physics 2 for electrical engineering ben gurion university of the negev

Physics 2 for Electrical EngineeringPhysics 2 for Electrical Engineering

Ben Gurion University of the Negevwww.bgu.ac.il/atomchip, www.bgu.ac.il/nanocenter

Lecturers: Daniel Rohrlich , Ron Folman Teaching Assistants: Ben Yellin, Yoav Etzioni

Grader: Gad Afek

Week 12. Outlook – Introduction • method of images • Poynting vector: examples • a paradox • transformers • Maxwell’s equations in matter • reflection and refraction • polarized light • spherical waves • interference Source: Halliday, Resnick and Krane, 5th Edition, Chap. 38;The Feynman Lectures on Physics, Chaps. 17, 21 and 27.

http://www.bgu.ac.il/atomchip

http://www.bgu.ac.il/nanocenter

Introduction

In the previous weeks in this course, we have offered you a basic understanding of classical electrodynamics, the theory electric and magnetic fields and forces. This theory – which Maxwell perfected – puts powerful tools in your hands. It does not include quantum theory but it is already consistent with Einstein’s special theory of relativity (although we have not talked about relativity).

Introduction

In the previous weeks in this course, we have offered you a basic understanding of classical electrodynamics, the theory electric and magnetic fields and forces. This theory – which Maxwell perfected – puts powerful tools in your hands. It does not include quantum theory but it is already consistent with Einstein’s special theory of relativity (although we have not talked about relativity).

In this lecture, we hope to entertain you with examples of how much you can already do with this theory, or can do with just one or two more tricks, and examples of everyday phenomena that you can now understand in detail.

Method of images

In electrostatics, E and B are constant in time; so we can write E as a gradient: Gauss’s law then becomes

Now suppose we have any function V0(r) with the property

that Any linear combination of V(r) and V0(r)

will be another solution of Gauss’s law. Which combination is the right one?

The answer depends on boundary conditions.

,0)( rE).()( rrE V

. )(

)(0

2

r

r V

.0)(02 rV

Method of images

We have solved electrostatic problems with a few charges, or with a lot of symmetry. Here is a different sort of electrostatic problem: A particle of charge q is at rest a distance R from an

infinite conducting plane. What is the potential V(r) everywhereto the right of the conductor?

Method of images


infinite conducting plane. What is the potential V(r) everywhereto the right of the conductor? Itcan’t be just the potential of the

particle…

xRq

z

Method of images


infinite conducting plane. What is the potential V(r) everywhereto the right of the conductor? It can’t be just the potential of the

particle, because the plane is an equipotential: V(0,y,z) = constant.

xRq

z

Equipotentialsurface

Method of images

Answer: Just put an “image” charge –q at on the x-axis at the point (–R,0,0) and write down the potential V(r) for x > 0:

xRq

z

–qR

Method of images


The Laplacian of the “image” potential vanishes for x > 0 and also V(0,y,z) = 0!

. ])([4

])([4

),,(

2/12220

2/12220

zyRx

q

zyRx

qzyxV

xRq

z

–qR

2

. ])([4

])([4

),,(

2/12220

2/12220

zyRx

q

zyRx

qzyxV

xq

z


R

Method of images


The Laplacian of the “image” potential vanishes for x > 0 and also V(0,y,z) = 0!

2

Method of images

This method also works for a hollow conducting sphere:

q

z

x

D

R


Method of images

This method also works for a hollow conducting sphere:

q

z

q' xR

D

R'


R

DR

qR

Dq

2'

'

Poynting vector

The relation between energy and momentum in radiation:

B

E

k

Poynting vector

The relation between energy and momentum in radiation:

Consider electromagnetic radiation normal to a metal sheet of size L2 and conductance σ. The electric field induces a current density J = σE. The magnetic field is B perpendicular to J, and like E lies in the metal sheet. It induces a force density J×B = σE×B in the direction of the radiation. The power delivered by the radiation to the metal sheet is I2R = (LJ)2/σ so the power per unit surface area is J2/σ = JE while the force per unit surface area is JB = JE/c. Although this result is for a special case, it must correspond to the energy and momentum content of the electromagnetic wave. Since the averaged Poynting vector Sav is the energy density flux in radiation, the

momentum density flux in radiation – i.e. the radiation pressure – must be Sav/c.

Poynting vector

Example 1: A great amount of debris – asteroids, meteors, dust – exists in interplanetary space. Although we might expect the smallest dust particles to be of molecular size, very little of the dust in our solar system is smaller than about 0.2 μm. Why?

Poynting vector


The sun radiates energy at the rate P = 3.90 × 1026 W. Assume each dust particle is spherical, with density ρ = 3.0 × 103 kg/m3 (density of the earth’s crust), and absorbs all energy incident on it. The mass of the sun is M = 1.99 × 1030 kg, and Newton’s gravitational constant is G = 6.67 × 10–11 m3/s2·kg. Assume the distance of the dust particle from the sun is R.

Poynting vector


The sun radiates energy at the rate P = 3.90 × 1026 W. Assume each dust particle is spherical, with density ρ = 3.0 × 103 kg/m3 (density of the earth’s crust), and absorbs all energy incident on it. The mass of the sun is M = 1.99 × 1030 kg, and Newton’s gravitational constant is G = 6.67 × 10–11 m3/s2·kg. Assume the distance of the dust particle from the sun is R. What is its radius r if the gravitational attraction of the sun just balances the radiation pressure from the sun?

Poynting vector


Answer: The attraction of the sun is

The energy flux from the sun per unit area is the Poynting vector S = P/4πR2. The radiation pressure S/c = (P/4πR2)/c if the dust particle absorbs all the energy. The radiation force is the cross-section of the particle times S/c: Frad = πr2(P/4πR2)/c.

If Frad = Fgrav then so r =

3P/16πρcGM. .

.3

4 32

rR

GMFgrav

3

22

2

3

4

4r

R

GM

cR

Pr

Poynting vector



If Frad = Fgrav then so r = 3P/16πρcGM

= 3(3.90 × 1026 W) / 16π (3.0 × 103 kg/m3) (3.00 × 108 m/s) × (6.67 × 10–11 m3/s2·kg) (1.99 × 1030 kg)

= 0.19 μm.

.3

4 32

rR

GMFgrav

3

22

2

3

4

4r

R

GM

cR

Pr

Poynting vector



If Frad = Fgrav then so r = 3P/16πρcGM

= 3(3.90 × 1026 W) / 16π (3.0 × 103 kg/m3) (3.00 × 108 m/s) × (6.67 × 10–11 m3/s2·kg) (1.99 × 1030 kg)

= 0.19 μm.

So why is interplanetary dust mostly larger than 0.2 μm?

.3

4 32

rR

GMFgrav

3

22

2

3

4

4r

R

GM

cR

Pr

Poynting vector

Example 2 (Halliday, Resnick and Krane, 5th Edition, Chap. 38, Prob. 12): The wire shown has length L, resistance R and radius a, and carries a current I. (a) Show that the Poynting vector S on the wire surface points inward. (b) Show that the integral of S over the wire surface is –I2R.

BI

a

Poynting vector


Answer: (a) E is parallel to the wire, hence S = E×B/μ0 points

into the wire.

BI

a

B

ES×

Poynting vector


Answer: (b) E = V/L = IR/L and B = μ0I/2πa, thus S = I2R/2πaL

and ∫ S·dA = 2πaLS = –I2R.

BI

a

B

ES× dA

Poynting vector


This looks strange at first – electromagnetic energy enters the wire from the sides? But maybe it is not so strange – heat energy leaves the wire from the sides….

BI

a

B

ES× dA

Lq

A paradox

Assume two uniformly (but oppositely) charged cylinders, both infinitely long. One fits inside the other but they have almost the same radius r. Outside, a distance L from their common axis, is a particle of charge q at rest. The two cylinders rotate at the same angular speed but in opposite directions. But mutual friction gradually stops them from rotating, starting at t = 0.

A paradox

Assume two uniformly (but oppositely) charged cylinders, both infinitely long. One fits inside the other but they have almost the same radius r. Outside, a distance L from their common axis, is a particle of charge q at rest. The two cylinders rotate at the same angular speed but in opposite directions. But mutual friction gradually stops them from rotating, starting at t = 0.

According to Faraday’s law, the decay of the magnetic flux in the cylinders creates an electric field that accelerates the charged particle.

Lq

A paradox

Let the magnitude of the magnetic field inside the cylinders be B(t). The force on the charged particle is qE(t) where

hence the final momentum of the charged particle is qr2B(0)/2L.

z

xLq

, )( )( 2 2 tBdt

drtLE

A paradox

Let’s integrate the initial Poynting vector, assuming L >> r. We need Ex(0,0,z) at the initial time:

so the integral of the Poynting vector is

which reduces (using z = L tan θ) to qr2B(0)/2ε0μ0L = qc2r2B(0)/2L.

z

xLq

, )(4

),0,0(2/322

0 zL

qLzEx

, )(4

)0( 2/322

00

2dz

zL

BrqL

A paradox

So the integral of the Poynting vector S is qr2B(0)/2ε0μ0L = qc2r2B(0)/2L.

Now S/c is energy density; and if we assume that S/c2 is momentum density, then the electromagnetic momentum was initially qr2B(0)/2L, exactly the final mechanical momentum.

Momentum is conserved, after all!

z

xLq

Transformers

Power transmission lines lose I2R = IV of power on their way to their destination. A way to reduce power losses is to reduce I while increasing V. 350-kV power lines are common. At the destination, where it is safer to use a smaller potential, a transformer reduces V. How does a transformer work?

Get yours today!

http://www.isomatic.co.uk/home.htm

Transformers

Power transmission lines lose I2R = IV of power on their way to their destination. A way to reduce power losses is to reduce I while increasing V. 350-kV power lines are common. At the destination, where it is safer to use a smaller potential, a transformer reduces V. How does a transformer work?

R2

S

N1 N2

ΔV1 Soft iron

Primary(input)

Secondary(output)

ΔV2

Transformers

With the switch S open, the potential across the primary coil (with N1 windings) is ΔV1 = – N1dΦB/dt, by Faraday’s law. If

the coils link with same magnetic flux ΦB then the potential on

the secondary coil is ΔV2 = – N2dΦB/dt. The relation between

the potentials is ΔV1 = ΔV2 N1/N2. If N2 > N1 then ΔV2 > ΔV1.

This is a step-up transformer.

R2

S

N1 N2

ΔV1 Soft iron

Primary(input)

Secondary(output)

ΔV2

Transformers

With the switch S open, the potential across the primary coil (with N1 windings) is ΔV1 = – N1dΦB/dt, by Faraday’s law. If

the coils link with same magnetic flux ΦB then the potential on

the secondary coil is ΔV2 = – N2dΦB/dt. The relation between

the potentials is ΔV1 = ΔV2 N1/N2. If N2 < N1 then ΔV2 < ΔV1.

This is a step-down transformer.

R2

S

N1 N2

ΔV1 Soft iron

Primary(input)

Secondary(output)

ΔV2

Transformers

With the switch S closed, the current in the secondary coil is I2 = ΔV2 /R2.

In an ideal transformer, there are no losses and I1ΔV1 = I2 ΔV2 .

R2

S

N1 N2

ΔV1 Soft iron

Primary(input)

Secondary(output)

ΔV2

Transformers

Example: A power station in Beersheba delivers 20 MW of power to a neighborhood 1 km away. The resistance in the transmission line is 2.0 Ω and the industrial price of electricity is₪ 0.30/kWh. The power station produces the electricity at 22 kV but steps it up to 230 kV before transmission. (a) What is the (approximate) daily transmission cost? (b) What would be the cost without stepping up the potential?

Transformers


Answer: (a) We approximate using dc relations. The current is power divided by potential: I = 20 MW / 230 kV = 87 A. The power lost in the line is I2R = (87 A)2(2.0 Ω) = 15 kW; in one day, the power loss is 360 kWh, worth ₪110.

Transformers


Answer: (b) We approximate using dc relations. The current is power divided by potential: I = 20 MW / 22 kV = 910 A. The power lost in the line is I2R = (910 A)2(2.0 Ω) = 1.65 MW; in one day, the power loss is 40 MWh, worth ₪12,000.

Maxwell’s equations in matter

Here are Maxwell’s equations, including sources (charges, currents) but not dielectric, diamagnetic, paramagnetic or ferromagnetic materials:

.

0

000

0

t

t

EJB

BE

B

E


Here are Maxwell’s equations, including sources (charges, currents) and many dielectric, diamagnetic, paramagnetic or ferromagnetic materials:

.

0

t

t

EJB

BE

B

E


We deduce the speed of electromagnetic waves in these materials just as we deduced the speed in the vacuum. First, we eliminate sources (charges, currents):

.

0

0

t

t

EB

BE

B

E


We find that the speed of electromagnetic waves in a medium with electric permittivity ε and magnetic permeability μ is .

EEE

E

BEE

BE

BE

22

22

2 )()(

0

0

ttt

t

t

t

Faraday’s law

Vector identity

Gauss’s law in a vacuum: •E = 0

Ampère’s law, as modified by Maxwell

/1

,02

22

t

EE

/1

, ),( tiiet rkErE


The plane wave E(r,t) = E cos (k·r – ωt + δ) is often written as a complex function:

where E is a constant vector that fixes the amplitude and the polarization of the wave, k = 2π/λ is the wave number of the wave, and ω is its angular frequency. (You can just ignore the imaginary part of the function.) Substituting this plane

wave into we obtain a solution if and

only if – k2 + μεω2 = 0, hence = ω/k = f λ is the speed

of the wave in the material.

),( as ),( tit rEkrE


Writing a plane wave as E(r,t) = Eeik·r – iωt is very convenient! It allows us to write , so we learn that E is transverse in this case as well. The factor of i just indicates a phase difference of π/2. It also allows us to write ik × E(r,t). Now from Faraday’s law we have

hence B(r,t) = k × E(r,t)/ω.

as ),( trE

, ),( ),(),(),( titt

tti rBrBrErEk

Reflection and refraction

Now let’s consider what happens when an electromagnetic wave crosses between two media, i.e. between two materials with different permittivities and permeabilities. Let the interface of the two materials be the plane x = 0, with a wave Ei(r,t) incident on the left. There will be also a reflected wave Er(r,t) on the left and a transmitted wave Et(r,t) on the right:

x

μiεi μtεt

kr kt

θi

θrθt

ki


On the x = 0 plane, the electric field must be continuous, which means that Ei(0,y,z,t) + Er(0,y,z,t) = Et(0,y,z,t); or in complex

wave notation (after cancelling the common factor e– iωt):

. ),,0(),,0(),,0( zyit

zyir

zyii

tri eee kkk EEE

x

μiεi μtεt

kr kt

θi

θrθt

ki


The equation

must hold for all y and all z. The only way it can hold for all y and all z is if ki · (0,y,z) = kr · (0,y,z) = kt · (0,y,z), i.e. ki , kr and

kt have the same y- and z-components:

),,0(),,0(),,0( zyit

zyir

zyii

tri eee kkk EEE

x

μiεi μtεt

kr kt

θi

θrθt

ki

ki sin θi = kr sin θr = kt sin θt .


The equation



),,0(),,0(),,0( zyit

zyir

zyii

tri eee kkk EEE


Since ki = kr we have

angle of incidence =angle of reflection!

x

μiεi μtεt

kr kt

θi

θrθt

ki


The equation



),,0(),,0(),,0( zyit

zyir

zyii

tri eee kkk EEE


Snell’s law ofrefraction!

x

μiεi μtεt

kr kt

θi

θrθt

ki

Polarized light

Our plane wave solution E(r,t) = Eeik·r – iωt implies that the electric field of a wave has a certain direction, perpendicular to propagation vector k. Is this a prediction we can test?

Yes, with a polarizer! A polarizer has a polarization axis, and passes only the component of light along the polarization axis.

We will not usually see an effect with one polarizer, because most light is unpolarized – it is mixture of waves with E pointing in any direction perpendicular to k. But with two polarizers we see an effect: the first polarizer polarizes the light emitted by a source, and as we turn the second polarizer relative to the first, we may see all or part or none of the polarized light.

Polarized light

Our plane wave solution E(r,t) = Eeik·r – iωt implies that the electric field of a wave has a certain direction, perpendicular to propagation vector k. Is this a prediction we can test?

Yes, with a polarizer! A polarizer has a polarization axis, and passes only the component of light along the polarization axis.

http://micro.magnet.fsu.edu/

http://micro.magnet.fsu.edu/

Polarized light

A plane wave E(r,t) = Eeik·r – iωt is linearly polarized, but there is also elliptically polarized light. Elliptically polarized light (including circularly polarized light) reduces to a combination of two linearly polarized waves, with orthogonal linear polarizations, and a relative phase difference:

Polarized light


Watch an animation!

http://www.nsm.buffalo.edu/~jochena/images/elliptic2.gif

Polarized light


Watch an animation!

http://www.nsm.buffalo.edu/~jochena/images/circular3.gif

Spherical waves

We found plane-wave solutions to the wave equation, but don’t assume these are the only solutions. Here a spherical wave:

where p(t) = p0 sin ωt represents an electric dipole oscillating on the z-axis at r = 0; p0 is a constant with dimensions of C·m,

pointing in the positive z-direction, and the dots denote time derivatives. E(r,t) is a bit more complicated.

At large r, both B(r,t) and E(r,t) drop as 1/r.

, 4

)/( )/()/(),(

320

rpp

rB

rc

crtcrcrtt

B

E

p

Interference

Spherical-wave solutions can help us understand what happens when a wave passes through a small hole in a screen. If the wave passes only through the hole, then on the other side of the screen it looks like a spherical wave with the hole as its source:

Watch this as an animation!

http://astro1.panet.utoledo.edu/~lsa/_color/14_1slitAnim.gif

Interference

If the wave arrives at a screen with two holes and passes through them, the result is an interference pattern where the crests of one spherical wave and the troughs of the other spherical wave add up to zero field:

Watch this as an animation!

http://astro1.panet.utoledo.edu/~lsa/_color/14_2slitAnim.gif

Interference

There are regions of destructive interference, where waves cancel each other,

L = difference in path length = (n + ½) λ

d = hole separation

D = distance to screen

sin θn = L/d = (n + ½) λ/d

Ld θn

Interference

and regions of constructive interference, where the waves add in phase to produce a stronger oscillation.

L = difference in path length = nλ

d = hole separation

D = distance to screen

sin θ'n = L/d = nλ/d

Ld

θ'n

physics 2 for electrical engineering ben gurion university of the negev

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