Niu (qn269) – Homework #5 – antoniewicz – (58855)
1
This
printout
should
have
17
questions.
Multiplechoice questions may continue on
the next column or page – find all choices
before answering.
001
10.0 points
A wire carrying a current 40 A has a length
0
.
1 m between the pole faces of a magnet at
an angle 60
◦
(see the figure). The magnetic
field is approximately uniform at 0
.
5 T.
We
ignore the field beyond the pole pieces.
ℓ
θ
I
B
What is the force on the wire?
Correct answer: 1
.
73205 N.
Explanation:
Let :
I
= 40 A
,
ℓ
= 0
.
1 m
,
θ
= 60
◦
,
and
B
= 0
.
5 T
.
we use
F
=
I ℓ B
sin
θ
, so
F
=
I ℓ B
sin
θ
= (40 A) (0
.
1 m) (0
.
5 T) sin 60
◦
=
1
.
73205 N
.
002
(part 1 of 2) 6.0 points
In the figure, a 40 cm length of conducting
wire that is free to move is held in place
between two thin conducting wires. All of the
wires are in a magnetic field.
When a 11 A
current is in the wire, as shown in the figure,
the wire segment moves upward at a constant
velocity.
The acceleration of gravity is 9
.
81 m
/
s
2
.
40 cm
11 A
11 A
11 A
a) Assuming the wire slides without friction
on the two vertical conductors and has a mass
of 0
.
13 kg, find the magnitude of the minimum
magnetic field that is required to move the
wire.
Correct answer: 0
.
289841 T.
Explanation:
Let :
ℓ
= 40 cm
,
I
= 11 A
,
m
= 0
.
13 kg
,
and
g
= 9
.
81 m
/
s
2
.
The magnetic and gravitational forces are
equal:
F
m
=
F
g
B I ℓ
=
m g
B
=
m g
I ℓ
=
(0
.
13 kg) (9
.
81 m
/
s
2
)
(11 A) (0
.
4 m)
=
0
.
289841 T
003
(part 2 of 2) 4.0 points
b) What is its direction?
1.
Into the page
2.
None of these
3.
Toward the left edge of the page
4.
Toward the top edge of the page
5.
Toward the bottom edge of the page
6.
Out of the page
correct
7.
Toward the right edge of the page
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Niu (qn269) – Homework #5 – antoniewicz – (58855)
2
Explanation:
Apply righthand rule; force directed out of
the palm of the hand, fingers in the direction
of the field, thumb in the direction of the
current.
Thumb points to the left, palm faces toward
the top of the page, and the fingers point out
of the page.
004
10.0 points
An alpha particle has a mass of 6
.
6
×
10
−
27
kg
and is accelerated by a voltage of 1
.
35 kV.
The charge on a proton is 1
.
60218
×
10
−
19
C.
If a uniform magnetic field of 0
.
073 T is
maintained on the alpha particle and perpen
dicular to its velocity, what will be particle’s
radius of curvature?
Correct answer: 0
.
102155 m.
Explanation:
Let :
B
= 0
.
073 T
,
V
= 1
.
35 kV = 1350 V
,
m
= 6
.
6
×
10
−
27
kg
,
and
q
= 2
e
= 3
.
20435
×
10
−
19
C
.
From Newton’s second law,
F
=
q v B
=
m v
2
r
v
=
q B r
m
.
The kinetic energy is
K
=
1
2
m v
2
=
q
2
B
2
r
2
2
m
=
qV ,
so the particle’s radius of curvature is
r
=
1
B
radicalBigg
2
V m
q
=
1
0
.
073 T
radicalBigg
2 (1350 V) (6
.
6
×
10
−
27
kg)
3
.
20435
×
10
−
19
C
=
0
.
102155 m
.
005
(part 1 of 3) 6.0 points
The toroid has its inner radius
a
, its outer
radius
b
, a height of
h
, and its number of
turns
N
. The rectangular crosssectonal area
of the hollow core is (
b
−
a
)
h
.
This is the end of the preview.
Sign up
to
access the rest of the document.
 Spring '09
 Antoniewicz
 Current, Work, Magnetic Field

Click to edit the document details