PH214_2010SPRING_EXAM1_PROFSOLN_[0] - Ph 214 Spring 2010...

Ph 214, Spring 2010
Exam # 1
Question
#1 SOLUTIONS
1. [30 pts] Mass spectroscopy: how heavy is that ion?
(a) [5 pts] Show that the deflection angle is given by θ =
q B
0
d
m v
.
The chamber is essentially a segment of a cyclotron and the particle
follows a circular trajectory while it is inside the chamber. The
radius of the trajectory is obtained by equating the centripetal
force with the Lorentz force:
mv
2
R
= qvB
0
→ R =
mv
qB
0
. [2 pts
for correctly determining radius of orbit]
We identify θ as the angle defined by the intersection of the line
representing the original (undeviated) trajectory and the line tan-
gent to the trajectory at the moment the particle leaves the cham-
ber (see figure). From the diagram it is clear to see that this
angle is the same as the angle subtended by the circular segment
of the particle’s trajectory. We can therefore write sin θ =
d
R
and, applying the small angle approximation, we can conclude that
θ ≈ sin θ =
d
R
=
qB
0
d
mv
. [3 pts: 1pt for defining θ , 1 pt for small angle approximation, 1 pt for final expression]
(b) [10 pts] Calculate the fractional energy loss
Δ E
E
0
where Δ E is the total energy radiated by each nucleus
while inside the chamber and E
0
=
1
2
m v
2
is the energy of one nucleus.
The particle is accelerated while in the chamber and therefore radiates electromagnetic waves. The total power
radiated is given by the Larmor formula: P
L
=
q
2
a
2
6 π ǫ
0
c 3
. The magnitude of the acceleration of our particle is
constant and is given by the Lorentz force: m | vectora | = qvB
0
→ | vectora | =
qvB
0
m
=
v
2
d
θ . Substituting this back into the
Larmor formula we obtain P
L
=
q
2
v
4
θ
2
6 π ǫ
0
c 3 d 2
. [3 pts for the correct expression for P
L
]
The energy lost due to radiation per nucleus is simply given by the product of the radiated power (energy per
unit time) and the length of time a single nucleus spends in the chamber. For small values of θ , the arc length
of the trajectory is approximately equal to the length of the chamber d . Therefore, the length of time a single
nucleus spends in the chamber is given by Δ t ≈
d
v
. This implies that Δ E = P
L
Δ t =
q
2
6 π ǫ
0
d
v
3
c 3
θ
2
. [5 pts for
correct expressions for Δ t and Δ E ]
This allows us to conclude that the fractional energy loss is
Δ E
E
=
Δ E
1
2
mv 2
=
q
2
3 π ǫ
0
d m v 2
v
3
c 3
θ
2
. [2 pts]
(c) [15 pts] For a single nucleus entering the chamber at t = 0, give complete expressions (amplitude, direction,
time dependence) for the electric field, magnetic field and Poynting vector of the radiation pulse detected by
observers P and Q , located a distance r >> d away from the chamber in the ˆ x and ˆ z directions, respectively.
1

Ph 214, Spring 2010
Exam # 1
Question
#1 SOLUTIONS
The radiated electric field from an accelerated point charge is given by
vector
E
rad
( vector r, t ) =
- q vectora
⊥
( t - r/c )
4 π ǫ
0
c
2
r
. For small angle deflection in the chamber we
obtain vectora = -
q v B
0
m
ˆ z = -
v
2
R
ˆ z .