terry (ect328) – homework 21 – Turner – (59130)
1
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printout
should
have
10
questions.
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before answering.
001
(part 1 of 3) 10.0 points
A metal strip 1
.
2 cm wide and 0
.
13 cm thick
carries a current of 22 A in a uniform magnetic
field of 1
.
4 T.
The Hall voltage is measured
to be 4
.
22
μ
V.
The charge on an electron is 1
.
6
×
10
−
19
C.
vector
B
1
.
2 cm
0
.
13 cm
I
a
b
Calculate the drift velocity of the electrons
in the strip.
Correct answer: 0
.
25119 mm
/
s.
Explanation:
Let :
V
H
= 4
.
22
μ
V = 4
.
22
×
10
−
6
V
,
B
= 1
.
4 T
,
and
w
= 1
.
2 cm = 0
.
012 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
.
22
×
10
−
6
V
(1
.
4 T) (0
.
012 m)
·
10
3
mm
1 m
=
0
.
25119 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: 3
.
50893
×
10
28
m
−
3
.
Explanation:
Let :
I
= 22 A
,
t
= 0
.
13 cm = 0
.
0013 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
=
22 A
(0
.
012 m) (0
.
0013 m)
×
1
(1
.
6
×
10
−
19
C) (0
.
00025119 m
/
s)
=
3
.
50893
×
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
correct
2.
V
a
< V
b
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
= 7 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
= 3 m, and
L
2
= 7 m and there
is a
B
= 5
.
75 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.
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terry (ect328) – homework 21 – Turner – (59130)
2
5 m
7 A
7 A
7 m
7 A
3 m
A
C
O
⊗
B
= 5
.
75 T
What is the magnitude of the force on the
wire due to the external magnetic field
B
?
Correct answer: 515
.
452 N.
Explanation:
Let :
R
= 3 m
,
I
= 7 A
,
L
1
= 5 m
,
L
2
= 7 m
,
and
B
= 5
.
75 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
)
.
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 Fall '09
 Turner
 Magnetic Field, Terry

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