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Physics 21
Fall, 2004
Solution to HW12
27
P29
An electron enters a uniform magnetic Feld 0.23 T
ata45
◦
to
B
. Determine the radius
r
and pitch
p
(distance
between loops) of the electron’s helical path assuming its
speed is 3
.
0
×
10
6
m/s.
Since
v
makes an angle of 45
◦
with
B
, the components of
v
parallel and perpendicular to
B
(
v
±
and
v
⊥
) both have the
value
v
0
/
√
2 (since sin
θ
=cos
θ
=1
/
√
2for
θ
=45
◦
), where
v
0
is the given speed 3
.
0
×
10
6
m/s. We get the radius of the
circular motion using
v
⊥
:
r
=
m
e
v
⊥
eB
=
(9
.
11
×
10
−
31
)(3
.
0
×
10
6
)
(1
.
6
×
10
−
19
)
√
2(0
.
23)
=52
.
5
µ
m
The time to make one loop of circular motion is
∆
t
=
2
πr
v
⊥
=
2
πm
e
v
⊥
eBv
⊥
=
2
πm
e
eB
.
The pitch is the distance travelled parallel to
B
(at speed
v
±
) in time ∆
t
(one loop):
p
=
v
±
∆
t
=
2
πm
e
v
±
eB
=3
.
30
×
10
−
4
m = 330
µ
m
27
P34
Show that the magnetic dipole moment
µ
of an
electron orbiting the proton nucleus of a hydrogen atom is
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 Fall '07
 Hickman
 Physics

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