Transcript Physics 2102 Spring 2002 Lecture 8

Physics 2102
Aurora Borealis
Jonathan Dowling
Lecture 14: TUE 09 MAR
Magnetic Fields II
Ch.28.6-10
“They are not supposed
to exist….”
QuickTime™ and a
decompressor
are needed to see this picture.
Second Exam Review:
10:40AM–12:00N THU 11 MAR
Nicholson 109 in Class
Second Exam (Chapters 24–28):
6–7PM THU 11 MAR
Lockett 6
Circular Motion:
v
F
Since magnetic force is perpendicular to motion,
the movement of charges is circular.
2
v
out
2
Fcentrifugal  ma  mr  m
r
in
magnetic
r
F
 qvB
FB  FC
mv 2
 qv B 
r

B into blackboard.



mv
Solve : r 
qB
In general, path is
a helix (component of v parallel to
field is unchanged).
v
F
C
.
.B
electron
mv
r
qB
Radius of Circlcular Orbit
r

v qB
 
r m
2r 2mv 2m
T


v
qBv
qB
Angular Frequency:
Independent of v
Period of Orbit:
Independent of v

1 qB
f  
T 2m
Orbital Frequency:
Independent of v
Example
Two charged ions A and B traveling with a
constant velocity v enter a box in which
there is a uniform magnetic field directed
out of the page. The subsequent paths
are as shown. What can you conclude?
A
v
B
v
(a) Both ions are negatively charged.
(b) Ion A has a larger mass than B.
(c) Ion A has a larger charge than B.
mv
r
qB
(d) None of the above.
Same charge q, speed v, and same B for both masses.
So: ion with larger mass/charge ratio (m/q) moves in circle of larger
radius. But that’s all we know! Don’t know m or q separately.
Examples of Circular Motion in
Magnetic Fields
Aurora borealis
(northern lights)
Synchrotron
Suppose you wish to accelerate charged
particles as fast as you can.
Linear accelerator (long).
Fermilab,
Batavia, IL (1km)
Magnetic Force on a Wire
L
L
q  it  i
vd

 
F  q vd  B


iL   
F  q B iLB
q

 
F iLB
Note: If wire is not straight,
compute force on differential
elements and integrate:

 
dF  i dL  B
Example
Wire with current i.
Magnetic field out of page.
What is net force on wire?
F1  F3  iLB
dF  iBdL  iBRd
By symmetry, F2 will only
have a vertical component,


0
0
F2   sin( )dF iBR  sin( )d 2iBR
Ftotal  F1  F2  F3  iLB  2iRB  iLB  2iB( L  R)
Notice that the force is the same as that for a straight wire,
and this would be true no
matter what the shape of
L
R
R
L
the central segment!.
Example 4: The Rail Gun
• Conducting projectile of length
2cm, mass 10g carries constant
current 100A between two rails.
• Magnetic field B = 100T points
outward.
• Assuming the projectile starts
from rest at t = 0, what is its
speed after a time t = 1s?
• Force on projectile: F= iLB
• Acceleration: a = F/m = iLB/m
• v = at = iLBt/m
rails
B
I
L
projectile
(from F = iL x B)
(from F = ma)
(from v = v0 + at)
= (100A)(0.02m)(100T)(1s)/(0.01kg) = 2000m/s
= 4,473mph = MACH 8!
Rail guns in the “Eraser” movie
"Rail guns are hyper-velocity weapons that shoot aluminum or clay rounds at just
below the speed of light. In our film, we've taken existing stealth technology one
step further and given them an X-ray scope sighting system," notes director
Russell. "These guns represent a whole new technology in weaponry that is still
in its infancy, though a large-scale version exists in limited numbers on
battleships and tanks. They have
incredible range. They can pierce
three-foot thick cement walls and
then knock a canary off a tin can with
absolute accuracy. In our film, one
contractor has finally developed an
assault-sized rail gun. We researched
this quite a bit, and the technology is
really just around the corner, which is
one of the exciting parts of the story."
Warner Bros., production notes, 1996.
http://movies.warnerbros.com/eraser/cmp/prodnotes.html#tech
Also: INSULTINGLY STUPID MOVIE PHYSICS: http://www.intuitor.com/moviephysics/
Rail guns in Transformers II
QuickTime™ and a
decompressor
are needed to see this picture.
Electromagnetic Slingshot
These Devices Can
Launch 1000kg Projectiles
At Mach 100 at a Rate of
1000 Projectiles Per Second.
Using KE = 1/2mv2
This corresponds to an output
about 1012 Watts = TeraWatt.
Uses: Put Supplies on Mars.
Destroy NYC in about 10
minutes.
QuickTime™ and a
H.264 decompressor
are needed to see this picture.
Principle behind electric motors.
Torque on a Current Loop:
Rectangular coil: A=ab, current = i
Net force on current loop = 0
But: Net torque is NOT zero!
F1  F3  iaB
F  F1 sin( )
Torque    Fb  iabB sin(  )
For a coil with N turns,
 = N I A B sin,
where A is the area of coil
Magnetic Dipole Moment
We just showed:  = NiABsin
N = number of turns in coil
A=area of coil.
Define: magnetic dipole moment 

  ( NiA)nˆ

 , n̂
Right hand rule:
curl fingers in direction
of current;
thumb points along 

  B


As in the case of electric dipoles, magnetic dipoles tend to align
with the magnetic field.
Electric vs. Magnetic Dipoles

  ( NiA)nˆ
+Q
p=Qa
-Q

QE
r r
E  p  E
r r
UE   p  E
r
QE
r
B   Br
r
U B   B
r
r
Magnetic Force on a Wire.
L

 
F iLB

 
dF  i dL  B
Magnetic Force on a Straight Wire in a Uniform
Magnetic Field
If we assume the more general case for which the
magnetic field B forms an angle  with the wire
the magnetic force equation can be written in vector
form as FB  iL  B. Here L is a vector whose
FB  iL  B
magnitude is equal to the wire length L and
has a direction that coincides with that of the current.
The magnetic force magnitude is FB  iLB sin  .
B
i
dF
.
dFB = idL  B
FB  i  dL  B

dL
Magnetic Force on a Wire of Arbitrary Shape
Placed in a Nonuniform Magnetic Field
In this case we divide the wire into elements of
length dL, which can be considered as straight.
The magnetic force on each element is
dFB = idL  B. The net magnetic force on the
wire is given by the integral FB  i  dL  B.
(28-12)
Principle behind electric motors.
Torque on a Current Loop:
Rectangular coil: A=ab, current = i
Net force on current loop = 0
But: Net torque is NOT zero!
F1  F3  iaB
F  F1 sin( )
Torque    Fb  iabB sin(  )
For a coil with N turns,
 = N I A B sin,
where A is the area of coil
Magnetic Dipole Moment
We just showed:  = NiABsin
N = number of turns in coil
A=area of coil.
Define: magnetic dipole moment 

  ( NiA)nˆ

 , n̂
Right hand rule:
curl fingers in direction
of current;
thumb points along 

  B


As in the case of electric dipoles, magnetic dipoles tend to align
with the magnetic field.
Electric vs. Magnetic Dipoles

  ( NiA)nˆ
+Q
p=Qa
-Q

QE
r r
E  p  E
r r
UE   p  E
r
QE
r
B   Br
r
U B   B
r
r