Transcript Chapter 28: Magnetic Field and Magnetic Forces

Chapter 28: Magnetic Field and Magnetic Forces
Iron ore found near Magnesia
Compass needles align N-S: magnetic Poles
North (South) Poles attracted to geographic North (South)
Like Poles repel, Opposites Attract
No Magnetic Monopoles
Magnetic Field Lines = direction of compass deflection.
Electric Currents produce deflections in compass direction.
=>Unification of Electricity and Magnetism in Maxwell’s
Equations.
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Magnetic Fields in analogy with Electric Fields
Electric Field:
– Distribution of charge creates an electric field E(r) in
the surrounding space.
– Field exerts a force F=q E(r) on a charge q at r
Magnetic Field:
– Moving charge or current creates a magnetic field B(r)
in the surrounding space.
– Field exerts a force F on a charge moving q at r
– (emphasis this chapter is on force law)
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Magnetic Fields and Magnetic Forces
Magnetic Force on a moving charge
– proportional to electric charge
– perpendicular to velocity v
– proportional to speed v (for a given geometry)
– perpendicular to Magnetic Field B
– proportional to field strength B (for a given geometry)
F=qvB
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F=qvB
F = |q| v B sinq
= |q| v B (v ^ B)
F
B
+
v
F
F=qvB
F = |q| v^ B
B
+
v^
v
F
F=qvB
F = |q| v B^
+
B^
v
B
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Magnetic Fields
Units of Magnetic Field Strength:
[B]
= [F]/([q][v])
= N/(C m s-1)
= Tesla
Defined in terms of force on standard current
CGS Unit 1 Gauss = 10-4 Tesla
Earth's field strength ~ 1 Gauss
Direction = direction of velocity which generates no force
Electromagnetic Force:
F=q(E+ vB)
= Lorentz Force Law
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Magnetic Field Lines and Magnetic Flux
Magnetic Field Lines
Mapped out with compass
Are not lines of force (F is not parallel to B)
Field Lines never intersect
Magnetic Flux
dFB = B . dA
 
dF B  B  dA
 
F B   B  dA
 
 B  dA  0 no magneti c charge! (no monopoles)
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• SI Unit of Flux:
– 1Weber = 1Tesla x 1 m2
– for a small area B = dFB /dA^
– B = “Magnetic Flux Density”
Flux through an open surface will play an important role
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Motion of Charged Particles in a Magnetic Field
Charged Particle moving perpendicular to the Magnetic Field
– Circular Motion!
– (simulations)
F
+
v
F
+
v
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Charged Particle moving perpendicular to a uniform
Magnetic Field
v
R
+
mv 2
F | q| vB 
R
mv
R
| q| B
v | q| B
 
R
m
 cyclotron frequency
+
v
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In a non-uniform field: Magnetic Mirror
v
F
B
Net component of force away from concentration of
field lines.
Magnetic Bottle
Van Allen Radiation Belts
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Work done by the Magnetic Field on a free particle:
 
dW  F  dx
  
 qv  B vdt
 0!
=> no change in Kinetic Energy!
Motion of a free charged particle in any magnetic field has
constant speed.
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Applications of Charged Particle Motion in a Magnetic Field
+
Recall:
Charged Particle moving perpendicular to a uniform
Magnetic Field
mv 2
F | q| vB 
R
v
R
mv
R
| q| B
+
v
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Velocity Selector
makes use of crossed E and B to provide opposing forces
+ +
- -
+ +
- -
+ + +
- - -
E
upwards
F=qvB
downwards
F = qE
No net deflection => forces exactly cancel:
|q| v B=|q| E
v = E/B
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J. J. Thomson’s Measurement of e/m
Electron Gun
+ +
- -
+ +
- -
+ + +
- - -
E
V
and velocity selector:
E
v
B
1 2
mv  eV
2
e
E2

m 2VB 2
e/m = 1.76x1011 C/kg
with Millikan’s measurement of e
=> mass of electron
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Example: Using an accelerating Potential of 150 V and a transverse
Electric Field of 6x106 N/C. Determine
a) the speed of the electrons,
b) the magnetic field magnitude required for no net deflection
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Mass Spectrometer
R2
+ +
+ +
+ + R1
- -
- -
- -
One method
velocity selector + circular
trajectory
E
E
v
B
mv
R
qB
RqB RqB 2
m

v
E
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Example: Vacuum System Leak Detector uses Helium
atoms. Ionized helium atoms (He +) are detected with
a mass spectrometer with a magnetic field strength of
.1 T. With a velocity selector tuned to 1x105 m/s,
where must the detector be placed to detect 4He +
ions?
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Magnetic Force on a Current Carrying Wire
B
I
A
vd
dl
Fi
 
F   Fi   qi v i  B




 Nqv d  B  n  volume  qv d  B
 


 nAdlqv d  B  JAdl  B
 
 Idl  B ( RHR )
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Example: A 1-m bar carries 50 A from west to east in a
1.2 T field directed 45° North of East. What is the
B
magnetic force on the bar?
I dl
Force will be directed upwards (out of the plane of the
page)
F  ILB sin q
 50 A 1m 12
. T sin 45
 42.4 N
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Torque on a Current Loop (from F = I l x B )
Rectangular loop in a magnetic field (directed along z axis)
short side length a, long side length b, tilted with short sides
at an angle with respect to B, long sides still perpendicular to
B.
B
Fb
Fa
Fb
Fa
Forces on short sides cancel: no net force or torque.
Forces on long sides cancel for no net force but there is a net
torque.
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Torque calculation: Side view
moment arm
a/2 sin q
q
Fb = IBb
Fb
t = Fb a/2 sin q + Fb a/2 sin q
 Iab B sin q = I A B sin q
magnetic moment
m
q
B
 I A B  m B
Magnetic Dipole ~ Electric Dipole
U = - m .B
Switch current direction every 1/2 rotation => DC motor
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y
Hall Effect
Bz
Conductor in a uniform magnetic field
x
z
+
Jx
Magnetic force on charge carriers F = q vd  B
Fz = qvdB Charge accumulates on edges
+
+
-
-
-
-
-
-
-
-
-
-
+
+
+
+
+
+
+
+
+
+
-
-
Jx
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Equilibrium: Magnetic Force = Electric Force
on bulk charge carriers
Bz
Ey
- - - - - - - - - - - w
t
+
+
+
+
+
+
+
+
+
+
+
+
Jx
Charge accumulates on edges Fz = 0 = qvdBy + q Ez
Ez
vd  By
Ez
J x  nqvd  - nq
By
nq 
- J x By
Ez
Hall EMF V H  E z w
I  J x tw
nq 
- IB y
VH t
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Negative Charge carriers:
velocity in negative x direction
magnetic force in positive z direction
=> resulting electric field has reversed polarity
Bz
+
-
-
+
-
+
-
+
-
+
-Ey
+
+
-
+
-
+
-
+
-
+
-
+
Jx
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Example: A ribbon of copper 2.0 mm thick and 1.5 cm
wide carries a 75 A current in a .40 T magnetic Field.
The resulting Hall emf is .81 mV. What is the density
of charge carrying electrons?
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