\FLPA}{\partial t^2}}=

\frac{1}{c^2}\,\frac{\partial^2}{\partial t^2}(\FLPcurl{\FLPA})=

\frac{1}{c^2}\,\frac{\partial^2\FLPB}{\partial t^2}.

\end{equation*}

Using these results, we get the following differential equation

for $\FLPB$:

\begin{equation}

\label{Eq:II:20:10}

\nabla^2\FLPB-

\frac{1}{c^2}\,\frac{\partial^2\FLPB}{\partial t^2}=\FLPzero.

\end{equation}

So each component of the magnetic field $\FLPB$ satisfies the

three-dimensional wave equation. Similarly, using the fact that

$\FLPE=-\FLPgrad{\phi}-\ddpl{\FLPA}{t}$, it follows that the electric

field $\FLPE$ in free space also satisfies the three-dimensional wave

equation:

\begin{equation}

\label{Eq:II:20:11}

\nabla^2\FLPE-

\frac{1}{c^2}\,\frac{\partial^2\FLPE}{\partial t^2}=\FLPzero.

\end{equation}

All of our electromagnetic fields satisfy the same wave equation,

Eq. (20.8). We might well ask: What is the most general

solution to this equation? However, rather than tackling that

difficult question right away, we will look first at what can be said

in general about those solutions in which nothing varies in $y$

and $z$. (Always do an easy case first so that you can see what is going

to happen, and then you can go to the more complicated cases.) Let’s

suppose that the magnitudes of the fields depend only upon $x$—that

there are no variations of the fields with $y$ and $z$. We are,

of course, considering plane waves again. We should expect to get

results something like those in the previous section. In fact, we will

find precisely the same answers. You may ask: “Why do it all over

again?” It is important to do it again, first, because we did not

show that the waves we found were the most general solutions for plane

waves, and second, because we found the fields only from a very

particular kind of current source. We would like to ask now: What is

the most general kind of one-dimensional wave there can be in free

space? We cannot find that by seeing what happens for this or that

particular source, but must work with greater generality. Also we are

going to work this time with differential equations instead of with

integral forms. Although we will get the same results, it is a way of

practicing back and forth to show that it doesn’t make any difference

which way you go. You should know how to do things every which way,

because when you get a hard problem, you will often find that only one

of the various ways is tractable.

We could consider directly the solution of the wave equation for some

electromagnetic quantity. Instead, we want to start right from the

beginning with Maxwell’s equations in free space so that you can see

their close relationship to the electromagnetic waves. So we start

with the equations in (20.1), setting the charges and

currents equal to zero. They become

\begin{equation}

\begin{alignedat}{3}

&\text{I.}&&\FLPdiv{\FLPE}&\;=&\;0\\[1ex]
&\text{II.}&&\FLPcurl{\FLPE}&\;=&\;-\ddp{\FLPB}{t}\\[1ex]
&\text{III.}&&\FLPdiv{\FLPB}&\;=&\;0\\[1ex]
&\text{IV.}&\quad c^2&\FLPcurl{\FLPB}&\;=&\;\ddp{\FLPE}{t}

\end{alignedat}

\label{Eq:II:20:12}

\end{equation}

We write the first equation out in components:

\begin{equation}

\label{Eq:II:20:13}

\FLPdiv{\FLPE}=\ddp{E_x}{x}+\ddp{E_y}{y}+\ddp{E_z}{z}=0.

\end{equation}

We are assuming that there are no variations with $y$ and $z$, so the

last two terms are zero. This equation then tells us that

\begin{equation}

\label{Eq:II:20:14}

\ddp{E_x}{x}=0.

\end{equation}

Its solution is that $E_x$, the component of the electric field in the

$x$-direction, is a constant in space. If you look at IV

in (20.12), supposing no $\FLPB$-variation in $y$ and $z$

either, you can see that $E_x$ is also constant in time. Such a field

could be the steady dc field from some charged condenser plates

a long distance away. We are not interested now in such an uninteresting

static field; we are at the moment interested only in dynamically

varying fields. For dynamic fields, $E_x=0$.

We have then the important result that for the propagation of plane

waves in any direction, the electric field must be at right

angles to the direction of propagation. It can, of course, still vary

in a complicated way with the coordinate $x$.

The transverse $\FLPE$-field can always be resolved into two

components, say the $y$-component and the $z$-component. So let’s

first work out a case in which the electric field has only one

transverse component. We’ll take first an electric field that is

always in the $y$-direction, with zero $z$-component. Evidently, if we

solve this problem we can also solve for the case where the electric

field is always in the $z$-direction. The general solution can always

be expressed as the superposition of two such fields.

How easy our equations now get. The only component of the electric

field that is not zero is $E_y$, and all derivatives—except those

with respect to $x$—are zero. The rest of Maxwell’s equations then

become quite simple.

Let’s look next at the second of Maxwell’s equations [II of

Eq. (20.12)]. Writing out the components of the curl

$\FLPE$, we have

\begin{alignat*}{4}

&(\FLPcurl{\FLPE})_x&&=\ddp{E_z}{y}&&-\ddp{E_y}{z}&&=0,\\[1.5ex]
&(\FLPcurl{\FLPE})_y&&=\ddp{E_x}{z}&&-\ddp{E_z}{x}&&=0,\\[1.5ex]
&(\FLPcurl{\FLPE})_z&&=\ddp{E_y}{x}&&-\ddp{E_x}{y}&&=\ddp{E_y}{x}.

\end{alignat*}

The $x$-component of $\FLPcurl{\FLPE}$ is zero because the derivatives

with respect to $y$ and $z$ are zero. The $y$-component is also zero;

the first term is zero because the derivative with respect to $z$ is

zero, and the second term is zero because $E_z$ is zero. The only

components of the curl of $\FLPE$ that is not zero is the

$z$-component, which is equal to $\ddpl{E_y}{x}$. Setting the three

components of $\FLPcurl{\FLPE}$ equal to the corresponding components

of $-\ddpl{\FLPB}{t}$, we can conclude the following:

\begin{align}

\label{Eq:II:20:15}

&\ddp{B_x}{t}=0,\quad\ddp{B_y}{t}=0.\\[1ex]
\label{Eq:II:20:16}

&\ddp{B_z}{t}=-\ddp{E_y}{x}.

\end{align}

Since the $x$-component of the magnetic field and the $y$-component of

the magnetic field both have zero time derivatives, these two

components are just constant fields and correspond to the

magnetostatic solutions we found earlier. Somebody may have left some

permanent magnets near where the waves are propagating. We will ignore

these constant fields and set $B_x$ and $B_y$ equal to zero.

Incidentally, we would already have concluded that the $x$-component

of $\FLPB$ should be zero for a different reason. Since the divergence

of $\FLPB$ is zero (from the third Maxwell equation), applying the

same arguments we used above for the electric field, we would conclude

that the longitudinal component of the magnetic field can have no

variation with $x$. Since we are ignoring such uniform fields in our

wave solutions, we would have set $B_x$ equal to zero. In plane

electromagnetic waves the $\FLPB$-field, as well as the $\FLPE$-field,

must be directed at right angles to the direction of propagation.

Equation (20.16) gives us the additional proposition that

if the electric field has only a $y$-component, the magnetic field

will have only a $z$-component. So $\FLPE$ and $\FLPB$ are at

right angles to each other. This is exactly what happened in the

special wave we have already considered.

We are now ready to use the last of Maxwell’s equations for free space

[IV of Eq. (20.12)]. Writing out the components, we have

\begin{equation}

\begin{alignedat}{4}

&c^2(\FLPcurl{\FLPB})_x&&=

c^2\,\ddp{B_z}{y}&&-c^2\,\ddp{B_y}{z}&&=\ddp{E_x}{t},\\[1ex]
&c^2(\FLPcurl{\FLPB})_y&&=

c^2\,\ddp{B_x}{z}&&-c^2\,\ddp{B_z}{x}&&=\ddp{E_y}{t},\\[1ex]
&c^2(\FLPcurl{\FLPB})_z&&=

c^2\,\ddp{B_y}{x}&&-c^2\,\ddp{B_x}{y}&&=\ddp{E_z}{t}.

\end{alignedat}

\label{Eq:II:20:17}

\end{equation}

Of the six derivatives of the components of $\FLPB$, only the

term $\ddpl{B_z}{x}$ is not equal to zero. So the three equations give us

simply

\begin{equation}

\label{Eq:II:20:18}

-c^2\,\ddp{B_z}{x}=\ddp{E_y}{t}.

\end{equation}

The result of all our work is that only one component each of the electric and

magnetic fields is not zero, and that these components must satisfy Eqs.

(20.16) and (20.18). The two equations can be combined

into one if we differentiate the first with respect to $x$ and the second with

respect to $t$; the left-hand sides of the two equations will then be the same

(except for the factor $c^2$). So we find that $E_y$ satisfies the equation

\begin{equation}

\label{Eq:II:20:19}

\frac{\partial^2E_y}{\partial

x^2}-\frac{1}{c^2}\,\frac{\partial^2E_y}{\partial t^2}=0.

\end{equation}

We have seen the same differential equation before, when we studied

the propagation of sound. It is the wave equation for one-dimensional

waves.

You should note that in the process of our derivation we have found

something more than is contained in Eq. (20.11).

Maxwell’s equations have given us the further information that

electromagnetic waves have field components only at right angles to the

direction of the wave propagation.

Let’s review what we know about the solutions of the one-dimensional

wave equation. If any quantity $\psi$ satisfies the one-dimensional

wave equation

\begin{equation}

\label{Eq:II:20:20}

\frac{\partial^2\psi}{\partial

x^2}-\frac{1}{c^2}\,\frac{\partial^2\psi}{\partial t^2}=0,

\end{equation}

then one possible solution is a function $\psi(x,t)$ of the form

\begin{equation}

\label{Eq:II:20:21}

\psi(x,t)=f(x-ct),

\end{equation}

that is, some function of the single variable $(x-ct)$. The

function $f(x-ct)$ represents a “rigid” pattern in $x$ which travels

toward positive $x$ at the speed $c$ (see Fig. 20–4). For

example, if the function $f$ has a maximum when its argument is zero,

then for $t=0$ the maximum of $\psi$, will occur at $x=0$. At some later

time, say $t=10$, $\psi$ will have its maximum at $x=10c$. As time goes

on, the maximum moves toward positive $x$ at the speed $c$.

Sometimes it is more convenient to say that a solution of the

one-dimensional wave equation is a function of $(t-x/c)$. However,

this is saying the same thing, because any function of $(t-x/c)$ is

also a function of $(x-ct)$:

\begin{equation*}

F(t-x/c)=F\biggl[-\frac{x-ct}{c}\biggr]=f(x-ct).

\end{equation*}

Let’s show that $f(x-ct)$ is indeed a solution of the wave

equation. Since it is a function of only one variable—the

variable $(x-ct)$—we will let $f’$ represent the derivative of $f$ with

respect to its variable and $f”$ represent the second derivative

of $f$. Differentiating Eq. (20.21) with respect to $x$, we

have

\begin{equation*}

\ddp{\psi}{x}=f'(x-ct),

\end{equation*}

since the derivative of $(x-ct)$ with respect to $x$ is $1$. The

second derivative of $\psi$, with respect to $x$ is clearly

\begin{equation}

\label{Eq:II:20:22}

\frac{\partial^2\psi}{\partial x^2}=f”(x-ct).

\end{equation}

Taking derivatives of $\psi$ with respect to $t$, we find

\begin{align}

&\ddp{\psi}{t}=f'(x-ct)(-c),\notag\\[1.5ex]
\label{Eq:II:20:23}

&\frac{\partial^2\psi}{\partial t^2}=+c^2f”(x-ct).

\end{align}

We see that $\psi$ does indeed satisfy the one-dimensional wave

equation.

You may be wondering: “If I have the wave equation, how do I know

that I should take $f(x-ct)$ as a solution? I don’t like this backward

method. Isn’t there some forward way to find the solution?”

Well, one good forward way is to know the solution. It is possible to

“cook up” an apparently forward mathematical argument, especially

because we know what the solution is supposed to be, but with an

equation as simple as this we don’t have to play games. Soon you will

get so that when you see Eq. (20.20), you nearly

simultaneously see $\psi=f(x-ct)$ as a solution. (Just as now when you

see the integral of $x^2\,dx$, you know right away that the answer

is $x^3/3$.)

Actually you should also see a little more. Not only is any function

of $(x-ct)$ a solution, but any function of $(x+ct)$ is also a

solution. Since the wave equation contains only $c^2$, changing the

sign of $c$ makes no difference. In fact, the most general

solution of the one-dimensional wave equation is the sum of two

arbitrary functions, one of $(x-ct)$ and the other of $(x+ct)$:

\begin{equation}

\label{Eq:II:20:24}

\psi=f(x-ct)+g(x+ct).

\end{equation}

The first term represents a wave travelling toward positive $x$, and

the second term an arbitrary wave travelling toward negative $x$. The

general solution is the superposition of two such waves both existing

at the same time.

We will leave the following amusing question for you to think

about. Take a function $\psi$ of the following form:

\begin{equation*}

\psi=\cos kx\cos kct.

\end{equation*}

This equation isn’t in the form of a function of $(x-ct)$ or

of $(x+ct)$. Yet you can easily show that this function is a solution of

the wave equation by direct substitution into Eq. (20.20).

How can we then say that the general solution is of the form of

Eq. (20.24)?

Applying our conclusions about the solution of the wave equation to

the $y$-component of the electric field, $E_y$, we conclude that $E_y$

can vary with $x$ in any arbitrary fashion. However, the fields which

do exist can always be considered as the sum of two patterns. One wave

is sailing through space in one direction with speed $c$, with an

associated magnetic field perpendicular to the electric field; another

wave is travelling in the opposite direction with the same speed. Such

waves correspond to the electromagnetic waves that we know

about—light, radiowaves, infrared radiation, ultraviolet radiation,

x-rays, and so on. We have already discussed the radiation of light in

great detail in Vol. I. Since everything we learned there applies to

any electromagnetic wave, we don’t need to consider in great detail

here the behavior of these waves.

We should perhaps make a few further remarks on the question of the

polarization of the electromagnetic waves. In our solution we chose to

consider the special case in which the electric field has only a

$y$-component. There is clearly another solution for waves travelling

in the plus or minus $x$-direction, with an electric field which has

only a $z$-component. Since Maxwell’s equations are linear, the

general solution for one-dimensional waves propagating in the

$x$-direction is the sum of waves of $E_y$ and waves of $E_z$. This

general solution is summarized in the following equations:

\begin{equation}

\begin{aligned}

\FLPE&=(0,E_y,E_z)\\[.5ex]
E_y&=f(x-ct)+g(x+ct)\\[.5ex]
E_z&=F(x-ct)+G(x+ct)\\[1ex]
\FLPB&=(0,B_y,B_z)\\[.5ex]
cB_z&=f(x-ct)-g(x+ct)\\[.5ex]
cB_y&=-F(x-ct)+G(x+ct).

\end{aligned}

\label{Eq:II:20:25}

\end{equation}

Such electromagnetic waves have an $\FLPE$-vector whose direction is

not constant but which gyrates around in some arbitrary way in the

$yz$-plane. At every point the magnetic field is always perpendicular

to the electric field and to the direction of propagation.

If there are only waves travelling in one direction, say the positive

$x$-direction, there is a simple rule which tells the relative

orientation of the electric and magnetic fields. The rule is that the

cross product $\FLPE\times\FLPB$—which is, of course, a vector at

right angles to both $\FLPE$ and $\FLPB$—points in the direction in

which the wave is travelling. If $\FLPE$ is rotated into $\FLPB$ by a

right-hand screw, the screw points in the direction of the wave

velocity. (We shall see later that the vector $\FLPE\times\FLPB$ has a

special physical significance: it is a vector which describes the flow

of energy in an electromagnetic field.)

## Physics – E\u0026M: Maxwell’s Equations (1 of 30) What are the Maxwell equations? Introduction

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In this video I will introduction to Maxwell’s equations.

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## Maxwell’s Equations: Crash Course Physics #37

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In the early 1800s, Michael Faraday showed us how a changing magnetic field induces an electromotive force, or emf, resulting in an electric current. He also found that electric fields sometimes act like magnetic fields, and developed equations to calculate the forces exerted by both. In the mid1800s, Scottish physicist James Maxwell thought something interesting was going on there, too. So he decided to assemble a set of equations that held true for all electromagnetic interactions. In this episode of Crash Course Physics, Shini talks to us about Maxwell’s Equations and how important they are to our understanding of Physics.

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## What’s a Tensor?

Dan Fleisch briefly explains some vector and tensor concepts from A Student’s Guide to Vectors and Tensors

## Divergence and Curl

Visualization of the Divergence and Curl of a vector field.

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นอกจากการดูบทความนี้แล้ว คุณยังสามารถดูข้อมูลที่เป็นประโยชน์อื่นๆ อีกมากมายที่เราให้ไว้ที่นี่: __ดูบทความเพิ่มเติมในหมวดหมู่Music of Turkey __

ขอบคุณที่รับชมกระทู้ครับ maxwell’s equations