Transcript PHYS_2326_040909

There will be a quiz next Tuesday, April 14
From now until the end of the semester, we will have 2 problem
solving session per week.
Next week, we will have one on Monday, April 13 at 1:00 PM and
one on Thursday, April 16 at 1:00 PM.
The electron spins on its axis, giving rise to a electron
current in the direction of rotation. Think of the electron
as a ball with charge distributed over its surface. When
the ball spins, that charge is set in motion around the
electron's spin axis, resulting in a magnetic field specific
to the electron.
The electron is like a magnetic dipole, a miniature
magnet, with a north end and a south end.
In most substances, electrons spin in random directions - magnetic fields cancel. For iron
and other magnetic substances, the spin magnetism is not canceled. Can be permanently
magnetized by placing in strong magnetic field and permanently aligning atoms - can be
demagnetized by dropping magnet and jostling atoms out of alignment.
Electromagnetic produced by wrapping coil around
iron bar - magnetic field produced that aligns atoms
in bar - more coils or more current - larger magnetic
field and greater atomic alignment
Magnetic Dipoles and How Magnets Work
The Direct-Current Motor
Magnetic Field of a Moving Charge
0 qv sin 
B
4 r 2

B
 
0 q v  r
4 r 2
Magnetic field of a point charge
moving with constant velocity

1

1
0
T

m
/
A
;



4

c
0

7
00 2
Example: Force between two moving protons
Find the ratio of electric and magnetic forces on the protons
1 q2
F
E
40 r2
0 qv
B
k
2
4
r

Magnetic field of the lower proton at
the position of the top one



F

qv
( )B
B

4
r
22

q
0 v
F
j
B
2
2
F
2 v
B


v 2
00
F
c
E
Magnetic Field of Current Element
The Biot-Savart law.
d
Q

n
q
A
d
l
f
l
o
w
w
i
t
h
v
e
l
o
c
i
t
y
v
d
For element of a (fine) wire:
 I dl  rˆ
dB  0
4 r 2
constant permeability of free space:
For the whole "circuit":
 I dl  rˆ
B 0  2
4
r
For arbitrary distribution of charge flow:
 j(1)  rˆ12
B(2)  0 
dV1
2
4
r12
(rˆ12 is from point 1 to point 2)
Magnetic field around a straight
wire
For the
fieldmagnitude
:
0I  sin dx
B

2

4  r
a
ad
[r 
; x  acot; dx 2 ] 
sin
sin 
0I 
0I

sin

d


4a 
2a
0
(where
ais thedistance
fromthewire)
Magnetic Field of Two Wires
Field at points on the x-axis
to the right of point (3)

I
0
B

;
1
2

(
x

d
)

I
0
B

;
2
2

(
x

d
)

I
d
0
B

B

B

t
o
t
a
l
2 1
2 2

(
x

d)
Magnetic field outside of a conductor pair falls off more rapidly
Magnetic field of a circular arc
For the field magnitude at O :
0 I
0 I
B
ds 
R
2 
2
4R
4R
0 I

4R
