Problem Set 1 Solutions

Lauren Pearce

January 9, 2011

1 Serway 3.1.1: Light as an Electromagnetic

Wave

PROBLEM: Classical Zeeman Effect or the Triumph of Maxwell’s Equa-tions! As pointed out in Section 3.1, Maxwell’s 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 ∮E · ds = −dΦB

dt

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 uniforminside of the circle, although it may be time dependent. Since ΦB is the fluxof the magnetic field through the circle, this reduces to:

ΦB =

∫B · da

=

∫B da

= B

∫da

= πr2B

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:

dΦB

dt= πr2dB

dt

Now 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 Lenz’s Law; the electric field will beinduced in a direction so that it would induce a magnetic field which opposesthe magnetic field creating it. The magnetic field is growing in the positivez direction; therefore the electric field will want to create a magnetic field inthe −z direction. By the right hand rule, the electric field will be in the −φ̂direction. Taking E = −Eφ̂, we then have E · ds = −E ds. Thus we have:∫

E · ds = −∫E · ds

= −E∫ds

= −2πrE

Setting the two sides of the equation equal, we find:∫E · ds = −dΦB

dt

−2πrE = −πr2dB

dt

E =r

2

dB

dt

E = −r2

dB

dtφ̂

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 −φ̂ direction.We will need this below.

1.2 B: Change of Speed

PROBLEM: Using F dt = mdv, calculate the change in speed, ∆v, of theelectron. Show that if r 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 don’tcare 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 theform dF dt = medv, using me for the mass of the electron. Then we have:

−eE dt = me dv

er

2

dB

dtφ̂ dt = me dv

er

2dB φ̂ = me dv

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 −φ̂ 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 = ∆v φ̂ to find:r

2dB φ̂ = me dvφ̂

r

2dB = me dv

Solving for dv, we find:

dv =er

2me

dB

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 + ∆v − v =er

2me

B

∆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, ∆ω, of the electronand calculate the numberical value of ∆ω for B = 1 T. Note that this is alsothe change in frequency of light emitted accordinate to Maxwell’s equations.Find the fractional change in frequency, ∆ω/ω, 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:

ω =v

r

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Since r is constant, this gives us:

∆ω =∆v

r

Substituting from the previous section, we have:

∆ω =eB

2me

Now we substitute the particular numbers, recalling that e is negative. Ifwe use MKS units, the result will be radians per second.

∆ω =(1.6 · 10−19 C

) 1 T

2· 1

9.1 · 10−31 kg

= 8.8 · 1010 rad

sec

To find the fractional change in the frequency, we need to find ω. We cando this from the information given, which is the wavelength of the associatedemission line. We use:

ω = 2πf = 2πc

λ

In the last equality, we have used the fact that λf = c. Substitutingnumbers, we have:

ω = 2π3.0 · 108 m/s

500 · 10−9 m

= 3.8 · 1015 rad

sec

Finally we have:∆ω

ω=

8.8 · 1010

3.8 · 1015= 2.3 · 10−5

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1.4 D: Emission Line Split into Three

PROBLEM: Actually, the original emission line at ω0 is split into threecomponents at ω0: ω0 − ∆ω, ω0, and ω0 + ∆ω. The line at ω0 + ∆ω isproduced by atoms with electrons rotating as shown in Figure P3.1, wherasthe lines at ω0 − ∆ω is produced by atoms with electrons rotating in theopposite sense. The line at ω0 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 doesn’t have to be lines up the magnetic field.The best way to picture this is that you have a bunch of atoms, and thecircular 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 doesn’t change). Similarly, the electric fieldwill be induced in the direction of ds, 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, anddoesn’t 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 signchange. Where does this come into the math? This is a very subtle point.The force hasn’t changed direction, so we still get the same VECTOR ∆v:

∆v =erB

2me

φ̂

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However, we now have v = −vv̂, and so the change in speed is given by:

v + ∆v = −vφ̂+erB

2me

φ̂

= −(v − erB

2me

)φ̂

We keep the negative sign out in front because that’s 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 ∆ω.

Now let us consider the case where the plane of the electron is parallel toB. Since the area element da is perpendicular to the plane of the electron,it is also perpendicular to B. Therefore B · da = 0, and following as in theabove sections, ∆ω = 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 Wein’s displacement law to find the maximum wavelength.Wein’s displacement law is:

λmaxT = 2.898 · 10−3 m ·K

We solve this for the maximum wavelength:

λmax =2.898 · 10−3 m ·K

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:

λmax =2.898 · 10−3

308m = 9.41 · 10−6 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-stant k = 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 accordingto 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:

ω =

√k

m=

√25 N/m

2.0 kg= 3.5 rad/sec

Now we find the frequency by dividing by 2π:

f =ω

2π=

3.5

2πHz = .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 satisfy En = nhf .We solve this for n:

n =En

hf

Now we substitute f and the energy from the first part of the problem,along with Planck’s constant, to find:

n =2.0 J

6.63 · 10−34 Js · .56 Hz= 5.4 · 1033

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 = 6.63 · 10−34 Js · .56 Hz = 3.7 · 10−34 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 Setan’s 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 Stefan’s Law, which relates the power perunit area, P/A, to the temperature in Kelvin through the constant σ. Specif-ically substituting numbers, we have:

P

A= σT 4

= 5.7 · 10−8 W

m2K4· (3000 K)4

= 4.6 · 106 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

4.6 · 106 W/m2

Substituting 75 W for the power gives us:

A =75 W

4.6 · 106 W/m2= 1.6 · 10−5 m2

We can also write this as 16 mm2, which is a reasonable size for a lightbulb.

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