Transcript Physics 2102 Spring 2002 Lecture 8

Physics 2102
Aurora Borealis
Jonathan Dowling
Physics 2102
Lecture 14
Ch28: Magnetic Forces on Current Wires
Star Quake on a
“I’ll be back….
Magnetar!
Crossed Fields E vs. B
q
v
B
E
y
L
FE=qE
FB=vqBFE
FE=ma => y=qEL2/(2mv2)
FB=FE => y=0 => v=E/B
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


0
0
By symmetry, F2 will only
have a vertical component,
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?
rails
B
L
I
projectile
• Force on projectile: F= ILB
(from F = iL x B)
• Acceleration: a = iLB/m
(from F = ma)
• v(t) = iLBt/m
(from v = v0 + at)
= (100A)(0.02m)(100T)(1s)/(0.01kg) = 2000m/s
= 4,473mph = MACH 8!
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  iabBsin( )
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

 , nˆ
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
QE
 
  p E


  B

