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Chapter 29
Magnetic Fields
Force on a Charge Moving in a
Magnetic Field, Formula

The properties can be summarized in a
vector equation:
FB  qv  B

FB is the magnetic force

q is the charge
v is the velocity of the moving charge
B is the magnetic field


Direction: Right-Hand Rule #1


The fingers point in the
direction of v
B comes out of your
palm


Curl your fingers in the
direction of B
The thumb points in the
direction of v  B which
is the direction of FB
q0
q0
Direction: Right-Hand Rule #2




Alternative to Rule #1
Thumb is in the
direction of v
Fingers are in the
direction of B
Palm is in the direction
of FB


On a positive particle
You can think of this as
your hand pushing the
particle
More Differences Between
Electric and Magnetic Fields

Work


The electric force does work in displacing a
charged particle
The magnetic force associated with a steady
magnetic field does no work when a particle is
displaced

This is because the force is perpendicular to
the displacement
 
  
WFB  FB  d r  (qv  B)  vdt  0


FB  v
Work in Fields, cont.


The kinetic energy of a charged particle
moving through a magnetic field cannot be
altered by the magnetic field alone
When a charged particle moves with a given
velocity through a magnetic field, the field can
alter the direction of the velocity, but not the
speed or the kinetic energy
1 2 1 2
 K  mv f  mvi  0  v f  vi
2
2
Wnet  K  Wnet  WFB  0,  K  0
자기장 내에서의 전자 운동
예제 29.1
v  8 106 m/s , q  1.6 1019 C, B  0.025T
z

 
FB  qv  B  qvB sin 60 (k̂ )
q
-

v
y
60°

FB
x

B : xy-plane
 (1.6 10 19 )(8 106 )(0.025) sin 60 (k̂ )
 2.8 10 14 (k̂ )( N)
Force on a Charged Particle
Equating the magnetic and centripetal forces:
2
mv



  v2


FB  qvB 
Fnet  ma, Fnet  FB  qv  B, a  (r̂ )
r
r


Solving for r:
mv
r
qB



FB  v , v  const
r is proportional to the linear momentum of the
particle and inversely proportional to the magnetic
field
More About Motion of Charged
Particle

The angular speed of the particle is
v qB
ω 
r
m


The angular speed, w, is also referred to as the
cyclotron frequency
The period of the motion is
2πr 2π 2πm
T


v
ω
qB
v, r 에 무관
Motion of a Particle, General


If a charged particle moves
in a magnetic field at some
arbitrary angle with respect
to the field, its path is a
helix
Same equations apply, with
v   v y2  v z2

Use the active figure to vary
the initial velocity and
observe the resulting
motion
PLAY
ACTIVE FIGURE
균일한 자기장에 수직으로 운동하는 양성자
예제 29.2
m  1.67 1027 kg, q  1.6 1019 C, B  0.35T, r  14cm

v
r
v?
qvB  mv2 /r
v  qBr/m
 (1.6 10-19 )(0.35)(0.14) / (1.67 10- 27 )
 4.7 106 (m / s)
예제 29.3 전자빔의 휘어짐
m  9.111027 kg, q  1.6 1019 C, r  7.5cm,V  350V

Bin
(A) B=?
mv2
mv
qvB 
B
r
qr
(9.1110-31 )(1.11107 )
B
(1.6 10-19 )(0.075)
r

v
4
 8.4 10 (T)
(A) ω=?
v 1.1110
ω 
 1.5 108 (rad / s)
r
0.075
7
V
v0
1 2
qV  mv
2
 v  2qV/m  1.11107 m / s
Particle in a Nonuniform
Magnetic Field



The motion is complex
For example, the
particles can oscillate
back and forth between
two positions
This configuration is
known as a magnetic
bottle
Velocity Selector, cont.


When the force due to the electric field is
equal but opposite to the force due to the
magnetic field, the particle moves in a
straight line
This occurs for velocities of value
v=E/B
FB  FE
FB  FE
FB  FE

Bin
FB  FE
v0  E/B
스크린
x1
t1  x1 /v0
전자
- q  1.6 1019 C
t 2  x2 /v0
x2
v y  a y t1 
a y x1
+++++++++++++++
v0
ay
-
-q
v0
qEx1

mv0
---------------
a y  qE/m
q
x1 , x2 ,v0 , E, B 
m
y2  v y t 2
y2
y1
1 2 qEx12
y1  a y t1 
2
2mv02

qEx1 x2
mv02
Cyclotron, 2



D1 and D2 are called
dees because of their
shape
A high frequency
alternating potential is
applied to the dees
A uniform magnetic
field is perpendicular to
them
qvB  mv2 /r
v  qBr/m
1 2πr 2πm
T 

f
v
qB
r
1 2
mv
2
q2 B2 2

r
2m
K
Force on a Wire, equation

The magnetic force is
exerted on each
moving charge in the
wire


F  qvd  B
The total force is the
product of the force
on one charge and
the number of
charges



F  qvd  B nAL

 
 

 nqvd AL  B  nqvd AL  B  IL  B
nqvd A  I
Force on a Wire, Arbitrary
Shape


Consider a small
segment of the wire, ds
The force exerted on
this segment is
dFB  I ds  B

The total force is
b
FB  I  ds  B
a
예제 29.4
반원형 도선에 작용하는 자기력
m  1.67 1027 kg, q  1.6 1019 C, B  0.35T, r  14cm
θ
dθ
r

B

ds
θ
I
반원에서 자기력

 
dFB  IdL  B  dFB  IdLB sin θ  I (ds sin θ ) B
FBcurve   IB (ds sin θ )  IB  ds sin θ  IB (2r )
직선에서 자기력

 
FB  IL  B  FBstr   ILB   IB (2r )

 curve  str
 FB  FB  FB  0
예제 29.4 반원형 도선에 작용하는 힘
반지름 R인 반원의 폐회로를 구성하고 있는 도선에 전류 I가 흐른다. 회로는 xy 평면에
놓여 있고, 균일한 자기장이 ＋y 방향을 따라 작용한다. 도선의 직선 부분과 곡선 부분에
작용하는 자기력을 구하라.
풀이
직선 부분에 작용하는 힘
곡선 부분에 작용하는 힘
F1  ILB  2IRB (+z방향)
b
FB  I  ds  B
a


0
0
F2  I  ( Rd ) B sin  IRB  sin d   2IRB
Torque on a Current Loop, 3


The forces are
equal and in
opposite
directions, but
not along the
same line of
action
The forces
produce a
torque around
point O
b
b
τ max=F2 +F4
2
2
b
b
=( IaB ) +( IaB )
2
2
=IabB
τ  τ max sin θ
Torque on a Current Loop,
Equation


The maximum torque is found by:
b
b
b
b
τ max  F2  F4  (I aB )  (I aB )
2
2
2
2
 I abB
The area enclosed by the loop is ab, so τmax =
IAB

This maximum value occurs only when the field is
parallel to the plane of the loop
Torque on a Current Loop,
General



Assume the magnetic
field makes an angle of
 < 90o with a line
perpendicular to the
plane of the loop
The net torque about
point O will be τ = IAB
sin 
Use the active figure to
vary the initial settings
and observe the
resulting motion
PLAY
ACTIVE FIGURE
Torque on a Current Loop,
Summary



The torque has a maximum value when the
field is perpendicular to the normal to the
plane of the loop
The torque is zero when the field is parallel to
the normal to the plane of the loop
  IA  B where A is perpendicular to the
plane of the loop and has a magnitude equal
to the area of the loop
τ  τ max sin θ
  IA  B
Magnetic Dipole Moment

The product IA is defined as
the magnetic dipole
moment,  , of the loop






μ  IA
Often called the magnetic
moment
SI units: A · m2
Torque in terms of magnetic
moment:
Analogous to  B
for
electric dipole   p  E
τ  τ max sin θ
  IA  B
   B
예제 29.5 코일의 자기 쌍극자 모멘트
A  5.4cm  8.5cm  45.9cm2 , N  25turns , I  15mA , B  0.35T
 
B⊥ A
(A) 코일의 자기 쌍극자 모멘트의 크기
μ=NIA=(25)(0.015 )(0.00459)
= 1.72 × 10-3 Am 2


(B) 코일에 작용하는 토크의 크기

 
τ = μ×B =Bsin90  = (1.72 × 10-3 )(0.35)
= 6.02 × 10-4 (Nm)

B