Magneto Statics

16
7/21/2019 Magneto Statics http://slidepdf.com/reader/full/magneto-statics-56d95cacc526e 1/16 Ch5 – 57  AAE 590E Magnetic Force Magnetic 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 magnetic fields is: Lorentz Force  ! F B  = q  !  v  ! ! B B  = ! F B  = qvB " sin  ! F  = ! F E  + ! F B  = q ! E  + !  v  ! ! B ( ) ! ! ! ! ! ! ! ! ! ! ! !  !  v +q  ! F B ! B

description

physics

Transcript of Magneto Statics

Page 1: 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

 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! 

"  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

 

!

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

""    =

µ 0

2# !I

ss d$ 

""    =

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

""    =  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

bc

!

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

""    =  µ 0Ienclosed

 

!B

 ! d

!

l

P ""   =

 µ 0Ienclosed

  = 0   Ienclosed

  = 0

 

!B !d

!

l

""    =  µ 0Ienclosed

#   Ienclosed

  =   J dA

a

"    =   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( )

 AL 

 AL 

 Amperian Loop (AL)