This

**preview**has**blurred**sections. Sign up to view the full version! View Full DocumentCHAPTER
28
The Magnetic Field
1* ·
When a cathode-ray tube is placed horizontally in a magnetic field that is directed vertically upward, the
electrons emitted from the cathode follow one of the dashed paths to the face of the tube in Figure 28-30. The
correct path is ____.
(
a
) 1
(
b
) 2
(
c
) 3
(
d
) 4
(
e
) 5
(
b
)
2
·
Why not define
B
to be in the direction of
F
, as we do for
E
?
One cannot define the direction of the force by fiat. By experiment,
F
is perpendicular to
B
.
3
·
Find the magnetic force on a proton moving with velocity 4.46 Mm/s in the positive
x
direction in a magnetic
field of 1.75 T in the positive
z
direction.
Use Equ. 28-1
F
= 1.25 pN
i
·
k
= –1.25 pN
j
4
·
A charge
q
= – 3.64 nC moves with a velocity of 2.75
×
10
6
m/s
i
. Find the force on the charge if the
magnetic field is (
a
)
B
= 0.38 T
j
, (
b
)
B
= 0.75 T
i
+ 0.75 T
j
, (
c
)
B
= 0.65 T
i
, (
d
)
B
= 0.75 T
i
+ 0.75 T
k
.
(
a
), (
b
), (
c
), (
d
) Use Equ. 28-1
(
a
)
F
= –3.8 mN
i
×
j
= -3.8 mN
k
(
b
)
F
= –7.5 mN
k
(
c
)
F
= 0
(
d
)
F
= 7.5 mN
j
5* ·
A uniform magnetic field of magnitude 1.48 T is in the positive
z
direction. Find the force exerted by the field
on a proton if the proton’s velocity is (
a
)
v
= 2.7 Mm/s
i
, (
b
)
v
= 3.7 Mm/s
j
, (
c
)
v
= 6.8 Mm/s
k
, and
(
d
)
v
= 4.0 Mm/s
i
+ 3.0 Mm/s
j
.
(
a
), (
b
), (
c
), (
d
) Use Equ. 28-1
(
a
)
F
= 0.639 pN
i
×
k
= –0.639 pN
j
(
b
)
F
= 0.876 pN
i
(
c
)
F
= 0
(
d
)
F
= 0.71 pN
i
– 0.947 pN
j
6
·
An electron moves with a velocity of 2.75 Mm/s in the
xy
plane at an angle of 60
°
to the
x
axis and 30
°
to
the
y
axis. A magnetic field of 0.85 T is in the positive
y
direction. Find the force on the electron.
Use Equ. 28-1
F
= –0.44(cos 60
o
i
+ cos 30
o
j
)
×
0.85
j
pN
= –0.187 pN
k
7
·
A straight wire segment 2 m long makes an angle of 30
°
with a uniform magnetic field of 0.37 T. Find the
magnitude of the force on the wire if it carries a current of 2.6 A.
Use Equ. 28-4
F
=
BI
l
sin
?
= 0.962 N
8
·
A straight wire segment
I
l
= (2.7 A)(3 cm
i
+ 4 cm
j
) is in a uniform magnetic field
B
= 1.3 T
i
. Find the

Chapter 28
The Magnetic Field
force on the wire.
Use Equ. 28-4
F
= –0.140 N
k
9* ·
What is the force (magnitude and direction) on an electron with velocity
v
= (2
i
– 3
j
)
×
10
6
m/s in a magnetic
field
B
= (0.8
i
+ 0.6
j
– 0.4
k
) T?
Use Equ. 28-4
F
=
–0.192 pN
i
–
0.128 pN
j
–
0.576 pN
k
;
F
= 0.621
pN
10 ··
The wire segment in Figure 28-31 carries a current of 1.8 A from
a
to
b
. There is a magnetic field
B
= 1.2
T
k
. Find the total force on the wire and show that it is the same as if the wire were a straight segment from
a
to
b
.
Use Equ. 28-4
If the wire is straight from
a
to
b
,
l
= 3 cm
i
+ 4 cm
j
F
= –0.0684 N
j
+ 0.0864 N
i
;
F
=
I
l
·
B
= 0.0864 N
i
– 0.0684 N
j
11 ··
A straight, stiff, horizontal wire of length 25 cm and mass 50 g is connected to a source of emf by light,
flexible leads. A magnetic field of 1.33 T is horizontal and perpendicular to the wire. Find the current necessary

This is the end of the preview. Sign up to
access the rest of the document.