CHAPTER
28
The Magnetic Field
1*
∙
When a cathoderay 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 2830. 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. 281
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. 281
(
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. 281
(
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. 281
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. 284
F
=
BI
!
sin
θ
= 0.962 N
8
∙
A straight wire segment
I
!
= (2.7 A)(3 cm
i
+ 4 cm
j
) is in a uniform magnetic field
B
= 1.3 T
i
. Find the
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Chapter 28
The Magnetic Field
force on the wire.
Use Equ. 284
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. 284
F
=
–0.192 pN
i
–
0.128 pN
j
–
0.576 pN
k
;
F
= 0.621 pN
10
∙∙
The wire segment in Figure 2831 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. 284
If the wire is straight from
a
to
b
,
!
= 3 cm
i
+ 4 cm
j
F
= –0.0684 N
j
+ 0.0864 N
i
;
F
=
I
!
×
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 to float the wire, that is, the current such that the magnetic force balances the weight of the wire.
F
= (
I
!
B – mg
)
k
= 0
I
=
mg/
!
B
= 1.48 A
12
∙∙
A simple gaussmeter for measuring horizontal magnetic fields consists of a stiff 50cm wire that hangs
from a conducting pivot so that its free end makes contact with a pool of mercury in a dish below. The mercury
provides an electrical contact without constraining the movement of the wire. The wire has a mass of 5 g and
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 Spring '08
 staff
 Magnetic Field, Ri, Equ.

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