Husain, Zeena – Exam 3 – Due: Apr 13 2004, 10:00 pm – Inst: Sonia Paban
1
This
printout
should
have
16
questions.
Multiplechoice questions may continue on
the next column or page – find all choices
before making your selection. The due time is
Central time.
001
(part 1 of 2) 10 points
At some time after the switch S is closed,
there is a current
I
flowing through the resis
tor and inductor which is increasing in time
d I
dt
>
0 (see the figure below). At this time
there is an amount of charge of magnitude
q
on each plate of the capacitor.
E
S
C
R
L
I
Which of the following equations is correct?
1.
none of these
2.
E
+
I R

L
d I
dt

q
C
= 0
3.
E 
I R

L
d I
dt

q
C
= 0
correct
4.
E 
I R
+
L
d I
dt

q
C
= 0
5.
E
+
I R

L
d I
dt
+
q
C
= 0
6.
E 
I R

L
d I
dt
+
q
C
= 0
7.
E
+
I R
+
L
d I
dt

q
C
= 0
8.
E
+
I R
+
L
d I
dt
+
q
C
= 0
9.
E 
I R
+
L
d I
dt
+
q
C
= 0
Explanation:
Apply Kirchoff’s loop rule to the above cir
cuit. Here we are told that
d I
dt
>
0; therefore
the
emf
induced by the inductor is

L
d I
dt
.
Summing the potential differences around the
loop gives
E 
I R

L
d I
dt

q/C
= 0
.
002
(part 2 of 2) 10 points
18 V
S
7
.
07
μ
F
4
.
18 MΩ
5
.
48 mH
In the circuit shown above, what is the
charge on the capacitor after the switch has
been closed for a long time?
Correct answer: 0
.
00012726 C.
Explanation:
Let :
E
= 18 V
,
R
= 4
.
18 MΩ
,
L
= 5
.
48 mH
,
and
C
= 7
.
07
μ
F
.
After a long time
t
the capacitor will reach
its maximum charge
q
max
.
When this occurs
the current will drop to zero. Setting
I
= 0 in
the loop equation from Part 1 gives
E 
q
max
C
= 0
,
or
q
max
=
C
E
= (7
.
07
μ
F) (1
×
10

6
F
/μ
F) (18 V)
= 0
.
00012726 C
,
003
(part 1 of 1) 10 points
Assume:
The induced emf for the closed
loop octagonal
CXDY C
is
E
.
A solenoid (with magnetic field
B
) pro
duces a steadily increasing uniform magnetic
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Husain, Zeena – Exam 3 – Due: Apr 13 2004, 10:00 pm – Inst: Sonia Paban
2
flux through its circular cross section.
A
octagonal circuit surrounds the solenoid as
shown in the figure.
The wires connecting
in the circuit are ideal, having no resistance.
The circuit consists of two identical light bulbs
(labeled
X
and
Y
) in series. A wire connects
points
C
and
D
.
The ratio of the solenoid’s
area
A
L
left of the wire
CD
and the solenoid’s
area
A
R
right of the wire
CD
is
A
L
A
R
= 4
.
B
B
B
B
Y
X
D
C
i
3
i
1
i
2
A
L
A
R
The equations for the (right) loop
CXDC
and the (left) loop
CDY C
are respectively
given by
1.
E
5
+
i
1
R
= 0
and
4
E
5

i
2
R
= 0
.
2.
4
E
5
+
i
1
R
= 0
and
E
5
+
i
2
R
= 0
.
3.
E
5

i
1
R
= 0
and
4
E
5

i
2
R
= 0
.
4.
4
E
5

i
1
R
= 0
and
E
5
+
i
2
R
= 0
.
5.
4
E
5

i
1
R
= 0
and
E
5

i
2
R
= 0
.
6.
E
5

i
1
R
= 0
and
4
E
5
+
i
2
R
= 0
.
7.
E
5
+
i
1
R
= 0
and
4
E
5
+
i
2
R
= 0
.
cor
rect
8.
4
E
5
+
i
1
R
= 0
and
E
5

i
2
R
= 0
.
Explanation:
By definition, the areas of the left and right
loops are related by
A
=
A
L
+
A
R
.
Since
A
L
A
R
= 4, we can solve for
A
L
and
A
R
in
terms of
A
.
A
L
=
4
A
5
A
R
=
A
5
.
Then we can compute the magnitude of the
induced emf around the right and left loops.
This is the end of the preview.
Sign up
to
access the rest of the document.
 Spring '08
 Turner
 Magnetic Field, Husain, Sonia Paban

Click to edit the document details