Transcript 2 Magnetic Force and Circular Motion

Magnetic
Force
and
Circular
Motion
Mrs. Coyle
AP Physics C
Force on a Charged Particle in a
Magnetic Field
For a charged particle moving
in an external magnetic field
with its velocity perpendicular
to the field:
• The force is always directed
toward the center of the circular
path
• The magnetic force causes a
centripetal acceleration,
changing the direction of the
velocity of the particle
Finding the radius of the circular
path
• Equate the magnetic and centripetal forces:
mv
FB  qvB 
r
• Solving for r:
mv
r
2
qB
– Note: r is proportional to the momentum of
the particle and inversely proportional to the
magnetic field
More About Motion of Charged
Particle
• The angular speed of the particle is
v qB
ω 
r
m
– The angular speed, w, is also referred to as
the cyclotron frequency
• The period of the motion is
2πr 2π 2πm
T


v
ω
qB
Force on a charge moving in a
magnetic field
Radius:
mv
r
qB
qBr
Velocity:
v
m
qB
Frequency: w 
m
Period:
2m
T
qB
If the angle between v and B
is not 90o .
• The path is a helix
• Same equations apply,
with
v  v v
2
y
2
z
Problem #29.
The magnetic field of the Earth at a certain location is
directed vertically downward and has a magnitude of 50.0
μT. A proton is moving horizontally toward the west in this
field with a speed of 6.20 × 106 m/s. (a) What are the
direction and magnitude of the magnetic force the field
exerts on this charge? (b) What is the radius of the circular
arc followed by this proton?
Ans: a)4.96x10-17 N South, b) 1.29km
Problem #32.
A proton moving freely in a circular path perpendicular to a
constant magnetic field takes 1.00 μs to complete one
revolution. Determine the magnitude of the magnetic field.
Ans: 6.56 x10-2 T
Problem #39.
A singly charged positive ion moving at 4.60 × 105 m/s
leaves a circular track of radius 7.94 mm along a direction
perpendicular to the 1.80-T magnetic field of a bubble
chamber. Compute the mass (in atomic mass units
1amu=1.66x10-27 kg) of this ion.
Ans: 4.97x10-27 kg=2.99amu