nguyen (jmn727) – homework 21 – Turner – (59070)
1
This
printout
should
have
10
questions.
Multiplechoice questions may continue on
the next column or page – find all choices
before answering.
001
(part 1 of 3) 10.0 points
A metal strip 1
.
8 cm wide and 0
.
09 cm thick
carries a current of 28 A in a uniform magnetic
field of 1
.
8 T.
The Hall voltage is measured
to be 4
.
35
μ
V.
The charge on an electron is 1
.
6
×
10
−
19
C.
vector
B
1
.
8 cm
0
.
09 cm
I
a
b
Calculate the drift velocity of the electrons
in the strip.
Correct answer: 0
.
134259 mm
/
s.
Explanation:
Let :
V
H
= 4
.
35
μ
V = 4
.
35
×
10
−
6
V
,
B
= 1
.
8 T
,
and
w
= 1
.
8 cm = 0
.
018 m
.
The Hall voltage as a function of the drift
velocity of the electrons in the strip is
V
H
=
v
d
B w ,
so
v
d
=
V
H
B w
=
4
.
35
×
10
−
6
V
(1
.
8 T) (0
.
018 m)
·
10
3
mm
1 m
=
0
.
134259 mm
/
s
.
002
(part 2 of 3) 10.0 points
The charge on the electron is 1
.
6
×
10
−
19
C.
Find the number density of the charge car
riers in the strip.
Correct answer: 8
.
04598
×
10
28
m
−
3
.
Explanation:
Let :
I
= 28 A
,
t
= 0
.
09 cm = 0
.
0009 m
,
and
q
= 1
.
6
×
10
−
19
C
.
The current is
I
=
n A q v
d
n
=
I
A q v
d
=
I
w t q v
d
=
28 A
(0
.
018 m) (0
.
0009 m)
×
1
(1
.
6
×
10
−
19
C) (0
.
000134259 m
/
s)
=
8
.
04598
×
10
28
m
−
3
.
003
(part 3 of 3) 10.0 points
Is point
a
or point
b
at the higher potential?
1.
V
a
< V
b
2.
V
a
> V
b
correct
3.
V
a
=
V
b
Explanation:
Apply a righthand rule to
I
vector
ℓ
and
vector
B
to
conclude that positive charge will accumulate
at
a
and negative charge at
b
, so
V
a
> V
b
.
004
(part 1 of 2) 10.0 points
A current
I
= 3 A flows through a wire
perpendicular to the paper and towards the
reader at
A
and back in the opposite direc
tion at
C
. Consider the wires below the plane
at
A
and
C
to be semiinfinite. In the figure,
L
1
= 5 m,
R
= 5 m, and
L
2
= 6 m and there
is a
B
= 9
.
73 T magnetic field into the paper
(not including the field due to the current in
the wire).
Caution:
It may be necessary to take
into account the contribution from the long
straight wire which runs
up to
and
down from
the underneath side of the page.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
nguyen (jmn727) – homework 21 – Turner – (59070)
2
5 m
3 A
3 A
6 m
3 A
5 m
A
C
O
⊗
B
= 9
.
73 T
What is the magnitude of the force on the
wire due to the external magnetic field
B
?
Correct answer: 433
.
941 N.
Explanation:
Let :
R
= 5 m
,
I
= 3 A
,
L
1
= 5 m
,
L
2
= 6 m
,
and
B
= 9
.
73 T
.
By the BiotSavart law,
d
vector
B
=
μ
0
4
π
I dvectors
×
ˆ
r
r
2
.
The contribution from the long straight wire
which runs
into
and
out of
the page is zero
since the external field and the current are
parallel.
The force on a current carrying wire from
point A, at
vectorr
1
, to point C, at
vector
r
2
, in a uniform
field is
vector
F
=
I
integraldisplay
vectorr
2
vectorr
1
(
dvectors
×
vector
B
)
.
Since
vector
B
is a constant, it can be taken out of
the integral and we can write (recalling that if
we change the order of the cross product, we
need to change the overall sign)
vector
F
=
−
I
vector
B
×
integraldisplay
vectorr
2
vectorr
1
dvectors
=
−
I
vector
B
×
(
vectorr
2
−
vectorr
1
)
.
This is the end of the preview.
Sign up
to
access the rest of the document.
 Spring '08
 Turner
 Work, Magnetic Field

Click to edit the document details