DC electrical circuits - University of Central Florida

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Transcript DC electrical circuits - University of Central Florida

Magnetic Fields
Chapter 29
(continued)
Force on a Charge in a
Magnetic Field
F  qv  B
or
F  qvBsin 
(Use “Right-Hand” Rule to
determine direction of F)
F
v
B
mq
Trajectory of Charged Particles
in a Magnetic Field
(B field points into plane of paper.)
+
+B
+
v+
+
+
+
+
+
+
+
+ F
+
+
+
+ F +
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
B
+
+
+
+
+
v
Magnetic Force is a centripetal force
Radius of a Charged Particle
Orbit in a Magnetic Field
+B
+
+
v+
+
Centripetal
Force
+
+
+
+
+
+
+
+
r
+
+
+
+
+
+
mv 2

 qvB
r
F
+
=
Magnetic
Force


mv
r
qB
Note: as Fv , the magnetic
force does no work.
Exercise
electron
v
B
v’
• In what direction does the magnetic field point?
• Which is bigger, v or v’ ?
Exercise: answer
electron
v
B
F
v’
• In what direction does the magnetic field point ?
Into the page [F = -e v x B]
• Which is bigger, v or v’ ?
v = v’ [B does no work on the electron, Fv]
Trajectory of Charged Particles
in a Magnetic Field
What if the charged particle has a velocity component along B?
v
x
z
y

B
Trajectory of Charged Particles
in a Magnetic Field
What if the charged particle has a velocity component along B?
v
Fz=0 so vz=constant
x

B
z
y
Result is a helix
The force is in the xy
plane. It acts exactly as
described before,
creating circular motion
in the xy plane.
Trajectory of Charged Particles
in a Magnetic Field
What if the charged particle has a velocity component along B?
v
Fz=0 so vz=constant
x

B
z
y
Result is a helix
The force is in the xy
plane. It acts exactly as
described before,
creating circular motion
in the xy plane.
The Electromagnetic Force
If a magnetic field and an electric field are simultaneously
present, their forces obey the superposition principle and
must be added vectorially:
F  qE  qv  B

+
+
+
+
+
+
+
q
+
B+
+

+v
+
+
+
+

E
The Lorentz force
The Electromagnetic Force
If a magnetic field and an electric field are simultaneously
present, their forces obey the superposition principle and
must be added vectorially:

+ FB +
+ +
q
+ +
+ B+
+ +
F  qE  qv  B

+v
+
+
+
+
FE

E
The Lorentz force
The Electromagnetic Force

v
+ +
+ +
FB+  +q
+ B+
+ +
+
+
+
+
+
When B, E, and v are mutually
perpendicular, as pictured here, FE
and FB point in opposite directions.
FE

E
The magnitudes do not have to be
equal, of course. But by adjusting E or
B you can set this up so the net force is
zero.
Set FE = qE equal to FB = qvB:
qE = qvB
Hence with the pictured orientation of fields and velocity, the
particle will travel in a straight line if v = E / B.
The Hall Effect
I
x
x
vd
x
x
x
x
B x
•Consider a conducting bar, carrying a current, with a
perpendicular magnetic field into the picture.
The Hall Effect
I
x
x
vd
x
x
x
x
B x
•Consider a conducting bar, carrying a current, with a
perpendicular magnetic field into the picture.
•The electrons drifting to the right tend to move down because
of the magnetic force.
The Hall Effect
+
I
+
x
x
-
+
-
+
vd
- -
+
+
x
-
+
+
x
-
-
+
x
-
-
+
+
+
x
B x
-
-
-
•Consider a conducting bar, carrying a current, with a
perpendicular magnetic field into the picture.
•The electrons drifting to the right tend to move down because
of the magnetic force.
•Thus you get a charge separation: a net negative charge along
the bottom edge, and positive along the upper.
The Hall Effect
+
I
+
x
x
-
+
-
+
vd
- -
+
+
x
-
+
+
x
-
-
+
x
-
-
+
+
+
x
B x
-
-
-
•Consider a conducting bar, carrying a current, with a
perpendicular magnetic field into the picture.
•The electrons drifting to the right tend to move down because
of the magnetic force.
•Thus you get a charge separation: a net negative charge along
the bottom edge, and positive along the upper.
•This charge separation sets up an electric field, top to bottom,
which pulls electrons up – opposing the magnetic force.
E
The Hall Effect
+
I
+
x
x
-
+
-
+
vd
- -
+
+
x
-
+
+
x
-
-
+
x
-
-
+
+
+
x
B x
-
-
-
•Consider a conducting bar, carrying a current, with a
perpendicular magnetic field into the picture.
•The electrons drifting to the right tend to move down because
of the magnetic force.
•Thus you get a charge separation: a net negative charge along
the bottom edge, and positive along the upper.
•This charge separation sets up an electric field, top to bottom,
which pulls electrons up – opposing the magnetic force.
•The charge separation builds up until the two forces are equal:
eE=evdB
E
The Hall Effect
+
I
+
x
x
-
+
-
+
vd
- -
+
+
x
-
+
+
x
-
-
+
x
-
-
+
+
+
x
B x
-
-
-
•The charge separation builds up until the two forces are equal:
eE=evdB
d
The Hall Effect
+
I
+
x
x
-
+
-
+
vd
- -
+
+
x
-
+
+
x
-
-
+
x
-
-
+
+
+
x
B x
-
-
d
-
•The charge separation builds up until the two forces are equal:
eE=evdB
•This means an electric potential difference develops between the
two edges: VH=Ed=vdBd
-the Hall voltage
The Hall Effect
+
I
+
x
x
-
+
-
+
vd
- -
+
+
x
-
+
+
x
-
-
+
x
-
-
+
+
+
x
B x
-
-
d
-
•The charge separation builds up until the two forces are equal:
eE=evdB
•This means an electric potential difference develops between the
two edges: VH=Ed=vdBd
-the Hall voltage
•This means that measuring the Hall voltage lets you work out the
drift velocity.
The Hall Effect
+
I
+
x
x
-
+
-
+
vd
- -
+
+
x
-
+
+
x
-
-
+
x
-
-
+
+
+
x
B x
-
-
d
-
•The charge separation builds up until the two forces are equal:
eE=evdB
•This means an electric potential difference develops between the
two edges: VH=Ed=vdBd
-the Hall voltage
•This means that measuring the Hall voltage lets you work out the
drift velocity.
•Moreover, using J=nevd and I=JA (with A the slab’s crosssectional area) gives vd=I/(Ane) and so VH=IBd/Ane . Measuring
the Hall voltage lets you find the density of conduction electrons.
The Hall Effect
+
I
+
+
x
x
-
-
+
+
vd
- -
+
x
-
+
+
x
-
+
x
-
-
-
+
+
+
x
B x
-
-
-
VH
•The Hall effect also lets you find the sign of the charge carriers
that make up the current. Above is the picture for electrons.
•But if the charge carriers actually had a positive charge, the
picture would look like this:
I
x
x
+
-
+
+
vd
+
-
x
+
-
-
x
+
+
-
-
x
+
x
+
+
-
-
B x
+
+
•The carriers would move to the bottom edge still, and the
Hall voltage would point in the opposite direction.
VH
Magnetic Force on a Current
• Force on one charge
F = q vd x B
• Forces on all charges in a length
L of a conductor:
F = n A L q vd x B
• Use I = n q vd A and define a
vector L whose length is L, and has
the same direction as the current I.
I
Then
F=I LxB
A
L
I = n q vd A
F

B
•
L
[F points out of the page]
Magnetic Force on a Current
Example: A current, I=10 A, flows through a wire, of
length L=20 cm, between the poles of a 1000 Gauss
magnet. The wire is at  = 900 to the field as shown.
What is the force on the wire?
N
I
S
L
Magnetic Force on a Current
Example: A current, I=10 A, flows through a wire, of
length L=20 cm, between the poles of a 1000 Gauss
magnet. The wire is at  = 900 to the field as shown.
What is the force on the wire?
F ILB
 ILB
(up)
N
I
S
L
Magnetic Force on a Current
Example: A current, I=10 A, flows through a wire, of
length L=20 cm, between the poles of a 1000 Gauss
magnet. The wire is at  = 900 to the field as shown.
What is the force on the wire?
F ILB
 ILB
(up)
N
 (10A)  (0.20m)  (0.1T)  0.2N
I
(up)
S
L
Magnetic Force on a Current Loop
A current loop is placed in a uniform magnetic field
as shown below. What will happen?
B
L
N
I
S
Magnetic Force on a Current Loop
No net force – but a
torque is imposed.
F=BIL

L
N
F’
S
B
F’
I
F=BIL
Magnetic Torque on a Current Loop
Simplified view:
F=BIL

L
d

B
I
F=BIL
Magnetic Torque on a Current Loop
Simplified view:
d

Torque  BIL   sin    2
2

F=BIL
 IAB sin 

L
d

B
I
F=BIL
Magnetic Torque on a Current Loop
  IA  B
F=BIL

L

d
for a current loop

B
A=vector with
magnitude A=Ld
and direction
given by a RH
rule.
I
F=BIL
Magnetic Force on a Current Loop
Torque & Magnetic Dipole
By analogy with electric dipoles, for which:
  p  E
The expression,
  IA  B
implies the a current loop acts as a magnetic dipole!
Here  IA is the magnetic dipole moment,
and


   B
(Torque on a
current loop)
Potential Energy of a Magnetic Dipole
By further analogy with electric dipoles:
UE  p  E
So for a magnetic dipole (a current loop)

UB   B
The potential energy is due to the fact that the magnetic field
tends to align the current loop perpendicular to the field.
Nonuniform Fields and Curved Conductors
• So far, we have considered only uniform fields and straight
current paths.
• If this is not the case, we must build up using calculus.
a
b
I
dF

B
Consider a small length,
dL, of current path.
The force on dL is:
dF = dL I x B
Nonuniform Fields and Curved Conductors
a

B
b
For a conductor of length L
F=LIxB
For a bit of length dL
I
To find the force exerted by a
non-uniform magnetic field on a
curved current we divide the
conductor in small sections dL and
add (integrate) the forces exerted
on every section dL.
dF = dL I x B
Then, for the total length of
the curved conductor in a
non-uniform magnetic field
F =  dF =  dL I x B
Nonuniform Fields and
Curved Conductors: Example
• What is the force on the current-carrying
conductor shown?
R
I
L
L

B
Nonuniform Fields and
Curved Conductors: Example

dF ( out of page )

d
R
I

L
L

B
Nonuniform Fields and
Curved Conductors: Example
 

dF  Id   B  I d B sin 

B
 
d
R
I

L
L
Nonuniform Fields and
Curved Conductors: Example
 

dF  Id   B  I d B sin 
d  Rd

B
 
d
R
I

L
L
Nonuniform Fields and
Curved Conductors: Example
 

dF  Id   B  I d B sin 
d  Rd

B
 
d
R
I

L

F  IBR  sin d   IBR cos
0
L

0
 2 IBR
Nonuniform Fields and
Curved Conductors: Example
 

dF  Id   B  I d B sin 
d  Rd

B
 
d
R
I

L

F  IBR  sin d   IBR cos
0
L

0
 2 IBR
Nonuniform Fields and
Curved Conductors: Example
 

dF  Id   B  I d B sin 
d  Rd

B
 
d
R
I

L

F  IBR  sin d   IBR cos
0
L

0
 2 IBR
Nonuniform Fields and
Curved Conductors: Example
 Ftot  2 IBR  2 IBL  IB( 2 L  2 R )

B
 
d
R
I

L

F  IBR  sin d   IBR cos
0
L

0
 2 IBR
Nonuniform Fields and
Curved Conductors: Example
 Ftot  2 IBR  2 IBL  IB( 2 L  2 R )

B
 
d
R
I

L
L
Equal to the force we would find for a straight wire of length 2(R+L)