Download B - LSU Physics

Physics 2113
Aurora Borealis
Jonathan Dowling
Lecture 25: FRI 24 OCT
Magnetic Fields III
“They are not supposed
to exist….”
Magnetic Force on a Wire.
L
!
! !
F =iL×B
!
! !
dF = i dL × B
Magnetic Force on a Wire
L
q = it = i
vd
!
! !
F = q vd × B
L
!
!
iL ! ! !
F = q ×B =iL×B
q
!
! !
F =iL×B
Note: If wire is not straight,
compute force on differential
elements and integrate:
i
!
! !
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 φ 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
!
B
i
!
dF
.
!
! !
dFB = idL × B
!
! !
FB = i ∫ dL × 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 φ .
φ
!
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)
28.8.2. A portion of a loop of wire passes between the poles of a magnet as shown.
We are viewing the circuit from above. When the switch is closed and a current
passes through the circuit, what is the movement, if any, of the wire between the
poles of the magnet?
a) The wire moves toward
the north pole of the magnet.
b) The wire moves toward the
south pole of the magnet.
c) The wire moves upward
(toward us).
d) The wire moves downward (away from us into board).
e) The wire doesn’t move.
i
Example
Wire with current i.
Magnetic field out of page.
What is net force on wire?
F1 = F3 = iLB
dF = iBdL = iBRdθ
By symmetry, F2 will only
have a vertical component,
π
π
0
0
F2 = ∫ sin(θ )dF =iBR ∫ sin(θ )dθ = 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!.
−y direction ↓
i
Fgrav = mg Fmag = iLB
−3
m / L = 46.6 × 10 kg/m
To balance:
Fgrav = Fmag
⇒ mg = iLB
mg ⎛ m ⎞ g
⇒B=
=⎜ ⎟
iL ⎝ L ⎠ i
i
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
Rail guns in Reality!
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.
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(θ )
Torque = τ = F⊥b = iabB sin(θ )
For a coil with N turns,
τ = N I A B sinθ,
where A is the area of coil
Magnetic Dipole Moment
We just showed: τ = NiABsinθ
N = number of turns in coil
A=area of coil.
Define: magnetic dipole moment m
!
µ = ( NiA)nˆ
!
µ , n̂
Right hand rule:
curl fingers in direction
of current;
thumb points along m
! ! !
τ = µ×B
As in the case of electric dipoles, magnetic dipoles tend to align
with the magnetic field.
Electric vs. Magnetic Dipoles
!
µ = ( NiA)nˆ
+Q
p=Qa
-Q
q
QE
!
! !
τE = p × E
! !
UE = − p ⋅ E
QE
!
! !
τ B = µ × B!
!
U B = −µ ⋅ B
Magnetic Dipole Moment
We just showed: t = NiABsinq
N = number of turns in coil
A=area of coil.
Define: magnetic dipole moment m
!
µ = ( NiA)nˆ
!
µ , n̂
Right hand rule:
curl fingers in direction
of current;
thumb points along µ
! ! !
τ = µ×B
! !
UE = − p ⋅ E
As in the case of electric dipoles, magnetic dipoles tend to align
with the magnetic field.
! ! !
τ =µ×B
τ = µ Bsin θ
τ1 = τ 2 = τ 3 = τ 4
τ is biggest
when B is at
right angles to μ 1 and 3 are “downhill”.
2 and 4 are “uphill”.
U1 = U4 > U2 = U3