mon. 10.1 - .2.1 potential formulation...

Fri. 7.3.1-.3.3 Maxwell’s Equations

Mon. 10.1 - .2.1 Potential Formulation HW8

Exercise: ‘very long’ solenoid, with radius a and n turns per unit length, carries time varying current, I(t). What’s an expression for the electric field a distance s from axis? Recall that inside a solenoid .

z

I

Using Faraday’s Law

t

BE

dt

d BEmfor area

B

tldE

or

zInB oˆ

Example: A slowly varying alternating current, , flows down a long, straight, thin wire and returns along a coaxial conducting tube of radius a. In what direction must the electric field point?


t

BE

dt

d BEmfor area

B

tldE

or

a

z

I t I0 cos t

I I

E

E

B

Lenz’ law says in the direction to drive current that would oppose changing flux, so down and up as the current varies up and down. z

What’s the electric field?

adBt

ldE

where B

0I

2 sˆ s a,

0 s a.

Calls for an Amperian loop

bottomouttopin

ldEldEldEldEldE

and I t I0 cos t

zsEzsEldE outin

a

sin

zsBdt

adBt

a

s

o

in

zsds

I

t 2z

s

aI

t in

o ln2

Example: A slowly varying alternating current, , flows down a long, straight, thin wire and returns along a coaxial conducting tube of radius a.


t

BE

dt

d BEmfor area

B

tldE

or

a

z

I t I0 cos t

I I

E

E

B

Lenz’ law says in the direction to drive current that would oppose changing flux, so down and up as the current varies up and down. z

What’s the electric field?

adBt

ldE

zsEsE outinz

s

aI

t in

o ln2

In what direction must the electric field point?

Right-hand-side is independent of how far out of loop sout is, so E is constant outside. But it should be 0 quite far away, so must be 0 everywhere outside.

insEin

o

s

aI

tln

2 in

oo

s

atI

tln

2

cos

insEin

oo

s

atIln

2

sin

Inductance

What is flux through path 2 due to current following path 1?

2

111

4 r

r

ldIB o

1

2 212 adB

22

112

4ad

ldIo

r

r22

11

4ad

ldI o

r

r

Equivilantly, can rephrase using product rules, or use A to get same result

212 adA11 AB

212 adB

Purely geometric factor

21 ldA

r11

14

ldIA o

r21

124

ldldI o

rr

r 2122

12,1

44

ldldad

ldM oo

Symmetric between two loops

2,11MI

12,12 IM

22,11 IM

Inductance

1

2 As with Resistance, sometimes it’s easiest to do the geometric integral, sometimes it’s easiest to find flux, factor out current, and thus find M.

rr

r 2122

1

44

ldldad

ld oo

2,1

2

1

1

2 MII

Example: Coaxial solenoids of radii a1 > a2 and windings per length n1 and n2.

a

z

a1 a2

B2 0n2I2

1

2

221 NaB

111 lnN

11

2

2221 lnaIno 21

2

212 Ilanno

M1,2

Overlapping volumes

Demo!

Faraday’s Law: time varying current in one solenoid induces Emf and drives current in other

22,1 IMdt

d

dt

d B 1.1Emf

Self Inductance

Current passing along the loop is itself responsible for flux through the loop

rr

r 1112

1

44

ldldad

ld oo

LI1

1

Time varying current along one segment of the loop produces field and Emf felt by other segments of the same loop.. z

a

nIB 0

NaB 2

nLN

nlanIo

2

1Ilano

22

L

volume

dt

d BEmf LIdt

d

Example: single solenoid

Consider driving charges around a solenoid. How much work would you have to do to get it going?

z

a

As you accelerate it up to speed, self inductance means a counter force is generated, so you must at least provide equal and opposite force.

q

ldF

Emfq

W

So per unit charge,

WqEmf

Then the rate at which work is done by the inductance’s emf is

PI

Pdt

dq

Emf

Emf

Or using the self-inductance relationship

PLIdt

dI

PLIdt

d 2

21

So bringing the current up to speed, you must oppose this, and invest energy at

2

21 LI

dt

dPyou

2

21 LIWyou

Energy to Generate Current (physical and special case derivation)

Energy to Generate Current (physical and special case derivation)

Consider driving charges around a solenoid. How much work would you have to do to get it going?

z

a

2

21 LIWyou

Rephrasing in terms of the corresponding field that’s generated, n

BI

0

2nL o

2

2

21

n

BnW

o

oyou

2

21 BW

oyou

Extrapolating to more general cases,

dBWo

2

21

(Griffiths does a more general derivation much like he did for the work of generating E field.)

“Where is the energy stored, field or current?” Neither / both – energy isn’t a substance (no “caloric fluid”) to be stored some where. It’s kinetic and potential energy, it’s “stored in” the motion of charges and their interactions situation of current flowing and field generated.

Energy to Generate Current

Exercise: Work to turn on co-axial solenoids of different wire density, n, and opposite current, I.

dBWo

2

21

a

z

a1 a2

For an individual solenoid

outside 0

inside 110

1

InB

Energy to Generate Current (mathematical and general case derivation)

Self-inductance should be a real phenomenon for any current path; the work to establish a current along any path should be

LIdt

d

dt

dEmf

fIP Em

PdtW

2

21 LI

dt

dLI

dt

dIP

dtLIdt

dW 2

21

2

21 LIW

LI adB

AB

adA

LI ldA

ldAIW

21 lIdA

21

dqv

dJAW

21

BJo

1

dBAWo

2

1

BAABBA

AB

dBABBWo

2

1

dBAdBWo

2

21

daBAdBWo

2

21

Rephrasing as sum over a volume containing current

Sending volume to cover all space, surface to infinity, where B is presumed to be 0

Correcting Ampere’s law Physical Motivation

J

J

1adJo

Io

a1 a2

2adJo

0

ldB

Mathematical Motivation It’s a mathematical fact that, the divergence of a curl of a vector field is 0

0B

JB

0So,

JB

0

Continuity Equation:

tJ Note: in the scenario above

this isn’t zero So,

tB 0

Correcting Ampere’s law The Fix

J

J

a1 a2

It’s a mathematical fact that, the divergence of a curl of a vector field is 0

0B

tsomethingB 00

So need

E

0

tE

too

0

tE

tB o 00 0

tt

EB o 00

Or rephrasing in terms of J again

Jt

EB o

00

Jt

EB o

00

In the scenario above, E is changing as the plates charge

Unfortunately permanent historically mistaken name: “Displacement Current”

t

EJ D

0

Conceptually, a stand-in for the effect of currents elsewhere

Corrected Maxwell-Ampere’s law

J

J

Jt

EB o

00

a) Find the electric field between the plates as a function of time t.

Example: Thin wires connect to the centers of thin, round capacitor plates. Suppose that the current I is constant, the radius of the capacitor is a, and the separation of the plates is w (<< a). Assume that the current flows out over the plates in such a way that the surface charge is uniform at any given time and is zero at t = 0.

zta

ItE

o

wire ˆ2

z

zt

tEo

ˆ

Approximating at infinite sheets, recall from Gauss’s law

or zAreatq

tEo

ˆ/)(

or since

a

tItqdt

tdqI wirewire )(

)(

Constant current and q(0)=0

and 2aArea

s

b) Using this as an Amperian Loop, find the magnetic field between the capacitor plate.

adJadt

EdB

000

2

2s

a

Iwireo

adJadt

EdB

000

None across this surface

adt

EdB

00

2

2s

a

Iwireo

sB 2Symmetry, as always, tells us B parallels our loop 2

2

0a

sIwire

ˆ2 2

0

a

sIB wire

so Just like field inside the wire!


J

J

Jt

EB o

00

a) Find the electric field between the plates as a function of time t.

Example: Thin wires connect to the centers of thin, round capacitor plates. Suppose that the current I is constant, the radius of the capacitor is a, and the separation of the plates is w (<< a). Assume that the current flows out over the plates in such a way that the surface charge is uniform at any given time and is zero at t = 0.

zta

ItE

o

wire ˆ2

z

a

c) Find the current along the surface of the capacitor plate.

adJadt

EdB

000

ˆ2 2

0

a

sIB wireb) Using this as an Amperian Loop, find the

magnetic field between the capacitor plate.

Compare Maxwell-Ampere’s Law for two, wisely-chosen surfaces bound by our Amperian loop.

Surface 1 Can lid

2.

0

2.

0

1.

0

1.

0

surfacesurfacesurfacesurface

adt

EadJdBad

t

EadJ

Surface 2 Can body

s

2.1.

0

surfacesurface

adJadt

E

platewire II

walllcylindricacapend

adJadJ..

2

2s

a

Iwire

21

as

wireplate II

Iwire

s

Iplate

ad

parallel

anti-parallel


Jt

EB o

00

Execise: Current flows down a long, straight, thin wire and returns along a thin, coaxial conducting tube of radius a. The electric field for the region s < a is

adJadt

EdB

000

a) Find an expression for the “displacement current” density.

I t I0 cos t

zts

aItsE ˆsinln

2, 00

t

E

0

a

z

I I

E

B

b) Integrate over a cross-section it pierces to find the “displacement current”.

adt

E

0

Fri. 7.3.1-.3.3 Maxwell’s Equations

Mon. 10.1 - .2.1 Potential Formulation HW8

E 0

E da Qenc

0

B 0 B da 0

E B

tad

t

B

tdE

a

B

Jt

EB

000

adJadt

EdB

000

Gauss’s Law

Gauss’s Law for Magnetism

Faraday’s Law

Maxwell – Ampere’s Law

mon. 10.1 - .2.1 potential formulation...

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