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Brannon (blb675) – HW #9 – Shih – (58250)
1
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beFore answering.
001
(part 1 oF 2) 10.0 points
Two parallel rails having negligible resistance
are connected by a resistor. The circuit also
contains two metal rods with resistance, one
with moving to the leFt and the other moving
to the right. At
t
= 0 seconds the rods are
4 cm From the 2 Ω resistor. The rods are
sliding along the rails and moving away From
the 2 Ω resistor at constant velocities as shown
in fgure below. A uniForm magnetic feld is
applied perpendicular to the plane oF the rails.
××××××
×
× ×
× ×
× ×
×
× ×
× ×
× ×
××××××××××
13Ω
2Ω
18Ω
B
=0
.
0152 T
5m
/
s
2m
/
s
10 cm
4 cm
4 cm
Note:
Drawing not to scale.
Determine the magnitude oF the current in
the 2 Ω resistor at
t
= 6 s.
Correct answer: 0
.
148919 mA.
Explanation:
R
L
E
C
I
L
I
R
I
Let :
R
L
= 13 Ω
,
R
C
= 2 Ω
,
R
R
= 18 Ω
,
v
L
= 2 m
/
s
,
v
R
= 5 m
/
s
,
d
= 10 cm
,
and
B
.
0152 T
.
Motional emF is
!
E
=
!v
×
!
B.
Ohm’s Law is
I
=
V
R
.
The 4 cm distances and the time interval
have no bearing on the solution to this prob
lem since the rods are traveling at a constant
velocity.
The motional emF induced in the two mov
ing rods are as Following, respectively For the
leFthand side, we have
E
L
=
B v
L
= (0
.
0152 T) (0
.
1 m) (2 m
/
s)
.
00304 V
,
and For the righthand side, we have
E
R
=
R
= (0
.
0152 T) (0
.
1 m) (5 m
/
s)
.
0076 V
.
The junction equation is
I
L
+
I
C

I
R
.
(1)
Using the leFthand current loop (in a counter
clockwise direction), we have
+
R
L
I
L

R
C
I
C
+0
I
R
=
E
L
.
(2)
Using the righthand current loop (in a clock
wise direction), we have
+0
I
L
+
R
C
I
C
+
R
R
I
R
=
E
R
.
(3)
Solving these three linear equations using the
determinant technique, we have
I
C
=
±
±
±
±
±
±
10

1
R
L
E
L
0
0
E
R
R
R
±
±
±
±
±
±
±
±
±
±
±
±
11

1
R
L

R
C
0
0
R
C
R
R
±
±
±
±
±
±
=
E
L
R
R
 E
R
R
L

R
R
R
C

R
L
R
C

R
L
R
R
=

0
.
04408 V Ω

296 Ω
2
.
000148919 A
.
148919 mA

I
C

=
0
.
148919 mA
.
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View Full DocumentBrannon (blb675) – HW #9 – Shih – (58250)
2
where
E
L
R
R
 E
R
R
L
= (0
.
00304 V) (18 Ω)

(0
.
0076 V) (13 Ω)
=

0
.
04408 V Ω
,
and

R
R
R
C

R
L
R
C

R
L
R
R
=

(18 Ω) (2 Ω)

(13 Ω) (2 Ω)

(13 Ω) (18 Ω)
=

296 Ω
2
.
Incorrect Solution:
In this problem, it’s
important to know that the current in the 2 Ω
resistor is the sum of
I
L
and
I
R
, which is due
to the two emfs induced in the moving rods
respectively.
From Lenz’s Law, we know that
I
L
has a
direction di±erent from that of
I
R
, namely,
I
L
is counterclockwise while
I
R
is clockwise. So
the current should be
I
C
=
I
R

I
L
.
Applying Ohm’s Law, we get
I
L
,
I
R
I
L
=
E
L
R
+
R
L
=0
.
000202667 A
.
I
R
=
E
R
R
+
R
R
.
00038 A
.
So the ²nal current ³ow through 2 Ω resistor
is
I
C
=
I
R

I
L
= (0
.
00038 A)

(0
.
000202667 A)
.
000177333 A
.
177333 mA
.
Since the two sides of the circuit are coupled,
they cannot be treated independently, conse
quently this
incorrect
solution is
WRONG
.
002
(part 2 of 2) 10.0 points
The current in the 2 Ω resistor is ³owing
1.
upward.
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 Spring '08
 Turner
 Physics, Resistance

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