Transcript Document

Magnetism
Magnetic Force
Magnetic Force Outline
•
•
•
•
•
•
Lorentz Force
Charged particles in a crossed field
Hall Effect
Circulating charged particles
Motors
Bio-Savart Law
Class Objectives
• Define the Lorentz Force equation.
• Show it can be used to find the magnitude
and direction of the force.
• Quickly review field lines.
• Define cross fields.
• Hall effect produced by a crossed field.
• Derive the equation for the Hall voltage.
Magnetic Force
• The magnetic field is defined
from
the



Lorentz Force Law,F  qE  q v  B
 
Magnetic Force
• The magnetic field is defined
from
the



Lorentz Force Law,F  qE  q v  B
• Specifically, for a particle with charge q
moving through a field B with a velocity v,

 
F  q vB
• That is q times the cross product of v and B.
 
 
Magnetic Force
• The
 cross product may be rewritten so that,
F  q vB sin 
• The angle  is measured from the direction


of the velocity v to the magnetic field B .
• NB: the smallest angle between the vectors!
vxB
B

v
Magnetic Force
Magnetic Force
• The diagrams show the direction of the
force acting on a positive charge.
• The force acting on a negative charge is in
the opposite direction.
B
F
-
v
B
+
v
F
Magnetic Force
• The direction of the force F acting on a
charged particle moving with velocity v
through a magnetic field B is always
perpendicular to v and B.
Magnetic Force
• The SI unit for B is the tesla (T) newton per
coulomb-meter per second and follows
from

F
 B.
the before mentioned equation
q v sin 
• 1 tesla = 1 N/(Cm/s)
Magnetic Field Lines
Review
Magnetic Field Lines
• Magnetic field lines are used to represent
the magnetic field, similar to electric field
lines to represent the electric field.
• The magnetic field for various magnets are
shown on the next slide.
Magnetic Field Lines
Crossed Fields
Crossed Fields
• Both an electric field E and a magnetic
field B can act on a charged particle. When
they act perpendicular to each other they
are said to be ‘crossed fields’.
Crossed Fields
• Examples of crossed fields are: cathode ray
tube, velocity selector, mass spectrometer.
Crossed Fields
Hall Effect
Hall Effect
• An interesting property of a conductor in a
crossed field is the Hall effect.
Hall Effect
• An interesting property of a conductor in a
crossed field is the Hall effect.
• Consider a conductor of width d carrying a
current i in a magnetic field B as shown.
d
i
x
x
x
x
x
x
x
x
x
x
x
x
x
Bx
x
x
Dimensions:
i
Cross sectional area: A
Length: x
Hall Effect
• Electrons drift with a drift velocity vd as
shown.
• When the magnetic field is turned on the
electrons are deflected upwards.
d
i
FB
x
x
x
x
x
x FB
x
x
x
x
x
x
x
Bx
x
x
vd
-
i
Hall Effect
• As time goes on electrons build up making
on side –ve and the other +ve.
x
d
i
-
x
+x
x
-
x
x
x FB
x
+x + x
vd
Bx
x
x
-
x
Low
-
+ x +
x
i
High
Hall Effect
• As time goes on electrons build up making
on side –ve and the other +ve.
• This creates an electric field from +ve to
–ve.
x
i
E
x
+x
x
-
x
x
x FB
x
+ x FE+ x
vd
Bx
x
x
-
x
Low
-
+ x +
x
i
High
Hall Effect
• The electric field pushed the electrons
downwards.
• The continues until equilibrium where the
electric force just cancels the magnetic
x
x
x
Low
force. x
-
i
E
x
+x
x
-
x FB
x
+ x FE+ x
vd
Bx
x
-
x
-
+ x +
x
i
High
Hall Effect
• At this point the electrons move along the
conductor with no further collection at the
top of the conductor and increase in E.
x
i
E
x
+x
x
-
x
x
x FB
x
+ x FE+ x
vd
Bx
x
x
-
x
Low
-
+ x +
x
i
High
Hall Effect
• The hall potential V is given by, V=Ed
Hall Effect
• When in balance, FB  FE
 eE  evd B
i
where vd 
neA
Hall Effect
• When in balance, FB  FE
 eE  evd B
i
where vd 
neA
dq
• Recall, i 
dt
dq  neAdx
dx
i  neA  neAvd
dt
dx
A
A wire
Hall Effect
• Substituting for E, vd into eE  evd B we get,
Bi
n
Vle
A
wherel 
d
A circulating charged particle
Magnetic Force
• A charged particle moving in a plane
perpendicular to a magnetic field will move
in a circular orbit.
• The magnetic force acts as a centripetal
force.
• Its direction is given by the right hand rule.
Magnetic Force
Magnetic Force
• Recall: for a charged particle
moving
in
a
2
mv
circle of radius R,FB 
2
mv
 qvB 
R
mv
R
qB
R
2m
• As so we can show that,T 
qB
qB
qB
,f 
, 
2m
m
Magnetic Force on a current carrying
wire
Magnetic Force
• Consider a wire of length L, in a magnetic
field, through which a current I passes.
x
x
x
x
B
x
x
x
x
I
Magnetic Force
• Consider a wire of length L, in a magnetic
field, through which a current I passes.
x
x
x
x
B
x
x
x
x
I
• The force actingon an element
of
the
wire

dl is given by,dFB  IdL  B
Magnetic Force
• Thus we can write the force acting on the
wire, dFB  BIdL
L
 FB  BI  dL
0
 FB  BIL
Magnetic Force
• Thus we can write the force acting on the
wire, dFB  BIdL
L
 FB  BI  dL
0
 FB  BIL
• In general, FB  BILsin 
Magnetic Force
• The force on a wire can be extended to that
on a current loop.
Magnetic Force
• The force on a wire can be extended to that
on a current loop.
• An example of which is a motor.
Interlude
Next….
The Biot-Savart Law
Biot-Savart Law
Objective
• Investigate the magnetic field due to a
current carrying conductor.
• Define the Biot-Savart Law
• Use the law of Biot-Savart to find the
magnetic field due to a wire.
Biot-Savart Law
• So far we have only considered a wire in an
external field B. Using Biot-Savart law we
find the field at a point due to the wire.
Biot-Savart Law
• We will illustrate the Biot-Savart Law.
Biot-Savart Law
• Biot-Savart law:
 0 I 
dB 
dl  rˆ
2
4r

 0 Idl sin 
dB 
2
4r

Biot-Savart Law
• Where0 is the permeability of free space.
0  4 107 Tm/ A
• And rˆ is the vector from dl to the point P.
Biot-Savart Law
• Example: Find B at a point P from a long
straight wire.
l
Biot-Savart Law
 0 I 
 0 Idl sin 
ˆ
dl  r  dB 
• Sol: dB 
2
4r
4r 2


l
Biot-Savart Law
• We rewrite the equation in terms of the
angle the line extrapolated from rˆ makes
with x-axis at the point P.
• Why?
• Because it’s more useful. l
Biot-Savart Law
 0 I 
 0 Idl sin 
ˆ
dl  r  dB 
• Sol: dB 
2
4r
4r 2


• From the diagram,
   180  
• And hence
    90

l

Biot-Savart Law
 0 I 
 0 Idl sin 
ˆ
dl  r  dB 
• Sol: dB 
2
4r
4r 2


• From the diagram,
   180  

• And hence
    90
 sin  sin  90
 cos
sinA  B  sin A cos B  sin B cos A
l

Biot-Savart Law
0 Idl cos
dB 
4r 2
• Hence,
• As well,
tan 
l
x
x
cos 
r
l
r  x2  l 2
 0 I cos d
• Therefore, dB 
4x

Biot-Savart Law
• For the case where B is due to a length AB,
B

 0i
B  dB 
cos d
4x

0


 0i
sin   sin 

4x
A
B
 
Biot-Savart Law
• For the case where B is due to a length AB,
B

 0i
B  dB 
cos d
4x

0


 0i
sin   sin 

4x
• If AB is taken to infinity,
 0i
B
2x
A
B
 