ragsdale (zdr82) – HW8 – ditmire – (58335)
1
This printout should have 21 questions.
Multiplechoice questions may continue on
the next column or page – fnd all choices
beFore answering.
001
10.0 points
A conductor consists oF an infnite number
oF adjacent wires, each infnitely long and
carrying a current
I
(whose direction is outoF
thepage), thus Forming a conducting plane.
A
C
IF there are
n
wires per unit length, what is
the magnitude oF
*
B
?
1.
B
=
μ
0
I
2
2.
B
=4
μ
0
I
3.
B
=2
μ
0
nI
4.
B
μ
0
5.
B
=
μ
0
I
6.
B
μ
0
I
7.
B
=
μ
0
4
8.
B
=
μ
0
I
4
9.
B
=
μ
0
2
correct
10.
B
=
μ
0
Explanation:
l
W
A
C
B
B
By symmetry the magnetic felds are equal
and opposite through point
A
and
C
and hori
zontally oriented. ±ollowing the dashed curve
in a counterclockwise direction, we calculate
±
*
B
·
d*s
, which by Ampere’s law is propor
tional to the current through the dashed loop
coming out oF the plane oF the paper.
In
this problem this is a positive current. Hence
*
B
along the horizontal legs points in the di
rection in which we Follow the dashed curve.
Ampere’s Law is
±
*
B
·
d*s
=
μ
0
I.
To evaluate this line integral, we use the rect
angular path shown in the fgure. The rectan
gle has dimensions
l
and
w
. The net current
through the loop is
nI l
. Note that since there
is no component oF
*
B
in the direction oF
w
, we
are only interested in the contributions along
sides
l
±
*
B
·
d*s
Bl
=
μ
0
nl I
B
=
μ
0
2
.
002
10.0 points
A superconducting solenoid has 5810 turns
/
m
and carries a current oF 2000 A.
What is the magnetic feld generated inside
the solenoid? The permeability oF Free space
is 1
.
25664
×
10

6
T
·
m
/
A.
Correct answer: 14
.
6021 T.
Explanation:
Let :
n
= 5810 turns
/
m
,
μ
0
=1
.
25664
×
10

6
T
·
m
/
A
,
and
I
= 2000 A
.
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View Full Documentragsdale (zdr82) – HW8 – ditmire – (58335)
2
The magnetic feld generated inside the
solenoid is
B
=
μ
0
nI
=
(
1
.
25664
×
10

6
T
·
m
/
A
)
×
(5810 turns
/
m) (2000 A)
=
14
.
6021 T
.
keywords:
003
(part 1 oF 2) 10.0 points
A capacitor oF capacitance
C
has a charge
Q
at
t
= 0. At that time, a resistor oF resistance
R
is connected to the plates oF the charged
capacitor.
±ind the magnitude oF the dis
placement current between the plates oF the
capacitor as a Function oF time.
1.
Q
RC
e

t/Q
2.
RC
Q
e
t/
(
RC
)
3.
Q
RC
e

t/
(
RC
)
correct
4.
Q
RC
e
t/
(
RC
)
5.
RC
Q
e

t/
(
RC
)
Explanation:
Basic Concept
RC circuits. Displacement Current.
The displacement current is defned to be
I
d
=
(
0
d
Φ
E
dt
.
The electric feld inside a capacitor is essen
tially uniForm and
E
=
q
(
0
A
. Since the charge
on a capacitor in a discharging
RC
circuit is
given by
q
(
t
)=
Qe

t/RC
, the displacement
current is Found by
I
d
=
(
0
d
Φ
E
dt
=
(
0
d
dt
±
q
(
0
A
A
²
=
dq
dt
=

Q
RC
e

t/
(
RC
)
.
Note that the displacement current equals the
actual current in the wires to the capacitor.
Thus, the AmpereMaxwell law tells us that
*
B
will be the same regardless oF which current
we evaluate.
004
(part 2 oF 2) 10.0 points
Given
C
=3
μ
±,
Q
= 37
μ
C,
R
= 384 kΩ,
and
(
0
=8
.
85419
×
10

12
C
2
/
N
·
m
2
, at what
rate is the electric ²ux between the plates
changing at time
t
=0
.
19 s?
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 Spring '10
 Turner
 Magnetic Field, ragsdale, PY PY PY, PX PX PX

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