Lesson 8 Ampère’s Law and Differential Operators.

Lesson 8Ampère’s Law and

Differential Operators

Section 1Visualizing Ampère’s Law

Amperian Loop•An Amperian loop is

–any closed loop

•Amperian loops include:–a circle–a square–a rubber band

•Amperian loops do not include:–a balloon–a piece of string (with two ends)

The Field Contour of a Wire

Take a wire with current coming out of the screen.


The field contour is made of half-planes centered on the wire.


We draw arrows on each plane pointing in the direction of the magnetic field.

Amperian Loops

We draw an Amperian loop around the wire.

Amperian Loops

We wish to count the “net number” of field lines pierced by the Amperian loop.

Amperian Loops

First, we put an arrow on the loop in an overall counterclockwise direction.

Counting Surfaces Pierced

To count the net number of surfaces pierced by Amperian loop, we add +1 when the loop is “in the direction’ of the plane and −1 when it is “opposite the direction” of the plane.

+1

+1

+1+1+1+1

+1

+1

+1

+1

+1+1 +1

+1

+1+1

+1

+1+1

−1

−1−1

−1+1


Note there are “+1” appears 20 times and “-1” appears 4 times.

+1

+1

+1+1+1+1

+1

+1

+1

+1

+1+1 +1

+1

+1+1

+1

+1+1

−1

−1−1

−1+1


The net number of surfaces pierced by the Amperian loop us therefore +16.

+1

+1

+1+1+1+1

+1

+1

+1

+1

+1+1 +1

+1

+1+1

+1

+1+1

−1

−1−1

−1+1

Other Amperian Loops

What is the net number of surfaces pierced by each of these Amperian loops?


The net numbers of surfaces pierced by each of these loops is 16.


What is the net number of surfaces pierced by this Amperian loop?


This time the net number of surfaces pierced by the loop is 0. Why?

Other Wires.

This is the same loop we saw earlier, but now only 8 surfaces are pierced, since there are only 8 surfaces extending outward from the wire.

Other Wires.

There are 8 surfaces coming from the wire because the current through the wire is half as much as it was before.

Ampère’s Law

The net number of perpendicular surfaces pierced by an Amperian loop is proportional to the current passing through the loop.

Sign Convention

•Always traverse the Amperian loop in a (generally) counterclockwise direction.

•If the number of surfaces pierced N>0, the current comes out of the screen.

•If N<0, the current goes into the screen.

Section 2Cylindrically Symmetric

Current Density

Current Density

•There are three kinds of charge density (ρ,σ,λ)

•There is one kind of current density (current/unit area)

A

Ij

Current Density

• The current passing through a small gate of area ΔA is

AjI

small gate

Current Density

• The total current passing through the wire is the sum of the current passing through all small gates.

dAjI

small gates

Cylindrically Symmetric Current Distribution

The current density, j, can vary with r only.

Below, we assume that the current density is greatest near the axis of the wire.

Outside the distribution, the field contour is composed of surfaces that are half planes, uniformly spaced.


Inside the distribution, it is difficult to draw perpendicular surfaces, as some surfaces die out as we move inward. – We need to draw many, many surfaces to keep them equally spaced as we move inward.


But we do know that if we draw enough surfaces, the distribution of the surfaces will be uniform, even inside the wire.


Let’s draw a circular Amperian loop at radius r.

r


Now we split the wire into two parts – the part outside the Amperian loop and the part inside the Amperian loop.

r r


The total electric field at r will be the sum of the electric fields from the two parts of the wire.


r r

Inside a Hollow Wire

The total number of perpendicular surfaces pierced by the Amperian loop is zero because there is no current passing through it.

r

How can we get zero net surfaces?

1. We could have all the surfaces pierced twice, one in the positive sense and one in the negative…

… but this violates symmetry!


2. We could have some surfaces oriented one way and some the other…

… but this violates symmetry, too!

3. Or we could just have no surfaces at all inside the hollow wire.


This is the only way it can be done!

If the current distribution has cylindrical symmetry, the magnetic field inside a hollow wire must be zero.

The Magnetic Field inside a Hollow Wire


Since the magnetic field inside a hollow wire is zero, the total magnetic field at a distance r from the center of a solid wire is the field of the “core,” the part of the wire within the Amperian loop.

r r

Outside the core, the magnetic field is the same as that of a thin wire that has the same current as the total current passing through the Amperian loop.

r


r

irB enc

2

)( 0

Inside a cylindrically symmetric current distribution, the magnetic field is:

r


Section 3Uniform Current Density

Example: Uniform Current Distribution

A wire of radius R with a uniform current distribution has a total charge i passing through it. What is the magnetic field at r < R ?



r

irB enc

2

)( 0

r


The current density is uniform, so:

enc

enc

A

i

A

ij


r

irB enc

2

)( 0

r


Therefore:

iA

Ai encenc


r

irB enc

2

)( 0

r


r

irB enc

2

)( 0


r

irB enc

2

)( 0 iA

Ai encenc


r

irB enc

2

)( 02

2

R

riienc


2

20

2)(

R

ri

rrB


20

2)(

R

irrB

Section 4The Line Integral

Line Integral

We know that the magnetic field is stronger where the perpendicular surfaces are closer together.

N

ksegmentlinefieldoflength

piercedsurfacesofnumberkB

Therefore, the number of surfaces pierced is

Bk

N1

Let

B

Line Integral

Therefore, the number of surfaces pierced is

kN

•Λ is called the “line integral”

•The line integral is proportional to the number of contours pierced.

•Λ=Bℓ if ℓ is a section of a field line and B is constant on ℓ.

Line Integral

•Λ is called the line integral because it is more generally given by the expression

dB

d

B

cos

dBdBd

Line Integral

• Note that this is similar to expression for work you learned in mechanics. Work is the line integral of force along the path an object follows.

dB

dFW

d

F

cos

dFdFdW

Line Integral

The line integral is a way of measuring the number of contour surfaces pierced by a line segment.

dB

d

B

cos

dBdBd

Line Integral

Roughly speaking, it is a measure of how much field lines along a path.

dB

d

B

cos

dBdBd

Ampère’s Law and the Line Integral

The net number of perpendicular surfaces pierced by an Amperian loop is proportional to the current passing through the Amperian loop.

Ampère’s Law and the Line Integral

The net number of perpendicular surfaces pierced by an Amperian loop is proportional to the current passing through the Amperian loop.

Therefore:

encidB 0

Class 23Today, we will use Ampere’s law to find the magnetic fields• inside and outside a long, straight wire with radial charge density• of a plane of wires• of a solenoid• of a torus

Section 5Applying Ampère’s Law

Ampère’s Law – the Practical Version

enciB 0


enciB 0

The number of surfaces pierced


enciB 0


The magnetic field on the Amperian loop – must be a constant over the whole loop.


enciB 0



The length of a the closed Amperian loop (or the part where the field is non-zero).


enciB 0



The length of a the closed Amperian loop (or the part where the field is non-zero).

The total current passing through the Amperian loop.


jdAiB enc 00

This is the line integral

This is NOT the line integral

Section 6Ampère’s Law and Cylindrical

Wires

A Typical Problem

• A wire of radius R has current density . Find the magnetic field inside the wire.

• “Inside the wire” means at some point P at a radius r < R.

r

P

2rj

R

Choosing the Amperian Loop

• What shape of Amperian Loop should we choose for current traveling through a cylindrical wire?

Choosing the Amperian Loop

• Choose a circular Amperian loop.

r

A wire of radius R has current density . Find the magnetic field inside the wire.What is the correct expression for the ℓ in the line integral?

2rj

r2

Integrating the Current Density

• How do we do slice the region inside the Amperian loop to integrate the current density?

Integrating the Current Density

• We slice the wire into rings.

•The current though each ring is dI = j dA

A wire of radius R has current density . Find the magnetic field inside the wire.What is the correct expression for ?

2rj

drrrir

enc 0

2 2

enci

4

)(3

0

rrB

A wire of radius R has current density . Find the magnetic field inside the wire.

2rj

Another Problem

• A wire of radius R has current density . . Find the magnetic field outside the wire.

• “Outside the wire” means at some point P at a radius r > R.

r

P

2rj

R

r

RrB

4)(

4

0

A wire of radius R has current density . Find the magnetic field outside the wire

2rj

drrriR

enc 0

2 2

Section 7Other Applications of

Ampère’s Law

Ampere’s Law – Plane of Wires

1

2

3

B1

P

The magnetic field from each wire forms circular loops around the wire.Therefore is perpendicular to .B

r


B2

1

2

3

d B1B3

P

Consider the magnetic field from three wires.


B2

1

2

3

d B1B3

P

When we add the magnetic fields from each wire, the vector sum points upward.


enciB 0

B

Now let’s look at the line integral of the magnetic field around the dotted Amperian loop.


enciB 0

d

B

Now let’s look at the line integral of the magnetic field around the dotted Amperian loop.


enciB 0

d

B

The bottom part of the Amperian loop pierces no contour surfaces – so the line integral here is zero.

We also know

but the B field and the path are perpendicular on this segment so Λ=0 for this part of the path.

dB

Λ=0 for this top of the path, too.


d

B

Conclusion: The line integral over the top and bottom segments of the Amperian loop is zero because the magnetic field is perpendicular to the path.

dB


enciB 0

d

B

The line integral over the rightside of the Amperian loop is

Bdright


enciB 0

d

B

The line integral over the leftside of the Amperian loop is

Bdleft


enciB 0

d

B

The total line integral is:

dB2The enclosed current is the number of wires in the loop times the current in each wire:

Niienc


enciB 0

d

B

d

Nnwhere

nii

d

NB

NidB

22

2

00

0

n is the number of wires per unit length.

Ampere’s Law – Two Planes of Wires

What would the field be like if there were two planes of wires with currents in opposite directions?


Field lines


Contour surfaces

What can you conclude about the magnetic field?


Field lines of the right plane


Field lines of the left plane


The field of the both planes


enciB 0

1

2

3

d

0B

B

What would the field be like if there were two planes of wires with currents in opposite directions?

nini

B 00

22

Ampere’s Law – SolenoidenciB 0

1

2

3

d

0B

A solenoid is similar to two planes of wires. The magnetic field inside this solenoid points downward. The magnetic fieldoutside the solenoid is zero.

B

Ampere’s Law – SolenoidenciB 0

1

2

3

d

0B

d

Nnwhere

niid

NB

NiBd

00

0

B

Right-hand Rule #3

The direction of the magnetic field inside a solenoid is given by a right-hand rule:

Circle the fingers of your right hand in the direction of the current. The magnetic field is inthe direction of your thumb.B

Ampere’s Law – TorusenciB 0

r

B

A torus is like a solenoidwrapped around a doughnut-shaped core.The magnetic field insideforms circular loops.The magnetic field outsideis zero.


r

NiB

NirB

2

2

0

0

r

B


r

NiB

NirB

2

2

0

0

r

B

The wires on the insideof the torus are closer together, so the fieldis stronger there.

Class 24Today, we will use direct integration to find• electric fields of charged rods and loops• electric potentials of charged rods and loops• magnetic fields of current-carrying wire segments and loop segments (Biot-Savart law)

Section 8Finding Fields by Direct

Integration

Geometry for Extended Objects

P

r

r

R

rRr

r

R

r origin to source

source to P

origin to P

The Basic LawsThe electric field and potential of a small charge dq:

The magnetic field of a current i in a small length of wire :

R

dqrdV

R

dqRrEd

03

0 4

1)(

4

1)(

30

4)(

R

RdirBd

d

Origin of the Basic Laws

Electric field and potential for slowly moving point charges – Coulomb’s Law:

r

qrV

rr

qr

r

qrE

0

30

20

4

1)(

4

1ˆ

4

1)(

The Basic Laws – for dq

Electric field and potential for dq.

R

dqrdV

R

dqRrEd

0

30

4

1)(

4

1)(


Remember the geometry from Lesson 2

motion of source

U

θ

head linethread

x

y

P P

tail line

hr

tr

S T

ray linerr

ψ

shr

An expression for the magnetic field of a point charge:

Ec

rEErcr

Errcr

Ercr

Erc

B

s

rhsh

rhsh

hh

h

1

toparallel is because 1

1

1ˆ

1


00

2

30

320

30

1

44

1)(

4

1)(

1)(

cas

rvr

qrv

r

q

crB

r

rq

crE

crB

ss

ss

An expression for the magnetic field of a point charge (moving slowly):


Origin of the Basic LawsMagnetic field for dq

rdr

irBd

diddt

dq

dt

ddqvdq

rvr

dqrBd

s

s

30

30

4)(

4)(

A little sleight of hand, but it’s the same as a more formal proof.

The Biot-Savart Law

30

4)(

R

RidrBd

A current i passes through the wire segment

The length of the wire segment is dℓ.

The direction of dℓ is the direction of the current.

is the vector from the origin to a field point.

is the vector from the segment (the origin) to a field point.

r

R

The formula for the magnetic field of a wire segment

Equations for Extended Objects

P

r

r

R

30

4)(

R

RidrBd

R

dqrdV

R

dqRrEd

0

30

4

1)(

4

1)(

Now, let’s work some problems…

A Charged Rod

−L +L

P

x

y

The rod has a linear charge density

L

q

2

Find the electric field at P.

A Charged Rod

−L +L

P

x

y

. and,, find toneed We

4

1)(

30

RRdq

RR

dqrEd

A Charged Rod

−L +L

P

y

1. Find r

r

A Charged Rod

−L +L

P

y

1. Find r

yyr ˆ

r

A Charged Rod

−L +L

P

1. Find

2. Choose a slice and find

r

yyr ˆ

r

r

r

A Charged Rod

−L +L

P

1. Find


Be sure to put primes on the slice variables!

r

yyr ˆ

r

r

r

xxr ˆ

A Charged Rod

−L +L

P

1. Find


3. Find the length and charge of the slice.

r

yyr ˆ

r

r

r

xxr ˆ

A Charged Rod

−L +L

P

1. Find



r

yyr ˆ

r

r

r

xxr ˆ

xd

xddq

A Charged Rod

−L +L

P

1. Find



4. Find

r

yyr ˆ

r

r

r

xxr ˆ

xddq rrR

R

A Charged Rod

−L +L

P

1. Find



4. Find

r

yyr ˆ

r

r

r

xxr ˆ

xddq rrR

R

xxyyR ˆˆ

A Charged Rod

−L +L

P

1. Find



4. Find

5. Find

r

yyr ˆ

r

r

r

xxr ˆ

xddq rrR

R

xxyyR ˆˆ

R

A Charged Rod

−L +L

P

1. Find



4. Find

5. Find

r

yyr ˆ

r

r

r

xxr ˆ

xddq rrR

R

xxyyR ˆˆ

R 22 yxR

A Charged Rod

1. Find



4. Find

5. Find

r

yyr ˆ

r xxr ˆ

xddq rrR

xxyyR ˆˆ

R 22 yxR

RR

dqrEd

3

04

1)(

Now just substitute!

A Charged Rod

1. Find



4. Find

5. Find

r

yyr ˆ

r xxr ˆ

xddq rrR

xxyyR ˆˆ

R 22 yxR

xxyyyx

xdR

R

dqrEd ˆˆ

4

1

4

1)(

2/3220

30

A Charged Rod

1. Find



4. Find

5. Find

r

yyr ˆ

r xxr ˆ

xddq rrR

xxyyR ˆˆ

R 22 yxR

L

L

L

L yx

xdxx

yx

xdyyrE

xxyyyx

xdR

R

dqrEd

2/3220

2/3220

2/3220

30

4ˆ

4ˆ)(

ˆˆ4

1

4

1)(

A Charged Rod

L

L

L

L yx

xdxx

yx

xdyyrE

xxyyyx

xdR

R

dqrEd

2/3220

2/3220

2/3220

30

4ˆ

4ˆ)(

ˆˆ4

1

4

1)(

I won’t expect you to evaluate these integrals!

yLyy

LrE ˆ

2)(

220

A Charged Rod

−L +L

P

x

y


L

q

2

Now find the electric potential at P.

A Charged Rod

1. Find



4. Find

5. Find

r

yyr ˆ

r xxr ˆ

xddq rrR

xxyyR ˆˆ

R 22 yxR

2/12200 4

1

4

1)(

yx

xd

R

dqrdV

A Charged Rod

1. Find



4. Find

5. Find

r

yyr ˆ

r xxr ˆ

xddq rrR

xxyyR ˆˆ

R 22 yxR

L

L yx

xdrV

yx

xd

R

dqrdV

2/1220

2/12200

4)(

4

1

4

1)(

A Charged Rod

L

L yx

xdrV

yx

xd

R

dqrdV

2/1220

2/12200

4)(

4

1

4

1)(

LyL

LyLrV

22

22

0

ln4

)(

You don’t need to evaluate this integral, either!

Current in a Wire Segment

−L +L

P

x

y

i

Current i travels to the left along a segment of wire.

Find the magnetic field at P.

−L +L

P

y

1. Find r

r


i

−L +L

P

y

1. Find r

yyr ˆ

r


i

−L +L

P

1. Find


r

yyr ˆ

r

r

r


i

−L +L

P

1. Find


Be sure to put primes on the slice variables!

r

yyr ˆ

r

r

r

xxr ˆ


i

−L +L

P

1. Find


3. Find

r

yyr ˆ

r

r

r

xxr ˆ


i

d

−L +L

P

1. Find


3. Find

r

yyr ˆ

r

r

r

xxr ˆ

xd

xxdxxdd ˆˆ


i

d

−L +L

P

1. Find


3. Find

4. Find

r

yyr ˆ

r

r

r

xxr ˆ

rrR

R


i

d xxdxxdd ˆˆ

−L +L

P

1. Find


3. Find

4. Find

r

yyr ˆ

r

r

r

xxr ˆ

rrR

R

xxyyR ˆˆ


i

d xxdxxdd ˆˆ

−L +L

P

1. Find


3. Find

4. Find

5. Find

r

yyr ˆ

r

r

r

xxr ˆ

rrR

R

xxyyR ˆˆ

R

i

d xxdxxdd ˆˆ


−L +L

P

1. Find


3. Find

4. Find

5. Find

r

yyr ˆ

r

r

r

xxr ˆ

rrR

R

xxyyR ˆˆ

R 22 yxR


i

d xxdxxdd ˆˆ

−L +L

P

1. Find


3. Find

4. Find

5. Find

6. Find

r

yyr ˆ

r

r

r

xxr ˆ

rrR

R

xxyyR ˆˆ

R 22 yxR


i

d xxdxxdd ˆˆ

Rd

−L +L

P

r

r

R


i

1. Find


3. Find

4. Find

5. Find

6. Find

r

yyr ˆ

r xxr ˆ

rrR

xxyyR ˆˆ

R 22 yxR

d xxdxxdd ˆˆ

Rd

zxydRd

Remember:

yxz

xzy

zyx

ˆˆˆ

ˆˆˆ

ˆˆˆ

yzx

xyz

zxy

ˆˆˆ

ˆˆˆ

ˆˆˆ

0ˆˆ

0ˆˆ

0ˆˆ

zz

yy

xx

2/322

03

0

4ˆ

4)(

yx

xydiz

R

RdirBd


1. Find


3. Find

4. Find

5. Find

6. Find

r

yyr ˆ

r xxr ˆ

rrR

xxyyR ˆˆ

R 22 yxR

d xxdxxdd ˆˆ

Rd

zxydRd

L

L yx

xdiyzrB

yx

xydiz

R

RdirBd

2/322

0

2/322

03

0

4ˆ)(

4ˆ

4)(


1. Find


3. Find

4. Find

5. Find

6. Find

r

yyr ˆ

r xxr ˆ

rrR

xxyyR ˆˆ

R 22 yxR

d xxdxxdd ˆˆ

Rd

zxydRd

A Charged Loop Segment

aPx

y


2/a

q

Find the electric field at P.


aPx

y

1. Find r


aPx

y

1. Find r

0r


aPx

y

1. Find

2. Slice and find

r

0r

r

r


aPx

y

1. Find

2. Slice and find

r

0r

r yaxar ˆsinˆcos

r


aPx

y

1. Find

2. Slice and find

3. Find the charge of the slice.

r

0r

r yaxar ˆsinˆcos

r

ad


aPx

y

1. Find

2. Slice and find


r

0r

r yaxar ˆsinˆcos

addq

r

ad


aPx

y

1. Find

2. Slice and find


4. Find

r

0r

r yaxar ˆsinˆcos

addqrrR

r

ad


aPx

y

1. Find

2. Slice and find


4. Find

r

0r

r yaxar ˆsinˆcos

addqrrR

yaxaR ˆsinˆcos

r

ad


aPx

y

1. Find

2. Slice and find


4. Find

5. Find

r

0r

r yaxar ˆsinˆcos

addqrrR

RyaxaR ˆsinˆcos

r

ad


aPx

y

1. Find

2. Slice and find


4. Find

5. Find

r

0r

r yaxar ˆsinˆcos

addqrrR

R aR yaxaR ˆsinˆcos

r

ad


1. Find

2. Slice and find


4. Find

5. Find

r

0r

r yaxar ˆsinˆcos

addqrrR


2/

03

03

0

ˆsinˆcos4

1

4

1)(

yaxaa

adR

R

dqrE


2/

03

03

0

ˆsinˆcos4

1

4

1)(

yaxaa

adR

R

dqrE

Here we only have to integrate sines and cosines, sothe integrals are easy!

)ˆˆ(4

)(0

yxa

rE

A Segment of a Current Loop

aPx

y i

Current i travels counterclockwise along a segment of a loop of wire.

Find the magnetic field at P.


aPx

y

1. Find r

i


aPx

y

1. Find r

0r

i


aPx

y

1. Find

2. Slice and find

r

0r

r

r

i


aPx

y

1. Find

2. Slice and find

r

0r

r yaxar ˆsinˆcos

r

i


aPx

y

1. Find

2. Slice and find

3. Find

r

0r

r yaxar ˆsinˆcos

r

adi

d


aPx

y

1. Find

2. Slice and find

3. Find

r

0r

r yaxar ˆsinˆcos

r

adi

d yadxadd ˆcosˆsin


aPx

y

1. Find

2. Slice and find

3. Find

4. Find

r

0r

r yaxar ˆsinˆcos

rrR

r

adi



aPx

y

1. Find

2. Slice and find

3. Find

4. Find

r

0r

r yaxar ˆsinˆcos

rrR

yaxaR ˆsinˆcos

r

adi



aPx

y

1. Find

2. Slice and find

3. Find

4. Find

5. Find

r

0r

r yaxar ˆsinˆcos

rrR

R

yaxaR ˆsinˆcos

r

adi



aPx

y

1. Find

2. Slice and find

3. Find

4. Find

5. Find

r

0r

r yaxar ˆsinˆcos

rrR


r

adi



aPx

y

1. Find

2. Slice and find

3. Find

4. Find

5. Find

6. Find

r

0r

r yaxar ˆsinˆcos

rrR


r

adi


Rd


aPx

y

1. Find

2. Slice and find

3. Find

4. Find

5. Find

6. Find

r

0r

r yaxar ˆsinˆcos

rrR


r

adi

d

zdazdazaRd ˆˆcosˆsin 22222

yadxadd ˆcosˆsin


1. Find

2. Slice and find

3. Find

4. Find

5. Find

6. Find

r

0r

r yaxar ˆsinˆcos

rrR


d

2/

03

20

30

4ˆ

4)(

a

daiz

R

RdirB

yadxadd ˆcosˆsin

zdazdazaRd ˆˆcosˆsin 22222


za

i

a

daizrB ˆ

84ˆ)( 0

2/

03

20

This is a really easy integral this time!

Class 25Today, we will:• learn the definition of divergence in terms of flux.• learn the definition of curl in terms of the line integral.• • find the gradient, divergence, and curl in terms of derivatives (differential operators) • write Gauss’s laws and Ampere’s law in differential form• work several sample problems

Section 9Gauss’s Law and

Divergence

Gauss’s Law

The net number of electric field lines passing through a Gaussian surface is proportional to the charge enclosed.

Gauss’s Law

This is true no matter how small the Gaussian surface is. But the number of field lines gets smaller as the volume gets smaller.

Divergence

Define divergence to be

Pv

PEdiv

vEdiv E

v

point around volumesmall a is

point at field E theof divergence theis

where

lim0

Divergence

Divergence is a scalar field – a scalar defined at every point in space – that tells us if diverging (or converging) field lines are being produced at that point. The larger the divergence, the more field lines are produced.

Divergence and Gauss’s Law

In a very small volume, charge density is nearly constant. (That is, until we get to the atomic scale where we can start seeing protons and electrons.)


In a very small volume, charge density is nearly constant.

0

)()(

1

000

vas

vr

dvrqenc

E


In a very small volume, charge density is nearly constant.

0

)()(

1

000

vas

vr

dvrqenc

E

00

)(lim

r

vEdiv E

v

Gauss’s Law in Differential Form

0

)(

r

Ediv


Electrical charge produces field lines that tend to spread from (or converge to) a point in space.

Divergence is a measure of how much field lines spread from (+) or converge to (-) a point of space. It is a measurement of “spreadingness.”

Section 10Ampère’s Law and Curl

Ampère’s Law

The net number of perpendicular surfaces pierced by an Amperian loop is proportional to the current passing through the loop.

This is true no matter how small the Amperian loop is. But the number of surfaces pierced gets smaller as the area of the loop gets smaller.

Ampere’s Law

Curl

Take a point in space and a line in the x direction passing through the point. The x-component of the curl (around the line) is defined to be:

arBcurl x

ax

0

lim)(

P x

Curl

Take a point in space and a line in the y direction passing through the point. The y-component of the curl (around the line) is defined to be:

arBcurl y

ay

0lim)(

P xy

Curl

Take a point in space and a line in the z direction passing through the point. The z-component of the curl (around the line) is defined to be:

arBcurl z

az

0lim)(

P xy

z

Curl

Take a point in space and a line in the x direction passing through the point. The curl (around the line) is defined by:

loopAmperian theof area theis

curl theofcomponent theis )(

where

lim)(0

a

xrBcurl

arBcurl

x

x

ax

Curl

Curl is a vector field, a vector defined a every point in space. The x component of curl tells us if something at the point is producing field lines that loop about a line going in the x direction and passing through the point.

In a very small area, current density is nearly constant.

Curl and Ampère’s Law

In a very small area, current density is nearly constant.

0

)()( 000

aas

arjdarji xxenc


In a very small volume, the density is nearly a constant.

)(lim)( 00

rja

rBcurl xx

ax

0

)()( 000

aas

arjdarji xxenc


More generally:

The curl points in the direction of the current at any point in space.

)()( 0 rjrBcurl


More generally:

Curl is a measure of how much field lines ccw (+) or cw to (-) around the direction of the current. It is a measurement of “loopiness.”

)()( 0 rjrBcurl


Electrical current produces magnetic field lines that form loops around the path of the moving charges.


Section 11Differential Operators

The Gradient

The gradient is a three-dimensional generalization of a slope (derivative). The gradient tells us the direction a scalar field increases the most rapidly and how much it changes per unit distance.

The Gradient

In terms of derivatives, the gradient is:

zz

yy

xx

ˆˆˆ

Electric Field and Electric Potential

VzyxE ),,(

z

Vz

y

Vy

x

VxVzyxE

ˆˆˆ),,(

The Divergence Operator

We can show that another way of representing divergence is in terms of derivatives.

z

E

y

E

x

EEEdiv zyx

The Curl Operator

We can also express the curl in terms of derivatives.

zyx

xyzxyz

BBBzyx

zyx

y

B

x

Bz

x

B

z

By

z

B

y

BxBBcurl

ˆˆˆ

ˆˆˆ

Gauss’s Laws and Ampère’s Law in Differential Form

0

E

0 B

jB

0

Some Problems

In a region of space, the electric potential is given by the expression

Find the electric field.

constant. a is where2 yxV

yxxxy

z

Vz

y

Vy

x

VxVzyxE

ˆˆ2

ˆˆˆ),,(

2

Some Problems

In a region of space, the electric potential is given by the expression

Find the charge density.

constant. a is where2 yxV

yxxxyzyxE ˆˆ2),,( 2

yy

E

x

EE

E

yx

0000

0

2

Some Problems

In a region of space, the magnetic field is given by the expression

Find β in terms of α.

constants. are and where

ˆ)3(ˆ)3(),( 2222

yyyxxxyxyxB

Some Problems ˆ)3(ˆ)3(),( 2222 yyyxxxyxyxB

0)33()33(

02)3(2)3(

0

2222

2222

yxyx

yyyxxxyx

y

B

x

BB yx

Some Problems

In a region of space, the magnetic field is given by the expression

Find the current density (magnitude and direction).

constant. a is where

ˆ)3(ˆ)3(),( 2222

yyyxxxyxyxB

BjjB

0

0

1

Some Problems

ˆ)3(ˆ)3(),( 2222 yyyxxxyxyxB

Bj

0

1

zxy

xyxyz

y

B

x

Bz

y

B

x

Bz

x

B

z

By

z

B

y

Bx

xy

xyzxyz

ˆ12

66ˆ

1ˆ

1ˆ

1ˆ

1ˆ

00

0

000

Lesson 8 Ampère’s Law and Differential Operators.

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Transcript of Lesson 8 Ampère’s Law and Differential Operators.