Announcements
• Quiz III: Median 87.5; highest yet
• Mean 83.2.
• Eight D’s and F’s!
• Quiz IV: April 21st
• Final Exam: May 2cd 9am
• Office Hrs 2-3pm today
Magnetic field from wires
I2
 0 I ds  rˆ
B
2

4
r
I1
In between one must add the fields from each wire, with the correct
direction vector .
Magnetic Field Integrated
around a wire
Current I out
of the plane
•What is the integral of the magnetic
field around wire on the path drawn?
0 I

2

a
 0 I
B

ds

2 a
 B  ds   I
0
Ampere’s Law
 B  ds   I
I
0
•Ampere’s law says that if we take the dot product of the field
and the length element and sum up (i.e. integrate) over a
closed loop, the result is proportional to the current through
the surface
•This is not quite the same as gauss’s law (ch32 for
magnetism)
Ampere’s Law:Symmetry
 B  ds   I
0
•Ampere’s law because easy (just as gauss’s law did) when
we have symmetry and a uniform field
•From a infinitely long wire:
Radial symmetric, so
B(2r)=0I
(same expression as from Biot-Savart law)
Quiz: Ampere’s Law
•Consider three wires with
current flowing in/out as shown
•Consider three different loops
surrounding the wires
X
Y
Which of the loops has the largest and
smallest integrals of the magnetic
field around the loops drawn?
A) X > Y > Z
C) Y > Z > X
B) X > Z > Y
D) Y > X > Z
2A
3A
1A
Z
Ampere’s Law: Wire
•We can use ampere’s law on problems difficult to solve with
the Biot-Savart law
We know that the current is uniformly distributed within the
wire, so Ienc=I(r2)/(R2)
B(2r)=0I(r2)/(R2)
B=0I(r)/(2R2)
Inside the field is proportional to the distance from the center
Outside the field falls over off
Magnetic Field From a
Plane of Current
What is the magnetic field?
•By symmetry, magnetic field
will be the same on both sides
L
•Draw a loop crossing the
boundary to use Ampere’s Law
N wires in
length L
 B  ds   I   NI
 B  ds  BL  BL  0  0
0 l
0
B
 0 NI
2L
Current I in
each wire
Magnetic Field of an Ideal Solenoid
length l
N turns
Length L
•If the length is much greater than the radius,
•Magnetic field far away is zero
•Magnetic field everywhere outside is small
•Draw a path for Ampere’s law that
crosses inside the solenoid
 B  ds   I
0 l
R
Current I
l
I l  NI
L
=Bl
B
 0 NI
L
Toroid
•Take a solenoid and bend into a torus
•Again we have symmetry!
B(2r)=0IN
•We do not have a constant B field!