Transcript Welcome to Physics 220!

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)
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 2
Cylindrically 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)
I
j
A
Current Density
• The current passing through a small gate of area ΔA is
I  jA
small gate
Current Density
• The total current passing through the wire is the sum of
the current passing through all small gates.
I   j dA
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.
Cylindrically Symmetric Current Distribution
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.
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!
How can we get zero net surfaces?
3. Or we could just have no surfaces at all inside the
hollow wire.
This is the only way it can be done!
The Magnetic Field inside a Hollow Wire
If the current distribution has cylindrical symmetry, the
magnetic field inside a hollow wire must be zero.
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
Cylindrically Symmetric Current Distribution
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
Inside a cylindrically symmetric current distribution, the
magnetic field is:
 0ienc
B(r ) 
2 r
r
Section 3
Uniform 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 ?
 0ienc
B(r ) 
2 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 ?
 0ienc
B(r ) 
2 r
r
The current density is
uniform, so:
ienc
i
j 
A Aenc
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 ?
 0ienc
B(r ) 
2 r
Therefore:
r
ienc
Aenc

i
A
Example: Uniform Current Distribution
 0ienc
B(r ) 
2 r
Example: Uniform Current Distribution
 0ienc
B(r ) 
2 r
ienc
Aenc

i
A
Example: Uniform Current Distribution
 0ienc
B(r ) 
2 r
r
i
2
R
2
ienc
Example: Uniform Current Distribution
0  r
i
B(r ) 
2
2 r  R
2
Example: Uniform Current Distribution
0  r
i
B(r ) 
2
2 r  R
2
Example: Uniform Current Distribution
 0ir
B (r ) 
2
2 R
Section 4
The Line Integral
Line Integral
We know that the magnetic field is stronger where the
perpendicular surfaces are closer together.
number of surfaces pierced
N
Bk
k
length of field line segment

Therefore, the number of surfaces pierced is
1
N  B
k
Let
  B
Line Integral
Therefore, the number of surfaces pierced is

N
k
•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
 
   B  d

d

B


d  B  d   B d cos 
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.
 
   B  d
 
W   F  d

d

F


dW  F  d   F d cos 
Line Integral
The line integral is a way of measuring the number of
contour surfaces pierced by a line segment.
 
   B  d

d

B


d  B  d   B d cos 
Line Integral
Roughly speaking, it is a measure of how much field
lines along a path.
 
   B  d

d

B


d  B  d   B d cos 
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:
 
   B  d   0ienc
Class 23
Today, 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 5
Applying Ampère’s Law
Ampère’s Law – the Practical Version
  B  0ienc
Ampère’s Law – the Practical Version
  B  0ienc
The number
of surfaces
pierced
Ampère’s Law – the Practical Version
  B  0ienc
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
  B  0ienc
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
  B  0ienc
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
  B   0ienc   0  jdA
This is the
line integral
This is NOT the
line integral
Section 6
Ampère’s Law and Cylindrical
Wires
A Typical Problem
• A wire of radius R has current density j
magnetic field inside the wire.
  r .2
Find the
• “Inside the wire” means at some point P at a radius r < R.
P
r
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 j   r 2. Find
the magnetic field inside the wire.
What is the correct expression for the ℓ in the line
integral?
  2 r
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 j   r 2. Find
the magnetic field inside the wire.
What is the correct expression for ienc ?
r
ienc    r 2 rdr
2
0
A wire of radius R has current density j   r 2. Find
the magnetic field inside the wire.
r
B(r )   0
4
3
Another Problem
• A wire of radius R has current density
magnetic field outside the wire.
j   .r 2 .
Find the
• “Outside the wire” means at some point P at a radius r > R.
P
r
R
A wire of radius R has current density j   r 2. Find
the magnetic field outside the wire
R
ienc    r 2 rdr
2
0
R
B(r )  0
4r
4
Section 7
Other Applications of
Ampère’s Law
Ampere’s Law – Plane of Wires
The magnetic field from each wire forms
circular loops
around
the
wire.


r
Therefore B is perpendicular to .
B1
1
2
3
P
Ampere’s Law – Plane of Wires
Consider the magnetic field from three
wires.
B3 B2 B1
d
1
2
3
P
Ampere’s Law – Plane of Wires
B3 B2 B1
d
1
2
3
P
When we add the magnetic
fields from each wire, the
vector sum points upward.
Ampere’s Law – Plane of Wires
  B  0ienc

B
Now let’s look at the line
integral of the magnetic field
around the dotted Amperian
loop.
Ampere’s Law – Plane of Wires
  B  0ienc

B
d
Now let’s look at the line
integral of the magnetic field
around the dotted Amperian
loop.
Ampere’s Law – Plane of Wires
  B  0ienc

B
d
The bottom part of the
Amperian loop pierces no
contour surfaces – so the line
integral here is zero.
 
We also know    B  d 
but the B field and the path are
perpendicular on this segment
so Λ=0 for this part of the path.
Λ=0 for this top of the path, too.
Ampere’s Law – Plane of Wires
 
   B  d

B
d
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.
Ampere’s Law – Plane of Wires
  B  0ienc

B
d
The line integral over the right
side of the Amperian loop is
 right  Bd
Ampere’s Law – Plane of Wires
  B  0ienc

B
d
The line integral over the left
side of the Amperian loop is
 left  Bd
Ampere’s Law – Plane of Wires
  B  0ienc

B
d
The total line integral is:
  B2d
The enclosed current is the number
of wires in the loop times the
current in each wire:
ienc  Ni
Ampere’s Law – Plane of Wires
  B  0ienc

B
d
B 2d   0 Ni
B
0 N
2 d
i
 0 ni
2
N
where n 
d
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
  B  0ienc
What would the field be like if
there were two planes of wires
with currents in opposite
directions?
d

B

B0
1
2
3
B  2
 0 ni
2
  0 ni
Ampere’s Law – Solenoid
  B  0ienc
d

B

B0
1
2
3
A solenoid is similar to two
planes of wires. The magnetic
field inside this solenoid points
downward. The magnetic field
outside the solenoid is zero.
Ampere’s Law – Solenoid
  B  0ienc
Bd   0 Ni
d

B

B0
1
2
3
N
B   0 i   0 ni
d
N
where n 
d
Right-hand Rule #3
The direction of the magnetic field
inside a solenoid is given by a
right-hand rule:

B
Circle the fingers of your right
hand in the direction of the
current. The magnetic field is in
the direction of your thumb.
Ampere’s Law – Torus
  B  0ienc
A torus is like a solenoid
wrapped around a
doughnut-shaped core.
The magnetic field inside
forms circular loops.
The magnetic field outside
is zero.
r

B
Ampere’s Law – Torus
  B  0ienc
B 2 r   0 Ni
r
 0 Ni
B
2 r

B
Ampere’s Law – Torus
  B  0ienc
B 2 r   0 Ni
r
 0 Ni
B
2 r

B
The wires on the inside
of the torus are closer
together, so the field
is stronger there.
Class 24
Today, 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 8
Finding Fields by Direct
Integration
Geometry for Extended Objects

r  origin to source

r origin to P

R source to P
  
r  R  r

r
P

R

r
The Basic Laws
The electric field and potential of a small
charge dq:

 
1 Rdq
dE (r ) 
40 R 3

1 dq
dV (r ) 
40 R
The magnetic field
 of a current i in a small
length of wire d  :
 
 
0 i d   R
dB(r ) 
4 R 3
Origin of the Basic Laws
Electric field and potential for slowly
moving point charges – Coulomb’s
Law:
 
E (r ) 

V (r ) 
q
1 q 
rˆ 
r
2
3
40 r
40 r
1
1
q
40 r
The Basic Laws – for dq
Electric field and potential for dq.

 
1 Rdq
dE (r ) 
40 R 3

dV (r ) 
1
dq
40 R
Origin of the Basic Laws
Remember the geometry from Lesson 2
y
head r
h
line

tail rt

θ
T

rh  s
thread


ray rr
line
ψ
line
S
P
P
U
motion of source
x
Origin of the Basic Laws
An expression for the magnetic field of
a point charge:
 1
 1  
B  rˆh  E 
rh  E
c
crh
 
1 

 s rh  rr  E
crh



1 

 s rh  E because E is parallel to rr
crh
1  
 s  E
c


Origin of the Basic Laws
An expression for the magnetic field
of a point charge (moving slowly):
 
  1    1 q s  r
B(r )   s  E (r ) 
c
c 40 r 3
 
1
q   0 q  
 B(r ) 
v r 
v r
2 3 s
3 s
40 c r
4 r
as c 
2
1
 0 0
Origin of the Basic Laws
Magnetic field for dq
 
 0 dq  
dB(r ) 
v r
3 s
4 r



d  dq 
 dq vs  dq

d  i d
dt dt
 
0 i  
dB(r ) 
d  r
3
4 r
A little sleight of hand, but it’s the same as a more formal proof.
The Biot-Savart Law
 
 
0 id   R
dB(r ) 
3
4 R
The formula for the magnetic field of a wire segment
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.

r is the vector from the origin to a field point.

R is the vector from the segment (the origin) to a field point.
Equations for Extended Objects

 
1 Rdq
dE (r ) 
3
40 R

dV (r ) 
1
dq
40 R
 
 
0 id   R
dB(r ) 
3
4 R
P

R

r

r
Now, let’s work some problems…
A Charged Rod
y
P
x
−L
+L
The rod has a linear charge density
q

2L
Find the electric field at P.
A Charged Rod
y
P
x
−L
+L
 
dE (r ) 
dq 
R
3
40 R
1

We need to find dq, R, and R.
A Charged Rod

r
−L
1. Find
P
y
+L

r
A Charged Rod

r
−L
1. Find
P
y
+L

r

r  yyˆ
A Charged Rod
P

r

r
−L
1. Find

r

r  yyˆ
2. Choose a slice and find
+L

r
A Charged Rod
P

r

r
−L
1. Find

r

r  yyˆ
2. Choose a slice and find
+L

r

r   xxˆ
Be sure to put primes on the slice variables!
A Charged Rod
P

r

r
−L
1. Find

r

r  yyˆ
2. Choose a slice and find
+L

r

r   xxˆ
3. Find the length and charge of the slice.
A Charged Rod
dx
−L
1. Find

r
P

r

r

r  yyˆ
2. Choose a slice and find
+L

r

r   xxˆ
3. Find the length and charge of the slice.
dq  dx 
A Charged Rod
P

r

R

r
−L
1. Find

r

r  yyˆ
2. Choose a slice and find
+L

r

r   xxˆ
3. Find the length and charge of the slice.
4. Find
  
R  r  r
dq  dx 
A Charged Rod
P

r

R

r
−L
1. Find

r

r  yyˆ
2. Choose a slice and find
+L

r

r   xxˆ
3. Find the length and charge of the slice.
4. Find
   
R  r  r  R  yyˆ  x xˆ
dq  dx 
A Charged Rod
P

r

R

r
−L
1. Find

r

r  yyˆ
2. Choose a slice and find
+L

r

r   xxˆ
3. Find the length and charge of the slice.
4. Find
5. Find
   
R  r  r  R  yyˆ  x xˆ
R
dq  dx 
A Charged Rod
P

r

R

r
−L
1. Find

r

r  yyˆ
2. Choose a slice and find
+L

r

r   xxˆ
3. Find the length and charge of the slice.
4. Find
5. Find
   
R  r  r  R  yyˆ  x xˆ
R
R  x 2  y 2
dq  dx 
A Charged Rod
 
dE ( r ) 
1. Find
dq 
R
3
40 R
1

r
Now just substitute!

r  yyˆ
2. Choose a slice and find

r

r   xxˆ
3. Find the length and charge of the slice.
4. Find
5. Find
   
R  r  r  R  yyˆ  x xˆ
R
R  x 2  y 2
dq  dx 
A Charged Rod
 
dE (r ) 
1. Find
dq 
1
dx 
R
3
40 R
40 x  2  y 2
1

r

r  yyˆ
2. Choose a slice and find
yyˆ  xxˆ 



r

r   xxˆ
3/ 2
3. Find the length and charge of the slice.
4. Find
5. Find
   
R  r  r  R  yyˆ  x xˆ
R
R  x 2  y 2
dq  dx 
A Charged Rod
 
dE ( r ) 
dq 
1
 dx 
R
3
40 R
4 0 x  2  y 2
1
 
y
E (r )  yˆ
40
1. Find

r

dx 
L
 x
L
2
y

2 3/ 2

r  yyˆ
2. Choose a slice and find

3/ 2

 xˆ
40

r
 yyˆ  xxˆ 
 x
L
5. Find
2
y

2 3/ 2

r   xxˆ
3. Find the length and charge of the slice.
4. Find
x dx 
L
   
R  r  r  R  yyˆ  x xˆ
R
R  x 2  y 2
dq  dx 
A Charged Rod
 
dE ( r ) 
dq 
1
 dx 
R
3
40 R
4 0 x  2  y 2
1

 
y
E (r )  yˆ
40
dx 
L
 x
L
2
y

2 3/ 2

3/ 2

 xˆ
40
 yyˆ  xxˆ 
x dx 
L
 x
L
2
y
I won’t expect you to evaluate these integrals!
 
E (r ) 
L
20 y y  L
2
2
yˆ

2 3/ 2
A Charged Rod
y
P
x
−L
+L
The rod has a linear charge density
q

2L
Now find the electric potential at P.
A Charged Rod

dV (r ) 
1. Find

r
dq
1
dx

40 R 40 x  2  y 2
1

r  yyˆ
2. Choose a slice and find


r

1/ 2

r   xxˆ
3. Find the length and charge of the slice.
4. Find
5. Find
   
R  r  r  R  yyˆ  x xˆ
R
R  x 2  y 2
dq  dx 
A Charged Rod

dV (r ) 

V (r ) 
1. Find

r
dq
1
 dx 

40 R 40 x  2  y 2
1

40

dx 
L
 x
L
2
y

1/ 2

2 1/ 2

r  yyˆ
2. Choose a slice and find

r

r   xxˆ
3. Find the length and charge of the slice.
4. Find
5. Find
   
R  r  r  R  yyˆ  x xˆ
R
R  x 2  y 2
dq  dx 
A Charged Rod
dq
1
dx

40 R 40 x2  y 2 1/ 2

 L
dx
V (r ) 
 2
40  L x  y 2 1/ 2

dV (r ) 
1




You don’t need to evaluate this integral, either!


V (r ) 
ln
40
L2  y 2  L
L2  y 2  L
Current in a Wire Segment
y
P
i
x
−L
+L
Current i travels to the left along a segment of
wire.
Find the magnetic field at P.
Current in a Wire Segment

r
−L
1. Find
P
y
i
+L

r
Current in a Wire Segment

r
−L
1. Find
P
y
i
+L

r

r  yyˆ
Current in a Wire Segment
P

r

r
−L
1. Find

r

r  yyˆ
2. Choose a slice and find
i
+L

r
Current in a Wire Segment
P

r

r
−L
1. Find
i

r

r  yyˆ
2. Choose a slice and find
+L

r

r   xxˆ
Be sure to put primes on the slice variables!
Current in a Wire Segment
P

r

r
−L
1. Find
i

r

r  yyˆ
2. Choose a slice and find

3. Find d 
+L

r

r   xxˆ
Current in a Wire Segment
dx
−L
1. Find

r
P

r
i

r

r  yyˆ
 
2. Choose a slice and find r  r   xx
ˆ


d   dx  xˆ   dx xˆ
3. Find d 
+L
Current in a Wire Segment
P

r

R

r
−L
1. Find
i

r

r  yyˆ
 
2. Choose a slice and find r  r   xx
ˆ


d   dx  xˆ   dx xˆ
3. Find d 
  
4. Find R  r  r 
+L
Current in a Wire Segment
P

r

R

r
−L
1. Find
i

r

r  yyˆ
 
2. Choose a slice and find r  r   xx
ˆ


d   dx  xˆ   dx xˆ
3. Find d 
  
4. Find R  r  r  R  yyˆ  x xˆ
+L
Current in a Wire Segment
P

r

R

r
−L
1. Find
i

r

r  yyˆ
 
2. Choose a slice and find r  r   xx
ˆ


d   dx  xˆ   dx xˆ
3. Find d 
  
4. Find R  r  r  R  yyˆ  x xˆ
5. Find R
+L
Current in a Wire Segment
P

r

R

r
−L
1. Find
i

r

r  yyˆ
 
2. Choose a slice and find r  r   xx
ˆ


d   dx  xˆ   dx xˆ
3. Find d 
  
4. Find R  r  r  R  yyˆ  x xˆ
5. Find R
R  x 2  y 2
+L
Current in a Wire Segment

R
−L 
1. Find r
P

r

r

r  yyˆ
i
 
2. Choose a slice and find r  r   xx
ˆ


d   dx  xˆ   dx xˆ
3. Find d 
  
4. Find R  r  r  R  yyˆ  x xˆ
5. Find
R
 
6. Find d   R
2
2

R x y
+L
Current in a Wire Segment

R
−L 
1. Find r
P

r

r

r  yyˆ
i
 
2. Choose a slice and find r  r   xx
ˆ


d   dx  xˆ   dx xˆ
3. Find d 
  
4. Find R  r  r  R  yyˆ  x xˆ
5. Find
R
 
6. Find d   R
2
2

R x y
 
d   R   ydx zˆ
+L
Remember:
xˆ  yˆ  zˆ
yˆ  zˆ  xˆ
zˆ  xˆ  yˆ
yˆ  xˆ   zˆ
zˆ  yˆ   xˆ
xˆ  zˆ   yˆ
xˆ  xˆ  0
yˆ  yˆ  0
zˆ  zˆ  0
Current in
a Wire Segment

 
 0 i d  R
0i
ydx
dB(r ) 
  zˆ
3
4 R
4 x 2  y 2

1. Find

r

r  yyˆ
5. Find
R
2
2

R x y

3/ 2
 
2. Choose a slice and find r  r   xx
ˆ


d   dx  xˆ   dx xˆ
3. Find d 
  
4. Find R  r  r  R  yyˆ  x xˆ
 
6. Find d   R
 
d   R   ydx zˆ
Current in a Wire Segment
 
 0 i d  R
0i
ydx 
dB(r ) 
  zˆ
3
4 R
4 x  2  y 2
 
 0 iy L
dx 
B(r )   zˆ
4 L x  2  y 2 3 / 2



3/ 2

1. Find

r

r  yyˆ
5. Find
R
2
2

R x y
 
2. Choose a slice and find r  r   xx
ˆ


d   dx  xˆ   dx xˆ
3. Find d 
  
4. Find R  r  r  R  yyˆ  x xˆ
 
6. Find d   R
 
d   R   ydx zˆ
A Charged Loop Segment
y
P
The rod has a linear charge density
q

a / 2
Find the electric field at P.
a
x
A Charged Loop Segment
y
1. Find

r
P
a
x
A Charged Loop Segment
y
1. Find

r

r 0
P
a
x
A Charged Loop Segment
y

r

r

1. Find
r 0

2. Slice and find r 
P
 x
a
A Charged Loop Segment
y

r

r
P
 x
a

1. Find
r 0


2. Slice and find r 
r   a cos   xˆ  a sin   yˆ
A Charged Loop Segment
y

r

r
P
ad 
 x
a

1. Find
r 0


2. Slice and find r 
r   a cos   xˆ  a sin   yˆ
3. Find the charge of the slice.
A Charged Loop Segment
y

r

r
P
ad 
 x
a

1. Find
r 0


2. Slice and find r 
r   a cos   xˆ  a sin   yˆ
3. Find the charge of the slice. dq  ad 
A Charged Loop Segment
y

r

r
P
ad 
 x
a

1. Find
r 0


2. Slice and find r 
r   a cos   xˆ  a sin   yˆ
3. Find the charge of the slice. dq  ad 
  
4. Find R  r  r 
A Charged Loop Segment
y

r

r
P
ad 
 x
a

1. Find
r 0


2. Slice and find r 
r   a cos   xˆ  a sin   yˆ
3. Find the charge of the slice. dq  ad 
  

4. Find R  r  r 
R  a cos   xˆ  a sin   yˆ
A Charged Loop Segment
y

r

r
P
ad 
 x
a

1. Find
r 0


2. Slice and find r 
r   a cos   xˆ  a sin   yˆ
3. Find the charge of the slice. dq  ad 
  

4. Find R  r  r 
R  a cos   xˆ  a sin   yˆ
5. Find R
A Charged Loop Segment
y

r

r
P
ad 
 x
a

1. Find
r 0


2. Slice and find r 
r   a cos   xˆ  a sin   yˆ
3. Find the charge of the slice. dq  ad 
  

4. Find R  r  r 
R  a cos   xˆ  a sin   yˆ
Ra
5. Find R
A Charged Loop Segment
 
E (r ) 
 /2

1
dq
1
ad 
 a cos  xˆ  a sin   yˆ 
R
3
3


40 R
40 0 a

r

1. Find
r 0


2. Slice and find r 
r   a cos   xˆ  a sin   yˆ
3. Find the charge of the slice. dq  ad 
  

4. Find R  r  r 
R  a cos   xˆ  a sin   yˆ
Ra
5. Find R
A Charged Loop Segment
 
E (r ) 
 /2

1
dq
1
ad 
 a cos  xˆ  a sin   yˆ 
R
3
3


40 R
40 0 a
Here we only have to integrate sines and cosines, so
the integrals are easy!
 
E (r )  

( xˆ  yˆ )
40 a
A Segment of a Current Loop
y
P
i
a
Current i travels counterclockwise along a
segment of a loop of wire.
Find the magnetic field at P.
x
A Segment of a Current Loop
y
1. Find

r
P
i
a
x
A Segment of a Current Loop
y
1. Find

r

r 0
P
i
a
x
A Segment of a Current Loop
i
y

r

r

1. Find
r 0

2. Slice and find r 
P
 x
a
A Segment of a Current Loop
i
y

r

r
 x
a

P
1. Find
r 0


2. Slice and find r 
r   a cos   xˆ  a sin   yˆ
A Segment of a Current Loop
i
y

r

r
ad 
 x
a

P
1. Find
r 0


2. Slice and find r 
r   a cos   xˆ  a sin   yˆ

3. Find d 
A Segment of a Current Loop
i
y

r

r
ad 
 x
a

P
1. Find
r 0


2. Slice and find r 
r   a cos   xˆ  a sin   yˆ

3. Find d 

d   ad  sin   xˆ  ad  cos   yˆ
A Segment of a Current Loop
i
y

r

r
ad 
 x
a

P
1. Find
r 0


2. Slice and find r 
r   a cos   xˆ  a sin   yˆ

3. Find d 
4. Find

d   ad  sin   xˆ  ad  cos   yˆ
  
R  r  r
A Segment of a Current Loop
i
y

r

r
ad 
 x
a

P
1. Find
r 0


2. Slice and find r 
r   a cos   xˆ  a sin   yˆ

3. Find d 
4. Find

d   ad  sin   xˆ  ad  cos   yˆ
  
R  r  r

R  a cos   xˆ  a sin   yˆ
A Segment of a Current Loop
i
y

r

r
ad 
 x
a

P
1. Find
r 0


2. Slice and find r 
r   a cos   xˆ  a sin   yˆ

3. Find d 
4. Find
5. Find

d   ad  sin   xˆ  ad  cos   yˆ
  
R  r  r
R

R  a cos   xˆ  a sin   yˆ
A Segment of a Current Loop
i
y

r

r
ad 
 x
a

P
1. Find
r 0


2. Slice and find r 
r   a cos   xˆ  a sin   yˆ

3. Find d 
4. Find
5. Find

d   ad  sin   xˆ  ad  cos   yˆ

  
R  r  r  R  a cos   xˆ  a sin   yˆ
R
Ra
A Segment of a Current Loop
i
y

r

r
ad 
 x
a

P
1. Find
r 0


2. Slice and find r 
r   a cos   xˆ  a sin   yˆ

3. Find d 
4. Find
5. Find

d   ad  sin   xˆ  ad  cos   yˆ

  
R  r  r  R  a cos   xˆ  a sin   yˆ
R
Ra
 
6. Find d   R
A Segment of a Current Loop
i
y

r

r
ad 
 x
a

P
1. Find
r 0


2. Slice and find r 
r   a cos   xˆ  a sin   yˆ

3. Find d 
4. Find
5. Find

d   ad  sin   xˆ  ad  cos   yˆ

  
R  r  r  R  a cos   xˆ  a sin   yˆ
R
Ra
 
2
2
2
2
2
6. Find d   R  a sin   zˆ  a cos   d  zˆ  a d  zˆ
A Segment of a Current Loop

   0i d  R
 0i  / 2 a 2 d 
B(r ) 
 zˆ


3
4
R
4 0 a 3

r

1. Find
r 0


2. Slice and find r 
r   a cos   xˆ  a sin   yˆ

3. Find d 
4. Find
5. Find

d   ad  sin   xˆ  ad  cos   yˆ

  
R  r  r  R  a cos   xˆ  a sin   yˆ
R
Ra
 
2
2
2
2
2
6. Find d   R  a sin   zˆ  a cos   d  zˆ  a d  zˆ
A Segment of a Current Loop
 
 0i  / 2 a 2 d   0i
B(r )  zˆ

zˆ

3
4 0 a
8a
This is a really easy integral this time!
Class 25
Today, 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 9
Gauss’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

E
div E  lim
v 0 v
where

div E is the divergence of the E field at point P
v is a small volume around point P
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.
E 
qenc
0

1
0
as v  0

  (r )dv 

 (r )
0
v
Divergence and Gauss’s Law
In a very small volume, charge density
is nearly constant.
E 
qenc
0

1
0

  (r )dv 

 (r )
as v  0

 (r )

E
 div E  lim

v  0 v
0
0
v
Gauss’s Law in Differential Form

div E 

 (r )
0
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 10
Ampère’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.
Ampere’s Law
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.
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:


 
x
curl B(r ) x  lim
a 0 a
P
x̂
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:


 
y
curl B(r ) y  lim
a 0 a
ŷ
P
x̂
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:


 
z
curl B(r ) z  lim
a 0 a
ŷ
P
ẑ
x̂
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:




 
x
curl B (r ) x  lim
a  0 a
where
 
curl B (r ) x is the x  component of the curl
a is the area of the Amperian loop
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.
Curl and Ampère’s Law
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.
   0ienc   0 
as a  0


j x (r ) da   0 j x (r )a
Curl and Ampère’s Law
In a very small volume, the density is
nearly a constant.
   0ienc   0 


j x (r ) da   0 j x (r )a
as a  0


 

x
curl B(r ) x  lim
 0 j x (r )
a 0 a
Curl and Ampère’s Law
More generally:
 
 
curl B(r )  0 j (r )
The curl points in the direction of the
current at any point in space.
Curl and Ampère’s Law
More generally:
 
 
curl B(r )  0 j (r )
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.”
Curl and Ampère’s Law
Electrical current produces magnetic
field lines that form loops around the
path of the moving charges.
Section 11
Differential 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:



  xˆ
 yˆ
 zˆ
x
y
z
Electric Field and Electric Potential

E ( x, y, z )  V

V
V
V
E ( x, y, z )  V   xˆ
 yˆ
 zˆ
x
y
z
The Divergence Operator
We can show that another way of
representing divergence is in terms of
derivatives.

  Ex  E y  Ez
div E    E 


x
y
z
The Curl Operator
We can also express the curl in terms
of derivatives.


  Bz  B y 
  Bx  Bz    B y  Bx 
curlB    B  xˆ 


 zˆ 

  yˆ 



y

z

z

x

x

y

 



xˆ


x
Bx
yˆ

y
By
zˆ

z
Bz
Gauss’s Laws and Ampère’s Law
in Differential Form
 
E 
0

 B  0


  B  0 j
Some Problems
In a region of space, the electric
potential is given by the expression
V   x 2 y where  is a constant.
Find the electric field.

V
V
V
E ( x, y, z )  V   xˆ
 yˆ
 zˆ
x
y
z
2
ˆ
 2xyx  x yˆ
Some Problems
In a region of space, the electric
potential is given by the expression
V   x 2 y where  is a constant.
Find the charge density.

2
ˆ
E ( x, y, z )  2xyx  x yˆ
 
E 
0

E y
E x
   0  E   0
 0
 2 0y
x
y
Some Problems
In a region of space, the magnetic field
is given by the expression

B( x, y )   ( x 2  3 y 2 ) xxˆ   (3x 2  y 2 ) yyˆ
where  and  are constants.
Find β in terms of α.
Some Problems

B( x, y )   ( x 2  3 y 2 ) xxˆ   (3 x 2  y 2 ) yyˆ
 Bx By
B 

0
x
y
 ( x  3 y )  x 2 x   (3x  y )  y 2 y  0
2
2
2
2
 (3x 2  3 y 2 )   (3x 2  3 y 2 )  0
   
Some Problems
In a region of space, the magnetic field
is given by the expression

B( x, y )   ( x 2  3 y 2 ) xxˆ   (3 x 2  y 2 ) yyˆ
where  is a constant.
Find the current density (magnitude and
direction).



 1
  B  0 j
j
 B
0
Some Problems

B( x, y )   ( x 2  3 y 2 ) xxˆ   (3x 2  y 2 ) yyˆ

 1
j
 B
0
1   Bz  B y 
1   B x  Bz 
1
ˆ
ˆ
ˆ
x 


 y 
z
0   y  z 
0   z
x 
0
1   B y  Bx 
 zˆ 

0   x  y 

12xy
 zˆ  6 xy  6 xy  
zˆ
0
0
  B y  Bx 




x

y


Some Problems

B( x, y )   ( x 2  3 y 2 ) xxˆ   (3x 2  y 2 ) yyˆ

 1
j
 B
0
1   Bz  B y 
1   B x  Bz 
1
ˆ
ˆ
ˆ
x 


 y 
z
0   y  z 
0   z
x 
0
1   B y  Bx 
 zˆ 

0   x  y 

12xy
 zˆ  6 xy  6 xy  
zˆ
0
0
  B y  Bx 




x

y

