W09D1:
Sources of Magnetic Fields:
Ampere’s Law
Today’s Reading Assignment Course Notes:
Sections 9.3-9.4, 9.7, 9.10.2
1
Announcements
Math Review Week Nine Tuesday from 9-11 pm in 26-152
PS 7 due Tuesday April 9 at 9 pm in boxes outside 32-082 or
26-152
Next reading Assignment W09D2 Faraday’s Law Course Notes:
Sections 10.1-10.4
2
Today: Biot-Savart vs. Ampere
BiotSavart
Law
Ampere’s
law
m0 I d s ´ r̂
B=
ò
2
4p
r
 
 B  d s   0 I enc
general
current source
ex: finite wire
wire loop
symmetric
current source
ex: infinite wire
infinite current sheet
3
Last Time:
Creating Magnetic Fields:
Biot-Savart
4
The Biot-Savart Law
Current element of length ds carrying current I
produces a magnetic field at the point P:
m0 I ds ´ r̂
dB =
4p r 2
m0
B(r) =
4p
ò
wire
Id s¢ ´ (r - r¢ )
r - r¢
3
5
The Biot-Savart Law: Infinite Wire
Magnetic Field of an Infinite
Wire Carrying Current I from
Biot-Savart:
m0 I
B=
k̂
2p y
See W06D3 Problem Solving
http://web.mit.edu/8.02t/www/materials/ProblemSolving/solution05.pdf
More generally:
6
3rd Maxwell Equation:
Ampere’s Law
(we will eventually add one more
term to this equation)
ò
closed path
B × d s = m0
òò
J × n̂ da
open surface
Open surface is bounded by closed path
7
Ampere’s Law: The Idea
In order to have a
B field around a
loop, there must be
current punching
through the loop
8
Concept Question: Line Integral
The integral expression
ò
B×ds
closed path
1) is equal to the magnetic work done around a closed
path.
2) is an infinite sum of the product of the tangent
component of the magnetic field along a small
element of the closed path with a small element of
the path up to a choice of plus or minus sign.
3) is always zero.
4) is equal to the magnetic potential energy between
two points.
5) None of the above.
9
C.Q. Answer: Line Integral
2. A line integral by definition is the sum
ò
closed path
i=N
B × d s = lim å B i × Dsi
N®¥
i=1
We need to make a choice of integration direction
(circulation) for the line integral. The small line
element Dsi is tangent to the line and points in the
direction of circulation. The dot product therefore is
the product of the tangent component of the
magnetic field in the direction of the line element.
So the answer depends on which way we circulate
around the path.
10
Current Enclosed
J Current density
I enc =
òò
J × n̂ dA
open
surfacce S
Current enclosed is the flux of the current
density through an open surface S bounded by
the closed path. Because the unit normal to an
open surface is not uniquely defined this
expression is unique up to a plus or minus sign.
11
Sign Conventions: Right Hand Rule
ò
closed path
B × d s = m0
òò
J × n̂ da
open surface
Integration direction
clockwise for line
integral requires that unit
normal points into page
for open surface integral
Current positive into
page, negative out of
page
12
Sign Conventions: Right Hand Rule
ò
closed path
B × d s = m0
òò
J × n̂ da
open surface
Integration direction
counterclockwise for line
integral requires that unit
normal points out of
page for open surface
integral
Current positive out of
page, negative into page
13
Concept Questions:
Ampere’s Law
14
Concept Question: Ampere’s Law
Integrating B around the loop shown gives us:
1. a positive number
2. a negative number
3. zero
15
C.Q. Answer: Ampere’s Law
Answer: 3. Total enclosed current is zero, so
I enc = 0 Þ
ò
B×ds = 0
closed path
16
Concept Question: Ampere’s Law
Integrating B around the loop in the clockwise
direction shown gives us:
1. a positive number
2. a negative number
3. zero
17
C.Q. Answer: Ampere’s Law
Answer: 2.
I enc < 0 Þ
ò
B×ds < 0
closed path
Net enclosed current is out of the page, so field is
counter-clockwise (opposite to circulation direction)
18
Applying Ampere’s Law
1) Identify regions in which to calculate B field.
2) Choose Amperian closed path such that by symmetry B is
zero or constant magnitude on the closed path!
ì"B times length"
B×ds = í
ò
îor "zero"
oriented closed path
3) Calculate
I enc = m0
4) Calculate current enclosed:
òò
J × n̂ da
open surface
5) Apply Ampere’s Law to solve for B: check signs
ò
closed path
B × d s = m0
òò
J × n̂ da
open surface
19
Infinite Wire
I
A cylindrical conductor has
radius R and a uniform
current density with total
current I. we shall find the
direction and magnitude of
the magnetic field for the
two regions:
(1) outside wire (r ≥ R)
(2) inside wire (r < R)
20
Worked Example: Ampere’s Law
Infinite Wire
I
B
I
Amperian Closed Path:
B is Constant & Parallel
Current penetrates surface
21
Example: Infinite Wire
Region 1: Outside wire (r ≥ R)
Cylindrical symmetry 
Amperian Circle
B-field counterclockwise
ò B × d s = B ò ds = B ( 2p r )
 0 I enc  0 I
22
Group Problem: Magnetic Field
Inside Wire
I
We just found B(r>R)
Now you find B(r<R)
23
Infinite Wire: Plot of B vs. r
 0 Ir
Bin 
2
2R
Bout
0 I

2r
24
Group Problem: Non-Uniform
Cylindrical Wire
A cylindrical conductor
has radius R and a nonuniform current density
with total current:
R
J = J 0 n̂
r
Find B everywhere
25
Other Geometries
26
Two Loops
http://web.mit.edu/viz/EM/visualizations/magnetostatics/MagneticFieldConfigurations/tworings/tworings.htm
27
Two Loops Moved Closer
Together
28
Multiple Wire Loops
29
Multiple Wire Loops –
Solenoid
http://youtu.be/GI2Prj4CGZI
30
Demonstration:
Long Solenoid
http://tsgphysics.mit.edu/front/?page=demo.php&letnum=G%2018&show=0
31
Magnetic Field of Solenoid
Horiz.
comp.
cancel
loosely wound
tightly wound
For ideal solenoid, B is uniform inside & zero outside
32
Magnetic Field of Ideal Solenoid
Using Ampere’s law: Think!
ìB ^ d s along sides 2 and 4
í
îB = 0 along side 3
ò B×d s = òB×d s + òB×d s + òB×d s + òB×d s

1
2
Bl
I enc  nlI

3
0

4
0
 0
n: # of turns per unit length
ò B × d s = Bl = m nlI
0
n  N / L : # turns/unit length
B=
m0 nlI
l
= m0 nI
33
Group Problem: Current Sheet
A sheet of current (infinite
in the y & z directions, of
thickness d in the x
direction) carries a
uniform current density:
J = Jk̂
Find the direction and
magnitude of B as a
function of x.
34
Ampere’s Law:
Infinite Current Sheet
B
I
B
Amperian Loops:
B is Constant & Parallel OR Perpendicular OR Zero
I Penetrates
35
Surface Current Density
A very thin sheet of current of width w carrying a
current I in the positive z-direction has a surface
current density
K = Kk̂
KI/w
For sheet of thickness d , width w, and current I
I  Jdw  Kw  J  K / d
36
Solenoid is Two Current Sheets
Consider two sheets each
of thickness d with current
density J. Then surface
current per unit length
K  Jd  nI
Use either Ampere’s Law
or superposition principle
B  0 K  0 Jd  0 nI
37
Biot-Savart vs. Ampere
BiotSavart
Law
Ampere’s
law
m0 I d s ´ r̂
B=
ò
2
4p
r
 
 B  d s   0 I enc
general
current source
ex: finite wire
wire loop
symmetric
current source
ex: infinite wire
infinite current sheet
38
Ampere’s Law:
ò
òò
B × d s = m0
closed path
J × n̂ da
open surface
B
Long
Circular
Symmetry
I
B
(Infinite) Current Sheet
X
X
X
X
X
X
X
X
X
X
X
X
X
X
B
X
X
Solenoid
=
2 Current
Sheets
X
X
X
X
X
X
X
X
X
X
X
X
Torus
39