Electromagnetism Applications of Particle
Deflection with Moving Charges in Magnetic and Electric Fields
Review of Previous Material
Magnitude of Deflecting Force
The deflecting force
on a charged particle moving through an external magnetic field is calculated using:
| Fm| = q v B sin θ
where: Fm = deflecting force B =magnetic field strength (T) q = charge (C) v = speed of particle (m/s) θ = angle between v and B
The maximum deflecting
force will occur when θ= 90o. Thus sin 90o= 1 and Fm= qvB.
Example: A 20 g particle with a charge of +2.0 C
enters 0.20 T a magnetic field at 90o to the field. If the speed of the particle is 40 m/s, what is the acceleration that is experienced by the particle in the diagram
below?
(2.0 )(40 / )(0.20 )
16
m
m
m
F qv B
F C m s T
F N
2 2
16
0.020
800 8.0 10 /
m net
m
F F
F Na
m kg
a m s out of the page
An alpha particle enters a 50 T field at 30°to the field at a speed of 500 m/s. What is the magnitude of the deflecting force experienced by the alpha particle? (An α+2 particle has a charge of 2 x 1.60 x 10-19C = 3.2 x 10-19C.)
19
15
sin
(3.20 10 )(500 / )sin30 (50 )
4.0 10
m
m
m
F qv B
F C m s T
F N
POS Checklist
explain, quantitatively, how uniform magnetic and electric fields affect a moving electric charge, using the relationships among charge, motion, field direction and strength, when motion and field directions are mutually perpendicular.
The Movement of
Charges Through
Electric and
Magnetic Fields
Simultaneously
The Movement of Charges Through Electric
and Magnetic Fields Simultaneously
-when a charge passes through a magnetic field which is perpendicular to an electric field, it can pass through undeflected
when no deflection of the charge occurs, the magnetic force is equal to the electrical force
em
e
e m
FE F qvB
q
F Eq
F F
Eq qvB
E vB
Ev
B
When the forces are equal, (Fm = Fe) the
speed of the charge can be determined
Fm is down by 3rd
LHR when e is in B
Fe is up, e is attracted to the positive plate
When Fe up = Fm down, e passes E and B undeflected
Speed of electron can be determined
Example Eg) An electron enters a magnetic field of 2.00 x
10-3 T at 90 . An electric field of 1000 N/C is perpendicular to the magnetic field. Determine the kinetic energy of the electron as it passes undeflected between the two fields.
e mF F
Eq qvB
E vB
Ev
B
3
5
2
31 5
19 2 2
1000 /
2.00 10 /
5.00 10 /
1
2
1(9.11 10 )(5.00 10 / )
2
1.14 10 ( / )
k
k
k
N Cv
Ns Cm
v m s
E mv
E kg m s
E J kgm s
Applications of
Magnetic Forces
ex) What is the velocity of a particle if the E-field is 500 N/C and the B-field is 0.75 T?
Step 1: Determine the direction of the E and B forces.
Step 2: Set FE equal to FB.
ex) Determine the charge to mass ratio of the particle using the data from previous example and given the particle deflects with r = 1.85 x 105 m.
Step 1: Set Fm = Fc.
Step 2: Solve for q/m.
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