Transcript Physics 2102 Spring 2002 Lecture 8

Physics 2102
Jonathan Dowling
Lecture 23: WED 11 MAR
Ampere’s law
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André Marie Ampère
(1775 – 1836)
Exam 02 and Midterm Grade
Exam 02 Average: 63/100
Q1&P1, Dr. Schafer, Office hours: MW 1:30-2:30 pm, 222B Nicholson
Q2&P2, Dr. Gaarde, Office hours: TTh 2:30-3:30 pm, 215B Nicholson
Q3&P3, Dr. Lee, Office hours: WF 2:30-3:30 pm, 451 Nicholson
Q4&P4, Dr. Buth, Office hours: MF 2:30-3:30 pm, 222A Nicholson
You can calculate your midterm letter grade, as posted on paws, via this handy formula used by all
sections:
[(MT01+MT02)*500/200+HWT/761]*30]/530*100 = Midterm Score
Here MT01 and MT02 are your scores on the first and second midterms and HWT is your total
homework points for HW01-07 only.
Once you have this Midterm Score you then can get your letter grade via:
A: 90-100
B: 80-89
C: 60-79
D: 50-59
F: below 50.
Details on the grading policy are at: http://www.phys.lsu.edu/classes/spring2009/phys2102/
I have posted both midterm exam grades and your midterm class score on WebAssign.
Forces Between Wires
Magnetic field due to wire 1
where the wire 2 is,
L I1
I2
F
B1 
0 I1
2 a
Force on wire 2 due to this field,
F21  L I 2 B1
a
Neil’s Rule:
Same Currents – Wires Attract!
Opposite Currents – Wires Repel!
0 LI1 I 2

2 a
Summary
• Magnetic fields exert forces on moving charges:
• The force is perpendicular to the field and the velocity.
• A current loop is a magnetic dipole moment.
• Uniform magnetic fields exert torques on dipole moments.
• Electric currents produce magnetic fields:
•To compute magnetic fields produced by currents, use BiotSavart’s law for each element of current, and then integrate.
• Straight currents produce circular magnetic field lines, with
amplitude B=0i/2r (use right hand rule for direction).
• Circular currents produce a magnetic field at the center (given by
another right hand rule) equal to B=0iF/4r
• Wires currying currents produce forces on each other: Neil’s
Rule: parallel currents attract, anti-parallel currents repel.
Remember Gauss Law for E-Fields?
Given an arbitrary closed surface, the electric flux through it is
proportional to the charge enclosed by the surface.
q
q
Flux = 0!
F

Surface
  q
E  dA 
0
New Gauss Law for BFields!
No isolated magnetic poles! The magnetic flux through any
closed “Gaussian surface” will be ZERO. This is one of the
four “Maxwell’s equations”.
B

d
A

0

There are no SINKS or SOURCES of B-Fields!
What Goes IN Must Come OUT!
Ampere’s law: Closed
Loops
r
—
 B  ds  0ienclosed
LOOP
The circulation of B
(the integral of B scalar
ds) along an imaginary
closed loop is
proportional to the net
amount of current
traversing the loop.
i4
i2
i3
i1
ds
 
 B  d s  0 (i1  i2  i3 )
loop
Thumb rule for sign; ignore i4
If you have a lot of symmetry, knowing the circulation of B
allows you to know B.
Sample Problem
• Two square conducting loops carry currents
of 5.0 and 3.0 A as shown in Fig. 30-60.
What’s the value of ∫B∙ds through each of
the paths shown?
Path 1: ∫B∙ds =
•(–5.0A+3.0A)
Path 2: ∫B∙ds
=
•(–5.0A–5.0A–3.0A)
Ampere’s Law: Example 1
• Infinitely long straight wire
with current i.
R
• Symmetry: magnetic field
consists of circular loops
centered around wire.
• So: choose a circular loop C
-- B is tangential to the loop
everywhere!
• Angle between B and ds =
0. (Go around loop in same
direction as B field lines!)
 
B

d
s


i
0

C
Bds

B
(
2

R
)


i
0

C
Much Easier Way to Get B-Field
Around A Wire: No Cow-Culus.
 0i
B
2R
Ampere’s Law: Example 2
• Infinitely long cylindrical
wire of finite radius R
carries a total current i
with uniform current
density
• Compute the magnetic
field at a distance r from
cylinder axis for:
– r < a (inside the wire)
– r > a (outside the wire)
Current out
of page,
circular field
lines
 
 B  ds  0i
C
i
Ampere’s Law: Example 2 (cont)
 
B

d
s


i
0

Current out of
page, field
tangent to the
closed
amperian loop
C
B(2r )  0ienclosed
 0ienclosed
Need Current Density J!
B
2r
2
i
r
2
2
ienclosed  J (r )  2 r  i 2
R
R
 0ir
For r>R, i =i, so
For r < R
B
2
B= i/2R = LONG WIRE!
2R
enc
0
Ampere’s Law: Example 2 (cont)
B
 0i
2 R
O
R
r
rR rR
B  r B 1/ r
rR
 0 ir
B
2
2 R
rR
0i
B
2 r
Outside Long Wire!
Solenoids
 
 B  ds  0ienc
 
 B  ds  0  B h  0  0
ienc  iN h  i ( N / L)h  inh
 
 B  ds  0ienc  Bh  0inh  B  0in
The n = N/L is turns per unit length.