Download magnetic field(磁场)

Chapter 28 Magnetic Field and
Magnetic Forces
Physicist: Hans Christian Orsted
Born: 14 August 1777
Died: 9 March 1851 Dane
What Orsted saw...
Magnetical Field
Intensity of
Magnetic field
Biot-Savart
law
Magnetic
Force
To
Moving
Circular
law
with
Magnetic
material
To Coil
Particle
Guass’s
law
Interaction
Hall
Effect
Key words
permanent magnet(永磁体)
magnetic monopole(磁单极)
magnetic field(磁场)
tesla(特斯拉)
magnetic field line(磁场线)
gauss(高斯)
magnetic flux(磁通量)
weber(韦伯)
magnetic flux density(磁通密
度)
Key words
cyclotron frequency(回旋频率)
mass spectronmeter(质谱仪)
isotope(同位素)
magnetic dipole(磁偶极子)
Hall effect(霍耳效应)
magnetic moment(磁矩)
solenoid(螺线管)
magnetic dipole moment(磁偶极矩)
28.1 Magnetism
• Permanent magnets.
• Magnetic poles: a north pole(N-pole) and a
south pole(S-pole).
• Compass.
• Opposite poles attract each, and like poles
repel each other.
• There are not magnetic monopoles.
28.2 Magnetic Field
• A moving charge or a current creates a
magnetic field in the surrounding space(in
addition to its electric field).
• The magnetic field exerts a force F on any
other moving charge or current that is
present in the field.
Magnetic field (cont)
• Magnetic field (B) is a vector field.
• At any position the direction of B is
defined as that in which the north pole of a
compass needle tends to point.
• For any magnet, B points out of its north
pole and into its south pole.
28.3 Magnetic Force on a
Moving Particle
• Consider a particle of charge q and velocity v at
a point in a magnetic field with induction B.
• The magnetic force exerted on the charge is
given by the definition of B,
F  q  B  q B sin 

 
F  qv  B
• Units are those of
force
N
1 1


NA
m  Tesla (T ).
1
charge  velocity C (ms )
1G  10 4 T
Magnetic force on a moving particle (cont)
Orders of Magnitude
•
•
•
•
•
•
•
Magnetic field at surface of Earth 10-4T.
Bar magnet 10-2T.
Write head for hard disk 10-1T.
Lab electromagnet 1T.
Superconducting magnet 101T.
Interior of atoms, up to 101T.
Surface of a pulsar(脉冲星) 108T.
28.4 Magnetic Field Lines
Magnetic flux
• Basic concept is as electric field lines but:
– they have the direction a compass needle
would point to at each location;
– tangent to the field lines gives the direction of
B at a point;
– they never cross and
– they form closed loops - a continuous field.
Magnetic flux
 
d B  B dA  B cos dA  B  dA
 
 B   B cos dA   B  dA
•
Magnetic flux is a scalar
quantity.
• This unit is the Weber
(1Wb)
1Wb=1T· m2=1N· m/A
28.5 Gauss’s Law for Magnetism
• The total magnetic flux through a closed
surface is always zero.
 
B

d
A

0

d B
B
dA
(magnetic flux through
any closed surface).
Magnetic field B is sometimes called
magnetic flux density.
28.6 Moving Charges in B Fields
•
•
•
•
Uniform field of B.
Non-uniform field of B.
Combined E and B fields.
Force on a current carrying conductor.
Basic Motion in Magnetic Field
• The velocity of a charged particle can be divided
into two components one parallel and one
perpendicular to the field.
• The perpendicular component gives circular
motion in a plane perpendicular to B.
• The parallel component gives constant linear
motion along B (no force).
• Combination is a helical path.
Uniform Field of B












F














B


  
F 
+


O



+









S

F
+




P







Uniform Field of B (cont)
• If v is perpendicular to B
– F=qvB the force is perpendicular to the velocity
giving circular motion.
– Modulus of force is qvB=mv2/r  r=(mv)/(qB).
– Define cyclotron frequency
1
v
qB
fc  

.
T 2r 2m
• Independent of speed and radius of orbit, particles of
same q/m have same fc.
Uniform Field of B (cont)
Uniform Field of B (cont)
• If v is not perpendicular to B.


 



F  q  B  q per   par  B  q per  B
• So we have circular motion from the
perpendicular component.
• The parallel component gives constant linear
motion along B (no force).
• Combination is the helix.
Non-Uniform field of B
• Effect can be to confine particles
• Examples magnetic bottles and van Allen belts.
Charged Particles from Sun Enter Earth’s
Magnetic Field
Example: Helical particle motion
q  1.60 1019 C
m  1.67 1027 kg
B  0.500T
At t=0, the proton has velocity
components vx=1.50x105m/s, vy=0,
and vz=2.00x105m/s
• At t=0, find the force on the proton ant its acceleration.
• Find the radius of helical path, the angular speed of
the proton, and the pitch of the helix (the distance
traveled along the helix axis per revolution).
Example: Helical particle motion (cont)
• The magnetic force on the proton is



 
F  q  B  q  x iˆ   z kˆ  Biˆ  q z Bˆj





 1.60 10 19 C 2.00 105 m s 0.500T  ˆj
14
 1.60 10 N ˆj
• The acceleration is

 F 1.60 1014 N ˆ
12
2 ˆ
a 
j

9
.
58

10
m
s
j
 27
m 1.67 10 kg


Example: Helical particle motion (cont)
• The force is perpendicular to the velocity, so the
speed of the proton does not change.
• The radius of the helical trajectory is given by



m z 1.67 10 27 kg 2.00 105 m s
3
R


4
.
18

10
m
19
qB
1.60 10 C 0.500T 


The angular speed is

qB
m

1.60 10


C 0.500T 
7

4
.
79

10
rad s
 27
1.67 10 kg
19
Example: Helical particle motion (cont)
• The time required for one revolution
(the period) is
2
2
7
T


1
.
31

10
s
7 1
 4.79 10 s
• The pitch is the distance traveled along the xaxis during this time, so



d   xT  1.50 105 m s 1.31107 s  0.0197m
28.7 Combined E and B Fields
 

  
F  q E B

• The velocity selector.
– Uniform, perpendicular B and E; v
perpendicular to both.
– Balance when q[E+(vB)]=0  v=E/B
– Undeflected particles have been velocity
selected.
Combined E and B fields (cont)
• e.g. Thomson's Experiment.
– First accelerate particles so ½mv2=eV.
– Then force straight motion so
E/B=v=(2eV/m)½.
– Thus gives e/m, which is constant!
Combined E and B fields (cont)
• e.g. The Bainbridge Mass
Spectrometer.
– First a velocity selector.
– Then uniform magnetic
field as r=(mv)/(qB) m/q.
28.8 Magnetic Force on Current
Carrying Conductor
Magnetic force on current carrying
conductor (cont)

 
F  q  B

 
so dF  ed  B
• Where vd is the drift velocity. If n conduction
electrons per unit volume, force summed over all

 
the electrons is
F  nAled  B
  
•Since I=nAvde,
F  Il  B.
•For variable magnetic fields, or curved wires
 

dF  Idl  B
Example
• The conductor has a straight segment with length
L perpendicular to the plane of the figure on the
right, with the current opposite to B.
• Find the total magnetic field force on these three
segments of wire.
Example (cont)
• There is no force on the segment on the right
perpendicular to the plane of the figure on the
right.

• For the straight segment on the
 left, L points
to the left, perpendicular to B . The force has
magnitude F=ILB, and its direction is up.

• Choose a segment dl with length dl  Rd on
the semicircle, so magnitude dF of the force on
the segment is dF  I ( Rd  ) B
Example (cont)
• The components of the force on segment are
dFx  IRd  cos dFy  IRd  sin 
• To Find the component of the total force, we
integral these expression.


0
0
Fx  IRB  cosd  0 Fy  IRB  sin d  2IRB
• Adding the forces on the straight and
semicircular segments, we find the total force:

Fx  0 Fy  IB L  2 R  or F  IBL  2 R  ˆj
28.9 Torque on a Current Loop
Torque on a current loop (cont)
• Consider a square current loop. Each side
– parallel to axis produces a force IaB,
– perpendicular to axis produces a force
IbBcos().
• Forces balance but parallel sides yield a torque 
=2(½bIaBsin())=IABsin()=|IAB|.
• So define  = B.
•  is called the magnetic dipole moment. Units?
Torque on a current loop (cont)

  IA

  NIA
  NIAB sin 
28.9 Hall Effect
Hall Effect
• Charges in a wire within a B field will be pushed to one
side of wire.
• Charge will build up until E and B forces match i.e.
FB=qvdB=FE=qE=qVH/h. (h is the height of the bar.)
• Recall that I=nqvdht, so VH=(IB)/(nqt). (t is the thickness
of the bar, so ht is the cross sectional area; vd is the drift
velocity, n the carrier density.)
• Measures field or charge carrier density.
• Sign of Hall voltage gives sign of charge carrier.
Electrostatic vs Magnetic Fields (I)
• Electrostatic Field
• Electric monopoles exist
(charges, +q, -q)
• Created by charges at
rest through E field
• The field exerts a force
(F=qE) on a static
charge.
• Magnetic field
• Magnetic monopoles do
not exist
• Created by moving
charge (or current)
through B field.
• The field exerts a force
(F=qvB) on a charge
moving at velocity v.
Electrostatic vs Magnetic Fields (II)
• E field is a vector (F is
parallel to E).
• The electric force does
work on a charged
particle.
• The energy of a charged
particle changes.
• B field is a vector (F is
perpendicular to v and
B).
• The magnetic force
does no work on a
charged particle.
• The energy of the
particle does not
change.
Electrostatic vs Magnetic Fields (III)
• E field lines start at +ve • B field lines are
and end on -ve charges.
continuous (form loops).
• Gauss's law tells us
• Since B lines have no end
points
 
 E  dA  q0 /  0 .
 
 B  dA  0.
 
• Since E  dl  0

the E field is conservative, • Wait and see…
hence definition of V.
Chapter 29 Sources of Magnetic
Andre Marie Ampere
Born: 20 January 1775
Died: 10 June 1836
French

C
 
B  dl  0 I enclosed
Key words
principle of superposition of
magnetic fields(磁场叠加原理)
law of Biot and Savart (毕奥－萨伐
定律)
source point(源点)
field point (场点)
ampere (安培)
Ampere’s law (安培定律)
toroidal solenoid(螺线管)
Bohr magneton(波尔磁子)
Key words
paramagnetic(顺磁性的)
relative permeability(相对渗透性)
permeability(渗透性)
magnetic susceptibility(磁化率)
diamagnetic(抗磁性的)
ferromagnetic(铁磁体)
magnetic domain(磁畴)
hysteresis(磁滞)
magnetization(磁化)
displacement current(位移电流)
29.1 Magnetic Field of a Moving Charge
ELECTRIC FIELD
Proportional to q.
Proportional to 1/r2.
MAGNETIC FIELD
Proportional to q.
Proportional to 1/r2.
AND to v.
AND to sin().
 is the angle between
 and .

r
Magnetic field of a moving charge (cont)
MAGNETIC FIELD
ELECTRIC FIELD
q moves with velocity v.
For a point charge q.
The field is circular, i.e in
Field is radial, i.e. the
direction of E is along r. the direction of vr.
Magnetic field of a moving charge (cont)

B


r
rˆ 
r
 0 q  sin 
B
4
r2
The field line
directions are given
by the right-hand
rule.


  0 q  rˆ  0 q  r
B

2
4 r
4 r 3

0
• In SI unit the numerical value of  0 is exactly
4 10
7
0
 110 7 T  m / A
4
c 
2
1
 00
Example: Forces between two moving protons
y
• Find the electric and
magnetic forces on the
upper proton, and
determine the ratio of
their magnitudes.


+
q
r

+ 
q
z
x
Example: Forces between two moving protons
(cont)
Solution:
By Coulomb’s law, the magnitude of electric force is
1 q2
FE 
40 r 2
The force on the upper proton is vertically upward.
The magnitude of B caused by the lower proton is
 0 q
B
4 r 2
From the right-hand rule, the direction of B is in
the +z direction.
Example: Forces between two moving
protons (cont)
The magnitude of the magnetic force on the upper
proton is
 0 q 2 2
FB  qB 
4 r 2
The ratio of the magnitude of the two forces is
2
FB  0 q 2 2 4r 2
2 
 2
  0  0  2
2
FE
q 40 r
c
c 
2
1
 00
So in non-relativity analysis, FB<<FE, FB can be ignored.
29.2 Magnetic Field of a Current Element

B
dQ  nqAdl I  n q d A
 0 n q  d Adl sin 
 0 dQ  d sin

dB 
2
4
r2
4
r
 0 Idl sin 

4
r2



  0 Idl  r̂  0 Idl  r
dB 

2
4 r
4 r 3
Law of Biot and Savart
29.3 Magnetic Field of a Straight Current
Carrying Conductor
y
a
 
dl
y  
r
P

O
I
a
dl  dy
Ixdy
 0 Idl sin   0

dB 
2
4 x 2  y 2 3 2
4
r
 0 I  a xdy
0 I
2a
B

3


a
2
2
2
4
x
4 x x 2  a 2


x

y
dB
When a>>x,
x2  a2  a
0 I
B
2x
r  x 2  y 2 sin   sin      x
x2  y 2
Magnetic field of a straight current
carrying conductor (cont)
For a long straight currentcarrying conductor, at all
points on a circle of radius r
around the conductor, the
magnitude B is
0 I
B
2r
y

Btotal
Example

B2

B1

B1
P1
Wire 1

Btotal
I
3d

B2
Wire 2
P2
d

d
I
2d
P3

B1

Btotal 
B2
• Find the magnitude and direction of B at points P1,
P2, and P3.
• Find the magnitude and direction of B at any point
on the x-axis to the right of wire 2 in terms of the xcoordinate of the point.
x
Example (cont)
Solution:
a) At pint 1:
0 I
0 I
0 I
0 I
0 I
B2 

B1 

Btotal  B1  B2 
2 4d  8d
2 2d  4d
8d
At pint 2:
0 I
B1  B2 
2d
At pint 3:
0 I
0 I
B1 

2 3d  6d
0 I
Btotal  B1  B2 
d
0 I
B2 
2d
0 I
Btotal  B2  B1 
3d
Example (cont)
Example (cont)
b) The magnitudes of the fields due to each wire
are
B1 
0 I
B2 
2 x  d 
0 I
2 x  d 
The total field is in the negative y-direction, and
has magnitude
Btotal  B2  B1 
0 I

0 I


 0 Id
2 x  d  2 x  d   x 2  d 2

At points very far from wires, so that x>>d, and
 Id
Btotal  0 2
x
0 I
B
2r
( a long, straight, currentcarrying conductor)
29.3 Force between Parallel Conductors
Force between parallel conductors (cont)
The magnitude of B field produced by the lower
conductor, at the position of the upper conductor, is
0 I
B
2r
The magnitude of the force exerted on a length
L of the upper conductor is
 0 II L
F  I LB 
2r
Force between parallel conductors (cont)
The force per unit length F/L is
F  0 II 

L
2r
(two long, parallel, currentcarrying conductors)
• Similar result for force on the lower
conductor.
• Like currents attract, unlike currents repel.
Definition of the Ampere
• One ampere is that unvarying current that, if
present in each of two parallel conductors of
infinite length and one meter apart in empty
space, cause each conductor to experience a force
of exactly 2x10-7 Newtons per meter of length.
29.4 Magnetic field of a circular current loop
 0 Idl
dB 
4 x 2  a 2 
 0 Idl
a
dBx  dB sin  
4 x 2  a 2  x 2  a 2 12
 0 Idl
x
dBy  dB cos 
4 x 2  a 2  x 2  a 2 12
 dl  2a

B 
4 x
Due to rotational symmetry about
x-axis, the y-component cancel.
Iadl
0
x
2
a
2

3
2
0
Ia
 0 Ia 2
(on the axis of a

dl 
3 
3
4 x 2  a 2  2
2 x 2  a 2  2 circular loop)
Magnetic field of a circular current loop
(cont)
• For N loop,
Bx 
N 0 Ia 2
(on the axis of
2 x 2  a 2  2 N circular
loop)
The maximum value of the field
is at x=0, the center of the loop.
3
 0 NI (at the center of N
Bx 
2a
circular loops)
Magnetic field of a circular current loop
(cont)
29.5 Ampere’s Law
• Recall Gauss's law vs. Coulomb's Law.
• Ampere's law is the alternative to the B-S law.
– Use to find B fields from highly symmetric
current distributions.
– Find current distributions for particular B
fields.
Ampere's Law (cont)
 
0 I
 B  dl   B// dl  B  dl  2r 2r   0 I
The line integral is
independent of the radius of
the circle and is equal to  0
multiplied by the current
passing through the area
bounded by the circle.
Ampere's Law (cont)


B and dlare antiparallel, so
 
 
B  dl   Bdl
 B  dl  0 I
The sign depends on the
direction of the current
relative to the direction of
integration.
Like directions is a positive
sign, unlike direction a
negative sign.
Ampere's Law (cont)
 
 B  dl   B// dl
 B1  dl  0 dl   B2  dl  0 dl
b
c
d
a
a
b
c
d
0 I
0 I
r1   0 
r2   0

2r1
2r2
0
Ampere's Law (cont)
 
B  dl  Bdl cos 
dl cos   rd
 Brd
 
0 I
 B  dl   2r rd 
0 I
d  2

d


2
 0 I
Ampere's Law (cont)
 
B  dl  Bdl cos 
dl cos   rd
 Brd
 
0 I
 B  dl   2r rd 
0 I
d  0

d


2
0
Ampere's Law (cont)
• It relates the tangential component of B summed
around a closed curve (C) to the current (I) which
passes through the curve.
 
B

d
l


I
0
enclosed

(Ampere’s Law)
• Holds for any curve provided the current is
continuous, i.e. it does not begin or end at any point.
Ampere's Law (cont)
 

• If  B  dl  0 , it does not necessarily mean that B  0
everywhere along the path, only that the total
current through an area bounded
by the path is

zero.

 B

dl

B

dl

B
I

dl

B
dl
dl

B

  I
B dl

 dl
B
 
 B  dl  0  I   I   0
Ampere's Law - The recipe
• Let geometric closed path pass through the
required field point and lie in a plane.
• And encompass the current
• Ensure B is constant along the path, and that B is
tangential to the line or
– it is perpendicular to the field lines (zero
contrib.)
– or it is in a region where B=0.
• To achieve this use the symmetry of the field
lines.
Example: inside a long cylindrical
conductor
• The current is uniformly
distributed over the
cross-section area of the
conductor.
• Find the magnetic field
as a function of the
distance r from the
conductor axis for points
inside (r<R) and outside
(r>r) the conductor.
Example: inside a long cylindrical
conductor (cont)
Solution:
i) r<R Choose integral path a circle with radius r<R.
 
 B  dl  B  dl  B2r 
 
 
2
I
Ir

 2
2
I enclosed  J r   2  r  2
R
 R 
 
By Ampere’s law,
B  dl   I

0 I
B2r   0 I  B 
2r
0 enclosed
Example: inside a long cylindrical
conductor (cont)
ii) r>R
Choose integral path a circle with radius r>R.
 
 B  dl  B  dl  B2r 
I enclosed  I
By Ampere’s law,
Ir 2
B2r    0 2
R
 
 B  dl  0 I enclosed
Example: inside a long cylindrical
conductor (cont)
 0 I r
 2 R 2 ,
B
 0 I ,
 2r
rR
rR
Example: Field of a solenoid
• The solenoid has n
turns of wire per unit
length and carries a
current I.
• Use Ampere’s law to
find the field at or near
the center of the
solenoid.
Example: Field of a solenoid (cont)
Solution:
Choose the integration path the
rectangle abcd.
Side ab, with length L, is
parallel to the axis of the
solenoid.
Side bc and da are taken to be
very long so that side cd is far
from the solenoid; the the field
at side cd is negligibly small.
  b  c  d  a 
 B  dl   B  dl   B  dl   B  dl   B  dl
a
abcd
b
c
 BL  0  0  0   0 nLI
d
 B  0 nI
Example: Field of a solenoid (cont)
Example: field of a toroidal(环行) solenoid
• A toroid is wound with N
turns of wire carrying a
current I.
• Find the magnetic field at
all points.
Example: field of a toroidal solenoid
(cont)
Solution:
To find the field using Ampere’s
law, we’ll use the integration
paths shown as black lines in the
right-hand picture.
i) Path 1

 B  dl  B2r   0 I enclosed  0
circle1
B0
Example: field of a toroidal solenoid
(cont)
ii) Path 2

 B  dl  B2r   0 I enclosed  0 NI
circle 2
 0 NI
B
2r
iii) Path 3

 B  dl  B2r   0 I enclosed  0 NI  0 NI  0  B  0
circle 3
29.5 Displacement Current
 
 B  dl  0 I enclosed
Where I enclosed  ic
or  0 ?
This is a clear
contradiction.

As the capacitor charges, the electric field E and
electric flux  e through the bulged-out surface
are increasing.
Displacement current (cont)
• The capacitor charge q is
A
q  C  Ed   EA   E
d
• As the capacitor charges, the rate of q is
the conduction current.
dq
d E
ic 

dt
dt
• We invent a fictitious current iD in the
region between the plates, defined as
d E
iD  
dt
(displacement current)
Displacement current (cont)
• We include displacement current iD, along with
conduction current iC, in Ampere’s law:
 
 B  dl  0 iC  iD 
enclosed
(generalized Ampere’s law)
Displacement current (cont)
• Apply Ampere’s law to
a circle of radius r,
 
 iD  2
 B  dl  2rB  0  R 2 r 
2
 ic  2
r
  0  2 r    0
ic
2
 R 
R
For r>R, B is the same as through the
wire were continues and plates not
present at all.
or B 
0 r
i
2 C
2 R
Summary
• The magnetic force exerted on the charge is given
by B, 
  
  
F  qv  B F  qE    B 
• The force on a straight segment
 of a conductor

current I in a uniformly B is F  Il  B
 

• For a infinitesimal segment dl the force is dF  Idl  B
 
• Gauss’s law for magnetism,  B  dA  0
Summary
• Magnetic field of a moving charge
• Magnetic field of a current
element
• Magnetic field of a long straight
current
 
• Ampere’s law  B  dl  0 iC  iD 
enclosed
  0 q  r
B
4 r 3
 
  0 Idl  r
dB 
4 r 3
0 I
B
2r
29.6 Magnetic Materials
• A extern magnetic field B0
• Put a magnetic material into the magnetic field.
• Interaction between the magnetic field and the
magnetic material.
• The accessional magnetic field by the magnetic
material is B¹.
 
• The total magnetic field: B  B  B
0
Magnetic materials (cont)
• Define the relative permeability:
B
r 
B0
• Paramagnetic(顺磁性) material: μr>1
• Diamagnetic(反磁性) material: μr<1
Molecule current
• The molecule current can
produce a magnetic
moment.
• Define the magnetization of
the material:
 total
M
V


e
I


• The total magnetic field B in the material is
 

B  B0  0 M
Molecule current (cont)
I
a
d
b
c
l
• Molecule current line density: js
p
m
 I s S  js Sl
p

M
m

js Sl
 js
Sl 
 V
 M  dl   M  dl  M  ab  Ml  jsl  I s
ab
Ampere’s law in material

 B  dl  0  I  I s   0

 
 I   M  dl

 
I s   M  dl

 
 B
    M   dl   I
 0


 B

• Define: H   M
0
 
• So:  H  dl   I
(Ampere’s law in material)
Boundary conditions of magnetic field

B1
S
1
2

B2
 
 B  dS  B2n S  B1n S  0
 B1n  B2 n
Boundary conditions of magnetic field
(cont)

H1
a
d
1
2
b
c

H2
 
 H  dl  H1t ab  H 2t cd
 H1t  H 2t