Download PHYS 222 Worksheet 14 Magnetic Field

PHYS 222
Worksheet 14 – Magnetic Field
Supplemental Instruction
Iowa State University
Leader:
Course:
Instructor:
Date:
Alek Jerauld
PHYS 222
Dr. Paula Herrera-Siklódy
2/21/12
Useful Equations
F  qv  B
F  q vB sin( )
F  q(E  v  B)
Magnetic force on a moving charged particle
Magnitude of the magnetic force. ∅ is the angle between the
velocity vector and the B-field vector
Total force exerted on a moving charged particle due to electric
and magnetic fields.
Magnetic field unit: Tesla [T] = 1 N/Am
or Gauss [G] = 10-4 T
Magnetic Interactions
 A moving charge or a current creates a magnetic field in the surrounding space
 The magnetic field exerts a force F on any other moving charge of current that is
present in the field
Diagrams
Right hand rule for magnetic force:
Related Problems
1) A particle with a charge of −1.20×10−8 C is moving with instantaneous velocity v =
(4.60 m/s) i + (-3.40 m/s) j. What is the force exerted by a magnetic field B = (1.40 T) i ?
(Book 27.1)
Hint:
F  qv  B
0
 4.60  1.40  Fx 
F  1.2(108 )  3.40    0   Fy 
0
 0   0  Fz  6.53(108 ) N
2) An electron moves at 2.60×106 m/s through a region in which there is a magnetic field of
unspecified direction and magnitude 7.60×10−2 T. What is the largest possible magnitude of
the acceleration of the electron due to the magnetic field? What is the smallest possible
acceleration? (Book 27.6)
When magnetic field is perpendicular to the velocity, acceleration is maximum:
F  ma
F  qvB sin( )
 ma  qvB sin(90 )
qvB
a
(1)  3.48(1016 ) m / s 2
m
When magnetic field is parallel to the velocity, acceleration is minimal:
 ma  qvB sin(0 )
a
qvB
(0)  0 m / s 2
m
3) In an experiment with cosmic rays, a vertical beam of particles that have charge of
magnitude 3e and mass 12 times the proton mass enters a uniform horizontal magnetic
field of 0.250 T and is bent in a semicircle of diameter 95.0 cm.
What is the speed of the particles? (27.22)
acentripetal
v2

d /2
v2
m
 3qvB sin(90 )
d /2
3qdB(1)
v
 2.84(106 ) m / s
(2)3me
4) A singly charged ion of 7Li (an isotope of lithium) has a mass of 1.16×10−26 kg. It is
accelerated through a potential difference of 260 V and then enters a magnetic field with
magnitude 0.726 T perpendicular to the path of the ion. What is the radius of the ion’s path
in the magnetic field? (Book 27.26)
First find the initial speed of the particle before entering the magnetic field:
W  KE 
1
1
1
mv12  mv0 2 , v0  0  W   mv12
2
2
2
W  qV
1
  mv12  qV
2
2qV
v12 
m
Now use centripetal acceleration equation and force equation to solve for R:
2qV
m
2
v
2qV
ac  1 
R
mR
mac  qvB sin(90)
v12 
m
2qV
2qV
1 2mV
q
BR
 8.46(103 ) m
mR
m
B
qe
5) If two deuterium nuclei (charge +e, mass 3.34 x 10-27kg) get close enough together, the
attraction of the strong nuclear force will fuse them to make an isotope of helium, releasing
vast amounts of energy. The range of this force is about 10-15 m. This is the principle
behind the fusion reactor. The deuterium nuclei are moving much too fast to be contained
by physical walls, so they are confined magnetically.
(a) How fast would two nuclei have to move so that in a head-on collision they would get
close enough to fuse? (Treat the nuclei as point charges, and assume that a separation
of 10-15 m is required for fusion.)
kq 2
kq 2
F  ma  2  a 
x
mx 2
d
da
kq 2
 v   2 dx
dx
mx

kqe 2
kq 2
v
 v0 , v  0  v0 
 8.3(106 ) m/ s
md
md
(b) What strength magnetic field is needed to make deuterium nuclei with this speed travel
in a circle of diameter 2.60 m?
For a circular motion, acceleration is only present in the centripetal direction:
v0 2
F  ma, a 
d /2
F  qv0 B sin(90)
mv0 2

 qv0 B sin(90)
d /2
mv0
B
 0.13 T
qd / 2