Transcript Magnetism Magnetic Fields Magnetic Forces •Certain objects and circuits produce magnetic fields •Magnetic fields, like electric fields, are vector fields •They have a magnitude and.

Magnetism
Magnetic Fields
Magnetic Forces
•Certain objects and circuits produce magnetic fields
•Magnetic fields, like electric fields, are vector fields
•They have a magnitude and a direction
•Denoted by B, or B(r)
FB  qv  B
•They have no effect on charges at rest
•They produce a force on moving charges given by
F  q vB sin
•Perpendicular to magnetic field
•Perpendicular to velocity
•Magnetic field strengths are measured in units called a tesla, abbreviated T
•A tesla is a large amount
F
N
N
B
T
of magnetic field
qv C  m/s A  m
The Right-Hand Rule
To figure out the direction of magnetic force, use the following steps: F  qv  B
1. Point your fingers straight out in direction of first vector v
2. Twist your hand so when you curl your fingers, they point in the
direction of B
3. Your thumb now points in the direction of v B
4. If q is negative, change the sign
vB
•Vectors in the plane are easy to draw
•Vectors perpendicular to the plane are hard
B
v
•Coming out of the plane
•Going into the plane
 ˆi

v  B  det  vx
B
 x
ˆj
vy
By
kˆ 

vz    v y Bz  vz By  ˆi   vz Bx  vx Bz  ˆj   vx By  v y Bx  kˆ
Bz 
Work and Magnetic Fields
How much work is being done on a point charge moving in a magnetic field?
W  F s
•Work = force  distance
•Divide the distance into little tiny steps, divide by time
F  qv  B
dW
ds
 F
 F  v  F  v cos  0
dt
dt
•But recall F  v
• = 90 and cos = 0
Magnetic fields do no
work on pure charges
F
B
q
v
Cyclotron Motion
Consider a particle of mass m and charge q
moving in a uniform magnetic field of strength B
B
v
v
F
F
F
q
F
F  qv  B
•Motion is uniform circular motion
•Centripetal force formula:
mv 2  qvB
mv  qRB
F
R
p  qRB
This version works even when
you take relativity into account
v
v
•Let’s find how long it takes to go around:
T
2 R
v
T
2 m
qB

2
T

qB
m
Motion in a Magnetic Field
•The particle may also move parallel to the magnetic fields
•No force F  qv  B
•Combined motion is a helix
•Net motion is along magnetic
q
field lines
•The Earth has magnetic field lines
•Charged particles from space follow them
•Hit only at magnetic poles
•aurora borealis
•aurora australis
B
Velocity Selector / Mass Spectrometer
•When we have both electric and magnetic fields, the force is F  q  E  v  B
•Magnetic field produces a force on the charge
vE B
0   E  vB
•Add an electric field to counteract the magnetic force
•Forces cancel if you have the right velocity
FB
v
detector
+
–
Now let it move into region with
magnetic fields only
•Particle bends due to cyclotron motion
•Measure final position
•Allows you to determine m/q
mv  qRB
FE
m RB 2

q
E
m RB

q
v
Sample Problem
An electron has a velocity of 1.00 km/s (in the positive x direction) and an
acceleration of 2.001012 m/s2 (in the positive z direction) in uniform
electric and magnetic fields. If the electric field has a magnitude of strength
of 15.0 N/C (in the positive z direction), determine the components of the
magnetic field. If a component cannot be determined, enter 'undetermined'.
F  q  E  v  B
1.602 1019 C
15.0kˆ  v  B  11.37kˆ
v  B  26.37kˆ
By  .02637 T
F ma
E  vB  
q
q
 ˆi
 3
ˆ
26.37k  det 10
B
 x
E  vB 
31
12
2
9.1094

10
kg
2.00

10
m/s



ˆj
0
By
kˆ 

0
Bz 

 10 kˆ By  ˆjBz
3

kˆ
Bz  0
Bx  undetermined
The Hall Effect
•Consider a current carrying wire in a magnetic field
•Let’s assume it’s actually electrons this time, because it usually is
I
FB
d
vd
V
t
I
•Electrons are moving at an average velocity of vd
vd 
•To the left for electrons
tdnq
I  AJ  td  nqvd 
•Because of magnetic field, electrons feel a force upwards
•Electrons accumulate on top surface, positive charge on bottom
•Eventually, electric field develops that counters magnetic force
EH   v d  B
0  F  q  EH  vd  B
IB
VH 
•This can be experimentally measured as a voltage
tnq
VH  EH d  vd Bd  IBd tdnq
Force on a Currenty-Carrying Wire
•Suppose current I is flowing through a wire of cross sectional area A and length L
•Think of length as a vector L in the direction of current
•Think of current as charge carriers with charge q and drift velocity vd
B
I
F  Nqvd  B  Vnqvd  B  ALJ  B  IL  B
F  IL  B
L
F
•What if magnetic field is non-uniform, or wire isn’t straight?
•Divide it into little segments
dF  I  ds  B
•Add them up
B
I
A
B
F  I  ds  B
A
Sample Problem
A loop of wire with length L and width W lies in the xy-plane with the
length L parallel to the x-axis and the nearest side a distance d away
from the x-axis. A current I runs clockwise around the loop. There is
a magnetic field in the plane given by the formula B  Akˆ y . What is
the force on each side of the loop, and the total force on the loop?
L
F  IL  B
F  I  ds  B
 
W
d
 A ˆ   ILA ˆj
ˆ
F
•Let’s do the bottom first F  IL  B  I iL   k 
d
I
d 
F
•All
points
have
y
=
d
F
ILA ˆ
F
j
•Top is the same, but y = d + W, direction opposite F  
W d
•Left
side
is
hard,
because
y
changes
B
x •But right side cancels it
ˆj
ILAW
ILA ˆ ILA ˆ
Ftot 
Ftot 
j
j
d d W 
d
W d
Force/Torque on a Loop
•Suppose we have a current carrying loop in a constant magnetic field
•To make it simple, rectangular loop size L  W
F  IL  B
L
Ft
W
I
t
•Left and right side have no force at all,
because cross-product vanishes
•Top and bottom have forces
•Total force is zero
•This generalizes to general geometry
F  I  ds  B  I
Fb
B
  ds   B
0
•There is, however, a torque on this loop
τ  r  F  Wˆj  Ft  ILWBˆj  kˆ  IABˆi
t  IAB
Ft  ILBkˆ
Fb  ILBkˆ
Torque on a Loop (2)
F
Wsin
•What if the loop were oriented differently?
•Torque is proportional to separation of forces
nˆ  t
t  FW sin   BIWL sin   BIA sin 
W
B
τ  IAnˆ  B
F
•Does this formula generalize to other shapes
Edge-on view of Loop
besides rectangles?
•Draw in imaginary wires to divide it into rectangles
•These carry equal and opposite current, so no contribution
to forces
•Now the whole thing is two rectangles
A1
A2
•Torque is sum of torques on each
B
I
t  t1  t 2  IA1B  IA2 B  IAB
Torque and Energy for a Loop
•Define A to be a vector perpendicular to the loop
with area A and in the direction of n-hat
A  Anˆ
•Determined by right-hand rule by current
•Curl fingers in direction current is flowing
τ  IA  B
•Thumb points in direction of A
•Define magnetic dipole moment of the loop as
μ  IA
τ  μB
τ  IAnˆ  B
A
R
I
t   B sin 
•Torque is like an angular force
•It does work, and therefore there is energy associated with it
U   t d    B sin  d    B cos 
•Loop likes to make A parallel to B
•Compare to formulas for electric dipole
U  p  E
τ  pE
 t
A
U  μ  B
B
Edge-on view of Loop
Comments on Magnetic Forces
τ  μB
U  μ  B
•Note that the simplest structures we can think of with
magnetic forces on them are dipoles
•As we will later discuss, there are only magnetic dipole sources
•No “fundamental magnetic charges” like electric forces
•We used work arguments to figure out the energy of a loop in a
magnetic field
•But we previously said magnetic fields do no work
This turns out to be surprisingly subtle:
•For current loops, the answer is that the battery that keeps the
current going is actually doing the work
•More on this later
How to Make an Electric Motor
•Have a background source of magnetic fields, like permanent magnets
•Add a loop of wire, supported so it can spin on one axis
•Add “commutators” that connect the rotating loop to outside wires
•Add a battery, connected to the commutators
•Current flows in the loop
τ  IA  B
•There is a torque on the current loop
•Loop flips up to align with B-field
•Current reverses when it gets there
t
•To improve it, make the
F
A
loop repeat many times
τ  INA  B
F
+
–