douglas c. giancoli chapter 28 - george mason...
TRANSCRIPT
Chapter 28
Sources of Magnetic Field
Physics for Scientists & Engineers, 3rd EditionDouglas C. Giancoli
© Prentice Hall
3P15-
Magnetic Fields
B q= ×F v B
Magnetic Dipoles Create and Feel B Fields:
Also saw thatmoving charges feel a force:
5P15-
Cyclotron Motion(1) r : radius of the circle
2mvqvBr
=mvrqB
⇒ =
(2) T : period of the motion
2 2r mTv qBπ π
= =
2 v qBfr m
ω π= = =(3) ω : cyclotron frequency
7P15-
B q= ×F v B
Magnetic Force on Current-Carrying Wire
( ) mcharges
= ×B
( )B I= ×F L B
charge ms
= ×B
9P15-
Magnetic Force on Current-Carrying Wire
Current is moving charges, and we know that moving charges feel a force in a magnetic field
17P15-
Electric Field Of Point ChargeAn electric charge produces an electric field:
r̂ 2
1 ˆ4 o
qrπε
=E r
r̂ : unit vector directed from q to P
18P15-
Magnetic Field Of Moving ChargeMoving charge with velocity v produces magnetic field:
2
ˆx4
o qr
µπ
=v rB
:r̂r̂
P
unit vector directed from q to P
permeability of free space70 4 10 T m/Aµ π −= × ⋅
19P15-
The Biot-Savart LawCurrent element of length ds carrying current I produces a magnetic field:
20 ˆ
4 rdI rsBd ×
=π
µ
(http://ocw.mit.edu/ans7870/8/8.02T/f04/visualizations/magnetostatics/03-CurrentElement3d/03-cElement320.html)
Figure 28-1
Figure 28-3
Figure 28-5
Figure 28-15
Figure 28-16
Figure 28-19
Figure 28-17
20P15-
The Right-Hand Rule #2
ˆ ˆˆ × =z ρ φ
23P15-
Example : Coil of Radius RConsider a coil with radius R and current I
II
IP
Find the magnetic field B at the center (P)
25P15-
Example : Coil of Radius RIn the circular part of the coil…
ˆd ⊥s r ˆ| d | ds→ × =s r
r̂sd
II
I0
2
ˆ4
dIdBr
µπ
×=
s rBiot-Savart:
024
I dsr
µπ
=
024
I R dR
µ θπ
=
0
4I d
Rµ θ
π=
24P15-
Example : Coil of Radius RConsider a coil with radius R and current I
II
IP
1) Think about it:• Legs contribute nothing
I parallel to r• Ring makes field into page
2) Choose a ds3) Pick your coordinates4) Write Biot-Savart
26P15-
Example : Coil of Radius RConsider a coil with radius R and current I
0
4I ddB
Rµ θ
π=
20
0 4I dB dB
R
π µ θπ
= =∫ ∫
( )2
0 0
0
24 4
I IdR R
πµ µθ ππ π
= =∫
II
Iθ
sd
0 into page2
IR
µ=B
27P15-
Example : Coil of Radius R
Notes:•This is an EASY Biot-Savart problem:
• No vectors involved•This is what I would expect on exam
II
PI
page into2
0
RIµ
=B
29P15-
Group Problem:B Field from Coil of Radius R
Consider a coil with radius R and carrying a current I
WARNING:This is much harder than what I just did! Why??
What is B at point P?
Figure 28-39
Figure 28-37
Figure 28-38
Figure 28-43
Figure 28-42
2P18-
Review:Right Hand Rules
1. Torque: Thumb = torque, fingers show rotation2. Feel: Thumb = I, Fingers = B, Palm = F3. Create: Thumb = I, Fingers (curl) = B4. Moment: Fingers (curl) = I, Thumb = Moment
20P18-
The Biot-Savart LawCurrent element of length ds carrying current Iproduces a magnetic field:
20 ˆ
4 rdI rsBd ×
=πµ
23P18-
Ampere’s Law: The IdeaIn order to have a B field around a loop, there must be current punching through the loop
24P18-
Ampere’s Law: The Equation
The line integral is around any closed contour bounding an open surface S.
Ienc is current through S:
∫ =⋅ encId 0µsB
encS
I d= ⋅∫ J A
Figure 28-6
Figure 28-7
26P18-
Biot-Savart vs. Ampere
Biot-Savart Law
general current sourceex: finite wire
wire loop
Ampere’s law
symmetriccurrent source
ex: infinite wireinfinite current sheet
02
ˆ4I d
rµπ
×= ∫
s rB
∫ =⋅ encId 0µsB
27P18-
Applying Ampere’s Law1. Identify regions in which to calculate B field
Get B direction by right hand rule2. Choose Amperian Loops S: Symmetry
B is 0 or constant on the loop!3. Calculate 4. Calculate current enclosed by loop S5. Apply Ampere’s Law to solve for B
∫ ⋅ sB d
∫ =⋅ encId 0µsB
28P18-
Always True, Occasionally Useful
Like Gauss’s Law, Ampere’s Law is always true
However, it is only useful for calculation in certain specific situations, involving highly symmetric currents. Here are examples…
Figure 28-10
29P18-
Example: Infinite Wire
I A cylindrical conductor has radius R and a uniform current density with total current I
Find B everywhere
Two regions:(1) outside wire (r ≥ R)(2) inside wire (r < R)
30P18-
Ampere’s Law Example:Infinite Wire
I
I
B
Amperian Loop:B is Constant & ParallelI Penetrates
31P18-
Example: Wire of Radius RRegion 1: Outside wire (r ≥ R)
d⋅∫ B s
ckwisecounterclo2
0
rI
πµ
=B
B d= ∫ s ( )2B rπ=
0 encIµ= 0Iµ=
Cylindrical symmetry Amperian CircleB-field counterclockwise
32P18-
Example: Wire of Radius RRegion 2: Inside wire (r < R)
2
0 2
rIR
πµπ⎛ ⎞
= ⎜ ⎟⎝ ⎠
ckwisecounterclo2 2
0
RIr
πµ
=B
Could also say: ( )222 ; rRIJAI
RI
AIJ encenc π
ππ====
d⋅∫ B s B d= ∫ s ( )2B rπ=
0 encIµ=
33P18-
Example: Wire of Radius R
20
2 RIrBin π
µ= r
IBout πµ2
0=
Figure 28-9
34P18-
Group Problem: Non-Uniform Cylindrical Wire
I A cylindrical conductor has radius R and a non-uniform current density with total current:
Find B everywhere
0RJr
=J
35P18-
Applying Ampere’s LawIn Choosing Amperian Loop:• Study & Follow Symmetry• Determine Field Directions First• Think About Where Field is Zero• Loop Must
• Be Parallel to (Constant) Desired Field• Be Perpendicular to Unknown Fields• Or Be Located in Zero Field
39P18-
Multiple Wire Loops
Figure 28-12
Figure 28-14
Figure 28-13
Figure 28-12
40P18-
Multiple Wire Loops –Solenoid
41P18-
Magnetic Field of Solenoid
loosely wound tightly wound
For ideal solenoid, B is uniform inside & zero outside
42P18-
Magnetic Field of Ideal Solenoid
d d d d d⋅ ⋅ + ⋅ + ⋅ + ⋅∫ ∫ ∫ ∫ ∫1 2 3 4
B s = B s B s B s B s
Using Ampere’s law: Think!
along sides 2 and 4
0 along side 3
d⎧ ⊥⎪⎨
=⎪⎩
B sB
n: turn densityencI nlI=
0d Bl nlIµ⋅ = =∫ B s
00
nlIB nIl
µ µ= =/ : # turns/unit lengthn N L=
0 0 0Bl= + + +
44P18-
Group Problem: Current Sheet
A sheet of current (infinite in the y & z directions, of thickness 2d in the x direction) carries a uniform current density:
Find B everywhere
ˆs J=J k
y
45P18-
Ampere’s Law:Infinite Current Sheet
I
Amperian Loops:B is Constant & Parallel OR Perpendicular OR Zero
I Penetrates
B
B
46P18-
Solenoid is Two Current SheetsField outside current sheet should be half of solenoid, with the substitution:
2nI dJ=
This is current per unit length (equivalent of λ, but we don’t have a symbol for it)
47P18-
=2 Current Sheets
Ampere’s Law: .∫ =⋅ encId 0µsB
IB
B
X XX
X
X
XX
X
XXX
X
X
XX
X
XXXXXXXXXXXX
B
LongCircular
Symmetry(Infinite) Current Sheet
Solenoid
Torus
49P18-
Maxwell’s Equations (So Far)
0Ampere's Law:
Currents make curling Magnetic Fields
encC
d Iµ⋅ =∫ B s
Magnetic Gauss's Law: 0
No Magnetic Monopoles! (No diverging B Fields)S
d⋅ =∫∫ B A
0
Gauss's Law:
Electric charges make diverging Electric Fields
in
S
Qdε
⋅ =∫∫ E A