basics of electricity and magnetism

7/25/2019 Basics of Electricity and Magnetism

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E&M Review

Jeremy Li

March 15, 2016



Abstract

All of electromagnetism is contained in the Maxwell equations: in differential form, they are

∇ · E = ρ

0(1)

∇ × E = −∂ B

∂t (2)

∇ × B = µ0J + µ00∂ E

∂t (3)

∇ · B = 0. (4)

In integral form, they are‹ ∂ Ω

E · dS = 1

0

˚ Ω

ρ dV (5)‹

∂ ΩB · dS = 0 (6)

∂ ΣE · dl = − d

dt

‹ Σ

B · dS (7) ∂ Σ

B · dl = µ0

¨ Σ

J · dS + µ00 dt

¨ Σ

E · dS. (8)

In statics, these can be separated into two groups — they are not interconnected. That is tosay, electricity and magnetism are distinct phenomena as long as charges and currents are static.Electrostatics has

∇ · E = ρ

0(9)

∇ × E = 0 (10)

and magnetostatics has

∇ × B = µ0J (11)

∇ · B = 0. (12)



Chapter 1

Statics

1.1 Important Notes

1. Recall what the potential actually is, and what the electric field actually is. The electric field

is the electric force per unit charge

, so that the amount of work needed per unit charge tomove it from a to b is

V = −W = − b

a

F · ds = − b

a

E · ds. (1.1)

1.2 Electrostatics

For static charges; that is, charges at rest, we start with Coulomb’s law, which says between twocharges q 1 and q 2 there is a force on charge q 1

F1 =

1

4π0

q 1q 2r2

12 e12 = −F2 (1.2)

where e12 is the unit vector in the direction from q 2 to q 1.Although Coulomb’s law and the principle of superposition together define all of electrostatics,

it is useful to introduce the concept of the electric field E, which is the force per unit charge onq 1 due to all other charges. In other words, the electric field at the position of charge q 1 is

E1 = 1

4π0

q 2r2

12

e12. (1.3)

Note that this field is a result of the other charges — regardless of whether q 1 is actually present,E1 remains the same. If there are many charges present, then the E field at any point is the sum

of the contributions from each of the other charges so that we have in total at location (1),

E(1) = j

1

4π0

q jr2

1 j

e1 j. (1.4)

When the charges are distributed on a small scale, we can introduce the charge density ρ(x,y ,z)which describes the amount of charge in a small volume ∆V :

∆q (r) = ρ(r)∆V R. (1.5)

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For convenience, we introduce the unit separation vector ˆ ≡ r−r

|r−r| , which describes the unit vector

vector pointing from a point r to the point r. Then the electric field is given by

E(r) = 1

4π0

R3

ρ(r)

r2 ˆ dV. (1.6)

1.2.1 Electric Flux and Gauss’s Law

The electric flux is the flow of the electric field E through some surface S . It is proportional to thenumber of electric field lines going through a surface—suppose we have some arbitrary surface S where da is the infinitesimal area—then the amount of E going through this infinitesimal area isequal to E · da; thus, the total flux through the surface S is given by

ΦE =

S

E · da. (1.7)

What Gauss’s Law says is in essence is that the flux through a closed surface is a measure of thetotal charge inside. That is, for any closed surface, it is true that

E · da =

V ρ dV =

1

0 Qenc. (1.8)

The differential version of this is just

∇ · E = ρ

0. (1.9)

The application of this is most evident when one uses it to find the field generated by a sourcewith some sort of symmetry. In other words, one wants to construct a Gaussian surface , whichis a closed surface where E is either of a constant magnitude or equal to 0. Then Gauss’s law justbecomes

E · da = Qenc

0(1.10)

E

da = Qenc

0(1.11)

=⇒ E = Qenc

0

da

(1.12)

and it becomes trivial to find the E-field.

Gaussian Surface Example

For example, consider an infinite sheet of charge with charge density σ on the x − y plane. Thenthe E-field is easily found by constructing a Gaussian surface which is a “pillbox”, like so:

2



so that the electric field can be found simply by E · da =

Qenc

0(1.13)

|E|2A = Aσ

0(1.14)

E = σ20

n (1.15)

where n is the unit normal vector of the plane.As for an cylinder of charge with radius R with a charge density proportional to the distance

from the axis, ρ = ks, we have the electric field to be different in two domains: inside the cylinder,one constructs a Gaussian surface which is a cylinder with a radius of s and length l centered aboutthe axis, so that we have

E · da = Qenc

0(1.16)

|E

|2πsl =

1

0 s

0 l

0

2π

0

ks

·s dφ dz ds (1.17)

|E|2πsl = 1

02πkl

s0

ks2

ds (1.18)

|E|2πsl = 1

0

2

3πlks3 (1.19)

E = ks2

30s. (1.20)

1.2.2 Electric Potential

The electric potential is related to the work done in carrying a charge from one point to another—there is some distribution of charge which produces an E-field. We are interested in the work

necessary to take a small charge from one point to another. It is interesting to note that the wayMaxwell’s equations are defined, it just so happens that E is an irrotational vector field: ∇ × E isidentically zero, and is therefore equal to the negative the gradient of some potential V . In otherwords, E is a conservative force field.

Then the electrostatic potential is defined

V (r) := − r

OE · dl (1.21)

where O is a standard reference point (usually taken to be ∞). Thus V (r) is negative the amountof work it takes to bring a charge from infinity to r. The potential difference is of more interest:

V (b) − V (a) = − bO E · dl +

aO E · dl (1.22)

= − ba

E · dl. (1.23)

And thus the relationship is established:

E = −∇V (1.24) E · ds = 0. (1.25)

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It follows that

∇ · E = ρ

0(1.26)

∇·−∇V = ρ

0(1.27)

∇2V = − ρ0. (1.28)

Given a stationary charge distribution ρ(r), the potential is found by the integral

V (r) = 1

4π0

ρ(r)

dτ . (1.29)

1.2.3 Boundary Conditions for Electrostatics

Crossing a surface charge σ affects the electric field and potential in different ways. Namely, whileE suffers a discontinuity at the surface, V is always continuous. Specifically,

Eabove − Ebelow = σ0n (1.30)

and

V above = V below (1.31)

∂V above

∂n − ∂ V below

∂n = − σ

0(1.32)

∂V

∂n = ∇V · n. (1.33)

1.2.4 Capacitors

When we have two conductors with opposite charges Q and −Q on them, there is a potentialdifference between the two. For two parallel plates each with σ of opposite charge a distance dapart, the potential difference is just

V = Ed = σ

0d =

d

A0Q. (1.34)

Clearly, V and Q are proportional to one another. The constant of proportionality C such that

Q = C V (1.35)

is called the capacitance of the system, and the system is called a capacitor . For the parallel

plate capacitor, we have

C = A0

d . (1.36)

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1.3 Magnetostatics

1.3.1 Current Density

The current density is the amount of charge crossing a unit area in a unit time; that is, it’s

j = ρv, (1.37)

and the total current through a surface S is the electric current I . Thus

I =

s j · dn dS. (1.38)

1.3.2 Magnetic Force on a Current

We’re interested in the force on a current-carrying wire in a magnetic field. The current consistsof charged particles moving with velocity v along the wire, so each charge feels a transverse force

F = q v

×B, (1.39)

and in total, the force on a wire of length ∆ L is

∆F = I × B∆L, (1.40)

so the force per unit length on a wire is I × B.

1.3.3 Ampere’s Law; The magnetic field of steady currents

Ampere’s original law gives us that the circulation of B around any closed curve is equal to thecurrent I through the loop times µ0:

C B · dl = µ0I enc. (1.41)

Similarly to how we constructed Gaussian surfaces to take advantage of Gauss’s law for the electricfield, we construct so called Amperian loops in order to be able to take B out of the line integral.Thus, along Amperian loops, we want |B| to be constant or zero.

Ampere’s Law for an Infinite Solenoid

An example of the usefulness of Ampere’s Law is the exercise of finding the magnetic field withinan infinite solenoid which has a current I running through it and has n turns per unit length. Thenby constructing an Amperian loop like so:

,

5



we see that B must be zero outside the solenoid and constant within: specifically, note that thecomponents of B normal to the surface of the solenoid are zero, so the only contribution to theintegral is the part inside:

C B · dl = µ0I enc (1.42)

|B|L = µ0nLI (1.43)

B = µ0nI z (1.44)

where z points in the axial direction.

1.3.4 Magnetic Vector Potential

Since B is defined to be a solenoidal vector field; that is, it has zero divergence, it can be writtenas the curl of a magnetic vector potential A:

B = ∇ × A. (1.45)

Similarly to how the electrostatic potential yields the same electric field up to addition of a constant,

so too does the magnetic vector potential yield the same magnetic field up to addition of a gradientfield; that is,

∇ × A = ∇ × (A + ∇ψ) = B. (1.46)

Out of convention, we restrict the choice of A with

∇ · A = 0. (1.47)

Vector potential of known currents

To find A from a distribution of known currents, we start with

∇ × B = µ0 j (1.48)∇ × (∇ × A) = µ0 j (1.49)

and apply the vector identity ∇ × (∇ × A) = ∇(∇ · A) − ∇2A to see

∇2A = −µ0 j (1.50)

or in components

∇2Ax = −µ0 jx, ∇2Ay = −µ0 jy, ∇2Az = −µ0 jz, (1.51)

each of which is mathematically identical to the electrostatic equation

∇2V =

−

ρ

0

. (1.52)

So a general solution for each coordinate i is

Ai(r) = 1

4π

ji(r)

dV, (1.53)

and so finally, from a general vector current density j, we can find A by

• finding each component of A by solving three imaginary electrostatic problems,

• and find B by taking various derivatives of A to obtain ∇ × A.

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Vector potential of a straight wire

Our goal is to find the field of a straight wire—take a long straight wire of radius a, carrying thesteady current I . Setting up coordinates like

we have the current density as

jx = bjy = 0 (1.54)

jz = I

πa2, (1.55)

and it follows immediately that Ax = Ay = 0, and ∇

2Az = −

µ0 I

πa2 ; the solution is found by

considering the equivalent electrostatic problem, where ρ = µ00I πa2 —outside an infinite charged

cylinder, the potential is

V = − λ

2π0ln r r =

x2 + y2 λ = πa2ρ. (1.56)

So in our case, we have

Az = −µ0I

2π ln r. (1.57)

Then B is just the curl of A:

Bx = −µ0I

2π

y

r2 (1.58)

By = µ0I

2π

x

r2 (1.59)

Bz = 0, (1.60)

which reveals that B just circles the wire and its magnitude is given by

|B| = µ0

4π

2I

r . (1.61)

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Biot-Savart Law

It is easy to extend this to just finding the magnetic field directly, via

B(r) = µ0

4π

I × ˆ

2 dl =

µ0

4πI

ˆ × dl

2 (1.62)

1.3.5 Magnetic Boundary Conditions

Just as the electric field suffers a discontinuity at a surface charge, so the magnetic field is dis-continuous at a surface current. However, this time it is the tangential component that changes:consider a surface current K : then by picking a Gaussian pillbox of infinitesimal width centered onthe boundary, we see that the condition ∇ · B = 0 means

B · da = 0 (1.63)

and so

B⊥above = B⊥below. (1.64)

On the other hand, if one takes the Amperian loop perpendicular to the surface current andextending to both sides of the sheet,

B · dl = Babovel − B

belowl = µ0I enc (1.65)

= µ0Kl (1.66)

=⇒ Babove − B

below = µ0K. (1.67)

So in summary, at a boundary where a surface current K is running,

Babove − Bbelow = µ0(K × n) (1.68)

Aabove = Abelow (1.69)

∂ Aabove

∂n − ∂ Abelow

∂n = −µ0K, (1.70)

where n is a unit vector perpendicular to the surface, pointing from “below” to “above”.

1.3.6 Multipole Expansion of the Vector Potential

The idea of a multipole expansion is to get approximations for the vector potential of a localizedcurrent that is valid at distant points—we write the potential in the form of a power series in 1/r

where r is the distance to the point in question; if r is sufficiently large, the series will be dominatedby the lowest nonvanishing contribution, and the higher terms can be ignored.Thus, the vector potential of a current loop can be written as (where α) is the angle between r

and r,

A(r) = µ0I

4π

1

dl (1.71)

= µ0I

4π

1

r

dl +

1

r2

r cos α dl +

1

r3

(r)2

3

2 cos2 α − 1

2

dl + . . .

. (1.72)

8



Clearly the first integral is zero since the vector displacement around a closed loop is identicallyzero. The dominant term is the dipole term,

Adip(r) = µ0

4π

m × r

r2 , (1.73)

where m is the magnetic dipole moment

m ≡ I

da = I a, (1.74)

where a is the vector area of the loop.

1.4 Magnetic Fields in Matter

When a magnetic field is applied to a piece of ordinary matter, a net alignment of the magneticdipoles within it occurs, and the medium becomes magnetically polarized, or magnetized . Thedirection of magnetization can be EITHER parallel or anti-parallel to B depending on the type

of material. In diamagnetism, the induced magnetic field is antiparallel to the direction of theexternally applied field, and in ferromagnetism, the induced magnetic field is parallel to the externalfield.

1.4.1 Torques and Forces on Magnetic Dipoles

For a loop of current in a uniform magnetic field, the aligning torque on the object from the fieldis given by

N = m × B (1.75)

where m = I a is the magnetic dipole moment of the loop. For an infinitesimal loop with dipole

moment m in a field B the force isF = ∇(m · B). (1.76)

1.4.2 Magnetization, Bound and Free Currents, Magnetic Susceptibility andPermeability

In the presence of a magnetic field, matter becomes magnetized : the state of magnetic polarizationis defined by the vector quantity

M ≡ magnetic dipole moment per unit volume. (1.77)

Interestingly, if we have a piece of magnetized material, its vector potential can be separated into

the sum of a volume and a surface integral, corresponding to the bound currents of the volumeand surface types, respectively. Namely, the potential of a magnetized object is the same as wouldbe produced by a volume current

Jb = ∇ × M (1.78)

throughout the material, plus a surface current

Kb = M × n (1.79)

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on the boundary.For most substances, the magnetization is proportional to the applied field, so that

M = χmH. (1.80)

χm is called the magnetic susceptibility , and it’s a dimensionless quantity. Thus we have

B = µ0(H + M) = µ0(1 + χm)H, (1.81)

and if we define

µ ≡ µ0(1 + χm), (1.82)

we get

B = µH. (1.83)

Incidentally, the volume bound current density in a homogeneous linear material is proportional tothe free current density:

Jb =

∇ ×M =

∇ ×(χmH) = χmJf , (1.84)

and therefore, unless free current actually flows through the material, all bound current will be atthe surface.

1.4.3 The Auxiliary Field H

Thus we see that the field of a magnetized object is the sum of two parts: the field due to magne-tization of the medium, and the field attributable to everything else—termed bound current and

free current respectively. In any event, the total current can be written as

J = Jb + Jf , (1.85)

and so plugging into Ampere’s law,

1

µ0(∇ × B) = Jf + (∇ × M) (1.86)

=⇒ ∇ ×

1

µ0B − M

= Jf . (1.87)

Now we define the auxiliary field H

H ≡ 1

µ0B − M, (1.88)

so that we have

∇ ×H = Jf , (1.89)

or H · dl = I f enc. (1.90)

So essentially H serves to let us write Ampere’s law in terms of the free current alone. Note,however, that this is not a perfect parallel—the divergence of H is in general not zero; in fact, it is

∇ · H = −∇ · M. (1.91)

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Chapter 2

Electrodynamics

2.1 Ohm’s Law, the Joule heating law

To make a current flow, a force must be applied to the charges. For most substances, the current

density J is proportional to the force per unit charge

f :

J = σf (2.1)

where σ is the conductivity of the medium—its reciprocal, ρ = 1/σ, is the resistivity. Usually, f is the electromagnetic force in its totality:

J = σ(E + v × B), (2.2)

but when the velocity of the charges is sufficiently small (which is usually the case), the secondterm can be ignored, so we have Ohm’s law

J = σE. (2.3)

As a result of all the collisions between electrons within a resistor, the work done by the electricalforce is converted into heat—work done per unit charge is V and the charge flowing per unit timeis I , so the power delivered is

P = V I = I 2R. (2.4)

NOTE: for steady currents and uniform conductivity, all unbalanced chargeresides on the surface.

When solving a problem, start off with J = σE, integrate over asurface to find the total current, then find V from E to cancel outthe line charge term, using V = IR.

2.1.1 Example of Ohm’s Law

1. Suppose there’s a cylindrical resistor of cross section A and length L made from a material of conductivity σ. Suppose further that the potential difference between the two ends is V : what

11



is the current that flows? Simply use Ohm’s Law:

J = σE (2.5)

AJ = I = σAE (2.6)

I = σAV

L

. (2.7)

2. Two long coaxial metal cylinders with radii a and b are separated by a material of conductivityσ; if they are maintained at a potential difference V , what current flows from one to the otherin a length L?

Starting from Ohm’s law, integrate to get total current

I =

J · da = σ

E · da =

σ

λL, (2.8)

and then find V from E:

E · da = λL

0(2.9)

=⇒ E = λ

2π0ss (2.10)

V =

ba

E · ds = λ

2π0ln

b

a

(2.11)

and rearrange to find λ to substitute into I ,

I = 2πσL

ln(b/a)V. (2.12)

2.2 Electromotive Force

In a typical electric circuit, say, a battery hooked up to a light bulb, there are in fact two forcesinvolved in driving the current around a circuit: the source, f s, which is confined to a part of the loop (say, the battery) and an electrostatic force, which serves to smooth out the flow andcommunicate the influence of the source to distant parts of the circuit:

f = f s + E. (2.13)

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The net effect of the source is called the electromotive force, or EMF —this is a bad name forit, since it’s not actually a force at all; it’s an integral of force per unit charge:

E ≡

f · dl =

f s · dl. (2.14)

It can be interpreted as the work done per unit charge by the source .

Example of emf problem

Suppose a battery of emf E and of an internal resistance r is hooked up to a variable “load”resistance, R. What resistance R should be chosen in order to maximize the amount of powerdelivered to the load?

We have that the emf of the battery and load are given by Ir and IR respectively, so that

E = I r + IR =⇒ I = E r + R

, (2.15)

and the power delivered is

P = I 2R = I 2(R) = E r + R

2

(R) (2.16)

=⇒ dP

dR = E 2

− 2R

(R + r)3 +

1

R + r

2

(2.17)

and the max power is thus

dP

dR = 0 = − 2R

(R + r)3 +

1

R + r

2

(2.18)

=⇒ 2R = R + r (2.19)

R = r (2.20)

for max power delivered.

2.2.1 Motional emf

Previously, we only considered possible sources of emfs like batteries—the most common, however,are ones exploited in generators. Specifically, generators exploit motional emfs, which arise asone moves a wire through a magnetic field. In other words, it is when the force driving the currentthrough a circuit is the magnetic force.

For example, in the below figure, the shaded area has a uniform B field pointing into the page,and the resistor R represents whatever it is we’re trying to drive current through. If the entire loopis pulled to the right, then the segment ab’s charges experience a magnetic force whose verticalcomponent drives current through the loop in a clockwise direction.

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Specifically, the emf is

E ≡

f mag · dl = vBh. (2.21)

Note that when one takes the integral, it is taken in a “snapshot” of time; that is, even though the

loop is moving to the right, dl still points straight up.Indeed, it turns out that the emf generated in a current loop is minus the rate of change of fluxthrough the loop, where the flux ΦB of B through the loop is

Φ ≡

B · da, (2.22)

so

E = −dΦ

dt . (2.23)

This is the flux rule for motional emf. It applies to a loop of any shape moving in arbitrarydirections through nonuniform magnetic fields; indeed, the loop needs not even maintain the same

shape. However, this rule is only useful when the current flows along a well-defined path.

Example of Motional emf

Suppose a metal bar of mass m slides frictionlessly on two parallel conducting rails a distance lapart, as in the below figure. A resistor R is connected across the rails and a uniform magneticfield B pointing into the page fills the entire region.

The questions of interest are as follows:

1. • If the bar moves to the right at speed v, what is the current in the resistor? Inwhat direction does it flow?

• The bar moving to the right induces an emf and hence a current, such that

E =

f mag · dl = lvB (2.24)

E = I R (2.25)

=⇒ I = lvB

R . (2.26)

2. • What is the magnetic force on the bar? In what direction?

• The magnetic force on the bar is simply

Fmag =

I dl × B = −I

B dl = I Bl = −

lvB

R

Bl = − l2B2v

R . (2.27)

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3. • If the bar starts out with speed v0 at time t = 0 and is left to slide, what is itsspeed at a later time t?

• We have

Fmag = mdv

dt =

−

l2vB2

R (2.28)

=⇒ dv

dt = − l2B2

Rm v v(0) = v0 (2.29)

=⇒ v(t) = v0 exp

− l2B2

Rm t

. (2.30)

4. • The initial kinetic energy of the bar was, of course, 12 m2

v0 . Check that the energy deliveredto the resistor is exactly 1

2 mv20.

• The total energy delivered to the resistor is equal to E = ∞

0 P dt where P is the powerdispensed into the resistor. We have power to be

P = I 2R = l2v2B2

R2 R = l2v2B2

R (2.31)

and so the total energy delivered to the resistor is

E =

∞0

P dt =

∞0

l2B2

R v2

0 exp

−2l2B2

Rm t

dt =

l2B2v20

R

− Rm

2l2B2 exp

−2l2B2

Rm t

t=∞

t=0

= −1

2m

(2.32)

as expected.

2.2.2 Electromagnetic Induction

Faraday in 1831 performed three experiments which revealed some remarkable facts:

• In the first experiment, he pulled a loop of wire to the right through a magnetic field, and acurrent flowed in the loop.

• In the second experiment, he moved the magnet to the left, holding the loop still, and againa current flowed in the loop.

• In the third experiment, with both the loop and magnet at rest, he changed the strength of the field, and once again current flowed in the loop.

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The curious thing here is that while in the first experiment, the moving loop means that it’s amagnetic force that sets up the emf, but if the loop is stationary, the force cannot be magnetic,since stationary charges experience no magnetic forces. Faraday then stumbled upon the incrediblefact that a changing magnetic field induces an electric field . In fact,

E = E · dl = −

dΦ

dt (2.33)

=⇒

E · dl = −

∂ B

∂t · da (2.34)

∇ × E = −∂ B

∂t . (2.35)

As one can see, in the case when the magnetic field is unchanging in time, Faraday’s law reducesto the old

E · dl = 0.

Lenz’s Law and the universal flux rule

We then arrive at a kind of universal flux rule :

Whenever (and for whatever reason) the magnetic flux through a loopchanges, an emf

E = −dΦ

dt (2.36)

will appear in the loop.

A rule to help understand which way induced currents flow is called Lenz’s Law , which says that

Nature abhors a change in flux. If an induced current flows, its directionis always such that it will oppose the change which produced it.

In a sense Faraday induction is a kind of “inertial” phenomenon: A conducting loop “likes” tomaintain a constant flux through it; if you try to change the flux, the loop responds by sending acurrent around in such a direction as to frustrate your efforts.

The Induced Electric Field

The induced electric field caused by a changing magnetic field acts precisely like magnetostaticfields that are determined by µ0J. More precisely, consider the case of an electric field which is dueexclusively to a changing B, with ρ = 0. Then the pure Faraday field is given by

∇ · E = 0, ∇ × E = −∂ B

∂t . (2.37)

This is precisely the same as in magnetostatics where

∇ · B = 0, ∇ × B = µ0J. (2.38)

Thus, just as we can take advantage of Ampere’s Law in integral form,

B· dl = µ0I enc, by drawingan Amperian loop to determine the magnetic field, we can do the same to determine the inducedelectric field:

E · dl = −dΦ

dt . (2.39)

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In other words, Faraday-induced electric fields are determined by an analog of the Biot-Savart law:

E = − 1

4π

∂ B∂t × ˆ

2 dτ (2.40)

=

−

1

4π

∂

∂t B × ˆ

2 dτ. (2.41)

First Example of an Induced Electric Field

Suppose a uniform magnetic field B(t) pointing straight up fills the shaded circular region depictedin the figure below. If B is time-dependent, what is the induced electric field?

Like a magnetic field caused by a straight wire with a current running through, the induced E-fieldwill be in the circumferential direction. If we draw an Amperian loop as in the figure, we have

E · dl = −dΦ

dt (2.42)

E 2πs = − ∂

∂t

B · da

(2.43)

= − ∂

∂t

B(t)πs2

(2.44)

=⇒ E = −s

2

∂ B

∂tφ. (2.45)

Hence if B is increasing, then E runs clockwise as viewed from above, and counterclockwise if decreasing.

Second Example of an Induced Electric Field

An infinitely long straight wire carries a slowly varying current I (t). Determine the induced electricfield as a function of the distance s from the wire.

17



Recall that the magnetic field caused by the current is just B · dl = B2πs = µ0I enc =⇒ B =

µ0I enc2πs

φ, (2.46)

and so Faraday’s law says, using the drawn Amperian loop, and realizing that E is zero on thevertical sides and of different magnitudes at different distances from the wire,

E

· dl =

−dΦ

dt

(2.47)

l(E (s0) − E (s)) = − d

dt

B · da

(2.48)

= − d

dt

µ0lI (t)

2πs · da

(2.49)

= − d

dt

µ0lI (t)

2π

ss0

1

s ds

(2.50)

= −dI

dt

µ0l

2π (ln s − ln s0) . (2.51)

It turns out that once we rearrange this to just find E (s), that the constant K (if we write it like)

E(s) =

µ0

2πdI dt

ln s + K z (2.52)

depends on the whole history of the function I (t). Also, we oddly find that |E| blows up as s → ∞—it turns out that this is because we’ve gone beyond the limits of the quasistatic approximation , inwhich we use the laws of magnetostatics when the current is changing . In the fully dynamic case,this issue disappears.

2.2.3 Inductance

Suppose we have two loops of wire at rest and run a current through one of them—a magnetic fieldB1 is created, and there is flux of B1 through both loops, albeit of different amounts, obviously.

Since B1 is proportional to the current being run through the first loop, so too is the flux throughthe second loop. In other words, if the current I 1 creates the magnetic field

B1 = µ0

4πI 1

dl1 × ˆ

2 , (2.53)

then the flux through the second loop is

Φ2 =

B1 · da2 = M 21I 1 (2.54)

18



where M 21 is the constant of proportionality—it is called the mutual inductance of the two loops.It can be derived via a clever use of the vector potential, and ends up to be

M 21 = µ0

4π

‹ dl1 · dl2

, (2.55)

which is called the Neumann formula. It’s not practical for actual calculations, but it reveals twoimportant things about the mutual inductance:

• M 21 is a purely geometrical quantity, having to do only with the shapes and positions andsizes of the loops.

• It is symmetric under exchange of which loop the current flows through—in other words,regardless of the shapes and positions of the loops, the flux through 2 when we run a currentI around 1 is identical to the flux through 1 when we send the same current I around 2.Thus, M 21 =12= M .

Example of Inductance Being Used to Simplify Flux Calculations

Suppose a short solenoid with length l and radius a with n1 turns per unit length lies on the axisof a long solenoid of radius b and n2 turns per unit length. A current I flows in the short solenoid:what is the flux through the long solenoid?

It is extremely difficult to calculate the flux of such a small solenoid: however, by taking intoaccount the fact that the flux through the outer solenoid when a current is running through thesmall one is equal to the flux through the small solenoid when the same current is running throughthe large one. The amount of flux through one turn of the small solenoid when the current I isrunning through the large solenoid is found thusly: recall that the magnetic field within a longsolenoid is constant, since

B · dl = µ0I enc =⇒ B = µ0In2, so the flux through a single turn

around the the small solenoid is

ϕs = Bπa2 = µ0In2πa2, (2.56)

and the total flux through the small solenoid is ϕ times the total number of turns in the smallsolenoid, n1l:

Φ = µ0In2πa2n1l. (2.57)

As stated before, this is also the flux through the large solenoid when the current I is runningthrough the small solenoid instead.

2.2.4 Self-Inductance

Now, if we vary the current in loop 1, then the flux of the B-field through the second loop willchange as well, thus inducing an emf; indeed,

E 2 = −dΦ2

dt = −M

dI 1dt

. (2.58)

Come to think of it, varying the current in the loop will induce an emf in the same loop! Indeed,the flux through the loop around which the current is flowing is proportional to the current:

Φ = LI , (2.59)

19



where L is the constant of proportionality called the self-inductance. As with M , it depends onthe geometry of the loop. Thus, if the current in the loop changes, the emf induced is

E = −LdI

dt. (2.60)

Now since L is defined to be a positive quantity, it acts as the analog of mass in mechanical

systems—the larger it is, the more difficult it is to change the current in a loop; that is, the emf induced by L, called the back emf , opposes the change in current.

Example of self-inductance of a toroidal coil

What is the self-inductance of a toroidal coil with a rectangular cross section (inner radius a, outerradius b, height h) that carries a total of N turns? Recall that the magnetic field inside the toroidis given by B = µ0NI

2πs φ where s is the radial distance from the axis around which the toroid issituated. Then the flux through a single turn is

ϕ =

µ0N Ih

2πs · da (2.61)

= µ0N Ih

2π

b

a

1

s ds (2.62)

= µ0N Ih

2π ln

b

a

, (2.63)

and the total flux is N ϕ, so the self-inductance is L = ΦI ,

L = µ0N 2h

2π ln

b

a

. (2.64)

Example of back emf

Suppose a battery which supplies a constant emf of E is connected to a circuit of resistance R andinductance L. Then the current that flows starts at 0 and slowly increases—what it is as a functionof time?

We know that the total emf is the sum of the battery’s and the emf produced by the increasingcurrent:

E = E 0 − LdI

dt, (2.65)

and by Ohm’s law is equal to I R; thus

E 0 − LdI

dt = I R (2.66)

=⇒ I (t) = E 0R + ke−(R/L)t, (2.67)

where k is a constant to be determined by initial conditions. If the switch is closed at time t = 0,then I (0) = 0, and k = −E 0/R, so

I (t) = E 0

R

1 − e−(R/L)t

. (2.68)

The quantity τ ≡ L/R is the time constant —it tells you how long the current takes to reach asubstantial fraction of its final value.

20



2.2.5 Energy in Magnetic Fields

Magnetic energy is the negative of the work necessary to establish a current against the back emf—that is, a circuit’s inductance opposes the change in current, and it takes a certain amount of workto go from o to I , and that amount of work gets stored in the magnetic field. When the currentgoes back to zero, the same amount of energy is recovered.

To be more precise, the amount of work done on a unit charge against the back emf is −E , andso the work per unit time is just −E I = LI dI dt . To find the total work it takes to establish thecurrent, we just integrate the RHS over time and get

W mag = 1

2LI 2 (2.69)

= 1

2

(A · J) dτ (2.70)

= 1

2µ0

all space

B2 dτ. (2.71)

Note that the energy per unit volume is just 1

2µ0

B2. Also note the similarity to the expression forthe energy stored in an electric field:

W elec = 1

2

V ρ dτ =

0

2

E 2 dτ. (2.72)

Example of Magnetic Energy

A long coaxial cable carries current I (which flows down the surface of the inner cylinder and upalong the outer cylinder); the radius of the inner cylinder is a, that of the outer is b. What’s themagnetic energy stored in a section of length l?

Recall that the B-field between the wires is (with it being zero elsewhere)

B = µ0I

2πsφ. (2.73)

Thus the magnetic energy per unit volume is

1

2µ0B2 =

1

2µ0

µ0I

2πs

2

= µ0I 2

8π2s2, (2.74)

21



and since it varies only along the radial axis, we first find the energy in a cylinder of infinitesimalthickness at a radius s, and integrate from a to b.

umag = −W =

ba

µ0I 2

8π2s2

2πls ds (2.75)

= µ0I 2

l4π

ba

1s

ds (2.76)

= µ0I 2l

4π ln

b

a

. (2.77)

Correspondingly, the self-inductance is

L = W

I 2 =

µ0l

2π ln

b

a

. (2.78)

22



Chapter 3

A Brief Summary of Maxwell’sEquations

It is often convenient to write Maxwell’s equations in a way that separates free and bound charges.

In other words, to rewrite the original four,

∇ · E = ρ

0(3.1)

∇ · B = 0 (3.2)

∇ × E = −dB

dt (3.3)

∇ × B = µ0J + µ00∂ E

∂t , (3.4)

as

∇ ·D = ρf (3.5)

∇ · B = 0 (3.6)

∇ × E = −∂B

∂t (3.7)

∇ × H = Jf + ∂ D

∂t . (3.8)

In the separated version, the total charge density can be separated into two parts:

ρ = ρf + ρb = ρf − ∇ · P (3.9)

where P is the electric polarization. Similarly, the current density can be split into three parts,

J = Jf + Jb + J p = Jf + (∇ × M) + ∂ P∂t

, (3.10)

and Gauss’s law can then be written as

∇ · D = ρf (3.11)

23



where D ≡ 0E + P. And Ampere’s law becomes, where H ≡ 1µ0

E + P,

∇ × B = µ0(Jf + Jb + J p) + µ00∂ E

∂t (3.12)

= µ0 Jf +

∇ ×M +

∂ P

∂t + µ00∂ E

∂t (3.13)

=⇒ ∇ × H = Jf + ∂ D

∂t . (3.14)

3.1 Constitutive Relations

These are the relationships between D, H and E, B. For linear media,

P = 0χeE M = χmH (3.15)

so

D = E H = 1

µB (3.16)

where ≡ 0(1 + χe) and µ ≡ µ0(1 + χm). The displacement current is

Jd ≡ ∂ D

∂t . (3.17)

24



Chapter 4

Conservation Laws

4.1 Charge and Energy

4.1.1 The Continuity Equation

The continuity equation describes the local conservation of charge—specifically, the charge in avolume V is

Q(t) =

V

ρ(r, t) dτ, (4.1)

and local conservation of charge says that

dQ

dt = −

S

J · ds, (4.2)

and so the continuity equation for charge is

∂ρ

∂t

=

−∇ ·J. (4.3)

4.1.2 Poynting’s Theorem

The total energy stored in electromagnetic fields per volume is

u = W e + W m = 1

2

0E 2 +

1

µ0B2

. (4.4)

Now suppose the charges move around within a volume V bounded by the surface S —what wasthe amount of work done by the electromagnetic forces to effect the move? The answer is given byPoynting’s theorem :

dW dt

= − ddt

V

12

0E 2 + 1

µ0B2

dτ − 1µ0

S

(E × B) · da. (4.5)

This says that the work done on the charges by the EM force is equal to the decrease in energyremaining in the fields, less the energy that flowed out the surface.

Then the energy flux density , or the energy per unit time, per unit area, transported by thefields is called the Poynting vector :

S ≡ 1

µ0(E × B). (4.6)

25



So Poynting’s theorem can be expressed as

dW

dt = − d

dt

V

u dτ − S

S · da. (4.7)

Then the “continuity” equation for energy , which corresponds to the flow of energy through free

space, is given by ∂u

∂t = −∇ · S. (4.8)

Note that in general EM energy itself is not conserved, since obviously energy constantly goesbetween charges and the fields.

4.2 Momentum

4.2.1 Maxwell’s Stress Tensor

Now consider the total electromagnetic force on the charges in volume V :

F =

V

(E + v × B)ρ dτ =

V

(ρE + J × B) dτ (4.9)

so that the force per unit volume isf = ρE + J × B. (4.10)

Now if we wish to express this in terms of fields alone, thus eliminating ρ and J, it turns out aftera lot of messy algebra that it is

f = ∇ ·↔T − 0µ0

∂ S

∂t , (4.11)

and so the total electromagnetic force on the charges in V is

F = S

↔T + · da. (4.12)

Here↔T is the Maxwell stress tensor , which characterizes the force per unit area (or stress)

acting on the surface. It is defined by

T ij ≡ 0

E iE j − 1

2δ ijE 2

+

1

µ0

BiB j − 1

2δ ijB2

. (4.13)

26



4.2.2 Conservation of momentum

Recall that in classical mechanics, the force on an object is given by

F = dpmech

dt . (4.14)

So the equation for the total magnetic force on the charges in volumeV

can be written in the form

dpp

dt = −0µ0

d

dt

V

S dτ +

S

↔T · da, (4.15)

where pmech is the mechanical momentum of the particles in volume V . Similarly to to Poynting’stheorem, it describes the conservation of momentum in electrodynamics: the first integral describesthe momentum stored in the fields , and the second integral represents the momentum per unit

time flowing in through the surface . Thus, if the mechanical momentum increases, either the fieldmomentum decreases , or else the fields are carrying momentum into the volume through the surface.The momentum density in the fields is evidently

g = µ00S = 0(E × B). (4.16)

Indeed, the momentum flux transported by the fields is −↔

T ; that is to say,

↔

T · da is the elec-tromagnetic momentum per unit time passing through the area da. If the amount of mechanicalmomentum in V isn’t changing, like in a region of empty space without charges, then g, mechanicalmomentum density, is conserved, and follows a continuity equation

∂ g

∂t = ∇ ·

↔T . (4.17)

In general, when there are indeed charges around, however, the field momentum and the mechanicalmomentum are not conserved, since charges and the fields will constantly exchange momentum—only the total is conserved.

Note that the Poynting vector S and the stress tensor↔T both play dual

roles. S is on its own the energy per unit area per unit time transported bythe EM fields, while µ00S is the momentum per unit volume stored in those

fields. Similarly,↔T is itself the electromagnetic stress (force per unit area)

acting on a surface, and −↔T describes the flow of momentum carried by the

fields.

4.2.3 Angular momentum

Interestingly, the conservation of momentum extends to angular momentum as well—recall thatEM fields, which started out as mediators of forces between charges, have taken on a life of theirown. They carry energy,

u = 1

2

0E 2 + 1

µ0 B2

, (4.18)

and momentum ,g = 0(E × B). (4.19)

Indeed, angular momentum too is conserved:

= r × g = 0[r × (E × B)] (4.20)

and indeed, even perfectly static fields can harbor momentum and angular momentum—it is onlywhen these field contributions are included that the conservation laws are sustained.

27



Chapter 5

Electromagnetic Waves

5.1 Waves in one dimension

5.1.1 The wave equation

Let us describe in a very hand-wavy manner what a wave is: a wave is a disturbance of a continuous medium that propagates with a fixed shape at constant velocity . This is the essence of wave motion:that some arbitrary function f (z, t), which might depend on z and t in any arbitrary manner,depends, in wave motion, only in the very special combination z − vt; when this is true, thefunction f (z, t) represents a wave of fixed shape traveling in the z direction at speed v.

The wave equation describes such motion. Suppose we have a very long string under tensionT , and that we shake it, creating a disturbance—these disturbances evolve in time according to thelinear PDE

∂ 2f

∂z 2 =

1

v2

∂ 2f

∂t2 , (5.1)

where v, which represents the speed of propagation, is

v =

T

µ. (5.2)

The solutions to this PDE are any functions of the form f (z, t) = g(z − vt); in fact, most generally,solutions to the wave equation are the sum of a left-moving and a right-moving wave:

f (z, t) = g(z − vt) + h(z + vt). (5.3)

5.1.2 Terminology

The one-dimensional waveform is

f (z, t) = A cos[k(z

−vt) + δ ]. (5.4)

A is the amplitude of the wave, the argument of the cosine is the phase, δ is the phaseconstant (which is normally 0 ≤ δ < 2π), k is the wavenumber, which is related to the wavelengthby the equation

λ = 2π

k . (5.5)

Thus, when z advances by 2π/k, the cosine executes one complete cycle. Note that at z = vt − δk ,

the phase is zero—this is the “central maximum”. Generally speaking, δ/k is the distance by whichthe central maximum is “delayed”.

28



As time passes, the entire wave train proceeds to the right at speed v, and at any fixed point z,the string vibrates up and down, undergoing one full cycle in a period

T = 2π

kv. (5.6)

The frequency is the number of oscillations per unit time,

ν = 1T

= kν 2π

= vλ

. (5.7)

It’s useful to use angular frequency for our purposes—in the analogous case of uniform circularmotion, it represents the number of radians swept out per unit time:

ω = 2πν = kv. (5.8)

Normally we write sinusoidal waves in terms of ω rather than v , so that we have

f (z, t) = A cos(kz − ωt + δ ). (5.9)

In complex notation , we apply Euler’s formula

eiθ = cos θ + i sin θ. (5.10)

Then the wave can be writtenf (z, t) =

Aei(kz−ωt+δ)

. (5.11)

The complex wavefunction is

f (z, t) := Aei(kz−ωt) A := Aeiδ. (5.12)

5.1.3 Polarization

Transverse waves, which describe displacement in directions perpendicular to that of propagation,

occur in two independent states of polarization : you can shake the string up and down, or left-and-right, or in any direction in the x-y plane. Namely, the polarization vector n defines theplane of vibration, so that

f (z, t) = Aei(kz−ωt)n. (5.13)

The polarization angle θ isn = cos θx + sin θy. (5.14)

Thus we can writef (z, t) = ( A cos θ)ei(kz−ωt)x + ( A sin θ)ei(kz−ωt)y. (5.15)

29



5.2 Electromagnetic waves in vacuum

In vacuum, Maxwell’s equations are

∇ · E = 0 ∇ × E = −∂ B

∂t (5.16)

∇ · B = 0 ∇ × B = µ00 ∂ E∂t

, (5.17)

and they constitute a set of coupled first-order PDEs subject to certain conditions. Specifically, weget

∇2E = µ00∂ 2E

∂t2 , ∇2B = µ00

∂ 2B

∂t2 , (5.18)

so each cartesian component of the fields satisfies the three-dimensional wave equation ,

∇2f = 1

v2

∂ 2f

∂t2 , (5.19)

which implies that the vacuum supports the propagation of EM waves traveling at a speed

v = 1√

0µ0= 3 × 108 m/s, (5.20)

which happens to be precisely the velocity of light, c. These waves are plane waves, because thefields are uniform over every plane perpendicular to the direction of propagation. Thus the fieldsare in the form

E(z, t) = E0ei(kz−ωt), B(z, t) = B0ei(kz−ωt), (5.21)

where E0 and B0 are the complex amplitudes, and ω = ck. They are transverse waves , andFaraday’s law demands that

B0 = k

ω(z

× E0). (5.22)

In other words, E and B are in phase and mutually perpendicular , and their real amplitudesare related by

B0 = k

ωE 0 =

1

cE 0. (5.23)

When we have a single wave of angular frequency ω, it is monochromatic.Consider an EM wave propagating in the z direction, and take E to point in the x direction.

Then B points in the y direction. Then the wave is said to be polarizated in the x direction out of convention, and the wave is described by the equations

E(z, t) = E 0 cos(kz − ωt + δ )x (5.24)

B(z, t) = 1c

E 0 cos(kz − ωt + δ )y. (5.25)

30



In general, monochromatic plane waves traveling in arbitrary direction can be described by theintroduction of the propagation vector k, which points in the direction of propagation and has amagnitude of the wavenumber k . Then k · r is the appropriate generalization of kz, so the complexwavefunctions are in general given by

E(r, t) = E 0ei(k·r−ωt)n (5.26)

B(r, t) = 1

ck × E (5.27)

and the actual fields are given by the real parts of these expressions.

5.2.1 Energy and momentum in EM waves

Since

u = 1

2

0E 2 +

1

µ0B2

, (5.28)

in the case of a monochromatic plane wave, B 2 = µ00E 2, and so the electric and magnetic contri-butions are equal :

u = 0E 2 = 0E 20 cos2(kz − ωt + δ ). (5.29)

Remarkably, as the wave travels, it carries this energy along with it . The energy flux densitytransported by the fields is given by the Poynting vector, which, in the case of monochromaticplane waves propagating in the z direction, is

S = c0E 20 cos2(kz − ωt + δ )z = cuz. (5.30)

EM waves therefore also carry momentum with them:

g = 1c2 = 1c uz. (5.31)

In the case of light , the wavelength and period is so short that we might as well just take theaverage value, since any macroscopic measurement will encompass many cycles. Since the averageof cosine-squared over a complete cycle is 1/2, we get the expressions

u = 1

20E 20 S =

1

2c0E 20 z g =

1

2c0E 20 z. (5.32)

31



The intensity of an EM wave is the average power per unit area it transports:

I := S = 1

2c0E 20 , (5.33)

and when light falls at normal incidence upon a perfect absorber , it delivers its momentum to the

surface. In a time ∆t, the momentum transfer is ∆p = gAc∆t, so the radiation pressure,which is the average force per unit area is

P = 1

A

∆ p

∆t =

1

20E 20 =

I

c. (5.34)

On a perfect reflector , the pressure is twice as great, since the momentum switches direction.

5.3 Electromagnetic waves in matter

5.3.1 Propagation in linear media

In matter, in regions where there’s no free charge or free current, Maxwell’s equations become

∇ · D = 0 ∇ × E = −∂ B

∂t (5.35)

∇ · B = 0 ∇ × H = ∂ D

∂t (5.36)

and in the case that the medium is linear , we have that

D = E H = 1

µB, (5.37)

and if we also require that the medium is homogeneous , so that and µ don’t change from pointto point, they reduce to

∇ · E = 0 ∇ × E = −∂ B

∂t (5.38)

∇ · B = 0 ∇ × B = µ∂ E

∂t , (5.39)

which differ from their vacuum analogues only by the replacement of µ00 with µ! This has re-markable physical implications, because it means that EM waves propagate through such materialsat a speed

v = 1√

µ =

c

n (5.40)

where we define the index of refraction

n :=

µ

0µ0. (5.41)

In most cases, µ ≈ µ0, so thatn ≈ √

r, (5.42)

where r is the relative permittivity of the material—i.e., /0, which describes the ability of thematerial to store electrical energy in an electric field. Since r is almost always greater than 1, lighttravels more slowly through matter.

32



The conclusions about energy and stuff also carry over, just with 0 → , µ0 → µ, and c → v .So we end up with an energy density of

u = 1

2

E 2 +

1

µB2

, (5.43)

and the Poynting vector becomesS =

1

µ(E × B). (5.44)

For monochromatic plane waves, the frequency and wavenumber are related by ω = kv and theamplitude of B is 1/v times the amplitude of E, and the intensity is

I = 1

2vE 20 . (5.45)

The question of what happens when a wave passes from one transparent medium to anotheris an interesting one, and is governed by electrodynamic boundary conditions earlier derived. Of course, we expect to get a reflected wave and a transmitted wave.

1E ⊥1 = 2E ⊥2 E1 = E

2 (5.46)

B⊥1 = B⊥

2

1

µ1B1 =

1

µ2B2. (5.47)

5.3.2 Reflection and Transmission at Normal Incidence

Suppose a plane wave of frequency ω travels in the positive z direction and is polarized in the xdirection, and encounters a boundary between two linear media at the xy plane, like shown below:

Then at the interface it gives rise to a reflected and transmitted wave. The reflected wave travelsin the −z direction and the transmitted wave continues in the +z direction. We can write the

33



incident, reflected, and transmitted waves, respectively, as

Incident

EI (z, t) = E 0I e

i(k1z−ωt)x

BI (z, t) = 1v1

B0I ei(k1z−ωt)y

(5.48)

Reflected ER(z, t) = E 0Rei(−k1z−ωt)x

BR(z, t) = 1v1 B0Rei(−k1z−ωt)y (5.49)

Transmitted

ET (z, t) = E 0T e

i(k2z−ωt)x

BT (z, t) = 1v1

B0T ei(k2z−ωt)y

(5.50)

Note that the reflected wave has a magnetic field component which has switched directions; thatis, it’s now pointing in the −y direction. This is consistent with the fact that the Poynting vectorof each wave (incident, transmitted, reflected) must point in the direction of propagation.

Recall that the electric field boundary conditions demand that the component of E parallel tothe interface not change when crossing, and that the magnetic field interface conditions demandthe the same (modulo the different permeabilities):

E1 = E

2 (5.51)

=⇒ E 0I + E 0R = E 0T (5.52)

1

µ1B =

1

µ2B2 (5.53)

=⇒ 1

µ1

1

v1E 0I −

1

v1E 0R

=

1

µ2

1

v2E 0T

, (5.54)

where the second condition can be rewritten

E 0I − E 0R = β E 0T β := µ1v1

µ2v2=

µ1n2

µ2n1. (5.55)

Then, solving for the reflected and transmitted waves in terms of the incident wave, we get

E 0g =

1 − β

1 + β

E 0I ,

E 0T =

2

1 + β

E 0I . (5.56)

Clearly, the reflected wave is with respect to the incident wave in phase if v2 > v1 and out of

phase if v1 > v2.The intensities (that is, the average power per unit area) of the waves clearly change as well.

Recalling that

I = 1

2vE 20 , (5.57)

then the ratios of the reflected and transmitted waves’ intensities to those of the incident wave’sare called the reflection , R, and transmission , T , coefficients. Of course, R + T = 1.

5.3.3 Reflection and Transmission at Oblique Incidence

Now suppose the wave is incident on the boundary at an oblique angle, θI , which is measured withrespect to the interface’s normal, as shown below:

34



Then the incident, reflected, and transmitted waves are all of the standard form

Eξ(r, t) = E0ξei(kξ·r−ωt) (5.58)

Bξ(r, t) = 1

vi(kξ × Eξ). (5.59)

Note that all three waves have the same frequency ω that is determined at the source of thelight. Thus

ωI =ωR = ωT (5.60)

kI v1 =kRv1 = kT v2 (5.61)

=⇒ kI = kR (5.62)

= v2

v1kT (5.63)

= n1

n2kT . (5.64)

Then as for the boundary conditions, they as always require that

(i) 1E ⊥1 = 2E ⊥2 (5.65)

(ii) E1 = E (5.66)

(iii) B⊥1 = B⊥

2 (5.67)

(iv) 1

µ1B1 =

1

µ2B2. (5.68)

Without doing any actual math we can immediately deduce some consequences: firstly, that all thex, y, and t dependence is confined to the exponential factors. Since a function of some variableshas, when one of those variables is held constant, dependence upon only its constituent parts which

35



depend on the remaining variables, the exponential factors in our cases must hold constant at all

points on the xy plane for all time . Thus if we hold z = 0, we see that

kI · r = kR · r = kT · r, (5.69)

which if expanded out shows that the x and y components of the wave directions must all be equal.

Thus we arrive at our first law:First Law: The incident, reflected, and transmitted wave vectors form a plane (calledthe plane of incidence ) which also includes the normal to the surface.

So in our case, if we choose that ky := 0 the plane of incidence is (for a given y) the xz plane.Then since (kξ)i = kξ sin θxi where the angle is taken with respect to the normal of the interface,we have the

Second Law: The angle of incidence is equal to the angle of reflection;

θI = θR. (5.70)

This is called the law of reflection .

As for the transmitted angle, this is the

Third Law:sin θT sin θI

= n1

n2. (5.71)

This is the law of refraction : Snell’s Law .

5.3.4 Fresnel’s Equations

Consider an EM wave incident at an oblique angle on an interface, and suppose it has polarizationthat is parallel to the plane of incidence; in other words, it is linearly polarized —the E-fieldand B-field are each confined to a given plane along the direction of propagation (in our case, the

electric field is confined to the xz plane). Then application of the EM interface conditions resultin expressions for the relationships between the field amplitudes on the two sides of the interface:

E 0R =

α − β

α + β

E 0I (5.72)

E 0T =

2

α + β

E 0I (5.73)

α := cos θT cos θI

(5.74)

β := µ1v1

µ1v2=

µ1n2

µ2n1. (5.75)

These are Fresnel’s equations for the case of polarization in the plane of incidence. Note thatthe transmitted wave is always in phase with the incident one, while the reflected one is either inphase (“right side up”) or 180 out of phase (“upside down”), depending on whether α > β orβ > α.

Clearly, we see that the amplitudes of the transmitted and reflected waves depend on the angleof incidence: since

α =

1 −

n1n2

sin θI

2

cos θI , (5.76)

36



we see that in the case of normal incidence, θI = 0, α = 1, and in the case of θI = 90, αdiverges—if the beam of light just grazes the interface, then there’s enormous reflection. Also,there’s an intermediate angle called Brewster’s angle θB at which the reflected wave is completelyextinguished—i.e., it’s completely transmitted. This occurs when

sin2

θB =

1

−β 2

n1n2

2 − β 2. (5.77)

The power per unit area striking the interface is S · z. Thus the incident intensities are

I I = 1

21v1E 20I cos θI (5.78)

I R = 1

21v1E 20R cos θR (5.79)

I T = 1

22v2E 20T cos θT . (5.80)

Note that these are the average power per unit area expressions with respect to the interface, andthat the interface is at an angle to the wavefront. The reflection and transmission coefficients

are given by

R := I RI I

=

E 0RE 0I

=

α − β

α + β

2

(5.81)

T := I T I I

= 2v2

1v1

E 0T E 0I

2 cos θT cos θI

= αβ

2

α + β

2

. (5.82)

basics of electricity and magnetism

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