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W13D2: Displacement Current, Maxwell’s Equations,

Wave Equations

Today’s Reading Course Notes: Sections 13.1-13.4

Announcements

Math Review Tuesday May 6 from 9 pm-11 pm in 26-152 Pset 10 due May 6 at 9 pm 2

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Outline

Displacement Current Poynting Vector and Energy Flow Maxwell’s Equations

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Maxwell’s Equations

E ⋅ n̂ da

S∫∫ =

1ε0

ρ dVV∫∫∫ (Gauss's Law)

B ⋅ n̂ da

S∫∫ = 0 (Magnetic Gauss's Law)

E ⋅ d s

C∫ = −

ddt

B ⋅ n̂ da

S∫∫ (Faraday's Law)

B ⋅ d s

C∫ = µ0

J ⋅ n̂ da

S∫∫ (Ampere's Law quasi - static)

Is there something missing?

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Maxwell’s Equations One Last Modification: Displacement Current

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Ampere’s Law: Capacitor Consider a charging capacitor: Use Ampere’s Law to calculate the magnetic field just above the top plate

What’s Going On?

B ⋅ ds∫ = µ0 Ienc

Ienc =

J ⋅ n̂ da

S∫∫

1) Surface S1: Ienc= I 2) Surface S2: Ienc = 0

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Displacement Current

dQdt

= ε0

dΦE

dt≡ Idis

We don’t have current between the capacitor plates but we do have a changing E field. Can we “make” a current out of that?

E = Q

ε0 A⇒ Q = ε0 EA = ε0ΦE

This is called the “displacement current”. It is not a flow of charge but proportional to changing electric flux

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Displacement Current:

Idis = ε0

ddt

E ⋅ n̂ da

S∫∫ = ε0

dΦE

dt

If surface S2 encloses all of the electric flux due to the charged plate then Idis = I

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Maxwell-Ampere’s Law

B ⋅ds

C∫ = µ0

J ⋅ n̂da

S∫∫ + µ0ε0

ddt

E ⋅ n̂da

S∫∫

= µ0 (Ienc + Idis )

Ienc =

J ⋅ n̂da

S∫∫

Idis = ε0

ddt

E ⋅ n̂da

S∫∫

“flow of electric charge”

“changing electric flux”

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Concept Question: Capacitor If instead of integrating the magnetic field around the pictured Amperian circular loop of radius r we were to integrate around an Amperian loop of the same radius R as the plates (b) then the integral of the magnetic field around the closed path would be

1.  the same. 2.  larger. 3.  smaller.

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Sign Conventions: Right Hand Rule

Integration direction clockwise for line integral requires that unit normal points into page for surface integral.

Current positive into the page. Negative out of page.

Electric flux positive into page, negative out of page.

B ⋅ds

C∫ = µ0

J ⋅ n̂da

S∫∫ + µ0ε0

ddt

E ⋅ n̂da

S∫∫

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Sign Conventions: Right Hand Rule

Integration direction counter clockwise for line integral requires that unit normal points out page for surface integral.

Current positive out of page. Negative into page.

Electric flux positive out of page, negative into page.

B ⋅ds

C∫ = µ0

J ⋅ n̂da

S∫∫ + µ0ε0

ddt

E ⋅ n̂da

S∫∫

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Concept Question: Capacitor Consider a circular capacitor, with an Amperian circular loop (radius r) in the plane midway between the plates. When the capacitor is charging, the line integral of the magnetic field around the circle (in direction shown) is

1.  Zero (No current through loop) 2.  Positive 3.  Negative 4.  Can’t tell (need to know direction of E)

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Concept Question: Capacitor

The figures above shows a side and top view of a capacitor with charge Q and electric and magnetic fields E and B at time t. At this time the charge Q is:

1.  Increasing in time 2.  Constant in time. 3.  Decreasing in time.

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Group Problem: Capacitor

A circular capacitor of spacing d and radius R is in a circuit carrying the steady current i shown. At time t = 0 , the plates are uncharged

1.  Find the electric field E(t) at P vs. time t (mag. & dir.) 2.  Find the magnetic field B(t) at P

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Energy Flow

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Poynting Vector

S =E ×B

µ0

Power per unit area: Poynting vector

P =S ⋅ n̂ da

opensurface

∫∫Power through a surface

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Energy Flow: Capacitor

S =E ×B

µ0

What is the magnetic field?

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Concept Question: Capacitor

The figures above show a side and top view of a capacitor with charge Q and electric and magnetic fields E and B at time t. At this time the energy stored in the electric field is:

1.  Increasing in time. 2.  Constant in time. 3.  Decreasing in time.

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Energy Flow: Capacitor

S = (

E×B) / µ0 = (Ek̂ × B θ̂) / µ0 = −(EB / µ0 ) r̂

!

P =S ⋅ n̂out da

cylindricalshell

∫∫ = −(EB / µ0 ) r̂ ⋅ r̂ dacylindricalbody

∫∫ = −(EB / µ0 )2πRh

B2πR = µ0ε0

dEdt

πR2 ⇒ B =µ0ε0

2dEdt

R

P = − EBµ0

⎛

⎝⎜⎞

⎠⎟2πRh

= − Eµ0

µ0ε0

2dEdt

R⎛⎝⎜

⎞⎠⎟

2πRh

= −ε0EdEdt

πR2h

= − ddt

12ε0E2⎛

⎝⎜⎞⎠⎟πR2h

= −duE

dt⎛⎝⎜

⎞⎠⎟

(Volume)

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Maxwell’s Equations

E ⋅ ⋅n̂ da

S∫∫ =

1ε0

ρ dVV∫∫∫ (Gauss's Law)

B ⋅ ⋅n̂ da

S∫∫ = 0 (Magnetic Gauss's Law)

E ⋅ d s

C∫ = −

ddt

B ⋅ n̂ da

S∫∫ (Faraday's Law)

B ⋅ ds

C∫ = µ0

J ⋅ n̂ da

S∫∫ + µ0ε0

ddt

E ⋅ n̂ da

S∫∫ (Maxwell - Ampere's Law)

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Electromagnetism Review E fields are associated with: (1) electric charges (Gauss’s Law )

(2) time changing B fields (Faraday’s Law)

B fields are associated with (3a) moving electric charges (Ampere-Maxwell Law)

(3b) time changing E fields (Maxwell’s Addition (Ampere-Maxwell Law)

Conservation of magnetic flux

(4) No magnetic charge (Gauss’s Law for Magnetism)

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Electromagnetism Review Conservation of charge:

E and B fields exert forces on (moving) electric charges: Energy stored in electric and magnetic fields

J ⋅ dA

closedsurface

∫∫ = −ddt

ρ dVvolumeenclosed

∫∫∫

Fq = q(

E + v ×

B)

U E = uE dV

all space∫∫∫ =

ε0

2E2 dV

all space∫∫∫

U B = uB dV

all space∫∫∫ =

12µ0

B2 dVall space∫∫∫

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Maxwell’s Equations in Vacua

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Maxwell’s Equations

1.E ⋅ dA

S∫∫ =

Qin

ε0

(Gauss's Law)

2.B ⋅ d

A

S∫∫ = 0 (Magnetic Gauss's Law)

3.E ⋅ d s

C∫ = −

dΦB

dt(Faraday's Law)

4.B ⋅ d s

C∫ = µ0 Ienc + µ0ε0

dΦE

dt(Ampere - Maxwell Law)

0

0

What about free space (no charge or current)?

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Maxwell: First Colour Photograph

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James Clerk Maxwell is sometimes credited as being the father of additive color. He had the photographer Thomas Sutton photograph a tartan ribbon on black-and-white film three times, first with a red, then green, then blue color filter over the lens. The three black-and-white images were developed and then projected onto a screen with three different projectors, each equipped with the corresponding red, green, or blue color filter used to take its image. When brought into alignment, the three images (a black-and-red image, a black-and-green image and a black-and-blue image) formed a full color image, thus demonstrating the principles of additive color.

http://upload.wikimedia.org/wikipedia/commons/7/7f/Tartan_Ribbon.jpg

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How Do Maxwell’s Equations

Lead to EM Waves?

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Wave Equation

Start with Ampere-Maxwell Eq and closed oriented loop

B ⋅ ds

C∫ = µ0ε0

ddtE ⋅ n̂ da∫

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Wave Equation

B ⋅ ds

C∫ = Bz (x,t)l − Bz (x + Δx,t)l

−

Bz (x + Δx,t) − Bz (x,t)Δx

= µ0ε0

∂Ey (x + Δx / 2,t)∂t

−∂Bz

∂x= µ0ε0

∂Ey

∂tSo in the limit that dx is very small:

Apply it to red rectangle:

B ⋅ ds

C∫ = µ0ε0

ddtE ⋅ n̂ da∫Start with Ampere-Maxwell Eq:

µ0ε0

ddt

E ⋅ n̂ da∫

= µ0ε0 lΔx∂Ey (x + Δx / 2,t)

∂t

⎛

⎝⎜

⎞

⎠⎟

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Group Problem: Wave Equation

Use Faraday’s Law

and apply it to red rectangle to find the partial differential equation in order to find a relationship between

∂Ey / ∂x and ∂Bz / ∂t

E ⋅ ds

C∫ = − d

dtB ⋅ n̂ da∫

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Group Problem: Wave Equation Sol.

E ⋅ ds

C∫ = − d

dtB ⋅ n̂ da∫

E ⋅ d s

C∫ = Ey (x + Δx,t)l − Ey (x,t)l

Ey (x + dx,t) − Ey (x,t)dx

= −∂Bz

∂t

∂Ey

∂x= −

∂Bz

∂t

− d

dt

B ⋅ n̂ da∫ = −ldx

∂Bz

∂t

Use Faraday’s Law:

So in the limit that dx is very small:

and apply it to red rectangle:

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1D Wave Equation for Electric Field

∂Ey

∂x= −

∂Bz

∂t (1) −

∂Bz

∂x= µ0ε0

∂Ey

∂t(2)

Take x-derivative of Eq.(1) and use the Eq. (2)

∂2 Ey

∂x2 = ∂∂x

∂Ey

∂x

⎛

⎝⎜

⎞

⎠⎟ =

∂∂x

−∂Bz

∂t⎛⎝⎜

⎞⎠⎟= − ∂

∂t∂Bz

∂x⎛⎝⎜

⎞⎠⎟= µ0ε0

∂2 Ey

∂t2

∂2Ey

∂x2 = µ0ε0

∂2Ey

∂t2

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1D Wave Equation for E

∂2Ey

∂x2 = µ0ε0

∂2Ey

∂t2

This is an equation for a wave. Let Ey = f (x − vt)

∂2Ey

∂x2 = f '' x − vt( )∂2Ey

∂t2 = v2 f '' x − vt( )

⎫

⎬⎪⎪

⎭⎪⎪

⇒ v = 1µ0ε0

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Definition of Constants and Wave Speed

c ≡ 299792458 m ⋅s−1

µ0 ≡ 4π ×10−7 N ⋅s2 ⋅C−2

Recall exact definitions of

The permittivity of free space is exactly defined by ε0

ε0 ≡

1c2µ0

≡ 8.854187817 ×10−12 C2 ⋅m-2 ⋅N−1

⇒ v = c =

1µ0ε0

in vacua

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Group Problem: 1D Wave Eq. for B

∂Bz

∂t= −

∂Ey

∂x

∂Bz

∂x= −µ0ε0

∂Ey

∂t

∂2Bz

∂x2 = 1c2

∂2Bz

∂t2

Take appropriate derivatives of the above equations and show that

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Wave Equations: Summary Both electric & magnetic fields travel like waves:

∂2Ey

∂x2 = 1c2

∂2 Ey

∂t2

∂2Bz

∂x2 = 1c2

∂2Bz

∂t2

with speed

But there are strict relations between them:

∂Bz

∂t= −

∂Ey

∂x∂Bz

∂x= −µ0ε0

∂Ey

∂t

c = 1

µ0ε0

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Electromagnetic Waves

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Electromagnetic Radiation: Plane Waves

http://youtu.be/3IvZF_LXzcc