Chapters 27--Examples
1
Problem
A electron that has a velocity v= (2x106
m/s) i + (3 x 106 m/s) j move through a
magnetic field, B=(0.030)i-(0.15)j.
a) Find the force on the electron
b) Recalculate for a proton.
2
F=qv x B

q=-1.602 x 10-19
x
y
z


v  B  2E 6
3E 6 0  2E 6 * 0.15  (0.030 * 3E 6) zˆ  0.39E 6 zˆ
0.030  0.15 0
F=(-1.602x10-19)*(-3.9 x 105) k
 F=(6.2x10-14 N) k
 If a proton, then q=1.602 x 10-19 so
 F=(-6.2x10-14 N) k (opposite direction of e)

3
Problem
A 150-g ball containing 4.0 x 106 excess
electrons is dropped into a 125-m vertical
shaft. At the bottom of the shaft, the ball
suddenly enters a uniform horizontal magnetic
field that has magnitude 0.250 T and direction
from east to west. If air resistance is negligibly
small, find the magnitude and direction of the
force that this magnetic field exerts on the ball
as it enters the field.
4
Prep Work

q=4 x 106 e = 4 x 106 * 1.602 x 10-19


q=6.408 x 10-11
No air resistance so mechanic energy,
E=KE+U is conserved.
(KE2-0)=-(0-U1)
 ½ m v2=mgy
 V2=2gy (y=125 m)


V=sqrt(2*9.8*125)=49.5 m/s
5
F=qv x B
||F||=qvB=6.408x10-11*49.5*0.250
 ||F||=7.93 x 10-10 N
 Direction

N
Down
v
F
E
W
B

Force is north
Up
S
6
Problem
In a certain cyclotron, a proton moves in a
circle of radius 0.5 m. The magnitude
of the magnetic field is 1.2 T
a) What is the cyclotron frequency?
b) What is kinetic energy of the proton in
electron volts?
7
Cyclotrons
v2
m  qvB
r
mv
R
qB
v qB
 
R m
  2f
qB 1.602  10 19 *1.2
7
f 


1
.
8

10
Hz
19
2m 2 (1.67  10 )
8
Part B)
v2
1
m  mv2  qvB
r
r
2
2
mv  qBR  mv  qBR 
mv
2
2
2

qBR 
1 2 1 qBR 

 KE  mv 
m
2

qBR 
KE 
2
1.6 10

19
2

m
2
*1.2 * 0.5
12

2
.
76

10
J
 27
2m
2 *1.67 10
7
KE  2.76 10 12 J 1.61eV
1 9  1.7  10 eV
10
9
Problem
A wire of 62.0 cm length and 13.0 g mass
is suspended by a pair of flexible leads in
a magnetic field of 0.44 T. What are the
magnitude and direction of the current
required to remove the tension in the
supporting leads?
X
X
X
X
X
X
X
X
X
X
X
X
X
X
X
X
X
X
X 0.62
Xm
X
X
X
X
X
X
X
X
X
X
X
X
X
X
X
B field into page
10
Free Body Diagram
X
X
X
X
X
X
X
X
X
X
X
X
X
F
X
X BX
X
X
X
X
X
X
X 0.62
Xm X
X
X
X
X
X
X
X
X
X
X
Tension vanishes
when FB =mg
mg
Based on the RH rule the current must go CCW, so i flows
F=iLB (if L perpendicular to B and it is)
mg=iLB or (mg)/(LB)=i
i=(.013*9.8)/(0.62*0.44)=0.467 A
11