magneto statics
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Ch5 – 57
AAE 590E
Magnetic ForceMagnetic Force
!Consider a charged particle moving through a magnetic field.
!The “magnetic force” acting on a moving charged particle is expresses as:
Lorentz Force Law:
!Net force on a charged particle due to the presence of electric and magneticfields is:
Lorentz Force
!
FB = q
!
v !!
B
F B =
!
FB = q vB " sin#
!
F =!
FE +
!
FB = q
!
E +!
v !!
B( )
! ! ! !
! ! ! !
! ! ! !
!
v
+q
!
FB
!
B
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Ch5 – 58
AAE 590E
Motion of Charged ParticlesMotion of Charged Particles
!Cyclotron Motion
!Newton’s Law:
!Cyclotron Formula:
!Momentum of Particle
!Radius
!Period
!Frequency
! Angular Speed/Frequency
!Helical Motion
!Controlling the radius of helix:
!Controlling the pitch of helix:
!
F = m
!
a =
!
Fcentripetal
q vB = ma = m
v2
r
p = mv = q rB
r =mv
qB=
p
qB
T =2! r
v=2! m
qB
f =1
T =
qB
2! m
! = 2" f =
qB
m
! ! !
! ! !
! ! !
! ! !
!
v
+q
!
v
r
v
! = v "sin#
v! = v !cos"
!
FB
q is in a plane to B.
!
B
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Ch5 – 59
AAE 590E
ExampleExample
!Cycloid Motion:
!Let’s consider the motion of a charged particle in a magnetic and electric field. We assume that the electric field is perpendicular to the magnetic field.
x
y
z
!
E
!
B +q
Charge is
initially at
rest! " Charged particle is at rest, v = 0 , F
mag = 0;
" Electric field accelerates charge particle, v is increasing in z–direction;
" Since the particle moves (at a certain speed) in a magnetic field, the particleexperiences a magnetic force to the right;
" The faster is goes, the stronger F mag
becomes;
" Eventually, it curves the particle back around towards the y-axis;
" Now particle moves against electric force and starts slowing down, v is
decreasing;
" With decreasing speed, the magnetic force decreases, electric force takes over till
the charge comes to rest at the y–axis a certain distance away form the origin.
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Ch5 – 60
AAE 590E
ExampleExample
!Equations of Motion for Cycloid Motion:
!Equations of Motion# Lorentz Force
# Newton’s Law Velocity
! Velocity
!Magnetic Field
!General Solution
!Boundary Conditions
!
FL =
!
FE +
!
FB = q
!
E +!
v !!
B( )
!FN = m
!a = m 0 "" y y ""z z( )
! v = 0 " y y "z z( )
!
B =ˆx 0 0( )
! v !
!B = B "z y "B " y z
!
FL =
!
FN
q E z +B !z y !B ! y z( ) = m !! y y + !!z z( )
!! y =qB
m!z = ! !z
!!z = !
E
B" ! y
# $ %
& ' (
y(t) = C1 cos! t + C
2 sin! t +
E
Bt + C
3
z(t) = C2 cos! t "
C1 sin! t + C4
! y(0) = !z(0) = 0, y(0) = z(0) = 0,
y(t) =E
! B! t " sin! t( ) = R ! t " sin! t( )
z(t) =E
! B1" cos! t
( )= R 1" sin! t
( )
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Ch5 – 61
AAE 590E
-0.04
-0.02
0
0.02
0.04
0 50 100 150 200
Acceleration ay
Acceleration az
c
e
o
y t
ExampleExample
!Equations of Motion for Cycloid Motion
-1
-0.5
0
0.5
1
0 50 100 150 200
Velocity Vy
Velocity Vz
e
o
t
y
y t
0
5
10
15
20
0 50 100 150 200
T
r
a
j
e
c
t
o
r
y
z
y t
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Ch5 – 62
AAE 590E
Work & Magnetic Field Work & Magnetic Field
! As we have done with the electric field, let’s examine the work done by the
magnetic field. Let’s consider a moving charge q at a velocity v moving
through a magnetic field.
!Definition of mag. Work:
!Conclusions:!Magnetic forces do NO work!
!Magnetic forces may alter the direction, in which a charged particle moves, but
they cannot alter the speed of the charged particle!
! ! !
! ! !
! ! !
! ! !
!
v
!
FB
d!
l
!
B
dW mag =
!
Fmag
!
d
!
l
!
Fmag
= q !
v "!
B( )d!
l =!
v dt
dW mag
= q !
v !!
B( ) "!
v dt = 0
= 0
!
v !!
B( ) " !
v
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Current
AAE 590E
In general, current is charge per unit time:
Average: I =Q
t A =
C
s
" # % &
Time varying: i =dq
dt ! Charge q = dq = i dt
0
t
! !
We distinguish between the following kinds of current:
! Line Current: Line Charge ! =Q
L traveling at velocity
!
v
Current:!
I = ! !
v
Magnetic Force:!
FB =
!
v ! !
B( )" dq =!
v ! !
B( )" # dl =!
I ! !
B( )" dl
!
FB= I d
!
l ! !
B( )"
! Surface Current Surface Charge Density ! =Q
A traveling at velocity
!
v
Surface Current Density!
K !d!
I
dl"
,!
K =! !
v
Magnetic Force:!
FB =
!
v ! !
B( )dq =!
v ! !
B( ) # da
!
FB=
!
K ! !
B( )" da
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Current
AAE 590E
With
!
J!
d!
I
da" , we can calculate the current crossing a surface as:
I = J da!
S
=
!
J # d!
a
S
.
Further according to Gauss’/Divergence Theorem, the total charge per unit time leaving a volume
can be expressed as:!
J ! d!
aS
" = # ! !
J( )V
" d$ .
Conservation of Charge: If total charge in a volume changes by !Q, then exactly that amount of
charge must have passed in or out through the surface of the volume.
Charge in a volume is defined as: Q(t) = ! (!
r,t)V
d#
Current flowing out through the surface boundary S is
!
J ! d
!
aS
:
dQ
dt= !
!
J " d!
aS
#
This minus sign reflects the fact that an outward flow
decreases the charge LEFT in the volume!!
Combining the above equations:dQ
dt=
d
dt!
V
" d# =$!
$ tV
" d#
Applying Gauss’ Theorem:
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Ch5 – 65
AAE 590E
Summary Summary
!Point Charge:
!Biot–Savart Law
!Biot-Savart Law describes the relationship between a moving charge and magnetic
field it induces.
!Lorentz Force:
!Magnetic Fields due to Currents:
!Straight Wire
! Wire Loop
!Magnetic Force on Current-Carrying “Structures”:
!
B =
µ 0
4!
"q
!
v # r
r 2 T $% &'
!
F =!
FE +
!
FB = q
!
E +!
v !!
B( )
!
Bz = B
z0 0 1( ) =
µ 0
2
IR2
R2+ z
2( )3
2
z
Bwire
=
µ 0
4!
"
L I
r " r 2+ (L 2)
2
Bwire
=
µ 0
4!
"
2 I
r
!
FB = I d
!
l !
!
B( )" !
FB = I
!
L !
!
B
!
FB =
!
K ! !
B( )" da
!
FB =
!
J ! !
B( )" d#
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Ch5 – 66
AAE 590E
DivergenceDivergence
!Biot-Savart gives the magnetic field
!at a point P
!due to a volume current at source S
!B is a function of (observation) point: (x, y, z).
! J is a function of (source) position: (xS, yS, zS).
Source
d!
S
x
y
zP
!
rscript
!
r
!
rS
!
J
!
r = x y z( )
!
rS = x
S y
SzS( )
!
B(!
r ) =µ 0
4!
!
J(
!
rS ) " rscript
r script
2d#
S$
!
J
!
B
d!
S = dx
Sdy
Sdz
S
!
rscript
= ( x ! xS)x + ( y ! y
S) y + (z ! z
S)z
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Ch5 – 67
AAE 590E
DivergenceDivergence
! As we have done it with the electric field, let’s examine divergence of
magnetic fields, where "* is with respect to coordinates of observation point.
! " !
B =
µ 0
4#
! " !
J $rscript
r script
2
%
& '
(
) * d+ S,
! " !
J #rscript
r script
2
$
% &
'
( ) =
rscript
r script
2 " ! #
!
J(!
rS)( ) *
!
J " ! #rscript
r script
2
$
% &
'
( )
=0, since J
does NOT depend
on observation
position (x,y,z).
=0!
J(!
rs
)
! "
!
B =
0
*! ="
" xx +
"
" y y +
"
"zz
Source
d!
S
x
y
zP
!
rscript
!
r
!
rS
!
J
d!
S = dx
Sdy
Sdz
S
!
rscript
= ( x ! xS)x + ( y ! y
S) y + (z ! z
S)z
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Ch5 – 68
AAE 590E
!To investigate the curl of a magnetic field, let’s consider current in a straight
line:
!Biot-Savart applied to long straight line:
! ! ! ! !
! ! ! ! !
! ! ! ! !! ! ! ! !
CurlCurl
!
I
dlS
P
!
rscript
!
!
!
I
!
B
P
!
s
B =
µ 0
4!
"
2I
s
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Ch5 – 69
AAE 590E
!Magnetic Field in Cylindrical Coordinates (s, ! , z):
!Stokes’ Theorem:
! Ampere’s Law:
CurlCurl
!B !d
!l
P
"" =
µ 0
2# !I
ss d$
P
"" =
µ 0I
2# d$
0
2#
" = µ 0I
!
B = µ 0I
2! sˆ
d!
l = ds s + sd" + dz z
!
B !d
!
lP ""
=
# $
!
B( ) !d!
a S"
= µ 0
!
J !d!
a S"
= µ 0Ienclosed
!"!
B = µ 0
!
J
!B !d
!
l
P
"" = µ 0Ienclosed
Integral Form Differential Form
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Ch5 – 70
AAE 590E
Application of Ampere’s Law Application of Ampere’s Law
!Magnetic Field of an infinite uniform Surface Current:
! Ampere’s Law
!NOTE: Magnetic field independent of distance from plane, just like electric field
of a uniform surface charge!
!B !d
!
l"" = 2Bl = µ 0Ienc
= µ 0K l # B =
1
2µ 0K
!
B =
+1
2µ 0K
!
y z < 0
!1
2µ 0K
!
y z > 0
"
#$$
%$$
x
y
z
!
K = K x
!
B
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Ch5 – 71
AAE 590E
Application of Ampere’s Law Application of Ampere’s Law
!Magnetic Field of an Ideal Very Long Solenoid:
!Magnetic Field of a Toroid
!
B =
µ 0N I
2! r
inside toroid
0 outside toroid
#
$%
&%
z
x #
N = Number of Windings of Toroid
Amperian Loop
Position 1
Position 2
!B !d
!l"" = BL = µ
0Ienc
L
a
bc
d
!
B !d!
l
d
a
" = BdaL = 0
!
B =
µ 0nI
!
z inside solenoid
0 outside solenoid
!"#$
n =Number of turns
enclosed by Amperian Loop
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Ch5 – 72
AAE 590E
ExampleExample
! Figure to the right shows the cross section of a long conducting cylinder with inner radius
a=2 cm and outer radius b=4 cm. The cylinder carries a current out of the page, and the
current density in the cross section is given by J=cr 2
, with c=3!
106
A/m2
and r in meters.! What is the magnetic field B at a point r < a, a < r < b (r=3 cm), and r > b from the central
axis of the cylinder?
! Ampere’s Law:
! Solution:
!r < a:
! a < r < b r=3 cm):
! r > b
!B !d
!
l
P
"" = µ 0Ienclosed
!B
! d
!
l
P "" =
µ 0Ienclosed
= 0 Ienclosed
= 0
!B !d
!
l
P
"" = µ 0Ienclosed
# Ienclosed
= J dA
a
r
" = cr 22$ r dr
a=2cm
r =3cm
" =
$ c
2r 4 % a4( )
Ienclosed
= 3.06 A
# B !2$ r = µ 0
$ c
2r 4 % a4( )
B =
µ 0c
4! r
4 % a4( )r
= 2 !10%5T
Ienclosed
= J dA
a
b
! =
" c
2b4 # a
4( ) =11.3 A $ B =
µ 0c
4% b
4 # a4( )
r
AL
AL
Amperian Loop (AL)