Lesson 8 Ampère’s Law and Differential Operators

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Lesson 8 Ampère’s Law and Differential Operators. Section 1 Visualizing 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). - PowerPoint PPT Presentation

Transcript of 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 WireTake a wire with current coming out of the screen.

The Field Contour of a WireThe field contour is made of half-planes centered on the wire.

The Field Contour of a WireWe draw arrows on each plane pointing in the direction of the magnetic field.

Amperian LoopsWe draw an Amperian loop around the wire.

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

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

Counting Surfaces PiercedTo 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

Counting Surfaces PiercedNote 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

Counting Surfaces PiercedThe 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 LoopsWhat is the net number of surfaces pierced by each of these Amperian loops?

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

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

Other Amperian LoopsThe net numbers of surfaces pierced by each of these loops is 16.

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

Other Amperian LoopsThis 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 LawThe 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)

AIj

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.

Cylindrically Symmetric Current Distribution

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.

Cylindrically Symmetric Current Distribution

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

Cylindrically Symmetric Current Distribution

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

r

Cylindrically Symmetric Current Distribution

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

r r

Cylindrically Symmetric Current Distribution

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

Cylindrically Symmetric Current Distribution

r r

Inside a Hollow WireThe 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!

How can we get zero net surfaces?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.

How can we get zero net surfaces?

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

Cylindrically Symmetric Current Distribution

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

Cylindrically Symmetric Current Distribution

rirB enc

2

)( 0

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

r

Cylindrically Symmetric Current Distribution

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 ?

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 ?

rirB enc

2

)( 0

r

Example: Uniform Current Distribution

The current density is uniform, so:

enc

enc

Ai

Aij

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 ?

rirB enc

2

)( 0

r

Example: Uniform Current Distribution

Therefore:

iA

Ai encenc

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 ?

rirB enc

2

)( 0

r

Example: Uniform Current Distribution

rirB enc

2

)( 0

Example: Uniform Current Distribution

rirB enc

2

)( 0 iA

Ai encenc

Example: Uniform Current Distribution

rirB enc

2

)( 02

2

Rriienc

Example: Uniform Current Distribution

2

20

2)(

Rri

rrB

Example: Uniform Current Distribution

2

20

2)(

Rri

rrB

Example: Uniform Current Distribution

20

2)(

RirrB

Section 4The Line Integral

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

Nk

segmentlinefieldoflengthpiercedsurfacesofnumberkB

Therefore, the number of surfaces pierced is

Bk

N 1

Let

B

Line IntegralTherefore, 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 IntegralThe line integral is a way of measuring the number of contour surfaces pierced by a line segment.

dB

d

B

cos

dBdBd

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

dB

d

B

cos

dBdBd

Ampère’s Law and the Line IntegralThe 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 IntegralThe 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

Ampère’s Law – the Practical Version

enciB 0

The number of surfaces pierced

Ampère’s Law – the Practical Version

enciB 0

The number of surfaces pierced

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

Ampère’s Law – the Practical Version

enciB 0

The number of surfaces pierced

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

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

Ampère’s Law – the Practical Version

enciB 0

The number of surfaces pierced

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

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.

Ampère’s Law – the Practical Version

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.

rP

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

0rrB

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

rRrB

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

Ampere’s Law – Plane of Wires

B2

1

2

3

d B1B3

P

Consider the magnetic field from three wires.

Ampere’s Law – Plane of Wires

B2

1

2

3

d B1B3

P

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

Ampere’s Law – Plane of Wires

enciB 0

B

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

Ampere’s Law – Plane of Wires

enciB 0

d

B

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

Ampere’s Law – Plane of Wires

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.

Ampere’s Law – Plane of Wires

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

Ampere’s Law – Plane of Wires

enciB 0

d

B

The line integral over the rightside of the Amperian loop is

Bdright

Ampere’s Law – Plane of Wires

enciB 0

d

B

The line integral over the leftside of the Amperian loop is

Bdleft

Ampere’s Law – Plane of Wires

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

Ampere’s Law – Plane of Wires

enciB 0

d

B

dNnwhere

niidNB

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?

Ampere’s Law – Two Planes of Wires

Field lines

Ampere’s Law – Two Planes of Wires

Contour surfaces

What can you conclude about the magnetic field?

Ampere’s Law – Two Planes of Wires

Field lines of the right plane

Ampere’s Law – Two Planes of Wires

Field lines of the left plane

Ampere’s Law – Two Planes of Wires

The field of the both planes

Ampere’s Law – Two Planes of Wires

The field of the both planes

Ampere’s Law – Two Planes of Wires

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?

niniB 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

dNnwhere

niidNB

NiBd

00

0

B

Right-hand Rule #3The 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.

Ampere’s Law – TorusenciB 0

rNiB

NirB

2

2

0

0

r

B

Ampere’s Law – TorusenciB 0

rNiB

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 :

RdqrdV

RdqRrEd

03

0 41)(

41)(

30

4)(

RRdirBd

d

Origin of the Basic LawsElectric field and potential for slowly moving point charges – Coulomb’s Law:

rqrV

rrqr

rqrE

0

30

20

41)(

41ˆ

41)(

The Basic Laws – for dqElectric field and potential for dq.

RdqrdV

RdqRrEd

0

30

41)(

41)(

Origin of the Basic LawsRemember the geometry from Lesson 2

motion of sourceU

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

1ˆ1

Origin of the Basic Laws

00

2

30

320

30

144

1)(

41)(1)(

cas

rvrqrv

rq

crB

rrq

crE

crB

ss

ss

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

Origin of the Basic Laws

Origin of the Basic LawsMagnetic field for dq

rdrirBd

diddtdq

dtddqvdq

rvrdqrBd

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)(

RRidrBd

A current i passes through the wire segmentThe 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)(

RRidrBd

RdqrdV

RdqRrEd

0

30

41)(

41)(

Now, let’s work some problems…

A Charged Rod

−L +L

P

x

y

The rod has a linear charge density

Lq

2

Find the electric field at P.

A Charged Rod

−L +L

P

x

y

. and,, find toneed We

41)( 3

0

RRdq

RRdqrEd

A Charged Rod

−L +L

Py

1. Find r

r

A Charged Rod

−L +L

Py

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

2. Choose a slice and 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

2. Choose a slice and 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

2. Choose a slice and find

3. Find the length and charge of the slice.

r yyr ˆ

r

r

r

xxr ˆ

xd

xddq

A Charged Rod

−L +L

P

1. Find

2. Choose a slice and find

3. Find the length and charge of the slice.

4. Find

r yyr ˆ

r

r

r

xxr ˆ

xddq rrR

R

A Charged Rod

−L +L

P

1. Find

2. Choose a slice and find

3. Find the length and charge of the slice.

4. Find

r yyr ˆ

r

r

r

xxr ˆ

xddq rrR

R

xxyyR ˆˆ

A Charged Rod

−L +L

P

1. Find

2. Choose a slice and find

3. Find the length and charge of the slice.

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

2. Choose a slice and find

3. Find the length and charge of the slice.

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

2. Choose a slice and find

3. Find the length and charge of the slice.

4. Find

5. Find

r yyr ˆ

r xxr ˆ

xddq rrR

xxyyR ˆˆ

R 22 yxR

RRdqrEd

3

041)(

Now just substitute!

A Charged Rod

1. Find

2. Choose a slice and find

3. Find the length and charge of the slice.

4. Find

5. Find

r yyr ˆ

r xxr ˆ

xddq rrR

xxyyR ˆˆ

R 22 yxR

xxyyyx

xdRRdqrEd ˆˆ

41

41)( 2/322

03

0

A Charged Rod

1. Find

2. Choose a slice and find

3. Find the length and charge of the slice.

4. Find

5. Find

r yyr ˆ

r xxr ˆ

xddq rrR

xxyyR ˆˆ

R 22 yxR

L

L

L

L yx

xdxxyx

xdyyrE

xxyyyx

xdRRdqrEd

2/3220

2/3220

2/3220

30

4ˆ)(

ˆˆ4

14

1)(

A Charged Rod

L

L

L

L yx

xdxxyx

xdyyrE

xxyyyx

xdRRdqrEd

2/3220

2/3220

2/3220

30

4ˆ)(

ˆˆ4

14

1)(

I won’t expect you to evaluate these integrals!

yLyy

LrE ˆ2

)(22

0

A Charged Rod

−L +L

P

x

y

The rod has a linear charge density

Lq

2

Now find the electric potential at P.

A Charged Rod

1. Find

2. Choose a slice and find

3. Find the length and charge of the slice.

4. Find

5. Find

r yyr ˆ

r xxr ˆ

xddq rrR

xxyyR ˆˆ

R 22 yxR

2/12200 4

14

1)(yx

xdRdqrdV

A Charged Rod

1. Find

2. Choose a slice and find

3. Find the length and charge of the slice.

4. Find

5. Find

r yyr ˆ

r xxr ˆ

xddq rrR

xxyyR ˆˆ

R 22 yxR

L

L yx

xdrV

yx

xdRdqrdV

2/1220

2/12200

4)(

41

41)(

A Charged Rod

L

L yxxdrV

yxxd

RdqrdV

2/1220

2/12200

4)(

41

41)(

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

Py

1. Find r

r

Current in a Wire Segmenti

−L +L

Py

1. Find r yyr ˆ

r

Current in a Wire Segmenti

−L +L

P

1. Find

2. Choose a slice and find

r yyr ˆ

r

r

r

Current in a Wire Segmenti

−L +L

P

1. Find

2. Choose a slice and find

Be sure to put primes on the slice variables!

r yyr ˆ

r

r

r

xxr ˆ

Current in a Wire Segmenti

−L +L

P

1. Find

2. Choose a slice and find

3. Find

r yyr ˆ

r

r

r

xxr ˆ

Current in a Wire Segmenti

d

−L +L

P

1. Find

2. Choose a slice and find

3. Find

r yyr ˆ

r

r

r

xxr ˆ

xd

xxdxxdd ˆˆ

Current in a Wire Segmenti

d

−L +L

P

1. Find

2. Choose a slice and find

3. Find

4. Find

r yyr ˆ

r

r

r

xxr ˆ

rrR

R

Current in a Wire Segmenti

d xxdxxdd ˆˆ

−L +L

P

1. Find

2. Choose a slice and find

3. Find

4. Find

r yyr ˆ

r

r

r

xxr ˆ

rrR

R

xxyyR ˆˆ

Current in a Wire Segmenti

d xxdxxdd ˆˆ

−L +L

P

1. Find

2. Choose a slice and find

3. Find

4. Find

5. Find

r yyr ˆ

r

r

r

xxr ˆ

rrR

R

xxyyR ˆˆ

R

i

d xxdxxdd ˆˆ

Current in a Wire Segment

−L +L

P

1. Find

2. Choose a slice and find

3. Find

4. Find

5. Find

r yyr ˆ

r

r

r

xxr ˆ

rrR

R

xxyyR ˆˆ

R 22 yxR

Current in a Wire Segmenti

d xxdxxdd ˆˆ

−L +L

P

1. Find

2. Choose a slice and find

3. Find

4. Find

5. Find

6. Find

r yyr ˆ

r

r

r

xxr ˆ

rrR

R

xxyyR ˆˆ

R 22 yxR

Current in a Wire Segmenti

d xxdxxdd ˆˆ

Rd

−L +L

P

r

r

R

Current in a Wire Segmenti

1. Find

2. Choose a slice and 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:

yxzxzyzyx

ˆˆˆˆˆˆˆˆˆ

yzxxyzzxy

ˆˆˆˆˆˆˆˆˆ

0ˆˆ0ˆˆ0ˆˆ

zzyyxx

2/322

03

0

4)(

yx

xydiz

RRdi

rBd

Current in a Wire Segment

1. Find

2. Choose a slice and 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

RRdi

rBd

2/322

0

2/322

03

0

4ˆ)(

4)(

Current in a Wire Segment

1. Find

2. Choose a slice and 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

The rod has a linear charge density

2/aq

Find the electric field at P.

A Charged Loop Segment

aPx

y

1. Find r

A Charged Loop Segment

aPx

y

1. Find r 0r

A Charged Loop Segment

aPx

y

1. Find

2. Slice and find

r 0r

r

r

A Charged Loop Segment

aPx

y

1. Find

2. Slice and find

r 0r

r yaxar ˆsinˆcos

r

A Charged Loop Segment

aPx

y

1. Find

2. Slice and find

3. Find the charge of the slice.

r 0r

r yaxar ˆsinˆcos

r

ad

A Charged Loop Segment

aPx

y

1. Find

2. Slice and find

3. Find the charge of the slice.

r 0r

r yaxar ˆsinˆcos

addq

r

ad

A Charged Loop Segment

aPx

y

1. Find

2. Slice and find

3. Find the charge of the slice.

4. Find

r 0r

r yaxar ˆsinˆcos

addqrrR

r

ad

A Charged Loop Segment

aPx

y

1. Find

2. Slice and find

3. Find the charge of the slice.

4. Find

r 0r

r yaxar ˆsinˆcos

addqrrR

yaxaR ˆsinˆcos

r

ad

A Charged Loop Segment

aPx

y

1. Find

2. Slice and find

3. Find the charge of the slice.

4. Find

5. Find

r 0r

r yaxar ˆsinˆcos

addqrrR

RyaxaR ˆsinˆcos

r

ad

A Charged Loop Segment

aPx

y

1. Find

2. Slice and find

3. Find the charge of the slice.

4. Find

5. Find

r 0r

r yaxar ˆsinˆcos

addqrrR

R aR yaxaR ˆsinˆcos

r

ad

A Charged Loop Segment

1. Find

2. Slice and find

3. Find the charge of the slice.

4. Find

5. Find

r 0r

r yaxar ˆsinˆcos

addqrrR

R aR yaxaR ˆsinˆcos

2/

03

03

0

ˆsinˆcos4

14

1)(

yaxaaadR

RdqrE

A Charged Loop Segment

2/

03

03

0

ˆsinˆcos4

14

1)(

yaxaaadR

RdqrE

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.

A Segment of a Current Loop

aPx

y

1. Find r

i

A Segment of a Current Loop

aPx

y

1. Find r

0r

i

A Segment of a Current Loop

aPx

y

1. Find

2. Slice and find

r

0r

r

r

i

A Segment of a Current Loop

aPx

y

1. Find

2. Slice and find

r

0r

r yaxar ˆsinˆcos

r

i

A Segment of a Current Loop

aPx

y

1. Find

2. Slice and find

3. Find

r

0r

r yaxar ˆsinˆcos

r

adi

d

A Segment of a Current Loop

aPx

y

1. Find

2. Slice and find

3. Find

r

0r

r yaxar ˆsinˆcos

r

adi

d yadxadd ˆcosˆsin

A Segment of a Current Loop

aPx

y

1. Find

2. Slice and find

3. Find

4. Find

r

0r

r yaxar ˆsinˆcos

rrR

r

adi

d yadxadd ˆcosˆsin

A Segment of a Current Loop

aPx

y

1. Find

2. Slice and find

3. Find

4. Find

r

0r

r yaxar ˆsinˆcos

rrR

yaxaR ˆsinˆcos

r

adi

d yadxadd ˆcosˆsin

A Segment of a Current Loop

aPx

y

1. Find

2. Slice and find

3. Find

4. Find

5. Find

r

0r

r yaxar ˆsinˆcos

rrR

RyaxaR ˆsinˆcos

r

adi

d yadxadd ˆcosˆsin

A Segment of a Current Loop

aPx

y

1. Find

2. Slice and find

3. Find

4. Find

5. Find

r

0r

r yaxar ˆsinˆcos

rrR

R aR yaxaR ˆsinˆcos

r

adi

d yadxadd ˆcosˆsin

A Segment of a Current Loop

aPx

y

1. Find

2. Slice and find

3. Find

4. Find

5. Find

6. Find

r

0r

r yaxar ˆsinˆcos

rrR

R aR yaxaR ˆsinˆcos

r

adi

d yadxadd ˆcosˆsin

Rd

A Segment of a Current Loop

aPx

y

1. Find

2. Slice and find

3. Find

4. Find

5. Find

6. Find

r

0r

r yaxar ˆsinˆcos

rrR

R aR yaxaR ˆsinˆcos

r

adi

d

zdazdazaRd ˆˆcosˆsin 22222

yadxadd ˆcosˆsin

A Segment of a Current Loop

1. Find

2. Slice and find

3. Find

4. Find

5. Find

6. Find

r

0r

r yaxar ˆsinˆcos

rrR

R aR yaxaR ˆsinˆcos

d

2/

03

20

30

4)(

adaiz

RRdirB

yadxadd ˆcosˆsin

zdazdazaRd ˆˆcosˆsin 22222

A Segment of a Current Loop

zai

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

Gauss’s LawThis is true no matter how small the Gaussian surface is. But the number of field lines gets smaller as the volume gets smaller.

DivergenceDefine divergence to be

PvPEdiv

vEdiv E

v

point around volumesmall a is point at field E theof divergence theis

where

lim0

DivergenceDivergence 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.)

Divergence and Gauss’s Law

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

0

)()(1

000

vas

vrdvrqencE

Divergence and Gauss’s Law

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

0

)()(1

000

vas

vrdvrqencE

00

)(lim

rv

Ediv E

v

Gauss’s Law in Differential Form

0

)(

rEdiv

Divergence and Gauss’s Law

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 LawThe 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

CurlTake 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

CurlTake 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

0

lim)(

P xy

CurlTake 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

0

lim)(

P xy

z

CurlTake 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

axrBcurl

arBcurl

x

x

ax

CurlCurl 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

Curl and Ampère’s Law

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

)(lim)( 00rj

arBcurl x

x

ax

0

)()( 000

aas

arjdarji xxenc

Curl and Ampère’s Law

More generally:

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

)()( 0 rjrBcurl

Curl and Ampère’s Law

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

Curl and Ampère’s Law

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

Curl and Ampère’s Law

Section 11Differential Operators

The GradientThe 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 GradientIn terms of derivatives, the gradient is:

zz

yy

xx

ˆˆˆ

Electric Field and Electric Potential

VzyxE ),,(

zVz

yVy

xVxVzyxE

ˆˆˆ),,(

The Divergence OperatorWe can show that another way of representing divergence is in terms of derivatives.

zE

yE

xEEEdiv zyx

The Curl OperatorWe can also express the curl in terms of derivatives.

zyx

xyzxyz

BBBzyx

zyx

yB

xB

zx

Bz

Byz

By

BxBBcurl

ˆˆˆ

ˆˆˆ

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

0

E

0 B

jB

0

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

Find the electric field.constant. a is where2 yxV

yxxxy

zVz

yVy

xVxVzyxE

ˆˆ2

ˆˆˆ),,(

2

Some ProblemsIn 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

Ex

EE

E

yx

0000

0

2

Some ProblemsIn 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

yB

xBB yx

Some ProblemsIn 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

01

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

Bj

0

1

zxyxyxyz

yB

xB

z

yB

xB

zx

Bz

Byz

By

Bx

xy

xyzxyz

ˆ1266ˆ

1ˆ1ˆ1ˆ

00

0

000

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

Bj

0

1

zxyxyxyz

yB

xB

z

yB

xB

zx

Bz

Byz

By

Bx

xy

xyzxyz

ˆ1266ˆ

1ˆ1ˆ1ˆ

00

0

000