Reilly (mar3978) – hw19 – turner – (56725)
1
This printout should have 17 questions.
Multiplechoice questions may continue on
the next column or page – fnd all choices
beFore answering.
001
10.0 points
Consider the set up shown in the fgure
where a solenoid has a steadily increasing
magnetic ±ux which generates identical in
duced
emf
s For the two cases illustrated.
B
B
B
B
A
#1
B
#2
i
i
Case 1:
Case 1: Two identical light bulbs are in
series.
The corresponding electrical power
consumed by bulb 1 and bulb 2 are
P
1
and
P
2
,
respectively.
B
B
B
B
A
#1
B
#2
D
C
O
i
2
1
Case 2:
Case 2: Let the points
C
and
D
be on the
symmetry line oF the diagram. Connect points
C
and
D
by a wire, which equally divides the
magnetic ±ux. The corresponding electrical
power consumed by bulb 1 and bulb 2 are
P
′
1
and
P
′
2
, respectively.
What is the ratio oF
P
′
1
P
1
? It may be helpFul
to frst fnd an expression For
P
1
and then
write down the loop equation For
ACODA
in
case 2.
1.
P
′
1
P
1
= 2
2.
P
′
1
P
1
= 0
3.
P
′
1
P
1
=
1
8
4.
P
′
1
P
1
= 8
5.
P
′
1
P
1
=
1
3
6.
P
′
1
P
1
=
1
2
7.
P
′
1
P
1
= 4
8.
P
′
1
P
1
= 3
9.
P
′
1
P
1
=
1
4
10.
P
′
1
P
1
= 1
correct
Explanation:
Let
E
and
R
be the induced
emf
and resis
tance oF the light bulbs, respectively.
²or case 1, since the two bulbs are in series,
the equivalent resistance is simply
R
eq
=
R
+
R
= 2
R
and the current through the bulbs is
I
=
E
2
R
.
Hence, For case 1, the power consumed by
bulb 1 is
P
1
=
p
E
2
R
P
2
R
=
E
2
4
R
.
²or case 2, the loop equation For
is
E
2
−
I
1
R
= 0
.
Solving For
I
1
yields
I
1
=
E
2
R
.
Hence the power dissipated by bulb 2 For case
2 is
P
′
1
=
p
E
2
R
P
2
R
=
E
2
4
R
.
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2
This is identical to the expression for
P
1
.
Therefore,
P
′
1
P
1
= 1
.
keywords:
002 (part 1 of 3) 10.0 points
Four long, parallel conductors carry equal cur
rents of
I
. An end view of the conductors is
shown in the ±gure. Each side of the square
has length of
ℓ
.
A
B
D
C
×
P
x
y
ℓ
Which diagram correctly denotes the direc
tions of the magnetic ±elds from each conduc
tor at the point
P
? The current direction is
out of the page at points
A
,
B
, and
D
indi
cated by the dots and into the page at point
C
indicated by the cross.
1.
P
B
A
B
B
B
C
B
D
2.
P
B
A
B
B
B
C
B
D
3.
P
B
A
B
B
B
C
B
D
4.
P
B
A
B
B
B
C
B
D
5.
P
B
A
B
B
B
C
B
D
correct
Explanation:
The direction of the magnetic ±eld due to
each wire is given by the right hand rule.
Place the thumb of the right hand along the
direction of the current; your ±ngers now curl
in the direction of the magnetic ±eld’s circular
path:
A
B
D
C
×
P
B
A
B
B
B
D
B
C
003 (part 2 of 3) 10.0 points
At point
P
, as far as the magnitudes are
concerned,
B
A
=
B
B
=
B
C
=
B
D
≡
B
i
,
where
B
i
is introduced to represent any of the
four magnetic ±elds.
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 Spring '10
 TURNER,J
 Force, Magnetic Field, Reilly

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