aljabr (faa335) – Hw16 – Ross – (89251)
1
This
printout
should
have
14
questions.
Multiplechoice questions may continue on
the next column or page – find all choices
before answering.
001
10.0 points
A rectangular loop located a distance from a
long wire carrying a current is shown in the
figure. The wire is parallel to the longest side
of the loop.
7
.
89 cm
4
.
89 cm
39
.
2 cm
0
.
0179 A
Find the total magnetic flux through the
loop.
Correct answer: 6
.
7682
×
10
−
10
Wb.
Explanation:
Let :
c
= 7
.
89 cm
,
a
= 4
.
89 cm
,
b
= 39
.
2 cm
,
and
I
= 0
.
0179 A
.
c
a
b
r
dr
I
From Amp`
ere’s law, the strength of the
magnetic field created by the currentcarrying
wire at a distance
r
from the wire is (see
figure.)
B
=
μ
0
I
2
π r
,
so the field varies over the loop and is directed
perpendicular to the page. Since
vector
B
is parallel
to
d
vector
A
, the magnetic flux through an area
element
dA
is
Φ
≡
integraldisplay
B dA
=
integraldisplay
μ
0
I
2
π r
dA .
Note:
vector
B
is not uniform but rather depends on
r
, so it cannot be removed from the integral.
In order to integrate, we express the area
element shaded in the figure as
dA
=
b dr
.
Since
r
is the only variable that now appears
in the integral, we obtain for the magnetic
flux
Φ
B
=
μ
0
I
2
π
b
integraldisplay
a
+
c
c
d r
r
=
μ
0
I b
2
π
ln
r
vextendsingle
vextendsingle
vextendsingle
a
+
c
c
=
μ
0
I b
2
π
ln
parenleftbigg
a
+
c
c
parenrightbigg
=
μ
0
(0
.
0179 A)(0
.
392 m)
2
π
ln
parenleftbigg
a
+
c
c
parenrightbigg
=
μ
0
(0
.
0179 A)(0
.
392 m)
2
π
(0
.
482285)
=
6
.
7682
×
10
−
10
Wb
.
002
(part 1 of 2) 10.0 points
A 0
.
0643 A current is charging a capacitor
that has square plates, 3
.
23 cm on a side.
The permitivity of free space is 8
.
85419
×
10
−
12
C
2
/
N m
2
.
If the plate separation is 2
.
35 mm find the
time rate of change of electric flux between
the plates.
Correct answer: 7
.
2621
×
10
9
V m
/
s.
Explanation:
Let :
I
= 0
.
0643 A
and
ǫ
0
= 8
.
85419
×
10
−
12
C
2
/
N m
2
.
The electric field between the two plates is
E
=
Q
ǫ
0
A
.
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aljabr (faa335) – Hw16 – Ross – (89251)
2
Then the rate of change of electric flux be
tween the plates is
d
Φ
E
dt
=
d
dt
(
E A
)
=
I
ǫ
0
=
0
.
0643 A
8
.
85419
×
10
−
12
C
2
/
N m
2
=
7
.
2621
×
10
9
V m
/
s
.
003
(part 2 of 2) 10.0 points
Find the displacement current between the
plates.
Correct answer: 0
.
0643 A.
Explanation:
The displacement current is
I
d
=
ǫ
0
d
Φ
E
dt
=
I
=
0
.
0643 A
.
004
(part 1 of 2) 10.0 points
A long, straight wire carries a current and lies
in the plane of a rectangular loops of wire, as
shown in the figure.
7
.
3 cm
12 cm
26 cm
(21 A) sin
bracketleftBig
(219 rad
/
s)
t
+
δ
bracketrightBig
→
140 loops (turns)
Determine the maximum
emf
,
E
, induced
in the loop by the magnetic field created by
the current in the straight wire.
Correct answer: 32
.
551 mV.
Explanation:
Let :
N
= 140
,
ω
= 219 rad
/
s
,
ℓ
= 26 cm
,
a
= 7
.
3 cm
,
b
= 12 cm
,
and
I
0
= 21 A
.
a
b
ℓ
r
dr
I
=
I
0
sin(
ω t
+
δ
)
Magnetic field near a long wire is
B
=
μ
0
I
2
π r
.
Faraday’s Law is
E
=

d
Φ
B
dt
.
The magnitude of the magnetic field is
B
=
μ
0
I
2
π r
.
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 Spring '10
 Gavrine
 Magnetic Field, Gauss’s Law

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