Chapter 27: Electric flux & Gauss’ law Chapter 29: Electric potential & work Chapter 30: Electric potential & field Chapter 28: Current & Conductivity Chapter 31: Circuits Chapter 32: Magnetic fields & forces
(exclude 32.6,32.8,32.10)
Exam 2 is Wed. Mar. 26, 5:30-7 pm, 2103 Ch: Adam(301,310), Eli(302,311), Stephen(303,306), 180 Science Hall: Amanda(305,307), Mike(304,309), Ye(308)
Wed. Mar. 26, 2008
Physics 208, Lecture 17 2
Electric current produces magnetic field
Current (flow of electric charges )in wire produces magnetic field.
That magnetic field aligns compass needle
Current
Magnetic field
Wed. Mar. 26, 2008
Physics 208, Lecture 17 3
Law of Biot-Savart
Each element of current produces a contribution to the magnetic field.
r Ids
€
dB =μo
4π
Ids × ˆ r
r2
B out of page
dI
dB
r
Wed. Mar. 26, 2008
Physics 208, Lecture 17 4
Magnetic field from long straight wire:Direction
What direction is the magnetic field from an infinitely-long straight wire?
I
€
dr B =
μo
4π
Idr s × ˆ r
r2
x
y
Wed. Mar. 26, 2008
Physics 208, Lecture 17 5
Current dependence
How does the magnitude of the B-field change if the current is doubled?
I
€
dr B =
μo
4π
Idr s × ˆ r
r2
x
y
A) Is halved
B) Quadruples
C) Stays same
D) Doubles
E) Is quartered
Wed. Mar. 26, 2008
Physics 208, Lecture 17 6
Distance dependence
How does the magnitude of the B-field at 2 compare to that at 1?
I
€
dr B =
μo
4π
Idr s × ˆ r
r2
x
y
1
2
A) B2=B1
B) B2=2B1
C) B2=B1/2
D) B2=4B1
E) B2=B1/4
Wed. Mar. 26, 2008
Physics 208, Lecture 17 7
Why?
Biot-Savart says
Why B(r) 1/r instead of 1/r2 ?
€
dr B =
μo
4π
Idr s × ˆ r
r2
€
rr
Large contribution from this current element.
Decreases as 1/r2
I
Small contribution from this current element.
~ independent of r
Wed. Mar. 26, 2008
Physics 208, Lecture 17 8
Long straight wire
All current elements produce B out of page
x
a
€
r = x 2 + a2
r
€
dB =μo
4π
Ids × ˆ r
r2
=μo
4π
I
r2sinθ
=μo
4π
I
r2
a
r=
μo
4πI
a
x 2 + a2( )
3 / 2
€
B =μoIa
4π
dx
x 2 + a2( )
3 / 2 =−∞
∞
∫ μoI
2πa
Add them all up:
Wed. Mar. 26, 2008
Physics 208, Lecture 17 9
Forces between currents
Which of these pairs of currents will attract each other?
A. A
B. A & C
C. B
D. A & B A B C
Wed. Mar. 26, 2008
Physics 208, Lecture 17 10
Force between current-carrying wires
Attractive for parallel currents.
Repulsive for antiparallel currents
Wed. Mar. 26, 2008
Physics 208, Lecture 17 11
Field from a circular loop
Each current element produce dB All contributions add as vectors Along axis, all
components cancelexcept for x-comp
Wed. Mar. 26, 2008
Physics 208, Lecture 17 12
Magnetic field from loopWhich of these graphs
best represents the magnetic field on the axis of the loop?
Bz
z
xy
Bz
Bz
Bz
A.
B.
C.
D.
z
z
z
z
Wed. Mar. 26, 2008
Physics 208, Lecture 17 13
Magnetic field from a current loop
One loop: field still loops around the wire.
Many loops: same effect
Wed. Mar. 26, 2008
Physics 208, Lecture 17 14
Solenoid electromagnet
Sequence of current loops can produce strong magnetic fields.
This is an electromagnet
Wed. Mar. 26, 2008
Physics 208, Lecture 17 15
Comparing Electric, Magnetic
Biot-Savart: calculate B-field from current distribution. Resulting B-field is a vector, and… complication: current (source) is a vector!
Coulomb: calculate E-field from charge distribution Resulting E is a vector
but charge (source) is not a vector
Wed. Mar. 26, 2008
Physics 208, Lecture 17 16
A shortcut: Ampere’s law Integral around closed
path proportional to current passing through any surface bounded by path.
Ampere’s law
€
B • ds∫ = μoI
closed path
surface bounded by path
I
Right-hand ‘rule’: Thumb in direction of
positive current Curled fingers show
direction integration
Wed. Mar. 26, 2008
Physics 208, Lecture 17 17
‘Testing’ Ampere’s law
Long straight wire , Br
€
B r( ) =μoI
2π r
r€
B • ds∫
I
B(r)
B||ds path has constant r
path length = 2πr
€
B • ds∫ =μoI
2πrds∫
€
B • ds∫ =μoI
2πrds∫ =
μoI
2πrds∫
€
B • ds∫ =μoI
2πrds∫ =
μoI
2πrds∫ =
μoI
2πr2πr = μoI
Circular path
Surface bounded by path
Wed. Mar. 26, 2008
Physics 208, Lecture 17 18
Using Ampere’s law
Could have used Ampere’s law to calculate B
r
I
B(r)€
B • ds∫ = Bds =∫ B ds∫ = B2π r= μoI ⇒ B =μoI
2π r
Circular path
Surface bounded by path
B||ds B constant on path
path length = 2πr
Wed. Mar. 26, 2008
Physics 208, Lecture 17 19
Quick Quiz
Suppose the wire has uniform current density. How does the magnetic field change inside the wire?
