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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 of a Wire
The field contour is made of half-planes centered on the wire.
The Field Contour of a 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
Counting Surfaces Pierced
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
Counting Surfaces Pierced
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?
Other Amperian Loops
What is the net number of surfaces pierced by each of these Amperian loops?
Other Amperian Loops
What is the net number of surfaces pierced by each of these Amperian loops?
Other Amperian Loops
The net numbers of surfaces pierced by each of these loops is 16.
Other Amperian Loops
What is the net number of surfaces pierced by this Amperian loop?
Other Amperian Loops
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.
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 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!
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
r
irB 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 ?
r
irB enc
2
)( 0
r
Example: Uniform Current Distribution
The current density is uniform, so:
enc
enc
A
i
A
ij
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
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 ?
r
irB enc
2
)( 0
r
Example: Uniform Current Distribution
r
irB enc
2
)( 0
Example: Uniform Current Distribution
r
irB enc
2
)( 0 iA
Ai encenc
Example: Uniform Current Distribution
r
irB enc
2
)( 02
2
R
riienc
Example: Uniform Current Distribution
2
20
2)(
R
ri
rrB
Example: Uniform Current Distribution
2
20
2)(
R
ri
rrB
Example: Uniform Current Distribution
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
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.
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
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
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?
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?
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.
Ampere’s Law – TorusenciB 0
r
NiB
NirB
2
2
0
0
r
B
Ampere’s Law – TorusenciB 0
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)(
Origin of the Basic Laws
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
Origin of the Basic Laws
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 Laws
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
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
RR
dqrEd
3
04
1)(
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
xdR
R
dqrEd ˆˆ
4
1
4
1)(
2/3220
30
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
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
The rod has a linear charge density
L
q
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
1
4
1)(
yx
xd
R
dqrdV
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
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
Current in a Wire Segment
i
−L +L
P
y
1. Find r
yyr ˆ
r
Current in a Wire Segment
i
−L +L
P
1. Find
2. Choose a slice and find
r
yyr ˆ
r
r
r
Current in a Wire Segment
i
−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 Segment
i
−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 Segment
i
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 Segment
i
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 Segment
i
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 Segment
i
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 Segment
i
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 Segment
i
d xxdxxdd ˆˆ
Rd
−L +L
P
r
r
R
Current in a Wire Segment
i
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:
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
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
R
RdirBd
2/322
0
2/322
03
0
4ˆ)(
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/a
q
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
1
4
1)(
yaxaa
adR
R
dqrE
A Charged Loop Segment
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.
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
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
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
4ˆ
4)(
a
daiz
R
RdirB
yadxadd ˆcosˆsin
zdazdazaRd ˆˆcosˆsin 22222
A Segment of a Current Loop
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.)
Divergence and Gauss’s Law
In a very small volume, charge density is nearly constant.
0
)()(
1
000
vas
vr
dvrqenc
E
Divergence and Gauss’s Law
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
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 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
Curl and Ampère’s Law
In a very small volume, the density is nearly a constant.
)(lim)( 00
rja
rBcurl xx
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 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
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
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