Download Part 2

University of Twente
Department Applied Physics
First-year course on
Electromagnetism
Part II: Magnetism: slides
© F.F.M. de Mul
1
FdM
Magnetic Force on Current Wire
dS
Suppose: total current = I ;
cross section S variable
eT
dl=dl.eT
B
| j | = dI/dS
j = n q v (n = part./m3)
Lorentz force on one charge: F = q v x B
….. per unit of volume : f = nq v x B = j x B
….. on volume dV : dF = j x B. dV
dF = j x B. dS.dl
j = j.eT
dF = j.dS. eT x B . dl
dF = I . dl x B
Straight conductor in homogeneous field:
B
FdM
F = I.eT x B  dl =
F = I. L. eT x B
2
Hall effect
l
I
G
P
Suppose n charge carriers / m3
I
a
d
H
B
v
Q
F
B-field causes deviation of path of charge carriers
Build up of electric field EHall between Q and P: Q+ ; P-
Stationary case: Fmagn = Felec  q v B = q EHall
IB
With j = nqv = I/(ad)  EHall  vB 
nqad
IB
Hall potential : VHall  VQ  VP  EHall . d 
nqa
FdM
3
Magnetic field of a line current
 0 I eT  er
Biot & Savart : dB 
dl
2
4 r
dl=dl.eT

er
dl=dz
Question: Determine B in P
z
O
r
e
R
P
I
(1) var  z : dB 
dB
0 I 1
4 R 2  z 2
Integration over z from - to +
Approach: Current line elements dl
Calculation: eT x er = e ;
tangential component only: dB
 0 I sin 
dB 
dz
2
4 r
R
dz
Result :
2
2
R z
0 I
1
Rd 
(2) var   : dB 
sin  2
2
2
4 R sin 
sin 
FdM
Integration over  from 0 to 
0 I
BP 
eφ
2R
4
Magnetic field of a circular circuit
 0 I eT  er
Biot & Savart : dB 
dl
2
4 r
I
l
er
dB
r
dl
P
dl = dl. eT
FdM
dBy
dB
r
R
Approach: Current line elements dl
R
a
a 
Question: Determine B in P

P
dBy
y
Calculation: eT x er = 1 ;
symmetry: y- component only:
I 1
dBy  0  2 dl cos 
4 l r
dBy 
0 I 1
a
2

a
4 a 2  R 2
a2  R2
0 I . a 2
BP  2 2 3 / 2 e y
2a  R 
5
Magnetic field of a circular solenoid
Radius: a ; Current I
Length: L
Coils: N , or per meter: n
y
P
Question: Determine B in P
Approach: Solenoid = set of circular circuits ;
and for each circuit:
0 I . a 2
BP  2
e
2 3/ 2 y
2a R


R is distance from
circuit to P
L
Each circuit: strip dy; current dI = n.dy.I
dy
L
O
R
y
yP
FdM
0
R=yp-y
a
P
B
 0 .nI .dy.R 2

2 R2  a

2 3/ 2
ey
Result for L   :
y
B = 0 n I ey
Result independent of6 a, L
Magnetic forces between currents
I1
Question: determine F12
I2
eT1
er1
r12 eT2
F12
B12
e 1
1
eT2  e 1  er1
FdM
dFL  I dl  B ; dl  dl.eT
Calculation
B1 2
 0 I1

e 1
2r12
F12   I 2 .eT2  B12 .dl2
2
If L1  L2
Relations:
0 I
B
eθ ; eθ  eT  er
2R
: F12  F21
F1 2
 0 I1 I 2

L2 (e r1 )
2 r12
7
Why is the wire moved by Lorentz force ?
(since inside the wire only the conduction electrons move)
Conductor:
- fixed ion lattice,
- conduction electrons
I
+ + + + + + +
+ + + + + + +
+ + + + + + +
I
electron
Magnetic field B
 plane of drawing
FL
B
v
B
-q
Hall effect: concentration of electrons
(- charge) at one side of conductor
Lattice ions feel a force FE upwards
FdM
This force (= FL ) is electrical !!
-------FE
+
8
Ampère’s Law (1)
Long thin straight wire; current I
I
l
Question: Determine the
“Circulation of B-field”
along circle l
B
r
e
 B  dl
0 I
0 I
l B  dl  l 2r e  e dl  2r 2r
 B  dl   I : Ampere
0
l
9
FdM
Ampère’s Law (2)
 B  dl   I : Ampere
I
0
l
l
B
r
dl
d
r
B
and again :
 B  dl   I
0
c
=r. d
FdM
 B  dl
0 I
c B  dl  c B.r.d  2r r.2
e
c
Question: Determine the
“Circulation of B-field”
along circuit c
Consequences:
1. More currents through c add up ;
2. Currents outside c do not contribute ;
3. Position of current inside c
10
is not important.
B-field from a thick wire
Cylinder: radius a
current : I   j  dS
S
I
 B  dl  0 I  0  j  dS
l
S
Options for current:
I: at surface II: in volume
(suppose: homogeneous)
Use Ampere-circuits (radius r):
0 I
2r
r  a B(r ).2r  0 I
B(r ) 
r  a ( I ) : B(r ).2r  0
B(r )  0
 0 I .r
B (r ) 
2a 2
r 2
( II ) : B(r ).2r  0 I 2
a
11
FdM
Magnetic Induction of a Solenoid
ez
I
er
G B
r
Br : Gauss-box G: total=0
B
c
c
top =bottom  wall=0  Br=0
B : Circuit c (radius r):
Ampere: B .2r=0  B =0
Bz
l 1
a
b
FdM
e
Radius: R ; Current: I
Length: L >> R
Coils: n per meter
Components: Bz Br B
2
Bz : Circuit 2: B(a)=B(b)=0
Circuit 1: Ampere: Bz l=0 n I l
Result: inside: B = 0 n I ez
outside: B = 0
12
Symmetries for Ampere’s Law
 B  dl  0 I  0  j  dS
l
Wire,
 long
Solenoid,
 long
S
Plane,
 extending
Toroid,
along core line
13
FdM
Magnetic Pressure
z
y
Plane layer L with current density j
L
x
BL
BL
Be
dFL
l
db
Suppose we add an external
field Be , with Be = BL , so that the
total field behind the layer = 0
Lorentz force on
: (FL=I.L.eT  B):
dFL= (j.db).dl  ½0 j ez
dFL= ½0 j2.db.dl ey
j
dl
B-field of the layer (circuit l ):
2 BL l = 0 j l  BL =  ½0 j ez
Be
j
Magnetic pressure: P = ½0 j2 = ½Be2 /0 .
Example: this situation is met at the wall of a (long) solenoid. Then
the pressure is outward, thus maximizing the cross section area.14
FdM
Vectorpotential A
Flow
tube
v’
dv’
r
er
P (x,y,z)
j
μ0
B
4

v'
B = rot A
x’,y’,z’
j  er
dv' 
2
r
μ0
4

v'
Question: determine A in P
from Biot-Savart for B
r = f (x,y,z, x’,y’,z’ )
μ0
1

 1
j     dv' 
    j.dv'

r
4 v '  r 

0   j    j 
u.dv  d (uv)  v.du  B       
dv'

4 v '   r  r 
  f ( x, y , z ) ; j  f ( x ' , y ' , z ' )    j  0
 0
 B   
 4
FdM
j 
dv'

r
v'

}
B
0
4
  j 
v'    r  dv'
0
j
 A
dv' in general,

4 v ' r
0 I
1
and A 
dl for a circuit
15

4 loop r
Magnetic Dipole (1): Far Field
r’
Far field: r’ << rP
1

r
1

rP
erP rP
r
O 
I
dl
1
2
rP
 r' 
r'
1     2 cos 
rP
 rP 
0 I
Thus : AP 
dl

4rP
FdM
0 I
A
4
dl
r
Goal: expression in rP in stead of all
r-values over circuit
1
1

2
r
rP  r '2 2rP r '.cos 
P
Monopole-term
=0
 r'

1  cos   .....
 rP

0 I

r '.cos  .dl
2 
4rP
Dipole-term
16
Magnetic Dipole (2): Dipole Moment
0 I
AP 
r '.cos  .dl
2 
4rP
z
en 
P
rP

erP
y
O
x
r’
Ax 

Assume: circular circuit,
with radius R << rP
Assume: Y-axis along OP’
 0   R sin  



r '.cos   er  r '   sin    R cos  
 cos 

0



P’
0 I
( R sin  cos  )( R.d )(  cos  ) 
2 
4rP
0 I 2
0 I 2
2
R
sin


cos

.
d


R sin  .
2
2

4rP
4rP
Ay  Az  0

FdM
 R sin  cos
Define: dipole moment:
m = I. Area. en = IR2.en
0
A
m  er
2
4rP
17
Magnetization and Polarization
Magnet = set of “elementary
circuits” ; n per m3
V=SL
Total surface current = Itot
Total magnetic moment = Itot S en
Def.: Magnetization M =
Polarization
L
en
S
magnetic moment / volume =
surface current / length
Magnetization
Dipole moment: p [Cm]
Dipole moment: m [Am2]
Polarization P = np [Cm-2]
= surface charge / m2
Magnetization M = nm [Am-1]
= surface current / m
18
FdM
Magnetic circuit: Torque and Energy
b
F
F
b.sin 
I
l
m

B
F

B
m
F
Torque: moment:  = F . b sin  = I B l b sin  = I B S sin 
Magnetic dipole moment: m = I S en
Torque: moment:  = m x B
Potential energy: min for  = 0 ; max for  =  
Potential energy: Epot = -MB. cos  = - m . B
19
FdM
Electret and Magnet
P
Electret
E = (D-P)/0
D = 0 E + P
 E  dl  0
 D  dS  Q
f
0
Magnet
E
D
H
B
+++
--M
H = B / 0 -M
B = 0 H + M
 H  dl  I  0
 B  dS  0
f
20
FdM
B- and H-fields at interface
1 1
B1
B2
Gauss box
Circuit:
2
Given: B1 ; 1 ; 2
Question: Calculate B2
Needed: “Interface-crossing relations”:
 H  dl  I free and  B  dA  0
2
Relation H and B: B = 0 r H
: B1.Acos 1 - B2.Acos 2 =0  B1 = B2
: no I : H1.Lsin 1 - H2.Lsin 2 =0
B1 1 A B2 1 A

H1 tan 1 L H 2 tan  2 L
 H1// = H2//
 r1 tan 1

 r 2 tan  2
21
FdM
Toroid with air gap
Core
line;
radius R;
L = 2R
I
Relations:
d
Result:
Suppose:
- toroid solenoid: R, L, N, r ;
- air gap, r =1 , width d <<R;
g = gap ; m = metal
Question: Determine Hg in gap
 H  dl  I
 B  dA  0
f
B  0 r H
Hg d + Hm (L-d) = N I
Bg =Bm
Bg=oHg ; Bm =0r Hm
Hg
 r NI
Hg 
and H m 
L  d (  r  1)
r
22
FdM
Induction: conductor moves in field
B
L1(t)
S1
L2 (t+dt )
S2
v
v.dt 
dl
   

top
bottom
wall

wall

Gauss box: top lid S1 in L1,
bottom lid S2 in L2
Area = v dt dl sin
 B  dS  0  
 (t )  (t  dt )  d(t )
 B  e [v.dt.dl.sin  ]   B  (v.dt  dl ) dt.  (v  B)  dl
n
wall
L
d(t )

   (v  B)  dl    E n  dl
dt
L
L
FdM
Suppose: circuit L moves with
velocity v through field B
Question: Show equivalence:
d
Vind   E  dl    B  dS
dt S
L
L
En = non-electrostatic field
23
Induction: Faraday’s Law
B
S
B
V0
c
Static:
I0
 E  dl 0
c
Vind
S
c
Iind
Dynamic: Suppose  changes 
Vind  0  Non-electrostatic field EN
d
d
Vind   E N  dl  
   B  dS
dt
dt S
c
Consequence: Let circuit c shrink, with keeping S.
c
For closed surface :
c
S
S
E
c
FdM
N
 dl  0   B  dS  0
S
24
Induction in rotating circuit frame

v
b
v
l
Bext


Bext
v
en
= t ; v= ½b
en
Induction potential difference :
(I) Using Lorentz force: Vind   E N  dl   (v  Bext )  dl 
=EN
2lvBext sin   2lvBext sin t
(II) Using flux change:
d
d
d
   Bext  dS   Bext [lb cos ] 
dt
dt
dt
Bext lb sin t  2 Bext lv sin t
Vind  
25
FdM
Electromagnetic brakes
Why is a conducting wire decelerated by a magnetic field?
Case I: switch open
B
v FL
P
Q
FE
B
vL
Q
I
Case II: switch closed
v FL
P
FL moves electrons: vL ,
which causes I and FL2 ,
which causes electric field,
FL2
Fions
FdM
Electrons feel FL
Potential difference VPQ :
P - , Q + (= Hall effect)
FL counteracted by FE
which will act on positive
metal ions: Fions.
26
Coupled circuits (1): M and L
1
2
I1
I2
Suppose: circuit 1 with current I1
Part of flux from 1 will pass
through 2 : 21
21 ~ I1
Definition: 21 = M21.I1
M : coefficient of mutual induction : “Mutual Inductance” :
[ H ] = [ NA-1m-1.m2.A-1] = [ NmA-2 ]
Suppose: Circuit 2 has current as well: I2
Flux through 2: 2= 21 + 22 = M21 I1 + M22 I2
M22 = L2 L : coefficient of self-induction “(Self) Inductance”:
M and L are geometrical functions (shape, orientation, distance etc.)
27
FdM
Coupled circuits (2): toroid
N1 , I1
r
N2
Question: determine M21
Cross
section S
Core line: L
Flux from 1 through S : S = BS = 0 r N1 I1 S / L
Linked flux from 1 through 2: 21 = N2 S = 0 r N1 N2 I1 S / L
Coefficient of mutual induction:
M21 = 0 r N1 N2 S / L
This expression is symmetrical in 1 and 2:
This result is generally valid: Mij = Mj i
FdM
M21 = M12
28
Coax cable : Self inductance
I
Radii: a and b (a<b)
Length: l
Current: I ; choice: inside = upward
B
Gauss
surface
0 r I
Ampere:
B(r ) 
B-field tangential
2r
Flux through circuit:
0  r I
0  r I . l b
 B   B  dS   dl 
dr 
ln
2r
2
a
S
0
a
l
b
Self-inductance (coefficient of self-induction): per unit of length:

b
L  0 r ln
(Compare with capacity of C  2  ln b  1
0 r
2
a

 a
coax cable, per meter:
29
FdM
Magnetic Field Energy
Circuit c in XY-plane
z
H-field lines:
circles around circuit.
I
H
y
A
A-surfaces: everywhere
 H (or B) -field lines
 through A:    B  dA
A
x
Magnetic energy:
Em= ½ L I2 =
 12  B  dA H  dl  12   ( B  H )(dA  dl )  12  B  H dv
A
c
dl
dA
A c
Compare: electric energy: Ee =
V
1
2
 D  E dv
v
30
FdM
Maxwell’s Fix of Ampere’s Law
Induction :
 En  dl  
c
d
B  dS

dt S
Changing B-field
causes E-field
Question: Does a changing E-field cause a B-field ?
D
L
Suppose: chargeing a capacitor using j
 B  dl    j  dS
L encloses S1 :
j
0
L
 B  dl    j  dS
L encloses S2 :
S1
0
L
S2
S3
S2
Volume enclosed  j  dS   dQ f   d D  dS
S
dt
dt 
by S2 and S3:
S
2
L encloses S3 :
3
 B  dl  0
L
FdM
S1
d
D  dS

dt S3
d
In general :  B  dl    j  dS    D  dS
dt S
L
S
31