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Maxwell'south Equations and Electromagnetic Waves

Michael Fowler, Physics Department, UVa

The Equations

Maxwell's four equations depict the electric and magnetic fields arising from distributions of electric charges and currents, and how those fields alter in time.   They were the mathematical distillation of decades of experimental observations of the electric and magnetic effects of charges and currents, plus the profound intuition of Michael Faraday.  Maxwell's ain contribution to these equations is just the final term of the last equation—merely the add-on of that term had dramatic consequences.  It made axiomatic for the get-go time that varying electric and magnetic fields could feed off each other—these fields could propagate indefinitely through space, far from the varying charges and currents where they originated.  Previously these fields had been envisioned as tethered to the charges and currents giving rise to them.   Maxwell'southward new term (called the displacement current) freed them to move through space in a self-sustaining manner, and fifty-fifty predicted their velocity—information technology was the velocity of lite!

Here are the equations:

1. Gauss' Law for electrical fields:

${\displaystyle \int \overrightarrow{E}}\cdot d\overrightarrow{A}=q/{\epsilon }_0.$

 (The integral of the outgoing electric field over an area enclosing a volume equals the total charge inside, in appropriate units.)

2. The respective formula for magnetic fields:

${\displaystyle \int \overrightarrow{B}\cdot d\overrightarrow{A}}=0.$

(No magnetic accuse exists: no "monopoles".)

3. Faraday's Police of Magnetic Induction:

${\displaystyle \oint \overrightarrow{\mathrm{Due\; east}}\cdot d\overrightarrow{\ell }}=-d/dt\left({\displaystyle \int \overrightarrow{B}\cdot d\overrightarrow{A}}\right).$

The first term is integrated round a closed line, commonly a wire, and gives the total voltage change around the excursion, which is generated by a varying magnetic field threading through the excursion.

four. Ampere'due south Police plus Maxwell's displacement electric current:

${\displaystyle \oint \overrightarrow{B}\cdot d\overrightarrow{\ell }}={\mu }_0\left(I+\frac{d}{dt}\left({\epsilon }_0{\displaystyle \int \overrightarrow{\mathrm{Due\; east}}}\cdot d\overrightarrow{A}\right)\right).$

This gives the full magnetic force around a excursion in terms of the current through the circuit, plus any varying electric field through the circuit (that'southward the "deportation current").

The purpose of this lecture is to review the first three equations and the original Ampere's law fairly briefly, every bit they were covered earlier in the grade, then to demonstrate why the displacement electric current term must be added for consistency, and finally to bear witness, without using differential equations, how measured values of static electrical and magnetic attraction are sufficient to determine the speed of lite.

 Preliminaries: Definitions of µ ₀ and ε ₀

Ampere discovered that two parallel wires carrying electrical currents in the same direction attract each other magnetically, the strength in newtons per unit of measurement length being given past

$F=2\left(\frac{{\mu }_0}{\mathrm{four}\pi }\right)\frac{I_1{I}_{\mathrm{two}}}{r},$

for long wires a distance $r$ apart. We are using the standard modern units (SI).  The constant ${\mu }_0/4\pi$ that appears here is exactly 10^-seven, this defines our present unit of electric current, the ampere.To echo: ${\mu }_0/4\pi$ is not something to measure out experimentally, it's just a funny style of writing the number 10^-seven!  That'due south not quite fair—it has dimensions to ensure that both sides of the above equation have the same dimensionality. (Of class, there's a historical reason for this strange convention, as we shall meet later on).  Anyway, if we bear in mind that dimensions take been taken intendance of, and only write the equation

$F=\mathrm{ii}\cdot {10}^{-7}\cdot \frac{I_1{I}_2}{r},$

it'south articulate that this defines the unit current—one ampere —as that current in a long straight wire which exerts a magnetic forcefulness of $2\times {10}^{-\mathrm{vii}}$ newtons per meter of wire on a parallel wire one meter away carrying the aforementioned current.

Still, after we have established our unit of electric current—the ampere— we have also thereby defined our unit of accuse, since current is a catamenia of charge, and the unit of charge must be the amount carried past a fixed point in unit of measurement time by unit current.  Therefore, our unit of measurement of charge—the coulomb —is divers past stating that a one amp current in a wire carries one coulomb per 2d past a fixed point.

 To be consequent, we must practice electrostatics using this aforementioned unit of charge. Now, the electrostatic force between 2 charges is $\left(1/\mathrm{four}\pi {\epsilon }_0\right)q_1{q}_2/r^2.$ The constant appearing here, at present written $1/4\pi {\epsilon }_0$, must be experimentally measured—its value turns out to be $\mathrm{nine}\times {10}^9$.

 To summarize:  to detect the value of $1/4\pi {\epsilon }_0$, two experiments have to be performed. We must first establish the unit of charge from the unit of electric current by measuring the magnetic forcefulness between two current-conveying parallel wires.  Second, nosotros must find the electrostatic force between measured charges. (We could, alternatively, have defined some other unit of current from the start, then nosotros would have had to find both ${\mu }_0$ and ${\epsilon }_0$ past experiments on magnetic and electrostatic attraction.  In fact, the ampere was originally divers equally the current that deposited a definite weight of silver per hr in an electrolytic cell).

Maxwell's Equations

We have so far established that the total flux of electric field out of a closed surface is just the total enclosed charge multiplied past $\mathrm{i}/{\epsilon }_0$,$$

${\displaystyle \int \overrightarrow{E}}\cdot d\overrightarrow{A}=q/{\epsilon }_0.$

This is Maxwell's first equation.  Information technology represents completely roofing the surface with a large number of tiny patches having areas $d\overrightarrow{A}$.  Nosotros correspond these small areas as vectors pointing outwards, because we tin can then take the dot product with the electric field to select the component of that field pointing perpendicularly outwards (information technology would count negatively if the field were pointing inwards)—this is the but component of the field that contributes to bodily period beyond the surface.  (Only as a river flowing parallel to its banks has no flow across the banks).

 The second Maxwell equation is the analogous one for the magnetic field, which has no sources or sinks (no magnetic monopoles, the field lines merely menses around in closed curves). Thinking of the strength lines every bit representing a kind of fluid flow, the then-called "magnetic flux", nosotros see that for a closed surface, as much magnetic flux flows into the surface every bit flows out.  This tin can perhaps be visualized most clearly by taking a group of neighboring lines of force forming a slender tube—the "fluid" within this tube flows circular and circular, so as the tube goes into the closed surface then comes out over again (maybe more than than once) information technology is like shooting fish in a barrel to see that what flows into the closed surface at one identify flows out at another. Therefore the net flux out of the enclosed book is cypher, Maxwell'southward second equation:

${\displaystyle \int \overrightarrow{B}\cdot d\overrightarrow{A}}=0.$

The first two Maxwell's equations, given higher up, are for integrals of the electric and magnetic fields over closed surfaces . The other 2 Maxwell'southward equations, discussed beneath, are for integrals of electric and magnetic fields around airtight curves (taking the component of the field pointing forth the curve). These represent the piece of work that would exist needed to take a accuse around a closed curve in an electric field, and a magnetic monopole (if i existed!) around a closed bend in a magnetic field.

 The simplest version of Maxwell'southward third equation is the electrostatic case:

The path integral ${\displaystyle \oint \overrightarrow{E}\cdot d\overrightarrow{\ell }}=0$ for electrostatics .

All the same, we know that this is only role of the truth, because from Faraday's Constabulary of Induction, if a closed circuit has a changing magnetic flux through information technology, a circulating current will arise, which ways at that place is a nonzero voltage around the circuit.

 The full version of Maxwell's third equation is:

${\displaystyle \oint \overrightarrow{\mathrm{Eastward}}\cdot d\overrightarrow{\ell }}=-d/dt\left({\displaystyle \int \overrightarrow{B}\cdot d\overrightarrow{A}}\right)$

where the area integrated over on the right mitt side spans the path (or circuit) on the left hand side, like a lather film on a loop of wire.

 It may seem that the integral on the right paw side is not very clearly defined, because if the path or circuit lies in a airplane, the natural pick of spanning surface (the "lather film") is flat, but how do you lot decide what surface to choose to practise the integral over for a wire bent into a circuit that doesn't lie in a airplane?   The reply is that it doesn't affair what surface yous choose, every bit long as the wire forms its purlieus.  Consider two different surfaces both having the wire equally a boundary (merely as both the northern hemisphere of the earth'southward surface and the southern hemisphere have the equator as a purlieus). If you add together these two surfaces together, they form a single airtight surface, and we know that for a airtight surface ${\displaystyle \int \overrightarrow{B}\cdot d\overrightarrow{A}}=0$.  This implies that ${\displaystyle \int \overrightarrow{B}\cdot d\overrightarrow{A}}$ for one of the 2 surfaces bounded by the path is equal to $-{\displaystyle \int \overrightarrow{B}\cdot d\overrightarrow{A}}$ for the other one, so that the ii will add to zero for the whole closed surface.  Only don't forget these integrals for the whole closed surface are defined with the little area vectors pointing outwards from the enclosed book. By imagining ii surfaces spanning the wire that are actually shut to each other, it is clear that the integral over one of them is equal to the integral over the other if we take the $d\overrightarrow{A}$ vectors to point in the aforementioned management for both of them, which in terms of the enclosed book would exist outwards for one surface, inwards for the other i. The lesser line of all this is that the surface integral ${\displaystyle \int \overrightarrow{B}\cdot d\overrightarrow{A}}$ is the same for any surface spanning the path, so information technology doesn't matter which nosotros choose.

The equation coordinating to the electrostatic version of the third equation given above, just for the magnetic field, is Ampere's law

${\displaystyle \oint \overrightarrow{B}\cdot d\overrightarrow{\ell }}={\mu }_0\cdot \text{(enclosed currents)}$

for magnetostatics, where the currents counted are those threading through the path nosotros're integrating around, and then if there is a soap motion picture spanning the path, these are the currents that punch through the picture (of course, we take to agree on a management, and subtract currents flowing in the contrary management).

 We must now consider whether this equation, like the electrostatic one, has limited validity. In fact, it was not questioned for a generation after Ampere wrote it down: Maxwell's bully contribution, in the 1860's, was to realize that it was not ever valid.

 When Does Ampere'southward Law Get Wrong?

 A unproblematic case to meet that something must be incorrect with Ampere'south Law in the general case is given by Feynman in his Lectures in Physics.  Suppose we use a hypodermic needle to insert a spherically symmetric blob of charge in the center of a large vat of solidified jello (which we assume conducts electricity).  Because of electrostatic repulsion, the charge volition dissipate, currents will menses outwards in a spherically symmetric way.Question: does this outward-flowing current distribution generate a magnetic field?  The answer must be no , considering since we have a completely spherically symmetric situation, it could merely generate a spherically symmetric magnetic field.  But the merely possible such fields are one pointing outwards everywhere and one pointing inward everywhere, both corresponding to non-existent monopoles. So, there can be no magnetic field.

 However, imagine we now consider checking Ampere's constabulary past taking equally a path a horizontal circumvolve with its middle in a higher place the point where we injected the charge (think of a halo above someone's caput.)  Evidently, the left mitt side of Ampere's equation is zero, since at that place tin can be no magnetic field.  (Information technology would have to be spherically symmetric, meaning radial.)  On the other manus, the right hand side is most definitely not nothing, since some of the outward flowing electric current is going to go through our circle.  So the equation must be wrong.

 Ampere's law was established as the result of large numbers of careful experiments on all kinds of current distributions.  Then how could information technology be that something of the kind we describe above was overlooked?  The reason is really similar to why electromagnetic consecration was missed for so long. No-one thought about looking at irresolute fields, all the experiments were done on steady situations.  With our ball of charge spreading outward in the jello, there is obviously a irresolute electric field.  Imagine yourself in the jello near where the charge was injected: at first, yous would feel a strong field from the nearby full-bodied charge, but equally the charge spreads out spherically, some of it going past y'all, the field will decrease with time.

Maxwell'southward Example

Maxwell himself gave a more practical example:  consider Ampere'southward law for the usual infinitely long wire carrying a steady current $I,$ but now break the wire at some point and put in 2 large circular metal plates, a capacitor, maintaining the steady current $I$ in the wire everywhere else, so that accuse is simply piling up on one of the plates and draining off the other.

 Looking now at the wire some distance away from the plates, the situation appears normal, and if we put the usual circular path around the wire, application of Ampere's law tells united states that the magnetic field at distance $r,$ from

${\displaystyle \oint \overrightarrow{B}\cdot d\overrightarrow{\ell }}={\mu }_{0}I$

is only

$B={\mu }_{0}I/2\pi r.$

Retrieve, withal, that we defined the electric current threading the path in terms of current punching through a soap film spanning the path, and said this was contained of whether the soap film was apartment, bulging out on one side, or any. With a single space wire, there was no escape— no contortions of this roofing surface could wriggle costless of the wire going through it (actually, if yous distort the surface enough, the wire could penetrate it several times, but yous have to count the net flow across the surface, and the new penetrations would come in pairs with the current crossing the surface in opposite directions, so they would cancel).

 Once we bring in Maxwell's parallel plate capacitor, nevertheless, at that place is a style to misconstrue the surface and so that no current penetrates it at all: we can run it betwixt the plates!

The question and then arises: can we rescue Ampere'south constabulary by adding another term merely as the electrostatic version of the third equation was rescued by adding Faraday'due south consecration term? The answer is of grade yes: although in that location is no electric current crossing the surface if we put it between the capacitor plates, at that place is certainly a irresolute electric field , because the capacitor is charging upwards equally the electric current $I$ flows in. Bold the plates are close together, we can take all the electric field lines from the accuse $q$ on i plate to menstruation across to the other plate, so the total electric flux across the surface betwixt the plates,

${\displaystyle \int \overrightarrow{E}}\cdot d\overrightarrow{A}=q/{\epsilon }_0.$

At present, the electric current in the wire $,I,$ is merely the rate of change of charge on the plate,

$I=dq/dt.$

Putting the above two equations together, we see that

$I=\frac{d}{dt}\left({\epsilon }_0{\displaystyle \int \overrightarrow{\mathrm{East}}}\cdot d\overrightarrow{A}\right).$

Ampere's law can at present be written in a way that is correct no matter where we put the surface spanning the path we integrate the magnetic field around:

${\displaystyle \oint \overrightarrow{B}\cdot d\overrightarrow{\ell }}={\mu }_0\left(I+\frac{d}{dt}\left({\epsilon }_0{\displaystyle \int \overrightarrow{E}}\cdot d\overrightarrow{A}\right)\right).$

This is Maxwell'southward fourth equation.

Find that in the case of the wire, either the current in the wire, or the increasing electric field, contribute on the correct hand side, depending on whether we take the surface simply cutting through the wire, or positioned betwixt the plates. (Actually, more complicated situations are possible—we could imaging the surface partly between the plates, and then cut through the plates to become out!  In this example, we would have to figure out the current really in the plate to get the right hand side, but the equation would still apply).

"Deportation Current"

Maxwell referred to the second term on the right hand side, the changing electric field term, equally the "displacement current".  This was an analogy with a dielectric material.  If a dielectric material is placed in an electric field, the molecules are distorted, their positive charges moving slightly to the right, say, the negative charges slightly to the left.  Now consider what happens to a dielectric in an increasing electric field.  The positive charges will be displaced to the right past a continuously increasing distance, so, equally long as the electric field is increasing in strength, these charges are moving: in that location is actually a displacement current .  (Meanwhile, the negative charges are moving the other way, but that is a current in the same management, so adds to the outcome of the positive charges' motion.)  Maxwell's picture of the vacuum, the aether, was that it likewise had dielectric backdrop somehow, so he pictured a similar move of charge in the vacuum to that we accept merely described in the dielectric.  The motion-picture show is wrong, but this is why the changing electric field term is often chosen the "deportation electric current", and in Ampere'south law (generalized) is simply added to the real current, to give Maxwell'southward fourth—and final—equation.

Some other Angle on the 4th Equation: the Link to Charge Conservation

Going back for a moment to Ampere's law, nosotros stated information technology every bit:

${\displaystyle \oint \overrightarrow{B}\cdot d\overrightarrow{\ell }}={\mu }_0\cdot \text{(enclosed currents)}$

for magnetostatics, where the currents counted are those threading through the path we're integrating around, so if there is a lather film spanning the path, these are the currents that punch through the film. Our mental picture show here is commonly of a few thin wires, perhaps twisted in various ways, conveying currents. More generally, thinking of electrolytes, or even of fat wires, we should exist envisioning a current density varying from indicate to bespeak in space. In other words, we accept a flux of electric current and the natural expression for the current threading our path is (analogous to the magnetic flux in the third equation) to write a surface integral of the current density $\overrightarrow{j}$ over a surface spanning the path, giving for magnetostatics

path integral ${\displaystyle \oint \overrightarrow{B}\cdot d\overrightarrow{\ell }}={\mu }_0{\displaystyle \int \overrightarrow{j}}\cdot d\overrightarrow{A}$(surface integral, over surface spanning path)

The question then arises as to whether the surface integral we have written on the right manus side above depends on which surface we choose spanning the path. From an argument exactly parallel to that for the magnetic flux in the third equation (come across higher up), this will be true if and only If ${\displaystyle \int \overrightarrow{j}}\cdot d\overrightarrow{A}=0$ for a airtight surface (with the path lying in the surface—this airtight surface is made upwardly past combining two unlike surfaces spanning the path).

 Now, ${\displaystyle \int \overrightarrow{j}}\cdot d\overrightarrow{A}$  taken over a closed surface is only the net current flow out of the enclosed book. Obviously, in a situation with steady currents flowing forth wires or through conductors, with no accuse piling up or draining away from anywhere, this is aught. Yet, if the total electric accuse $q,$ say, enclosed by the closed surface is changing as time goes on, then apparently

${\displaystyle \int \overrightarrow{j}}\cdot d\overrightarrow{A}=-dq/dt,$

where nosotros put in a minus sign because, with our convention, $d\overrightarrow{A}$ is a footling vector pointing outwards, so the integral represents net menstruum of accuse out from the surface, equal to the charge per unit of decrease of the enclosed total charge.

 To summarize: if the local accuse densities are changing in time, that is, if charge is piling up in or leaving some region, so ${\displaystyle \int \overrightarrow{j}}\cdot d\overrightarrow{A}\ne 0$ over a airtight surface around that region. That implies that ${\displaystyle \int \overrightarrow{j}}\cdot d\overrightarrow{A}$ over one surface spanning the wire will be different from ${\displaystyle \int \overrightarrow{j}}\cdot d\overrightarrow{A}$ over some other surface spanning the wire if these two surfaces together make upward a closed surface enclosing a region containing a changing amount of charge.

 The fundamental to fixing this up is to realize that although ${\displaystyle \int \overrightarrow{j}}\cdot d\overrightarrow{A}=-dq/dt\ne 0,$ the right-paw side can be written as another surface integral over the aforementioned surface, using the beginning Maxwell equation, that is, the integral over a closed surface

${\displaystyle \int \overrightarrow{\mathrm{Due\; east}}}\cdot d\overrightarrow{A}=q/{\epsilon }_0$

where $q$ is the full charge in the book enclosed past the surface.

Past taking the time rate of change of both sides, we detect

$\frac{d}{dt}{\displaystyle \int \overrightarrow{E}}\cdot d\overrightarrow{A}=\frac{1}{{\epsilon }_0}\frac{dq}{dt}$

Putting this together with ${\displaystyle \int \overrightarrow{j}}\cdot d\overrightarrow{A}=-dq/dt$ gives:

${\epsilon }_0\frac{d}{dt}{\displaystyle \int \overrightarrow{E}}\cdot d\overrightarrow{A}+{\displaystyle \int \overrightarrow{j}}\cdot d\overrightarrow{A}=0$

for whatever closed surface, and consequently this is a surface integral that must be the same for whatever surface spanning the path or excursion!  (Because two different surfaces spanning the same circuit add up to a airtight surface. Nosotros'll ignore the technically trickier instance where the two surfaces intersect each other, creating multiple volumes—there one must treat each created volume separately to get the signs right.)

Therefore, this is the way to generalize Ampere's constabulary from the magnetostatic situation to the case where charge densities are varying with time, that is to say the path integral

${\displaystyle \oint \overrightarrow{B}\cdot d\overrightarrow{\ell }}={\mu }_0{\displaystyle \int \left(\overrightarrow{j}+{\epsilon }_0\frac{d\overrightarrow{\mathrm{Due\; east}}}{dt}\right)}\cdot d\overrightarrow{A}$

and this gives the same result for whatsoever surface spanning the path.

A Sheet of Electric current: A Elementary Magnetic Field

As a preliminary to looking at electromagnetic waves, we consider the magnetic field configuration from a sail of uniform current of large extent.  Think of the canvass as perpendicular to this sheet of paper, the current running vertically upwards.  Information technology might be helpful to visualize the sail as many equal parallel fine wires uniformly spaced close together:

...................................................................................... (wires)

The magnetic field from this current canvass tin be found using Ampere's law applied to a rectangular profile in the plane of the paper, with the current sail itself bisecting the rectangle, so the rectangle's top and bottom are equidistant from the current sheet in reverse directions.

 Applying Ampere's constabulary to the in a higher place rectangular contour, there are contributions to ${\displaystyle \oint \overrightarrow{B}\cdot d\overrightarrow{\ell }}$ only from the top and lesser, and they add together to give $2B\mathrm{50}$ if the rectangle has side $L.$ The full electric current enclosed by the rectangle is $I\mathrm{Fifty},$ taking the current density of the sheet to be $I$ amperes per meter (how many little wires per meter multiplied past the electric current in each wire).

Thus, ${\displaystyle \oint \overrightarrow{B}\cdot d\overrightarrow{\ell }}={\mu }_0\cdot \text{(enclosed currents)}$ immediately gives:

$B={\mu }_{0}I/2$

a magnetic field strength independent of altitude $d$ from the sheet. (This is the magnetostatic analog of the electrostatic result that the electric field from an infinite sheet of accuse is independent of distance from the sheet.)  In real life, where at that place are no infinite sheets of anything, these results are good approximations for distances from the canvass pocket-sized compared with the extent of the canvass.

Switching on the Sheet: How Fast Does the Field Build Up?

Consider now how the magnetic field develops if the electric current in the sail is suddenly switched on at time $t=0.$ We volition presume that sufficiently shut to the sheet, the magnetic field pattern plant above using Ampere's law is rather speedily established.

In fact, we will assume further that the magnetic field spreads out from the sheet similar a tidal wave, moving in both directions at some speed $v,$ then that after time $t$ the field within distance $\mathrm{five}t$ of the sail is the same every bit that institute above for the magnetostatic case, but across $\mathrm{five}t$ there is at that instant no magnetic field nowadays.

Let us now employ Maxwell's equations to this guess to see if it can make sense. Certainly Ampere's law doesn't work by itself, because if we take a rectangular path every bit we did in the previous section, for $d<vt$ everything works as before, merely for a rectangle extending beyond the spreading magnetic field, $d>\mathrm{five}t,$ there volition be no magnetic field contribution from the top and bottom of the rectangle, and hence

${\displaystyle \oint \overrightarrow{B}\cdot d\overrightarrow{\ell }}=0$

but at that place is definitely enclosed electric current!

We are forced to conclude that for Maxwell'southward fourth equation to exist correct, at that place must as well be a irresolute electric field through the rectangular contour.

Let us now endeavour to nail down what this electric field through the contour must wait like.  First, it must be through the contour, that is, accept a component perpendicular to the airplane of the profile, in other words, perpendicular to the magnetic field.  In fact, electrical field components in other directions won't touch on the fourth equation we are trying to satisfy, so we shall ignore them.  Detect start that for a rectangular contour with $d<vt,$  Ampere'southward police works, and so nosotros don't want a changing electrical field through such a contour (but a constant electric field would be ok).

Now apply Maxwell's quaternary equation to a rectangular contour with $d>5t,$

It is:  path integral ${\displaystyle \oint \overrightarrow{B}\cdot d\overrightarrow{\ell }}={\mu }_0{\displaystyle \int \left(\overrightarrow{j}+{\epsilon }_0\frac{d\overrightarrow{\mathrm{Eastward}}}{dt}\right)}\cdot d\overrightarrow{A}$ (over surface spanning path).

For the rectangle shown above, the integral on the left hand side is zero because $\overrightarrow{B}$ is perpendicular to $d\overrightarrow{\ell }$ along the sides, so the dot product is zero, and $\overrightarrow{B}$ is zilch at the elevation and bottom, because the outward moving "wave" of magnetic field hasn't gotten there even so. Therefore, the right hand side of the equation must too be zero.

Nosotros know ${\displaystyle \int \overrightarrow{j}}\cdot d\overrightarrow{A}=\mathrm{Fifty}I$, so we must have:${\epsilon }_0\frac{d}{dt}{\displaystyle \int \overrightarrow{E}}\cdot d\overrightarrow{A}=-LI.$

Finding the Speed of the Outgoing Field Forepart: the Connection with Light

So, equally long as the outward moving front of magnetic field, travelling at $v,$ hasn't reached the top and bottom of the rectangular profile, the electric field through the contour increases linearly with fourth dimension, merely the increment drops to zero (because Ampere's law is satisfied) the moment the front reaches the top and bottom of the rectangle. The simplest way to get this behavior is to have an electric field of strength $\mathrm{Due\; east},$ perpendicular to the magnetic field, everywhere there is a magnetic field, and then the electric field also spreads outwards at speed $v.$  Note that, unlike the magnetic field, the electric field must point the same way on both sides of the current canvas, otherwise its cyberspace flux through the rectangle would be zero.

Later on fourth dimension $t,$ so, the electric field flux through the rectangular contour ${\displaystyle \int \overrightarrow{E}}\cdot d\overrightarrow{A}$ will exist just field x area = $E\cdot 2\cdot \mathrm{five}tL,$ and the charge per unit of alter will exist $2E\mathrm{five}\mathrm{50}.$ (Information technology's spreading both ways, hence the ii).

Therefore

${\epsilon }_0\mathrm{East}\cdot 2\cdot 5tL=-LI,$

the electric field is downwards and of strength $\mathrm{Eastward}=I/\left(2{\epsilon }_{0}v\right).$

Since $B={\mu }_{0}I/\mathrm{two},$ this implies:

$B={\mu }_0{\epsilon }_{0}5E.$

Just nosotros take another equation linking the field strengths of the electric and magnetic fields, Maxwell'southward tertiary equation:

${\displaystyle \oint \overrightarrow{E}\cdot d\overrightarrow{\ell }}=-d/dt\left({\displaystyle \int \overrightarrow{B}\cdot d\overrightarrow{A}}\right)$

Nosotros tin apply this equation to a rectangular contour with sides parallel to the $E$ field, one side existence within $vt$ of the current canvass, the other more distant, so the only contribution to the integral is $\mathrm{East}\mathrm{50}$  from the first side, which nosotros take to have length $L.$ (This contour is all on 1 side of the current sheet.) The area of the rectangle the magnetic flux is passing through volition be increasing at a rate $Lv$ (square meters per 2d) as the magnetic field spreads outwards.

It follows that

$E=vB.$

Putting this together with the effect of the quaternary equation,

$B={\mu }_0{\epsilon }_{0}v\mathrm{Eastward},$

we deduce

$${\mathrm{five}}^2=\frac{1}{{\mu }_0{\epsilon }_0}.$$

Substituting the divers value of ${\mu }_0,$ and the experimentally measured value of ${\epsilon }_0,$  we find that the electric and magnetic fields spread outwards from the switched-on current sheet at a speed of three x 10⁸ meters per second.

This is how Maxwell discovered a speed equal to the speed of light from a purely theoretical argument based on experimental determinations of forces betwixt currents in wires and forces between electrostatic charges. This of form led to the realization that light is an electromagnetic wave, and that at that place must be other such waves with different wavelengths.  Hertz detected other waves, of much longer wavelengths, experimentally, and this led directly to radio, telly, radar, etc.

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Source: http://www.phys.virginia.edu/classes/109N/more_stuff/Maxwell_Eq.html

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