Anderson, Matthew – Homework 12 – Due: Apr 24 2008, noon – Inst: Vitaly
1
This
printout
should
have
17
questions.
Multiplechoice questions may continue on
the next column or page – fnd all choices
beFore answering.
The due time is Central
time.
001
(part 1 oF 3) 15 points
A 20
.
8 mA current is carried by a uniFormly
wound aircore solenoid with 233 turns as
shown in the fgure below.
The
permeability
oF
Free
space
is
1
.
25664
×
10

6
N
/
A
2
.
20
.
8
mA
23
.
2 mm
14
.
9 cm
Compute the magnetic feld inside the
solenoid.
Correct answer: 4
.
08736
×
10

5
T.
Explanation:
Let :
N
= 233
,
‘
= 14
.
9 cm
,
I
= 20
.
8 mA
,
and
μ
0
= 1
.
25664
×
10

6
N
/
A
2
.
I
d
‘
The magnetic feld inside the solenoid is
B
=
μ
0
n I
=
μ
0
µ
N
‘
¶
I
= (1
.
25664
×
10

6
N
/
A
2
)
µ
233
0
.
149 m
¶
×
(0
.
0208 A)
=
4
.
08736
×
10

5
T
.
002
(part 2 oF 3) 15 points
Compute the magnetic ±ux through each
turn.
Correct answer: 1
.
72786
×
10

8
Wb.
Explanation:
Let :
B
= 4
.
08736
×
10

5
T
,
and
d
= 23
.
2 mm = 0
.
0232 m
.
The magnetic ±ux through each turn is
Φ =
B A
=
B
‡
π
4
d
2
·
= (4
.
08736
×
10

5
T)
π
4
(0
.
0232 m)
2
=
1
.
72786
×
10

8
Wb
.
003
(part 3 oF 3) 15 points
Compute the inductance oF the solenoid.
Correct answer: 0
.
193554 mH.
Explanation:
The inductance oF the solenoid is
L
=
N
Φ
I
=
(233) (1
.
72786
×
10

8
Wb)
0
.
0208 A
=
0
.
193554 mH
.
keywords:
004
(part 1 oF 5) 15 points
A circuit is set up as shown in the fgure.
L
R
1
R
2
E
S
I
1
I
2
I
This preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentAnderson, Matthew – Homework 12 – Due: Apr 24 2008, noon – Inst: Vitaly
2
The switch is closed at
t
= 0. The current
I
through the inductor takes the form
I
=
E
R
x
‡
1

e

t/τ
x
·
where
R
x
and
τ
x
are to be determined.
Find
I
immediately after the circuit is
closed.
1.
I
=
E
R
1
+
R
2
2.
I
=
E
R
2
3.
I
= 0
correct
4.
I
=
E
R
1
Explanation:
Before the circuit is closed, no current is
±owing. When we have just closed the circuit
we are at “
t
= 0
+
”, a mathematical nota
tion meaning a very short time
²
after
t
= 0.
(Nothing happens in the circuit at
t
= 0, only
immediately after when the switch is, indeed,
closed. However, this is just a mathematical
detail.)
There are two loops in the prob
lem, one with
E
,
R
1
,
R
2
and one with
E
,
R
1
,
L
.
So at
t
= 0
+
, the battery “wants to”
drive a current through both loops. The ²rst
loop presents no problem; since there is no in
ductance “working against us,” a current will
immediately be set up. The second loop, how
ever, has an inductor which tries to prevent
any change in the current going through it,
and so goes up smoothly from
I
= 0, as can
be seen in the given solution (just put
t
= 0
to ²nd
I
= 0). Therefore, at this instant, the
inductor
L
carries no current, and we can ne
glect it when we ²nd the current through
R
2
.
The equivalent resistance is
R
eq
=
R
1
+
R
2
so, from
E
=
R
eq
I
we ²nd
I
2
=
E
R
1
+
R
2
005
(part 2 of 5) 15 points
Find
I
2
immediately after the circuit is closed.
This is the end of the preview.
Sign up
to
access the rest of the document.
 Spring '08
 Opyrchal
 Inductance, Work, Resistor, Inductor, Electrical resistance, RL circuit

Click to edit the document details