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Kapoor (mk9499) – oldhomework 18 – Turner – (60230)
1
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001
10.0 points
What is the maximum torque on a 450turn
circular coil oF radius 0
.
85 cm that carries
a current oF 1 mA and resides in a uniForm
magnetic feld oF 0
.
2 T?
Correct answer: 2
.
04282
×
10
−
5
N
·
m.
Explanation:
Let :
N
= 450
,
B
= 0
.
2 T
,
r
= 0
.
85 cm = 0
.
0085 m
,
and
I
= 1 mA = 0
.
001 A
.
μ
=
N I A
=
N I π r
2
τ
max
=
μ B
=
N I π r
2
B
= (450) (0
.
001 A)
π
×
(0
.
0085 m)
2
(0
.
2 T)
=
2
.
04282
×
10
−
5
N
·
m
.
002
(part 1 oF 3) 10.0 points
A particle with charge
q
and mass
m
is un
dergoing circular motion with speed
v
. At
t
= 0, the particle is moving along the nega
tive
x
axis in the plane perpendicular to the
magnetic feld
V
B
, which points in the positive
z
direction (up From the plane oF the fgure).
x
y
z
Vv
V
B
±ind the magnitude oF the centripetal ac
celeration (
b
Va
b ≡
a
).
1.
a
=
q v B
m
correct
2.
a
=
q B m
v
3.
a
=
q m
v B
4.
a
=
v B
q m
5.
a
=
q v B m
6.
a
=
q v m
B
Explanation:
Basic Concepts:
Newton’s 2nd Law:
F
=
m a .
Centripetal acceleration:
F
c
=
m v
2
r
±orce on charge
q
in magnetic feld (no electric
feld):
V
F
B
=
qVv
×
V
B .
Solution:
The trajectory oF the particle will
be bent into a circle by the magnetic feld.
±rom this we understand there has to be a
centripetal acceleration
a
to keep the particle
in this circle (just like a string is needed to
provide the tension to keep a ball whirling in
a circle). Since
F
=
m a
and
F
B
=
q v B
, the
particle moves perpendicularly to the mag
netic feld, so that
m a
=
q v B ,
a
=
q v B
m
.
003
(part 2 oF 3) 10.0 points
±ind the period
T
oF oscillation;
i.e.
, the time
it takes For the particle to complete one revo
lution.
1.
T
=
q B
m
2.
T
=
2
π m
q B
correct
3.
T
= 2
π
m B
q
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View Full DocumentKapoor (mk9499) – oldhomework 18 – Turner – (60230)
2
4.
T
= 2
π
q B
m
5.
T
=
m
q B
6.
T
=
m B
q
Explanation:
The centripetal force
F
c
is provided by
F
B
,
so
m v
2
r
=
q v B ,
v
r
=
q B
m
.
the period of oscillation is
T
=
2
π r
v
,
which is intuitive since the particle traverses
a distance 2
π r
in a revolution and, moving at
speed
v
, takes the time
T
to do so. We know
the ratio
v
r
, so
T
=
2
π r
v
=
2
π
m
q B
.
004
(part 3 of 3) 10.0 points
Find the direction of the instantaneous accel
eration
h
a
at
t
= 0 if
q
is negative.
1.
h
a
=
−
ˆ
j
correct
2.
h
a
=
ˆ
k
3.
h
a
=
ˆ
i
4.
h
a
=
−
ˆ
k
5.
h
a
=
−
ˆ
i
6.
h
a
=
ˆ
j
Explanation:
The particle is moving along the negative
x
axis in this instant:
Vv
=
v
(
−
ˆ
i
) ;
since it is moving in a circle, we need to talk
about instantaneous direction.
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 Fall '08
 Turner
 Physics, Work

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