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Problem Set 1 Solutions
Lauren Pearce
January 9, 2011
1 Serway 3.1.1: Light as an ElectromagneticWave
PROBLEM: Classical Zeeman Effect or the Triumph of Maxwells
Equa-
tions! As pointed out in Section 3.1, Maxwells equations may be
used topredict the change in emission frequency when gas atoms are
placed in amagnetic field. Consider the situation shown in Figure
P3.1. Note thatthe application of a magnetic field perpendicular to
the orbital plane of theelectron induces an electric field, which
changes the direction of the velocityvector.
1.1 A: Magnitude of Electric Field
PROBLEM: Using Eds=
dBdt
show that the magnitude of the electric field is given by
E=r
2
dB
dt
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SOLUTION: From the diagram, we take the magnetic field to be
uniform
inside of the circle, although it may be time dependent. Since B
is the fluxof the magnetic field through the circle, this reduces
to:
B =
B da
=
B da
=B
da
=r2B
Here we have used the fact that the infinitesimal area element
of the surfaceis oriented perpendicular to the surface (using the
right hand rule); from thediagram we see that it is then parallel
to the magnetic field. This is shownin the diagram:
Since the radius is held constant, the time derivative of the
magnetic fluxis:
dB
dt
=r2dB
dtNow we want to consider the left hand side of the equation.
From electro-
magnetism, we know that the magnitude of the electric field will
be constantalong the circle, and it will be parallel (or
antiparallel) to the direction of
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the circle. To find the direction, we use Lenzs Law; the
electric field will be
induced in a direction so that it would induce a magnetic field
which opposesthe magnetic field creating it. The magnetic field is
growing in the positivezdirection; therefore the electric field
will want to create a magnetic field inthezdirection. By the right
hand rule, the electric field will be in the direction. Taking E=E,
we then have Eds= E ds. Thus we have:
Eds=
Eds
=E
ds
=2rE
Setting the two sides of the equation equal, we find: Eds=
dBdt
2rE=r2dB
dt
E=r
2
dB
dt
E= r2 dBdt
We have found the magnitude of the field as requested in the
problem;however, it is also important to note that it is in the
negative direction.We will need this below.
1.2 B: Change of Speed
PROBLEM: UsingF dt= m dv, calculate the change in speed, v, of
the
electron. Show that ifr remains constant,
v=erB
2me
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SOLUTION: First we ask what force we should be concerned with
here.
Let us think about the magnetic force, which is qv B. If we do
the crossproduct, we see that v B will be pointed out from the
center of the cir-cle. However, we assume that the electron is
forced to go along in a circle;therefore there must be some
restoring force that counteracts this. Anyway,because we assume the
electron goes in a circle, this means that we dontcare about the
magnetic force, which leaves the electric force.
If the radius remains constant, then the magnitude of the
electric fieldalso remains constant. Therefore, the magnitude of
the force on the particle,F = |eE|, is also constant. We will use e
for the MAGNITUDE of theelectron; therefore it is a positive
number. Thus we have F =eE.
To be absolutely clear about signs, we will use the equation
above in theformdFdt = medv, using me for the mass of the electron.
Then we have:
eEdt= medv
er
2
dB
dt dt= medv
er
2 dB= medv
This shows us that the component of the velocity that will
change is thecomponent that is going around in the circle, as we
expect. Now we need tobe exceedingly careful with the signs. We are
asked to find the change in thevelocity of the electron. While the
velocity of the electron is by definitionalways postivie, the
change in the velocity can be positive or negative; thevelocity can
increase or decrease. Let us first consider whether it will
increaseor decrease.
Now let us think of the direction of the force acting on the
electron. Wealready determined that the electric field was pointing
in the direction,
which is the opposite direction as how the electron is
travelling. The force,as we saw above, is in the positive direction
because (since the electron isnegatively charged), it is in the
opposite direction as the electric field. Sincethe force is then
parallel to the direction that the electron is moving, it
shouldaccelerate the electron. Then we expect v to be positive.
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Now let us figure this out in the mathematics. We set v= vto
find:
r2
dB = medv
r
2dB =medv
Solving for dv, we find:
dv= er
2medB
We want to integrate this; we must take care with our limits of
integration.
The particle is initially moving with speed v and after the
magnetic field isturned on, it has speed v + v. Initially, the
magnetic field is zero, but afterit is turned on, it has the value
B . We assume that r is constant and take itoutside of the
integral. Thus we have:
v+v
v
dv= er
2me
B
0
dB
v+ vv = er
2meB
v=erB
2me
As expected, this will be positive. Note how important it is
that we takee to be the magnitude of the electric charge.
1.3 C: Change in Angular Frequency
PROBLEM: Find the change in angular frequency, , of the
electronand calculate the numberical value of for B = 1 T. Note
that this is alsothe change in frequency of light emitted
accordinate to Maxwells equations.Find the fractional change in
frequency, /, for an ordinary emission lineof 500 nm.
SOLUTION: First we recall how angular velocity is related to the
speedthat the particle is moving at. This is just from circular
motion:
=v
r
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Sincer is constant, this gives us:
=v
r
Substituting from the previous section, we have:
= eB
2me
Now we substitute the particular numbers, recalling that e is
negative. Ifwe use MKS units, the result will be radians per
second.
=
1.61019 C
1 T2 1
9.11031 kg
= 8.81010 rad
sec
To find the fractional change in the frequency, we need to find
. We cando this from the information given, which is the wavelength
of the associatedemission line. We use:
= 2f= 2c
In the last equality, we have used the fact that f = c.
Substitutingnumbers, we have:
= 23.0108 m/s
500109 m
= 3.81015 rad
sec
Finally we have:
=
8.81010
3.81015 = 2.3105
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1.4 D: Emission Line Split into Three
PROBLEM: Actually, the original emission line at 0 is split into
threecomponents at 0: 0 , 0, and 0 + . The line at 0 + isproduced
by atoms with electrons rotating as shown in Figure P3.1, wherasthe
lines at 0 is produced by atoms with electrons rotating in
theopposite sense. The line at 0 is produced by atoms with
electronic planesof rotation oriented parallel to B. Explain.
SOLUTION: We first note that the electron is just orbiting the
nucleusin the atom in any old way; it doesnt have to be lines up
the magnetic field.The best way to picture this is that you have a
bunch of atoms, and the
circular orbits of the electrons are aligned more-or-less
randomly. Some willbe lined up with the magnetic field as shown in
the image; then everythingwe derived will be correct.
Some will be lined up going around in the opposite direction.
There areseveral ways of accounting for this change; the important
thing is that youpick up the correct number of sign changes. In the
way I look at it, thischanges the direction of ds and da, since the
orientation of the circle isdefined by the motion of the electron.
Now we need to carefully count howmany negative sign changes we
find. The flux of the magnetic field willchange by a sign (because
B doesnt change). Similarly, the electric fieldwill be induced in
the direction ofds, instead of the opposite direction, andthis will
pick up a sign. Thus we find the same result for E, which we
shouldhave expected. (The electric field is just induced by the
magnetic field, anddoesnt have anything to do with the
electron.)
However, now we need to think about the next part. Now we have
theelectric field is parallel to the direction that the electron is
going. Since thecharge of the electron is negative, this means that
the force is antiparallelto the direction that the electron is
going. Therefore it is going to slow theelectron down, instead of
speeding it up. So we expect it to have a sign
change. Where does this come into the math? This is a very
subtle point.The force hasnt changed direction, so we still get the
same VECTOR v:
v=erB
2me
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However, we now have v= vv, and so the change in speed is given
by:
v + v= v + erB2me
=
v
erB
2me
We keep the negative sign out in front because thats the
direction theelectron is moving. Therefore, we see that the change
in the SPEED (themagnitude of the velocity) is negative. Of course,
this then changes the signof .
Now let us consider the case where the plane of the electron is
parallel toB. Since the area elementda is perpendicular to the
plane of the electron,it is also perpendicular to B. Therefore Bda=
0, and following as in theabove sections, = 0.
2 Serway 3.2.2
PROBLEM: The temperature of your skin is approximately 35 C.
Whatis the wavelength at which the peak occurs in the radiation
emitted fromyour skin?
SOLUTION: We begin by modelling the human body as a
blackbody.Then we can use Weins displacement law to find the
maximum wavelength.Weins displacement law is:
maxT = 2.898103 mK
We solve this for the maximum wavelength:
max =2.898103 mK
T
We see that we want to substitute the temperature in Kelvin, not
Celsius.We can convert by adding 273, which gives us the temperatue
308 K. Wefind:
max =2.898103
308 m = 9.41106 m
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In nanometers, this is 9410 nm.
3 Serway 3.2.3
PROBLEM: A 2.0 kg mass is attached to a massless spring of force
con-stantk = 25 N/m. The spring is stretched.40 m from its
equilibrium positionand released.
3.1 A: Classical Energy and Frequency of Oscillation
PROBLEM: Find the total energy and frequency of oscillation
according
to classical calculations.
SOLUTION: We recall that the total energy of a simple harmonic
oscil-lator is:
E=kA2
2
where A is the amplitude of the oscillation. We assume that the
spring isreleased from rest; then the amplitude is .40 m.
Substituting this, we have:
E=25 N/m
2 (.40 m)2 = 2.0 J
Next we consider the frequency of oscillation. We first find the
angularfrequency:
=
k
m=
25 N/m
2.0 kg = 3.5 rad/sec
Now we find the frequency by dividing by 2:
f=
2 =
3.5
2 Hz =.56 Hz
3.2 B: Quantum Number
PROBLEM: Assume that the energy is quantized and find the
quantumnumber,n, for the system.
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SOLUTION: If the energy is quantized then it has to satisfyEn
=nhf.
We solve this for n:n=
Enhf
Now we substitute fand the energy from the first part of the
problem,along with Plancks constant, to find:
n= 2.0 J
6.631034 Js.56 Hz= 5.41033
This is a very large number.
3.3 C: Energy in a Quantum Change
PROBLEM: How much energy would be carried away in a
1-quantumchange?
SOLUTION: To find the amount of energy in a one quantum change,
wesimply need to evaluate hf:
E= hf= 6.631034 Js.56 Hz = 3.71034 J
This is a very small number.
4 Serway 3.2.4
4.1 A: Total Power Radiated Per Unit Area by Tung-
sten Filament
PROBLEM: Use Setans law to calculate the total power radiated
perunit area by a tungsten filament at a temperature of 3000 K.
(Assume thatthe filament is an ideal radiator.)
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SOLUTION: First we recall Stefans Law, which relates the power
per
unit area,P/A, to the temperature in Kelvin through the constant
. Specif-ically substituting numbers, we have:
P
A=T4
= 5.7108 W
m2K4 (3000 K)4
= 4.6106 W/m2
4.2 B: Surface Area of Filament in a Lightbulb
PROBLEM: If the tungsten filament of a lightbulb is rated at 75
W, whatis the surface area of the filmanet? (Assume that the main
energy loss is dueto radiation.)
SOLUTION: Rearranging the equation found above, we have:
A= P
4.6106 W/m2
Substituting 75 W for the power gives us:
A= 75 W
4.6106 W/m2 = 1.6105 m2
We can also write this as 16 mm2, which is a reasonable size for
a lightbulb.
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