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Ch. 29
•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)
•They have no effect on charges at rest
•They produce a force on moving charges given by FB  qv  B
•Perpendicular to magnetic field
F  q vB sin 
•Perpendicular to velocity
•Magnetic field strengths are measured in units called a tesla,
abbreviated T
•A tesla is a large amount of magnetic field
B
F
qv
N
C  m/s
N
T
Am
Warmup 12
F  qv  B
The Right Hand Rule
To figure out the direction of magnetic force, use the following steps:
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
vB
3. Your thumb now points in the direction of v B
4. If q is negative, change the sign
B
•Vectors in the plane are easy to draw
v
•Vectors perpendicular to the plane are hard
•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
Bz 
  vx By  v y Bx  kˆ
Finding the direction
proton
What is the direction of the force for each
F  qv  B
of the situations sketched?
A)
B)
C)
D)
E)
F) None of the above
B
p
v
F  qv  B
F  q vB sin 
Ca+2
4. If q is negative, change the sign
ion
electron
e
v
Ca
B
v
B
Ex- (Serway 29-11) A proton moves perpendicular to a uniform magnetic field B at 1.00
x 10 7m/s and experiences an acceleration of 2.00 x 1013 m/s2 in the positive x direction
when its velocity is in the positive z direction. Determine the magnitude and direction of
the field.
Solve on
Board
Warmup 12
Warmup 13
Cyclotron Motion
Consider a particle of mass m and charge q moving in a uniform
F  qv  B
magnetic field of strength B
B
v
v
F
F
F
q
F
•Motion is uniform circular motion
•Centripetal force formula:
mv 2
mv  qRB
 qvB
F
R
p  qRB
v
v
•Let’s find how long it takes to go around:
2 R
T 
v
2 m
T
qB
2

T
qB

m
CT –1 Cosmic rays (atomic nuclei stripped bare of their electrons) which come from all
directions, would continuously bombard Earth’s surface if most of them were not
deflected by Earth’s magnetic field. Given that Earth is, to an excellent approximation, a
magnetic dipole, the intensity of cosmic rays bombarding its surface is greatest at the
A. poles.
B. mid-latitudes.
C. equator.
The Earth has magnetic field lines
Charged particles from space follow them
Hit only at magnetic poles
aurora borealis
aurora australis
Concept Question
Two particles with the same mass are
moving in the same magnetic field, but
particle X is circling in less time than
particle Y. What can account for this?
A) Particle X is moving faster (only)
B) Particle Y is moving faster (only)
C) Particle X has more charge (only)
D) Particle Y has more charge (only)
E) A and C could both account for this
F) B and C could both account for this
•Moving faster doesn’t help
•Higher speed means bigger radius
•Higher charge does help
•You turn corners faster
mv  qRB
2 m
T
qB
B
X
Y
Velocity Selector / Mass Spectrometer
•When we have both electric and magnetic fields, the force is
•Magnetic field produces a force on the charge F  q  E  v  B 
•Add an electric field to counteract the magnetic force
•Forces cancel if you have the right velocity 0  E  vB v  E B
FB
v
FE
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
m RB 2

q
E
m RB

q
v
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
E  vB 
F ma
E  vB  
q
q
31
12
2
9.1094

10
kg
2.00

10
m/s



kˆ
1.602 1019 C
v  B  26.37kˆ
15.0kˆ  v  B  11.37kˆ
By  .02637 T
ˆ
ˆ
ˆ
 i
j
k 
 3

3 ˆ
ˆ

10
kBy  ˆjBz
26.37k  det 10
0
0
Bz  0
B

B
B
y
z 
B  undetermined
 x

x

Force on a Current-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  Nqv d  B  Vnqv d  B
 ALJ  B  IL  B
L
F
F  IL  B
Ex- (Serway 29-39) A wire having a mass per unit length of 0.500 g/cm carries a 2.00 A
current horizontally to the south. What are the direction and magnitude of the minimum
magnetic field needed to lift this wire vertically upward?
Solve on
Board
Warmup 13
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
•Left and right side have no force at all,
because cross-product vanishes
I
Ft
•Top and bottom have forces
W
ˆ
t
Ft  ILBk
Fb
B
Fb   ILBkˆ
•Total force is zero
•This generalizes to general geometry
F  I  ds  B  I
  ds   B  0
•There is, however, a torque on this loop
τ  r  F  Wˆj  Ft   ILWBˆj  kˆ  IABˆi
t  IAB
•What if the loop were oriented differently?
•Torque is proportional to separation of forces
t  FW sin   BIWL sin   BIA sin 
Wsin
Torque on a Loop (2)
F
n̂  t
W
B
F
Edge-on view of Loop
ˆ B
τ  IAn
This formula generalizes to other shapes besides rectangles?
•It is true for circular loops, or oddly shaped loops of current
Torque and Energy for a Loop
ˆ B
•Define A to be a vector perpendicular to the loop τ  IAn
with area A and in the direction of n-hat
ˆ
A  An
•Determined by right-hand rule by current
A
•Curl fingers in direction current is flowing
R
•Thumb points in direction of A
τ  IA  B
I
•Define magnetic dipole moment of the loop as
τ  μB
μ  IA
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 
U  μ  B
 t
A
•Loop likes to make A parallel to B
B
Edge-on view of Loop
CT – 2 A rectangular loop is placed in a uniform magnetic field with the plane of the loop
perpendicular to the direction of the field. If a current is made to flow through the loop in
the sense shown by the arrows, the field exerts on the loop:
A.
B.
C.
D.
a net force.
a net torque.
a net force and a net torque.
neither a net force nor a net torque.
CT - 3 A rectangular loop is placed in a uniform magnetic field with the plane of the loop
parallel to the direction of the field. If a cur-rent is made to flow through the loop in the
sense shown by the arrows, the field exerts on the loop:
A.
B.
C.
D.
a net force.
a net torque.
a net force and a net torque.
neither a net force nor a net torque.
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
F
A
the loop repeat many
times
F
τ  INA  B
+
–