Version 069 – TEST02 – tsoi – (57210)
1
This printout should have 16 questions.
Multiplechoice questions may continue on
the next column or page – fnd all choices
beFore answering.
001
10.0 points
AcircularcoilismadeoF
N
turns oF copper
wire as shown in the fgure. A resistor
R
is in
serted in the copper wire. Initially, a uniForm
magnetic feld oF magnitude
B
i
points hor
izontally From leFttoright through the per
pendicular plane oF the coil.
When viewed From the right the coil is
wound counterclockwise.
R
Magnetic
±ield
B
(
t
)
During a time interval
t
the feld uniFormly
changes at a constant rate, until a reversed
feld is reached equal in magnitude to the
initial feld.
The electrons in the resistor,
R,
shown in
the fgure
1.
move in a direction that is undetermi
nated From the inFormation given.
2.
do not move in a preFerred direction, since
they have thermal kinetic energy.
3.
move leFttoright.
4.
move righttoleFt.
correct
Explanation:
As the leFttoright magnetic feld decreases
(and eventually ²ipping sign and increasing
in magnitude) it Follows From Lenz’s law (op
position to the change in magnetic feld will
tend to keep the current constant and ²owing
in the same direction) that the induced emF
will produce an leFttoright magnetic feld
arising From induced currents in the coil.
By the right hand rule, the induced current
²ows counterclockwise when viewed From the
right. The coils are wound counterclockwise
as the wire goes From the right to leFt termi
nals. The current must enter the loop From
the right terminal and exit at the leFt termi
nal.
Since the current is continuous, the current
must ²ow through the resistor in the leFtto
right direction, which means that electrons
are ²owing
righttoleFt
in the resistor. That
is, the electrons ²ow in the opposite direction
From the current since they have a negative
charge.
002
10.0 points
Calculate the resistance in an
RL
circuit in
which
L
=1
.
67 H and the current increases
to 85
.
2% oF its fnal value in 4
.
57 s.
1. 2.13199
2. 1.15712
3. 1.35266
4. 1.30311
5. 0.698163
6. 2.93158
7. 7.1511
8. 3.30293
9. 6.53223
10. 6.25451
Correct answer: 0
.
698163 Ω.
Explanation:
Let :
L
.
67 H
,
R
=85
.
2%
,
and
t
=4
.
57 s
.
At time
t
the current is
I
(
t
)=
I
0
±
1

e

Rt/L
²
=
I
0
R
.
So the resistance is
R
=

L
t
ln(1
R
)
=

1
.
67 H
4
.
57 s
ln(1

0
.
852)
=
0
.
698163 Ω
.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentVersion 069 – TEST02 – tsoi – (57210)
2
003
10.0 points
The fgure below shows a cylindrical coaxial
cable oF radii
a
,
b
,and
c
in which equal, uni
Formly distributed, but antiparallel currents
i
exist in the two conductors.
O
i
out
±
i
in
⊗
F
E
D
C
r
1
r
2
r
3
r
4
c
b
a
Which expression gives the magnitude
B
(
r
3
)a
t
D
oF the magnetic feld in the re
gion
b<r
3
<a
?
1.
B
(
r
3
)=
μ
0
ir
3
2
πb
2
2.
B
(
r
3
μ
0
i
2
πr
3
3.
B
(
r
3
μ
0
i
(
a
2

r
2
3
)
2
3
(
a
2

b
2
)
correct
4.
B
(
r
3
μ
0
i
(
a
2
+
r
2
3

2
b
2
)
2
3
(
a
2

b
2
)
5.
B
(
r
3
μ
0
i
(
a
2

b
2
)
2
3
(
r
2
3

b
2
)
6.
B
(
r
3
μ
0
3
2
πc
2
7.
B
(
r
3
μ
0
3
2
πa
2
8.
B
(
r
3
μ
0
i
3
9.
B
(
r
3
μ
0
i
(
r
2
3

b
2
)
2
3
(
a
2

b
2
)
10.
B
(
r
3
)=0
Explanation:
Ampere’s Law
states that the line inte
gral
±

B
·
d

,
around any closed path equals
μ
0
I
,where
I
is the total steady current pass
ing through any surFace bounded by the closed
path.
This is the end of the preview.
Sign up
to
access the rest of the document.
 Spring '10
 TSOI
 Physics, Magnetic Field, Correct Answer

Click to edit the document details