This preview shows pages 1–3. Sign up to view the full content.
Chapter26 The Magnetic Field:
(Additional Questions)
1)
An electron that has velocity
(
29
(
29
6
6
ˆ
ˆ
2 10 m/s
3 10 m/s
v
i
j
=
×
+
×
r
moves through the magnetic field
(
29
(
29
ˆ
ˆ
0.03 T
0.15 T
B
i
j
=

r
.
(a)
Find the force on the electron.
(b)
Repeat your calculation for a proton having the same velocity.
Solution:
(a)
The force on the electron is
(
29
(
29
(
29
(
29
(
29
(
29
(
29
(
29
(
29
19
6
6
14
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
1.6 10
C
2 10 m/s
0.15 T
3 10 m/s
0.03 T
ˆ
6.2 10
N
B
x
y
x
y
x
y
y
x
F
qv
B
q v i
v j
B i
B j
q v B
v B k
k
k


=
×
=
+
×
+
=

= 
×
×


×
=
×
r
r
r
Thus, the magnitude of
B
F
r
is
14
6.2 10
N

×
, and
B
F
r
points in the positive
z
direction.
(b)
This amount to repeating the above computation with a change in the sign in the change.
Thus,
B
F
r
has the same magnitude but points in the negative
z
direction.
2)
An electron has a velocity of 1.20 × 10
4
m/s (in the positive
x
direction), and an
acceleration of 2.00 × 10
12
m/s
2
(in the positive
z
direction) in uniform electric and magnetic
fields. If the electric field has a magnitude of 20.0 N/C (in the positive
z
direction), what can
you determine about the magnetic field in the region? What can you not determine?
Solution:
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document 3)
An alpha particle (
2 ,
4 u
q
e m
= +
=
) travels in a circular path of radius 4.5 cm in a
uniform magnetic field with
1.2
B
=
T. Calculate
(a)
its speed,
(b)
its period of revolution,
(c)
This is the end of the preview. Sign up
to
access the rest of the document.
This note was uploaded on 04/03/2012 for the course ENGINEERIN 12 taught by Professor Who during the Spring '09 term at Kadir Has Üniversitesi.
 Spring '09
 who

Click to edit the document details