25
EM INDUCTION AND EM WAVES
Q25.1. Reason: With the assumption that the loop is stationary, no current will be induced in the loop
because the magnetic flux through it is not changing.
Assess: It doesn’t matter how strong the magnetic field is (or how big the loop is), if the field through the loop
isn’t changing then there won’t be any current induced in it. Lenz’s law tells us this.
Q25.2. Reason: When the mayfly (and hence the magnet) is close to the coil, there is a magnetic flux
through the coil determined by Φ = B A, where B is the magnetic field strength at the site of the coil and A is the
area of the coil. As the mayfly moves, the flux experienced by the coil varies and this induces a current that can
be detected. The more rapid the movement, the greater the time rate of change of the flux and the greater the
induced current.
Assess: Some small magnets are quite strong. This coupled with the rapid vibration can induce a detectable
current. If necessary we can amplify the current.
Q25.3. Reason: In each case we apply Lenz’s law: “The direction of the induced current is such that the
induced magnetic field opposes the change in the flux.”
(a) The magnetic field is out of the page and the flux through the rectangular loop is increasing because the loop
is growing. A current will be induced that will create a magnetic field to oppose that change. The induced
magnetic field must be into the page; this requires (by the right-hand rule) that the induced current be clockwise.
(b) The magnetic field is into the page and the flux through the rectangular loop is increasing because the loop is
growing. A current will be induced that will create a magnetic field to oppose that change. The induced magnetic
field must be out of the page; this requires (by the right-hand rule) that the induced current be counterclockwise.
(c) The magnetic field is parallel to the plane of the page and so the flux through the rectangular loop is zero and
not changing. Since the flux through the loop isn’t changing then the induced current is zero.
(d) The magnetic field is out of the page and the flux through the triangular loop is decreasing because the loop is
shrinking. A current will be induced that will create a magnetic field to oppose that change. The induced magnetic
field must be out of the page; this requires (by the right-hand rule) that the induced current be counterclockwise.
(e) The magnetic field is into the page but the flux through the rectangular loop is constant because the loop is
not growing or shrinking as the two rails move at the same velocity. Since the flux through the loop isn’t
changing then the induced current is zero.
(f) The magnetic field is into the page but the flux through the rectangular loop is constant because the moving
wire is outside the region of the magnetic field; as the loop grows it doesn’t enclose any more flux. Since the flux
through the loop isn’t changing then the induced current is zero.
Assess: Flux though a loop of wire can change for various reasons, either because the area of the loop changes,
or the strength of the field changes, or the orientation of the loop changes. Only if the flux through the loop
changes is a current induced.
Q25.4. Reason: In order to compare one or more things, you need to pick a reference and then compare to that
reference. Let’s set B = 1 as the reference for the magnetic field and call it B1. Let’s pick a reference area as A1 =
1 × 1. We can now determine the flux through each loop in terms of these reference values and then compare
them:
Φ1 = (4 A1 ) B1 = 4 A1B
Φ 3 = (2 A1 ) B1 = 2 A1B1
Φ 2 = A1 (2 B1 ) = 2 A1B1 ,
Φ 4 = A1 (2 B1 ) + A1 0 = 2 A1B1
Comparing these values, we can rank the magnetic flux through the coils as follows: Φ 1 > Φ 2 = Φ 3 = Φ 4 .
Assess: This question is rather straightforward once we establish a reference area and magnetic field strength
and then determine the flux through each coil in terms of these. This strategy makes it easy to make comparisons.
25-1
25-2
Chapter 25
Q25.5. Reason: The flux is defined in Equation 25.9: Φm = AB cosθ .
For circular loops A = π r 2.
We compute the flux for each case.
Φ1 = (π 12 ) B cos0 ° = π B
Φ 2 = (π 12 ) B cos 45 ° = 0 .707 Φ1
Φ 3 = (π 22 ) B cos 45 ° = 2 . 83 Φ1
Φ 4 = (π 22 ) B cos90 ° = 0
So the final ranking is Φ 3 > Φ1 > Φ 2 > Φ4 .
Assess: We see that both the area A and the cosine of the angle matter in the calculation of the flux, even
where B is uniform and constant.
Q25.6. Reason: In this question, we are interested in the value of Φ at various angles and also the manner in
which Φ is changing at various angles. We can answer the question easily by examining the function which
allows us to determine the magnetic flux: Φ = AB cosθ . A plot of this function is shown.
Examining the function and its graphical representation we see that:
(a) When the angle φ = 0°, 180 °, and 36 0°, the magnetic flux through the coil has the maximum value Φ = BA.
(b) When the angle φ = 90 ° and 270° the magnetic flux through the coil is zero. At these angles the magnetic
field does not go through the coil: Φ = 0 .
(c) The flux is changing most rapidly at φ = 90° and 270 °.
Assess: The plot makes it easy to examine the expression Φ = AB cosθ .
Q25.7. Reason: The magnetic field lines generated by a long current-carrying wire are concentric circles
around the wire. The strength of that field changes continuously in the AC power lines.
If you orient your loop so its axis is parallel to the power line then you won’t ever get any flux through it and
certainly no changing flux through it. So that is a bad idea.
To get the most flux through your loop you want to place it close to the power line (where the field is strongest) and in an
orientation so the plane of the loop also contains the power line. Not only will this get you the greatest value of the flux,
but in this case it also produces the greatest change in the flux and therefore the greatest induced emf in your loop as the
field goes from a maximum to zero and then to a maximum in the other direction.
Assess: We also don’t want the loop to be flat on the power line (with two opposite points of the loop touching
the line) because some field lines would be going through the loop one way and other lines going through the
opposite way leaving you with very little flux through the loop.
Q25.8. Reason: An induced current will be present in the coil any time the magnetic flux is changing. From
Figure Q25.8, we note that the magnetic flux is changing during the following time intervals: 0 to t1, t2 to t3, and
t4 to t5.
EM Induction and EM Waves 25-3
Assess:
The magnetic flux must be changing in order to induce a current.
Q25.9. Reason: The original field is into the page within the loop and is changing strength. The induced,
counterclockwise current produces a field out of the page within the loop that is opposing the change. This
implies that the original field must be increasing in strength so the flux into the loop is increasing.
Assess: If the original flux were decreasing then the induced current would be in the opposite direction from
that drawn to produce an induced field into the page to oppose the decrease. That isn’t the case, so we know the
field isn’t decreasing.
Q25.10. Reason: As the magnets move in the region near the aluminum vanes, the changing magnetic flux
experienced by the vanes induces eddy currents. These eddy currents generate a magnetic field which (according
to Lenz’s law) first opposes the magnets attached to the cars as they enter the region near the vanes and then
opposes their leaving the region near the vanes. This will create a strong braking effect as the cars both enter and
exit the area of the vanes.
Assess: Magnetic braking is fairly common in tool technology.
Q25.11. Reason: There is no mystery about the clear plastic tube. The magnet falls as expected in free fall.
However, the aluminum tube, while not magnetic, is a good electrical conductor. Think of the tube as a stack of coaxial
conducting rings. As the magnet falls each ring in succession sees a changing flux through it; therefore currents are
induced around the ring that then create magnetic fields to oppose the change. These induced magnetic fields act on the
magnet to slow it. The magnet reaches a slow terminal speed very soon after being dropped.
Assess: This is a fairly impressive demonstration, especially if you can get a very strong magnet and a tube
whose inner diameter isn’t too much bigger than the magnet. Making the resistance of the tube lower helps too;
you can do this by cooling it. To show that the currents are induced around the tube you can cut various slits
lengthwise staggered around the tube at various places along the length and the magnet will fall much faster
because the cut “rings” can’t carry a current.
Q25.12. Reason: As you shove the coil into the region of the magnetic field, the magnetic flux through the
coil will be increasing. This changing magnetic flux will create an induced clockwise current (see the figure
below). This induced current interacts with the existing magnetic field in such a manner as to oppose what
caused it. Since the induced current was caused by you pushing the coil into the region of the magnetic field, as
the induced current interacts with the existing magnetic field the resulting force will oppose its motion into the
region of the field. The figure below shows the magnet, magnetic field, coil, induced current in the coil, the force
you are exerting, and the subsequent magnetic force. As shown in the figure, you need to push the loop in against
a repulsive force.
Assess: This problem is an excellent example of Lenz’s law. As the coil is pushed into the region of the field,
there is an increasing upward magnetic flux. The induced current creates an induced magnet flux that is directed
downward.
Q25.13. Reason: Any induced current must oppose the change that created it. As a result, the current induced
in the inner loop must move in the a direction opposite that in the outer loop. This is shown in the figure.
25-4
Chapter 25
Assess: The induced current will always oppose the change that created it.
Q25.14. Reason: (a) The flux in the upper loop is zero and not changing. Since the flux isn’t changing no
current is induced.
(b) The lower loop has a counterclockwise current (looking down) that makes a magnetic field directed up, and it
is increasing. So the induced magnetic field will need to be down; this is accomplished by an induced current in
the top loop in the clockwise direction.
(c) Because the flux through the upper loop is no longer changing there is no current induced.
(d) The field due to the lower loop is directed up, but it is decreasing, so the induced field will need to be up; this
is accomplished by an induced current in the top loop in the counterclockwise direction.
Assess: As always, we must carefully consider the change in flux and the induced current will be in a direction
to create an induced field to oppose the change.
Q25.15. Reason: (a) As the north magnetic pole of the magnet approaches the coil, there is an increasing
upward magnetic flux through the coil. The induced current must be in such a direction as to oppose this increase
in magnetic flux. If the induced current is clockwise (as viewed from above), the induced magnetic field will be
downward. As the coil is pushed into the field, the upward magnetic flux increases, the induced clockwise
current increases, the induced magnetic field increases and the induced magnetic flux directed in the downward
direction increases.
(b) When the magnet is being pushed toward the coil, the magnetic situation at the site of the coil is as shown in
the following figure. The magnetic field at the site of the coil may be resolved into a vertical and horizontal
component.
Recall that the current is clockwise as indicated by the tail of an arrow representing the current at the left and the
head of an arrow at the right. Notice that as the induced current interacts with the magnetic field of the
approaching magnet, the vertical component of the magnetic field results in forces that try to collapse the coil
and the horizontal component of the magnetic field results in forces that try to push the coil vertically upward.
Assess: If the magnetic field associated with the magnet is strong enough and magnet is thrust toward the coil
fast enough, the resulting induced forces will support the coil.
Q25.16. Reason: While there are field lines through the loop there are equal numbers going in each
direction, so the net flux through the loop is zero when the magnet is in that orientation, even as the magnet
moves toward the loop.
Assess: It is the net flux that matters. Field lines in one direction cancel field lines in the opposite direction.
EM Induction and EM Waves 25-5
Q25.17. Reason: As the magnetic field rapidly increases, the magnetic flux through the coil increases. This
changing magnetic flux inside the coil will induce a current in the coil. The induced current will be the same for
both bulbs and both bulbs will be equally bright (assuming the induced current is sufficient).
Assess: The arrangement as described acts as a generator of electric current.
Q25.18. Reason: (a) The original magnetic field B is into the page and increasing, so Binduced is out of the
page. This leads to a counterclockwise induced current. By the right-hand rule, with a current going up through
the sliding wire and Bnet into the page, the force exerted by the field on the wire is to the left. So the wire moves to
the left.
(b) The original magnetic field B is into the page and decreasing, so Binduced is into the page. This leads to a
clockwise induced current. By the right-hand rule, with a current going down through the sliding wire
and Bnet into the page, the force exerted by the field on the wire is to the right. So the wire moves to the right.
However, the field extends beyond the rail so the wire will break contact. At that point the force on the wire will
drop to zero.
Assess: Just as a moving wire can induce a current as shown in the text, a changing field can induce a current
that creates a force on the wire and causes it to move.
Q25.19. Reason: The light we are interested in blocking is the light reflected off water and other horizontal
surfaces. This light is partially polarized parallel to the plane of the reflecting surface, hence the electric field vector
we wish to block is oscillating parallel to the horizontal surface. If the long axis of the polarizing molecules is also
parallel to the horizontal surface, the electrons in the molecules will be forced into oscillations along the long
molecules by the electric field. During this process, the polarizing coating on the surface of the sunglasses absorbs
the energy of the electric field and as a result the electric field does not pass through the sunglasses.
Assess: Polarized sunglasses are a good example of better living through a combination of chemistry and physics.
Q25.20. Reason: Because no light gets through until the plastic between the two polarizers is stressed, the
stressed plastic must change the plane of polarization of the light by rotating the plane of polarization.
Assess: This technique is used to find out how much the plastic is stressed, in, say, a plastic model of a building.
Q25.21. Reason: The metal bars of the antenna have different orientations and different lengths. The signal
coming from a television station is unpolarized. If the antenna had only vertical bars, it would “pick up” only the
signal that was polarized in this direction. Since different stations transmit at different frequencies, their signals
have different wavelengths. The length of the metal bar in the antenna is related to the wavelength of the
transmitted signal. Having numerous bars oriented in several directions and of several different lengths will
result in “picking up” a larger signal from more stations.
Assess: The length and orientation of the metal bars in the antenna is critical to reception.
Q25.22. Reason: (a) As the oscillating magnetic field of the radio wave goes through the coil, it induces a little
current in the coil because the flux through it is changing. This small current is amplified and then fed to the speakers.
(b) The coil and ferrite bar need to be horizontal since the magnetic field will be oscillating in a horizontal plane
(perpendicular to the vertical plane of the electric field). If the station is north of you then the magnetic field will
oscillate in an east-west direction, and that is the way you should orient your ferrite bar to detect the largest
changes in flux.
Assess: Look at some household AM radios. The ferrite bar on the back is horizontal, but whether it points
east-west is up to you and where your favorite station broadcasts from.
Q25.23. Reason: (a) The energy of a photon is related to the wavelength by E = hc /λ. This says that the energy
is inversely proportional to the wavelength: the longer the wavelength the less the energy of the photon. This
allows us to rank the energy of the photons as follows: Since λ1 < λ2 < λ3 we may write that E1 > E2 > E3 .
Since power is energy per unit time and the energy comes from individual photons, we have
P=
Energy NEphoton
=
t
t
(b) The number of photons delivered per second is
25-6
Chapter 25
N
P
P
Pλ
=
=
=
t
Ephoton hc / λ hc
This says that the number of photons needed per second is inversely proportional to the energy of the individual
photons and directly proportional to the wavelength associated with the photons. This allows us to rank the
number of photons of each wavelength needed per second to deliver the same power as follows:
Since λ1 < λ2 < λ3 we may write that N 3 > N 2 > N1.
Assess: Since photons with a greater wavelength have less energy, it will require more of them.
Q25.24. Reason: Only D is true. More intense light means more photons per second.
Assess: We know this experimentally from the photoelectric effect: More intense light ejects more
photoelectrons.
Both the energy and wavelength are related to the frequency, but we were told the frequency was unchanged.
The intensity does not affect the speed. The speed of all photons in a vacuum is the same.
Q25.25. Reason: The intensity of a photon beam may be determined as follows:
I=
P E / t NEphoton ⎛ N ⎞⎛ hf ⎞
=
=
= ⎜ ⎟⎜ ⎟
A
A
At
⎝ t ⎠⎝ A ⎠
Notice that the intensity of the photon beam (in this case visible light) is proportional to the number of photons hitting the
surface each second ( N / t ) and the frequency of the photons in the beam. If the frequency is increased, each photon now
has more energy ( Ephoton = hf ). If the frequency increases and the intensity of the beam remains the same, then fewer
photons are delivered per second ( N / t ). This allows us to say that statements B and C are correct.
Assess: This question was answered by first having sufficient command of the fundamental principles to obtain
the above expression and then having the skill to interpret what the expression had to offer. Physics will help you
develop these skills.
Q25.26. Reason: The arc is so hot that the thermal emission spectrum has a significant amount in the UV
range. The higher the temperature the shorter the wavelength of the peak of the spectrum.
Assess: Red hot is hot, but blue hot is hotter. Objects that appear blue to us due to their extreme temperature
also give off lots of UV. Even the sun, which is only yellow hot or white hot, gives off lots of UV that our ozone
layer protects us from.
Q25.27. Reason: The flux is given by Equation 25.9. We are given A = 0 .30 m 2, B = 0 .40 T, and θ = 45 °.
Φ m = AB cosθ = (0 . 30 m 2 )(0 . 40 T)cos 45 ° = 8. 5 × 10−2 T ⋅ m 2
The correct choice is A.
Assess: A bigger loop, a stronger field, or a smaller angle would increase the flux.
Q25.28. Reason: The figure indicates that the magnetic field is uniform and we are going to assume that the
coil is rotating at a constant angular frequency. As a result the magnetic flux in the coil is constant. Since the
magnetic flux in the coil is not changing, no electric current will be induced. The correct choice is C.
Assess: A changing magnetic flux is needed in order to induce an electric current.
Q25.29. Reason: We know that while the loop is entirely in the field the flux isn’t changing and so the
induced current is zero. But all the choices have the initial current at zero. We also know that when the loop is
completely out of the field the flux will be zero and constant so the induced current will be zero. This eliminates
choices A, which doesn’t return to zero. Regardless of which current direction is positive, the sign won’t change;
as the loop exits the field the flux is always decreasing, never increasing. So the graph won’t change sign. This
eliminates choice E. As the loop moves with a constant velocity, the amount of area still in the field is
decreasing, but, due to the shape of the loop, not at a constant rate. The area in the field decreases faster where
the loop is wider (in the middle). This leads us to conclude that B is the right choice.
The correct choice is B.
Assess: Choice B also provides the correct direction of current. The field is into the page but decreasing so the
induced field will also be into the page; this is produced by a clockwise-induced current.
EM Induction and EM Waves 25-7
If the diamond-shaped loop were rotated 45° so it looked like a square, then choice C would be correct, as the
rate of decrease of area in the field would be constant.
Q25.30. Reason: The magnetic flux through the coil is Φ = BA = B( L2 /2). As the magnetic field changes the
magnetic flux changes and the induced emf, which is the rate of change of the magnetic flux, is determined by
E=
ΔΦ Δ( BA) ΔB
ΔB 2
0.02 T
=
=
A=
( L /2) =
(0.2 m) 2 /2 = 8.0 mV
Δt
Δt
Δt
Δt
5.0 × 10−2 s
The correct choice is A.
Assess: This is a small but reasonable value for the induced emf.
Q25.31. Reason: The question asks for the emf generated by a coil where the flux is changing. This makes
us think of Faraday’s law: Equation 25.12.
E=N
ΔΦ per coil
Δt
We are given N = 100 turns, Δ t = 0 . 50 s, and ΔΦ = 2Φ because as the coil flips the flux goes from the original
maximum to zero and then to a maximum in the other direction, so the change is two times the original. Φ max =
AB cosθ = (0 . 010 m 2 )(0 . 050 mT) cos0° = 0 . 0005 mT ⋅ m2.
E=N
ΔΦ per coil
Δt
= 100
2(0 .0005 mT ⋅ m 2 )
= 0 .20 mV
0 . 50 s
The correct choice is C.
Assess: The rate of change of flux through the coil is not constant, so the induced emf is not constant; however,
we were only asked for an average over the time interval, and we computed that correctly.
Note the units: T = kg/(A ⋅ s 2 ) and V = kg ⋅ m 2 /(A ⋅ s3 ), so the units in the question work out.
T ⋅ m2
=V
s
Q25.32. Reason: The frequency and wavelength of an electromagnetic wave are related by c = f λ . Using
this we can determine the wavelength of the various frequency radio waves.
λ = c /f = (3.0 × 108 m/s)/(87.9 × 106 Hz) = 3.41 m
and
λ = c /f = (3.0 × 108 m/s)/(107.9 × 106 Hz) = 2.78 m
From this calculation we see that the wavelength range associated with the frequency range 87.9 MHz to 107.9
MHz is 3.41 m to 2.78 m. The correct choice is C.
Assess: According to Figure 25.35, radio wavelengths are in the 1 m to 10 m range.
Q25.33. Reason: The time of travel for the radio signal is t = d /c = 2h /c, where h is the height of the
spacecraft above the earth.
t = d / c = 2h / c = 2(1.0 × 104 m)/(3.0 × 108 m/s) = 67 μs
The correct choice is D.
Assess: This is a reasonable amount of time for the signal to travel this distance at the speed of light.
Q25.34. Reason: The intensity of light through a polarizing filter is given by Malus’s law in Equation 25.20.
Intensity, power, and cross-sectional area of the beam are related by I = P / A. Putting this last expression into
Malus’s Law obtain
25-8
Chapter 25
I trans = I incident cos 2 θ ⇒
Ptrans Pincident
=
cos 2 θ ⇒ Ptrans = Pincidemnt cos 2 θ = (6 . 0 mW)cos 2 75 ° = 0 . 40 mW
A
A
The cross-sectional area of the incident and transmitted beam are the same.
The correct choice is A.
Assess: We got a fairly small answer (compared to the incident power), but that is because 75 ° is close to 90°
for which none of the light would get through, so we expected a small answer.
Remember that cos 2 θ = (cosθ ) 2 , not cos(θ ) 2 , nor cos(cosθ ).
Q25.35. Reason: The wavelength of an electromagnetic wave is related to the frequency by λ = c /f = (3.0 ×
108 m/s)/(76 Hz) = 3900 km. The correct choice is D.
Assess: This rather long wavelength is expected due to the low frequency.
Q25.36. Reason: First we’ll compute the energy emitted in the 5 s if the power is 2.8 mW. Then we’ll follow
the strategy in Example 25.8 to find the energy of one photon of 635 nm wavelength. The last step will be to
compute N from those two quantities.
We use energy = power × time, given P = 2. 8 mW and Δ t = 5 . 0 s.
E = PΔ t = (2 .8 mW)(5 . 0 s) = 14 mJ
The frequency of the photon is
f =
c
λ
=
3 . 0 × 108 m/s
= 4 . 72 × 1014 Hz
635 × 10−9 m
The energy of one such photon is
Ephoton = hf = (6 . 63 × 10−34 J ⋅ s)(4. 62 × 1014 Hz) = 3 . 13 × 10−19 J
The number of 650 nm photons in 14 mJ of light energy is
N=
14 × 10−3 J
= 4 . 5 × 1016 photons
3.13 × 10 –19 J/photon
The correct choice is D.
Assess: The answer is a very large number, but not as large as the answer in Example 25.8, which is as we
expect because our laser pointer has less power than the 40 W bulb.
Problems
P25.1. Prepare: Please refer to Figure P25.1. The magnetic field is perpendicular to the wire. The positive
charges experienced a magnetic force to the left. We will assume the magnetic field to be uniform.
Solve: This is a straightforward use of Equation 25.3. We have
B=
E
0.050 V
=
= 0.10 T
vl (5.0 m/s)(0.10 m)
By the right-hand rule the field must be out of the page, so that the magnetic force on the positive charges is to the left.
Assess: This is reasonable. Laboratory fields are typically up to a few teslas in magnitude.
P25.2. Prepare: The motional emf induced may be determined by E = Blv.
Solve: The motional emf is E = Blv = (50 × 10−6 T)(0.85 m)(1.5 m/s) = 64 μ V.
Assess: This is large enough potential difference for the hammerhead to navigate.
P25.3. Prepare: The wire is pulled with a constant force in a magnetic field. This results in a motional emf
and produces a current in the circuit. From energy conservation, the mechanical power provided by the puller
must appear as electrical power in the circuit.
EM Induction and EM Waves 25-9
Solve:
(a) Using Equation 25.6,
P = Fpullv ⇒ v =
P
4.0 W
=
= 4.0 m/s
Fpull 1.0 N
(b) Using Equation 25.6 again,
P=
Assess:
v 2l 2 B 2
⇒B=
R
RFpull
vl
2
=
(0.20 Ω)(1.0 N)
= 2.2 T
(4.0 m/s)(0.10 m) 2
This is a reasonable field for the circumstances given.
P25.4. Prepare: The motional emf induced may be determined by E = Blv, the current is determined
by I = E / R, the magnetic force and hence the force required to keep the rod moving at a constant speed is
determined by F = BIL.
Solve: (a) Combing the first two equations and solving for the current, obtain
I = E / R = Blv / R = (1.4 T)(0.15 m)(3.5 m/s) / (0.65 Ω) = 1.13 A ≈ 1.1 A
(b) The magnetic force and hence the force required to keep the rod moving at a constant speed is
F = BIL = (1.4 T)(1.13 A)(0.15 m) = 0.237 N ≈ 0.24 N
Assess: This is a reasonable current and force for this situation.
P25.5. Prepare: We neglect air resistance for this case. The terminal speed will be achieved when the
downward force of gravity on the bar is equal in magnitude to the upward force due to the induced current in the
bar in the horizontal magnetic field. We’ll use Equation 25.5 for the force needed to produce a constant speed.
Known
l = 0.12 m
B = 0.060 T
R = 1.0 Ω
m = 0.050 kg
Find
v
We use the given mass m = 0. 050 kg to compute the gravitational force w = mg = (0. 050 kg)(9 . 8 m/s 2 ).
Solve:
See Equation 25.5. For terminal speed
25-10 Chapter 25
Fgrav = Fmag
mg = IlB
⎛ vlB ⎞
mg = ⎜
⎟ lB
⎝ R ⎠
vl 2 B 2
mg =
R
Solve for v.
v=
mgR (0.050 kg)(9.8 m/s 2 )(1.0 Ω)
=
= 9500 m/s
l 2B2
(0.12 m) 2 (0.060 T) 2
Assess: The answer is quite large, but that is because most real magnetic fields are quite weak (as is this one).
If we hadn’t neglected air resistance the bar would have reached terminal speed due to air resistance long before
it does due to the magnetic force.
P25.6. Prepare: The motional emf induced may be determined by E = Blv.
Solve: The motional emf induced is
E = Blv = (5.0 × 10−5 T)(2.8 m)(75 km/hr)(103 m/km)(hr/(3.6 × 103 s)) = 2.9 mV.
Using the right-hand rule we see that top of the panels will be negative compared to the bottom,
Assess: This is a small but reasonable motional emf.
P25.7. Prepare: Refer to Figure P25.7. The direction of magnetic field is shown as left to right. The direction
of the vector representing the area of the coil is perpendicular to the coil. The angle θ is the angle between the
r
r
direction of the magnetic field B and the area A . The magnetic flux is obtained by Φ = BA cosθ .
Solve: An expression for the magnetic flux for any angle θ is as follows:
Φ = BA cosθ = Bπ r 2 cosθ = (0.05T )π (0.050m) 2 cosθ = (3.93 × 10−4 Wb)cosθ
For the angle
θ = 0° obtain Φ = 3.93 × 10-4 Wb ≈ 3.9 × 10-4 Wb
θ = 30° obtain Φ = 3.40 × 10-4 Wb ≈ 3.4 × 10-4 Wb
θ = 60° obtain Φ = 1.97 × 10-4 Wb ≈ 2.0 × 10-4 Wb
θ = 90° obtain Φ = 0 Wb
Assess: Given the strength of the magnetic field and the area of the coil, these are reasonable values for the
magnetic flux.
P25.8. Prepare: Please refer to Figure P25.8. We will assume that the field changes abruptly at the
boundary between the two sections. The directions of the fields are opposite, so some flux is positive and some
negative. The total flux is the sum of the flux in the two regions.
r
Solve: The field is constant in each region so we will use Equation 25.9. Take A to be into the page. Then, it is
parallel to the field in the left region so the flux is positive, and it is opposite to the field in the right region so the flux is
negative. The total flux is Φ = AL BL cosθ L + AR BR cosθ R = (0.20 m) 2 (2.0 T) − (0.20 m) 2 (1.0 T) = 0.04 0 Wb.
Assess: The flux is positive because the areas are equal and the stronger field is parallel to the normal of the surface.
P25.9. Prepare: Please refer to Figure P25.9. Consider the solenoid to be long so the field is constant inside
and zero outside. The field of a solenoid is along the axis. The flux through the loop is only nonzero inside the
solenoid. Since the loop completely surrounds the solenoid, the total flux through the loop will be the same in
both the perpendicular and tilted cases.
r
Solve: The field is constant inside the solenoid so we will use Equation 25.9. Take A to be in the same direction as
the field. The magnetic flux is
2
Φ = Aloop Bloop cosθ = Asol Bsol cosθ = π rsol
Bsol cosθ = π (0.010 m) 2 (0.20 T) = 6.3 × 10−5 Wb
r
r
When the loop is tilted, the component of B in the direction of A is less, but the effective area of the loop surface
through which the magnetic field lines cross is increased by the same factor.
Assess: Because the field created is along the length of the solenoid, we used θ = 0° .
EM Induction and EM Waves 25-11
P25.10. Prepare: The magnetic flux through the book may be determined by Φ = ( B cos φ ) A, where
( B cos φ ) .is the component of the earth’s magnetic field perpendicular to the book and A is the area of the book.
Solve:
The magnetic flux through the book is
Φ = ( B cos φ ) A = ( B cos φ )( LW ) = (5.0 × 10−5 T)(cos 25 °)(0.28 m)(0.22 m) = 9.9 μ Wb
Assess: Since both the earth’s field and the surface area of the book are small, we expect a small magnetic flux.
P25.11. Prepare: Please refer to Figure P25.11. Assume the plane of the loop is perpendicular to the field
direction. The flux is due to the field through the area of the triangle. Only the left half gives a contribution as the
field strength is zero on the right half.
r
Solve: (a) The flux given by Equation 25.9 is Φ = AB cosθ . Take A to be into the page, perpendicular to the
r
triangle, and thus parallel to B. In this case Φ = AB where A is the area of half of the triangle. This smaller
triangle has a base of 10 cm and height 20sin 60° cm = 17.32 cm. Thus,
1
(0.10 m)(0.1732 m) × 0.1 T = 8.7 × 10−4 Wb
2
(b) The flux is directed into the loop. According to Lenz’s law, the induced current will try to prevent the
decrease of flux. To do this, the field of the induced current will have to point into this loop. This requires the
induced current to flow clockwise.
Φ = AB =
P25.12. Prepare: Please refer to Figure P25.12. If the changing field produces a changing flux in the loop
there will be a corresponding induced emf and current. We will use Equations 25.9 and 25.11 and Lenz’s law.
Solve: (a) The induced emf is ε = |ΔΦ/Δt| and the induced current is I = E /R . The field B is changing, but the
r
r
area A is not. Take A to be out of the page and parallel to B, so from Equation 25.9 Φ = AB. Thus,
E= A
I=
ΔB
ΔB
= π r2
= |π (0.050 m) 2 (0.50 T/s)| = 3.93 × 10−3 V = 4 mV
Δt
Δt
E 3.93 × 10−3 V
=
= 3.93 × 10−2 A = 40 mA
R
0.10 Ω
The field is increasing out of the page. To prevent the increase, the induced field needs to point into the page.
Thus, the induced current must flow clockwise.
(b) As in part a, E = A(ΔB/Δt) = 4.0 × 10−3 V and I = 40 mA. Here the field is into the page and decreasing. To
prevent the decrease, the induced field needs to point into the page. Thus the induced current must flow
clockwise.
r
r
(c) Now A (left or right) is perpendicular to B and so AB cosθ = 0 Wb. That is, the field does not penetrate the
plane of the loop. If Φ = 0 Wb, then E = ΔΦ / Δt = 0 V and I = 0 A. There is no induced current.
Assess: Note that the induced field opposes the change.
P25.13. Prepare: Equation 25.11 gives the induced emf for a loop of wire in a changing magnetic field.
E=
ΔΦ m
Δt
Since the loop is not changing size, and the orientation stays perpendicular, this boils down to
E=
Δ( AB cosθ )
A cos0 °(Δ B)
ΔB
=
=A
Δt
Δt
Δt
Since A is (positive and) constant, the ranking of E will be the same as the ranking for Δ B / Δ t .
Solve:
Plug in the respective values of the data and compute Δ B / Δt for each case.
25-12 Chapter 25
A.
ΔB 1 T − 0 T 1
=
= T/s
6s
6
Δt
B.
4 T −1 T 3
ΔB
=
= T/s
2s
2
Δt
C.
4T−4T
ΔB
=
= 0 T/s
1 min
Δt
D.
3T−4T 1
ΔB
=
= T/s
4s
4
Δt
E.
0 T −3T
ΔB
=
= 3 T/s
1s
Δt
The ranking from greatest to least is | EE |>| EB |>| ED |>| EA |>| EC | .
Assess: Since we wanted only the magnitude of the emf, the absolute value in the equation meant it didn’t
matter whether the magnetic field was increasing or decreasing.
P25.14. Prepare: The emf produced by a changing magnetic flux is related to the changing magnetic flux by
E = ΔΦ
. Magnetic flux is related to the strength of the magnetic field, the area and the angle between the
Δt
area and magnetic field vectors by Φ = BA cosθ . In order to calculate the maximum emf generated, let’s use the
angle θ = 0 °.
Solve: Since the cross-sectional area of the eyeball does not change, the emf generated is
E=
ΔΦ
ΔB
ΔB
=A
= π r2
= π (12.5 × 10−2 m) 2 (50 T/s) = 2.5 V
Δt
Δt
Δt
This amount of emf is more than adequate to trigger an action potential.
Assess: A large value is expected due to the large change in magnetic field per unit of time.
P25.15. Prepare: Assume the field strength is changing at a constant rate. The changing field produces a
changing flux in the coil and there will be a corresponding induced emf and current. We will use Equations 25.9
r
r
and 25.12. When B is parallel to A , Φ = BA cos0 ° = B A.
Solve:
The induced emf of the coil is
E=N
Assess:
⎛ 0.10 T ⎞
ΔΦ
Δ( AB )
ΔB
ΔB
=N
= NA
= Nπ r 2
= (103 )π (0.01 m) 2 ⎜
⎟ = 3.1 V
−3
Δt
Δt
Δt
Δt
⎝ 10 × 10 s ⎠
This seems to be a reasonable emf as there are many turns.
P25.16. Prepare: Please refer to Figure P25.16. Assume the field is uniform across the loop. There is a
current in the loop so there must be an emf that is due to a changing flux. With the loop fixed, the area is constant
so the change in flux must be due to a changing field strength.
Solve: The induced emf is E = |ΔΦ/Δt| and the induced current is I = E / R. The B field is changing, but the
r
r
area A is not. Take A as being into the page and parallel to B, so Φ = AB and ΔΦ/Δt = A(B/Δt). We have
EM Induction and EM Waves 25-13
E=
ΔΦ
ΔB
Δ B IR (150 × 10−3 A)(0.10 Ω)
=A
⇒
=
=
= 2.3 T/s
Δt
Δt
Δt
A
(0.080 m) 2
The original field and flux is into the page. The induced counterclockwise current produces an induced field and
flux that is out of the page. Since the induced field opposes the change, the field must be increasing.
Assess: This range of changing magnetic field is rather large, but still reasonable.
P25.17. Prepare: The induced emf is determined by Einduced = ΔΦ / Δt = A(ΔB / Δt). The induced emf is in the
opposite direction of the applied voltage (the 9.0 volts from the battery, so the net emf is Enet = Eapplied − Einduced .
The net emf and current are related by Enet = IR. Combining these expressions and solving for the current we
obtain: I = [Eapplied − A(ΔB / Δt )]/ R.
Solve:
The current in the circuit and hence the resistor is
I = [Eapplied − A( ΔB / Δ t )]/ R = [9.0 V − (25 × 10−4 m 2 )(0.50 T/10 × 10−3 s)]/(20 Ω) = 0.44 A
Assess:
The current in the circuit is decreased slightly due to the fact that the induced emf opposes the applied voltage.
P25.18. Prepare: The emf is related to the changing magnetic field and cross-sectional area of the loop by
ΔΦ
ΔB
=A
. The current in the loop is related to the emf generated by the changing magnetic field and the
Δt
Δt
resistance of the loop by I = E / R .
Solve: Combining these results and solving for the current, obtain
E=
I=
A ⎛ Δ B ⎞ π r 2 ⎛ ΔB ⎞ π (2.5 × 10−2 m) 2 ⎛ Δ B ⎞
⎛ ΔB ⎞
=
=
= (1.64 × 10−3 m 2 /Ω) ⎜
⎜
⎟
⎜
⎟
⎜
⎟
⎟
1.2 Ω
R ⎝ Δt ⎠
R ⎝ Δt ⎠
⎝ Δt ⎠
⎝ Δt ⎠
Next note that Δ B / Δt is the slope of the B versus t plot. The slope has a nonzero value only between 0.1 and 0.3
seconds and during this time interval Δ B / Δ t = 15 T/s . As a result, current exists only during this time interval
and it is
⎛ ΔB ⎞
−3
2
I = (1.64 × 10−3 m 2 / Ω) ⎜
⎟ = (1.64 × 10 m / Ω)(15 T/s) = 25 mA
⎝ Δt ⎠
A plot of the current generated as a function of time is shown in the figure.
Assess: Current is generated only when there is a changing magnetic field. In this case that is the time interval
between 0.1 and 0.3 seconds.
P25.19. Prepare: The electric and magnetic field amplitudes of an electromagnetic wave are related as
E0 = cB0 .
Solve: The electric field amplitude of the electromagnetic wave is
E0 = cB0 = (3.0 × 108 m/s)(2.0 × 10−3 T) = 6.0 × 105 V/m
Assess: Because the magnetic field amplitude is much larger than the earth’s magnetic field, we expected a
large electric field amplitude.
P25.20. Prepare: The electric and magnetic field amplitudes of an electromagnetic wave are related as E0 = cB0 .
Solve:
The magnetic field amplitude of the electromagnetic wave is
25-14 Chapter 25
B0 =
Assess:
E0
10 V/m
=
= 3.3 × 10−8 V ⋅ s/m 2 = 3.3 × 10−8 T
c 3.0 × 108 m/s
The magnetic field is quite small, about 1/1000 of the magnetic field of the earth.
P25.21. Prepare: The intensity of the microwaves is related to the amplitude of the oscillating electric field
by I = cEo Eo2 / 2 and the amplitude of the oscillating magnetic field is related to the amplitude of the oscillating
electric field by Bo = Eo / c.
Solve: The amplitude of the oscillating electric field is
Eo = (2 I /(cEo ))1/ 2 = [2(2500 W/m 2 ) / ((3.0 × 108 m/s)(8.85 × 10−12 C 2 / (N ⋅ m 2 ))]1/ 2 = 1.37 × 103 V/m ≈ 1.4 × 103 V/m
The amplitude of the oscillating magnetic field is
Bo = Eo / c = (1.37 × 103 V/m) / (3.0 × 108 m/s) = 4.6 × 10−6 T
Assess: These are reasonable values for the amplitude of the oscillating electric and magnetic fields of
microwaves.
P25.22. Prepare: The intensity of the microwaves is related to the amplitude of the oscillating electric field
by I = cEo Eo2 / 2.
Solve: The amplitude of the oscillating electric field is
Eo = (2 I / (cEo ))1/ 2 = [2(5.0 mW/m 2 ) / ((3.0 × 108 m/s)(8.85 × 10−12 C 2 / (N ⋅ m 2 ))]1/ 2 = 1.9 V/m
Assess: For such a small intensity we expect the amplitude of the oscillating electric field to be small.
P25.23. Prepare: The intensity of the microwaves is related to the amplitude of the oscillating electric field
by I = cEo Eo2 / 2 and to the power and cross-sectional area of the beam by I = P / A . The amplitude of the
oscillating magnetic field is related to the amplitude of the oscillating electric field by Bo = Eo / c.
Solve: The amplitude of the oscillating electric field is
Eo = (2 I / (cEo ))1/ 2 = [2 P / (π r 2cE)]1/ 2
[2(10−3 W/m 2 ) / π (5.0 × 10−4 m) 2 (3.0 × 108 m/s)(8.85 × 10−12 C2 / (N ⋅ m 2 ))]1/ 2 = 9.6 × 105 V/m
The amplitude of the oscillating magnetic field is
Bo = Eo / c = (9.6 × 105 V/m) / (3.0 × 108 m/s) = 3.2 × 10−3 T
Assess: These are reasonable values for the amplitude of the oscillating electric and magnetic fields of
microwaves.
P25.24. Prepare: Electromagnetic waves are sinusoidal and travel with the speed of light. The magnetic
field is Bz = B0 sin(kx − ωt ), where B0 = 3.0 μΤ and k = 1.00 × 107 m–1.
Solve: (a) The wavelength is
λ=
2π
2π
=
= 6.28 × 10−7 m = 628 nm
k 1.00 × 107 m −1
(b) The frequency is
f =
c
λ
=
3.0 × 108 m/s
= 4.77 × 1014 Hz
6.28 × 10−7 m
(c) The electric field amplitude is
E0 = cB0 = (3.0 × 108 m/s)(3.0 × 10–6 T) = 900 V/m
P25.25. Prepare: Electromagnetic waves are sinusoidal and travel with the speed of light. The electric field
is Ey = E0 cos (kx – ωt), where E0 = 20 V/m and k = 6.28 × 108 m–1.
EM Induction and EM Waves 25-15
Solve:
(a) The wavelength is
λ=
2π
2π
=
= 1.00 × 10−8 m = 10.0 nm
k
6.28 × 108 m −1
(b) The frequency is
f =
c
λ
=
3.0 × 108 m/s
= 3.00 × 1016 Hz
1.00 × 10−8 m
(c) The magnetic field amplitude is
B0 =
Assess:
E0
20 V/m
=
= 6.67 × 10−8 T
vem 3.0 × 108 m/s
This field is approximately 500 times smaller than the earth’s magnetic field, and is thus reasonable.
P25.26. Prepare: A radio signal is an electromagnetic wave. We will use Equation 25.17 to find intensity
corresponding to the amplitude of the detected signal’s amplitude.
Solve: From Equation 25.17, the intensity of an electromagnetic wave is
I=
cE0 2 (3.0 × 108 m/s)(8.85 × 10−12 C 2 /(N ⋅ m 2 ))
E0 =
(300 × 10−6 V/m) 2 = 1.2 × 10−10 W/m2
2
2
Assess: Extremely small signals may be picked up with an antenna system and them amplified.
P25.27. Prepare: The laser beam is an electromagnetic plane wave. We will assume that the energy is
uniformly distributed over the diameter of the laser beam. Equation 25.17 connects power of a wave that
impinges on an area A with the wave’s electric field amplitude.
Solve: (a) Using Equation 25.17, the light intensity is
I=
P cε 0 2
200 × 106 W
(3.0 × 108 m/s)(8.85 × 10−12 C 2 /(N ⋅ m 2 )) 2
=
E0 ⇒
=
E0 ⇒ E0 = 2.19 × 1011 V/m
π (1.0 × 10−6 m) 2
A
2
2
which will be reported as 2.2 × 1011 V/m.
(b) The electric field between the proton and the electron is
E=
q (9 × 109 N ⋅ m 2 /C 2 )(1.60 × 10−19 C)
=
= 5.13 × 1011 V/m
4πε 0 r 2
(0.053 × 10−9 m) 2
1
The ratio of the two electric field amplitudes is thus
E0 / E = (2.19 × 1011 V/m) /(5.13 × 1011 V/m) = 0.43
Assess: The laser beam’s electric field is approximately half the electric field that keeps the electron in its
orbit. These electric field amplitudes are very large!
P25.28. Prepare: A radio wave is an electromagnetic wave. The power or energy transported per second by the
radio wave is 25 kW, or 25 × 103 J/s. This energy is carried uniformly in all directions. Equation 25.17 connects power of
a wave that impinges on an area A (defined as its intensity) with the wave’s electric field amplitude.
Solve:
(a) From Equation 25.17, the light intensity is
I=
P
P
25 × 103 W
=
=
= 2.21 × 10−6 W/m 2
A 4π r 2 4π (30 × 103 m) 2
(b) Using Equation 25.17 again,
cε
(3.0 × 108 m/s)(8.85 × 10−12 C2 /(N ⋅ m 2 )) 2
I = 0 E02 ⇒ 2.21 × 10−6 W/m 2 =
E0 ⇒ E0 = 0.041 V/m
2
2
Assess: These values are reasonable.
25-16 Chapter 25
P25.29. Prepare: The energy transported per second by the wave is 10 W or 10 J/s. This energy is carried
uniformly in all directions. Equation 25.17 connects power of a wave that impinges on an area A with the wave’s
electric field amplitude.
Solve: (a) From Equation 25.17, the light intensity is
I=
P
cE
= 0 E02 ⇒ r =
2
4π r
2
2P
=
4π cE0 E02
2(10 W)
599.45 24.48
=
=
−12
2
2
2
4π (3.0 × 10 m/s)(8.85 × 10 C /(N· m )) E0
E02
E0
8
For E0 = 100 V/m, r = 0.25 m.
(b) For E0 = 0.010 V/m, r = 2500 m.
Assess: Greater distances result in smaller electric fields. This is exactly what we expect.
P25.30. Prepare: We will use Malus’s law for the polarized light. From Equation 25.20 the relationship between the
incident and the transmitted polarized light is Itransmitted = Iincident cos2θ, where θ is the angle between the electric field and
the axis of the filter. In our case θ will be the angle between the filter axis and the plane of polarization.
Solve: From Equation 25.20 the intensity is
(a) Itransmitted = Iincident cos2θ = (10 W/m2) cos2 (0 °) = 10 W/m2
(b) I
=I
cos2θ = (10 W/m2) cos2 (30 °) = 7.5 W/m2
transmitted
incident
(c) Itransmitted = Iincident cos2θ = (10 W/m2) cos2 (45 °) = 5 W/m2
(d) Itransmitted = Iincident cos2θ = (10 W/m2) cos2 (6 0 °) = 2.5 W/m2
(e) Itransmitted = Iincident cos2θ = (10 W/m2) cos2 (90 °) = 0 W/m2
Assess: The final intensity must be between 0 W/m2 and 10 W/m2 and all values are in this range.
P25.31. Prepare: We will use Malus’s law for the polarized light. From Equation 25.20 the relationship
between the incident and the transmitted polarized light is Itransmitted = I0 cos2θ, where θ is the angle between the
electric field and the axis of the filter.
Solve:
From Equation 25.20,
0.25 I0 = I0 cos2θ ⇒ cosθ = 0.50 ⇒ θ = 60 °
Assess:
Note that θ is the angle between the electric field and the axis of the filter.
P25.32. Prepare: We will use Malus’s Law for polarized light I transmitted = I o cos 2 θ where θ is the angle
between the electric field and the axis of the filter and the relationship between power and intensity I = P / A.
Solve: The following figure helps us visualize the situation and get the angles correct. The incident beam is polarized
horizontally and the transmitted beam is polarized at an angle of 25 ° wrt the vertical and 65 ° wrt the horizontal.
EM Induction and EM Waves 25-17
The incident intensity and power are related by I o = Po / A and the transmitted intensity and power are related by
IT = PT / A . No subscript is used for the A since it is the same for the incident and transmitted beam. Combining
these results and recalling that θ is the angle between the electric field of the incident beam (horizontal) and the
axis of the filter (25 ° wrt the vertical and 65 ° wrt the horizontal) obtain
I T = I o cos 2 65 ° ⇒
PT Po
= cos 2 65 ° ⇒ PT = Po cos 2 65 ° = (200 mW)cos 2 65 ° = 36 mW
A A
Assess: We are expecting that the power of the beam will be decreased as it passes through the filter and this is
the case. Note that it is important to know that θ is the angle between the electric field and the axis of the filter,
which is 90 ° − 25 ° = 65 °.
P25.33. Prepare: Malus’s Law for polarized light ( I transmitted = I o cos 2 θ ) provides the relationship between the
incident and transmitted light intensity. Note that θ is the angle between the electric field and the axis of the filter.
Solve: The intensity of the transmitted beam while the laser is cold is I TC = I o cos 2 30 °.
The intensity of the transmitted beam while the laser is warm is
ΔI
I −I
I (cos 2 60 ° − cos 2 30 °)
Percent =
(100) = TW TC (100) = o
(100) = − 67%
ITC
ITC
I o cos 2 30 °
Assess: The minus reminds us that indeed the light intensity has decreased.
P25.34. Prepare: Use the photon model of light as given by Equation 25.21.
Solve:
The energy of the photon is
⎛ 3.0 × 108 m/s ⎞ ⎛
1e V
⎞
= (6.63 × 10−34 J ⋅ s) ⎜
= 2.5 eV
⎟⎜
−9
−19 ⎟
×
×
500
10
m
1.60
10
J
λ
⎝
⎠
⎝
⎠
Assess: The energy of a single photon in the visible light region is small.
Ephoton = hf = h
c
P25.35. Prepare: We will use the photon model of light as given by Equation 25.21. We will also use f = c/λ
in Equation 25.21.
Solve: The energy of the x-ray photon is
⎛ 3.0 × 108 m/s ⎞
⎛c⎞
−16
E = hf = h ⎜ ⎟ = (6.63 × 10−34 J ⋅ s) ⎜
⎟ = 1.99 × 10 J
−9
⎝λ⎠
⎝ 1.0 × 10 m ⎠
In eV, the energy is (1.99 × 10−16 J)
(
1 eV
1.6 × 10−19 J
) = 1200 eV.
Assess: This is a small amount of energy, but it is larger than the energy of a photon in the visible wavelength.
P25.36. Prepare: We will use the photon model of light. The energy of a photon with wavelength λ1
is E1 = hf1 = hc / λ1. Similarly, E2 = hc / λ2 .
Solve:
Since E2 be equal to 2E1,
hc
λ2
=2
hc
λ1
⇒ λ2 =
λ1
2
=
600 nm
= 300 nm
2
Assess: A photon with λ = 300 nm has twice the energy of a photon with λ = 600 nm. This is an expected
result, because energy is inversely proportional to the wavelength.
P25.37. Prepare: The energy of a photon is given by Equation 25.21: Ephoton = hf .
We first compute f from Equation 25.16: c = λ f .
Solve:
(a)
f =
c
λ
=
3 . 0 × 108 m/s
= 6 . 0 × 1019 Hz
0 .0050 × 10−9 m
Ephoton = hf = (6. 63 × 10−34 J ⋅ s)(6 . 0 × 1019 Hz) = 4. 0 × 10−14 J
25-18 Chapter 25
We want to know the answer in eV as well.
⎛
⎞
1 eV
5
4 .0 × 10−14 J ⎜
⎟ = 2 . 5 × 10 eV = 250 keV
−19
⎝ 1. 60 × 10 J ⎠
(b) The energy just found divided by the energy required to break a hydrogen bond between two water molecules
gives the number of bonds it can break: 250 keV/(0. 24 eV) = 1,000,000 bonds.
Assess: This is actually a very large energy for one photon, compared to visible light photons (which have
energies of just a few eV), or radio wave photons (which have even less energy than visible photons).
P25.38. Prepare: Wavelength and energy of photons are related by E = hc /λ .
Solve:
The maximum wavelength that can cause the desired transition is
λ = hc / E = [(6.63 × 10−34 J ⋅ s)(3.0 × 108 m/s)]/[(1.8 eV)(1.6 × 10−19 J/eV)] = 690 nm
Assess:
As expected, this wavelength is in the visible range.
P25.39. Prepare: We will use Equation 25.21 and the relationship f = c/λ.
Solve:
The energy of the single photon is
⎛ c ⎞ (6.63 × 10−34 J ⋅ s)(3.0 × 108 m/s)
= 1.99 × 10−19 J
Ephoton = hf = h ⎜ ⎟ =
1.0 × 10−6 m
⎝λ⎠
⇒ Emol = N A Ephoton = (6.023 × 1023 )(1.99 × 10−19 J) = 1.2 × 105 J
Assess: Although the energy of a single photon is very small, a mole of photons has a significant amount of energy.
P25.40. Prepare: The relationship between energy and wavelength for a photon is E = hc / λ .
Solve: The energy of the photons is E = hc / λ = (6.63 × 10−34 J ⋅ s)(3.0 × 108 m/s) / (9.5 × 10−6 m) = 2.1 × 10−20 J.
Assess: We expect the energy of a single photon of this wavelength to be small.
P25.41. Prepare: We follow somewhat the strategy outlined in Example 25.8. Along the way we’ll use
these relationships: intensity = power/area and energy = power × time.
We’ll first find the energy per photon for the given wavelength from E = hf where f = c / λ .
We are given λ = 680 × 10−9 m, A = 1 . 00 m 2, Δ t = 1 . 00 s, and the intensity I = 1370 W/m 2 .
Solve:
c 3 .00 × 108 m/s
f = =
= 4 .412 × 1014 Hz
λ 680 × 10−9 m
The energy per photon is
Ephoton = hf = (6 . 63 × 10−34 J ⋅ s)(4 . 412 × 1014 Hz) = 2. 925 × 10−19 J
Now we estimate the number of photons by applying the relationships.
N≈
total energy
power × time
(intensity area) × time
=
=
energy/photon energy/photon
energy/photon
N≈
IAΔ t
(1370 W/m 2 )(1 .00 m 2 )(1 . 00 s)
=
= 4 . 68 × 1021 photons
E /photon
2. 925 × 10−19 J/photon
Assess: Don’t confuse the usage of I as intensity here with its usage elsewhere for current.
The photons emitted by the sun will span a range of energies, because the light spans a range of wavelengths, but
the average photon energy is assumed to be 680 nm.
EM Induction and EM Waves 25-19
P25.42. Prepare: The intensity, power, and area are related by I = P / A. Power, energy, and time are related
by P = ΔE / Δ t . The total amount of energy is related to the energy of each photon by E = NE photon . The energy of
a photon is related to its wavelength by Ephoton = hc / λ .
Solve:
Combining these results we have
I=
P (Δ E / Δt ) NEphoton ⎛ N ⎞ ⎛ Ephoton ⎞ ⎛ N ⎞⎛ hc / λ ⎞ ⎛ N ⎞⎛ hc ⎞
=
=
= ⎜ ⎟⎜
⎟ = ⎜ ⎟⎜
⎟ = ⎜ ⎟⎜
⎟
A
A
AΔ t
⎝ Δt ⎠ ⎝ A ⎠ ⎝ Δ t ⎠⎝ A ⎠ ⎝ Δt ⎠⎝ Aλ ⎠
Solving this expression for the number of photons per second, obtain
2
−11
2
−3
2
−7
⎛ N ⎞ IAλ I (π r )λ (1.6 × 10 W/m )π (3.5 × 10 m) (5.50 × 10 m)
=
=
=
= 1.7 × 103 photons/s
⎜ Δt ⎟ hc
hc
(6.63 × 10−34 J ⋅ s)(3.0 × 108 m/s)
⎝ ⎠
Assess: It is impressive that these photons traveled all the way from the distant star to your eye. The expression
we developed cannot be found in the text. The purpose of your text is not to give you the answers but rather to
give you the tools to find the answers.
P25.43. Prepare: Wein’s law, given in Equation 25.23, tells us the relationship between the temperature and
the peak wavelength, although the answer will be in kelvin and we’ll then convert to °C.
λpeak (in nm) =
2 .9 × 106 nm ⋅ K
T
We are given λpeak = 1200 nm.
Solve:
Solve the equation for T .
T=
2 . 9 × 106 nm ⋅ K
λpeak
=
2. 9 × 106 nm ⋅ K
= 2417 K
1200 nm
Now we must convert the answer in kelvin to °C by subtracting 273: 2417 K − 273 ° = 2144 °C.
This should be rounded to 2100 °C.
Assess: Regular tungsten filaments are usually a bit hotter than this, but we are in the right ballpark.
P25.44. Prepare: The wavelength and the maximum intensity are related by λpeak (in nm) = (2.9 × 106 nm ⋅ K)/T
(in K). According to the Stefan-Boltzmann law (Δ Q / Δt = eσ AT 4 ) , the thermal radiation is proportional to the
temperature raised to the fourth power (T 4 ).
Solve: The temperature for the given wavelengths is
T1800nm = (2.9 × 106 nm ⋅ K)/λ = (2.9 × 106 nm ⋅ K)/(1.8 × 103 nm) = 1.6 × 103 K
and
T1600nm = (2.9 × 106 nm ⋅ K)/ λ = (2.9 × 106 nm ⋅ K)/(1.6 × 103 nm) = 1.8 × 103 K
The resulting temperature change is ΔT = T1600 nm − T1800 nm = 200 K = 200 °C.
The positive sign informs us that the temperature increases.
The energy radiated for each wavelength case is
4
3
4
12
4
(ΔQ / Δ t )1800 nm = eσ AT1800
nm = eσ A(1.6 × 10 K) = eσ A(6.5 × 10 K )
and
4
3
4
12
4
(Δ Q / Δt )1600 nm = eσ AT1600
nm = eσ A(1.8 × 10 K) = eσ A(10.5 × 10 K )
( ΔQ / Δ t )1600 nm
( ΔQ / Δ t )1800 nm
=
eσ A(10.5 × 1012 K 4 )
= 1.6
eσ A(6.5 × 1012 K 4 )
25-20 Chapter 25
This informs us that as the peak wavelength is decreased from 1800 nm to 1600 nm, the temperature increases by
200 °C and the thermal radiation increases by a factor of 1.6.
Assess: Looking at Figure 25.41, notice that as the peak wavelength goes down the temperature goes up and the
intensity of the radiation increases. This is consistent with our calculation.
P25.45. Prepare: For both cases we will use Ephoton = hf to find the frequency, then c = λ f to arrive at the
wavelength.
In the second case the energy of the photons is E = (2. 5)(25 keV) = 62 . 5 keV.
Solve: For the 25 KeV photons:
25000 eV ⎛ 1.6 × 10−19 J ⎞
E
18
=
⎜
⎟ = 6 . 0 × 10 Hz
1 eV
h 6.63 × 10−34 J ⋅ s ⎝
⎠
c 3.0 × 108 m/s
−11
λ= =
= 5 . 0 × 10 m = 50 pm
f 6.0 × 1018 Hz
f =
For the 62.5 keV photons:
f =
62500 eV ⎛ 1 . 6 × 10−19 J ⎞
E
19
=
⎜
⎟ = 1 .5 × 10 Hz
1 eV
h 6 . 63 × 10−34 J ⋅ s ⎝
⎠
λ=
c 3 . 0 × 108 m/s
=
= 2 . 0 × 10−11 m = 20 pm
f 1 . 5 × 1019 Hz
Assess: We see that the more energetic photons have a shorter wavelength, as we expected. These wavelengths
are in the x-ray range.
P25.46. Prepare: Please refer to Figure P25.46. To calculate the flux we need to consider the orientation of the
normal of the surface relative to the magnetic field direction. We will consider the flux through the surface in the two
parts corresponding to the two different directions of the surface normals. The top and left surface areas are equal.
Solve: The flux is
Φ = Φ top + Φ left = Atop B cos 45 ° + Aleft B cos 45 ° = 2 × (0.050 m × 0.10 m)(0.050 T)cos 45 ° = 3.5 × 10−4 Wb
Assess: Since both the area and the magnetic field are small, we expect a small value for the total flux through
the square.
P25.47. Prepare: The flux measures how much of the field penetrates the chosen surface. We will break the
surface up because the magnetic field has different strengths over different parts of the surface of the loop.
Solve: For convenience we choose the normal to the loop to be into the page so it is in the same direction as
the magnetic field. It is clear that the field is uniform inside the rectangular region and zero outside. The total
flux thus is Φ = BAinside = B ab.
rectangle
Assess:
All flux in the loop is contained inside the rectangle.
P25.48. Prepare: The magnetic flux through a coil is Φ = BA cos φ .
Solve: (a) When the plane of the coil is perpendicular to the earth’s magnetic field, the magnetic flux through
the coil is
Φ = BA cos φ = Bπ r 2 cos φ = (50 × 10−6 T)π (0.10 m) 2 cos0 ° = 1.6 × 10−6 Wb
(b) If the coil is rotated 30°, the magnetic flux through the coil is
Φ = BA cos φ = Bπ r 2 cos φ = (50 × 10−6 T)π (0.10 m) 2 cos30 ° = 1.4 × 10−6 Wb
Assess:
A small value is expected for the magnetic flux since the earth’s magnetic field and the coil are both small.
P25.49. Prepare: The flux is given by Equation 25.9. In this case B = 1. 6 T, and A = πr2 = π (d/2)2 = π (0.55 m)2
= 0. 95 m 2.
Solve:
EM Induction and EM Waves 25-21
Φ m = AB cosθ = (0. 95 m 2 )(1 . 5 T)cos0 ° = 1 . 4 T ⋅ m 2 = 1 . 425 Wb
Induced emf is
E=
Δφ 1.425 Wb
=
= 1.2 V
Δt
1.2 s
Assess: We put in cos 0° because the field created by the solenoid will be oriented along the length of the
solenoid. The length of the solenoid was unneeded information because we were given the total magnetic field
created by the solenoid. We kept a couple of extra significant figures on the intermediate calculation but rounded
the final answer to two.
P25.50. Prepare: The induced current will be in such a direction as to oppose what created it.
Solve: As the magnet is pushed into the coil, the coil experiences a changing magnetic flux. According to the
right-hand rule, this changing magnetic flux will create a counterclockwise current. This counterclockwise
current is considered to be positive since it enters the ammeter at the positive terminal. This current creates a
magnetic flux that opposes the incoming flux associated with the magnet. While the magnet is stationary, there is
no flux change in the coil and hence no induced current. As the magnet is pulled out of the coil, a clockwise
current is created. This current is considered to be negative since it enters the ammeter at the negative terminal.
This current creates a magnetic flux that adds to the decreasing flux associated with the magnet. The current in
the circuit is shown in the figure.
Assess: When the motion of the magnet is reversed, the direction of the induced current is reversed. When the
magnet has no motion relative to the coil, there is no induced current.
P25.51. Prepare: Equation 25.11 gives the induced emf for a loop of wire in a changing magnetic field.
E=
ΔΦ m
Δt
Since the loop is not changing size and the orientation stays perpendicular, this boils down to
E=
Δ( AB cosθ )
A cos30 °( ΔB )
ΔB
=
= A cos30 °
Δt
Δt
Δt
We are given A = 0 . 020 m 2, Δ B = 0 T − 0 . 30 T = −0 . 30 T, and Δ t = 45 × 10−3 s.
Solve:
E = A cos30 °
ΔB
− 0 . 30 T
= (0 . 020 m 2 )cos30 °
= 0 .11V
Δt
45 × 10−3 s
Assess: This is a reasonable value for the induced emf and checking the units the units. T = kg/(A ⋅ s 2 ) and
V = kg ⋅ m 2 /(A ⋅ s3 ) so the units in the question work out.
T ⋅ m2
= V.
s
25-22 Chapter 25
P25.52. Prepare: Assume that B changes uniformly with time. The magnetic strength is changing so the
flux is changing and this will create an induced emf. The magnetic field is at an angle θ = 60° to the normal of
the plane of the coils.
Solve: The flux for a single loop of the coil is Φ = BA cosθ . The radius and angle don’t change with time, but
B does. According to Faraday’s law, the induced emf is
E=N
ΔΦ
ΔB
ΔB
= NA cosθ
= N π r 2 cosθ
Δt
Δt
Δt
= 100π (0.010 m) 2 (cos60 °)
Assess:
1.50 T − 0.50 T
= 2.62 × 10−2 V = 26 mV
0.60 s
This is a reasonable induced emf for a 100-turn coil and a relatively fast increase in the magnetic field.
P25.53. Prepare: We will assume that B changes uniformly with time. The magnetic strength is changing so
the flux is changing and this will create an induced emf. The magnetic field is at an angle θ = 45° to the normal
to the plane of the coils.
Solve: The flux for a single loop of the coil is Φ = BA cosθ . The radius and angle don’t change with time, but B
does. According to Faraday’s law, the induced emf is
ΔΦ
ΔB
ΔB
E=N
= NA cosθ
= N π r 2 cosθ
Δt
Δt
Δt
= 25π (0.050 m) 2 (cos 45 °)
Assess:
0.20 T − 0.80 T
= 4.17 × 10−2 V = 42 mV
2.0 s
This is a reasonable induced emf for a 25-turn coil and a rather fast decrease in the magnetic field.
P25.54. Prepare: Magnetic flux is determined by Φ = BA cosθ , where θ is the angle between the direction of
the area A and the magnetic field B. The emf generated is related to the changing magnetic flux by E = ΔΦ / Δt .
A typical tongue has an area of approximately 25 × 10−4 m 2 .
Solve: The emf generated is
E=
ΔΦ
Φ − Φ initial
AB cosθ − AB
AB(cosθ − 1)
(25 × 10−4 )(3 T)(cos30o − 1)
= final
=
=
=
= 6.7 mV
Δt
Δt
Δt
Δt
0.15 s
This emf is less than half that needed to trigger an action potential and as a result does not support the notion that
the induced emf is responsible for the effect.
Assess: This is a reasonable emf even though less than that needed to trigger an action potential.
EM Induction and EM Waves 25-23
P25.55. Prepare: The circumference of the loop and the radius of the loop are related by C = 2π rloop . The emf
generated in the loop is related to the changing magnetic field by
2
ΔB
ΔB
⎛ C ⎞ ΔB ⎛ C 2 ⎞ ΔB
2
= π rloop
=π⎜
=⎜
⎟
⎟
Δt
Δt
⎝ 2π ⎠ Δ t ⎝ 4π ⎠ Δ t
The emf generated is related to the induced current, the resistivity of the wire, the cross-sectional area of the wire
C
C
= Iρ 2 .
and the length (circumference) of the wire by ε = IR = I ρ
Awire
π rwire
Solve: Combining these two expressions for the emf and solving for the current obtain
E= A
⎛ Cr 2
I = ⎜ wire
⎝ 4ρ
⎞ ΔB ⎛ (0.2 m)(1.6 × 10−4 m) 2 ⎞ ⎛ 0.55 T ⎞
=⎜
= 31 mA
⎟
⎟⎜
−6
−3 ⎟
⎠ Δt ⎝ 4(1.5 × 10 Ω ⋅ m) ⎠ ⎝ 15 × 10 s ⎠
Assess: This is a reasonable induced current.
P25.56. Prepare: Since this problem deals with numerous concepts, we will start by listing all concepts involved.
Cross-sectional area of a loop coil or wire A = π r 2
Circumference of a circle C = 2π r
Resistance R = ρ l / A
Induced emf E = Aloop
ΔB
Δt
Current I = E / R
Power P = W / Δt = I 2 R
Mass m = ρV
Heat energy Q = mcΔ T
Solve: Now let’s determine the value of each quantity and then put the pieces together.
2
= π (4.0 × 10−2 m) 2 = 5.03 × 10−2 m 2
Cross-sectional area of the muscle coil Acoil = π rcoil
2
= π (5.0 × 10−3 m) 2 = 7.85 × 10−5 m 2
Cross-sectional area of the muscle Amuscle = π rmuscle
Length (circumference) of the wire l = 2π rloop = 2π (4.0 × 10−2 m) = 2.51 × 10−1 m
Resistance of the muscle Rmuscle = ρ
Induced emf E = Aloop
l
Amuscle
= (13 Ω ⋅ m)
(2.51 × 10−1 m)
= 4.16 × 104 Ω
(7.85 × 10−5 m 2 )
ΔB
(1.6 T)
= (5.03 × 10−3 m 2 )
= 2.68 × 10−2 V
Δt
(0.3 s)
E (2.68 × 10−2 V)
=
= 6.44 × 10−7 A
R (4.16 × 104 Ω)
(a) The energy dissipated in the muscle loop is
Induced current I =
W = PΔt = I 2 RΔt = (6.44 × 10−7 A) 2 (4.16 × 104 Ω)(0.3 s) = 5.15 × 10−9 J = 5.2 × 10−9 J
Mass of muscle m = ρV = ρ Amusclelmuscle = (1.1 × 103 kg/m3 )(7.85 × 10−5 m 2 )(2.51 × 10−1 m)
W
(5.15 × 10−9 J)
=
= 6.6 × 10−11 K.
−2
mc (2.16 × 10 kg)(3.6 × 103 J/(kg ⋅ K))
Assess: Due to the small current and small time involved, we expect a small amount of energy to be dissipated
and hence a very small temperature change.
(b) The change in temperature of the muscle tissue is Δ T =
P25.57. Prepare: We will assume that the field is uniform in space over the coil. We want an induced
current so there must be an induced emf created by a changing flux. The magnetic field is perpendicular to the
plane of the coil so the flux for a single loop of the coil is Φ = BA , if we take the normal to the coil to be in the
same direction as the field.
Solve: To relate the emf and the current we need to know the resistance. From Equation 22.8,
25-24 Chapter 25
R=
ρCu lwire
Awire
=
ρCu N 2π r 2(100)(1.7 × 10−8 Ω ⋅ m)(0.040 m)
=
= 2.18 Ω
2
(2.5 × 10−4 m) 2
π rwire
Using Faraday’s law,
E = IR = NA
ΔB
ΔB
ΔB
IR
(2.0 A)(2.18 Ω)
= Nπ r 2
⇒
=
=
= 8.7 T/s
Δt
Δt
Δt
Nπ r 2 100π (0.040 m) 2
Assess: We wanted rather large induced emf of magnitude (2.0 A)(2.18 Ω) = 4.36 V, which in a 100-turn coil
requires a large rate of increase of B.
P25.58. Prepare: Please refer to Figure P25.58. We assume that the field is uniform. The motion of the loop
changes the flux through it. This results in an induced emf and current. The area A is changing, but the field B is
r
r
not. Take A as being out of the page and parallel to B, so Φ = AB and ΔΦ / Δt = B (ΔA / Δ t ) The flux is through
that portion of the loop where there is a field, that is, A = l x.
Solve: The induced emf is E = ΔΦ / Δt and the induced current is I = E / R. The emf and current are
ΔA
Δ (lx)
=B
= Blv = |(0.20 T)(0.050 m)(50 m/s)| = 0.50 V
Δt
Δt
E 0.50 V
= 5.0 A
I= =
R 0.10 Ω
E=B
The field is out of the page. As the loop moves the flux increases because more of the loop area has field through
it. To prevent the increase, the induced field needs to point into the page. Thus, the induced current flows
clockwise.
Assess: This seems reasonable since there is rapid motion of the loop.
P25.59. Prepare: We will assume that the field is uniform in space over the loop. The moving slide wire in
a magnetic field develops a motional emf and a corresponding current in the wire and the rails. The current
through the resistor will cause energy dissipation and subsequent warming.
Solve: (a) The induced emf depends on the changing flux, not on where the resistance in the loop is located.
From Equation 25.3, we have Eloop = vlB = (10 m/s)(0.20 m)(0.10 T) = 0.20 V.
The total resistance around the loop is due entirely to the carbon resistor, so the induced current is I = Eloop /R =
(0.20 V)/(1.0 Ω) = 0.20 A.
(b) To move with a constant velocity the acceleration must be zero, so the pulling force must balance the retarding
magnetic force on the induced current in the slide wire. Thus from Equation 25.5, Fpull = Fmag = IlB = (0.20 A)
(0.20 m)(0.1 T) = 4.0 × 10–3 N.
(c) The current in the resistor will result in power being dissipated. The power is P = I 2 R = (0.20 A)2 (1 Ω) =
0.040 W = 0.040 J/s.
During a 10-second period Q = 0.40 J of energy is dissipated by the current, increasing the internal energy of the
carbon resistor and raising its temperature. This is, effectively, a heat source. The heat is related to the
temperature rise by Q = mcΔT, where c is the specific heat of carbon. Thus,
ΔT =
Assess:
Q
0.40 J
=
= 11 K = 11 °C
mc (5.0 × 10−5 kg)(710 J/(kg °C))
This is a significant change in the temperature of the resistor.
EM Induction and EM Waves 25-25
P25.60. Prepare: The induced emf is related to the area of the ring and the changing magnetic field by
ΔB
l
2π r
E= A
. The resistance of the ring may be determined by R = ρ = ρ
. The current in the ring is related
Δt
A
A
to the emf and resistance by I = E / R . Finally, gold rings are usually an alloy of gold, silver, and copper, called
jeweler’s gold and the resistivity is ρ = 13.2 × 10−10 Ω ⋅ m.
Solve: (a) The emf generated in the ring is
E= A
ΔB
ΔB
(2.5 T)
= π r2
= π (1.0 × 10−2 m) 2
= 3.93 V ≈ 3.9 V
Δt
Δt
(2.0 × 10−4 s)
l
2π r
2π (1 × 10−2 m)
=ρ
= (13.2 × 10−10 Ω ⋅ m)
= 2.07 × 10−5 Ω.
A
A
4.0 × 10−6 m 2
3.93 V
E
(b) The current induced in the ring is I = =
= 1.9 × 105 A.
R 2.07 × 10−5 Ω
The current would cook the technician’s finger.
Assess: Considering the magnitude of the changing magnetic flux, the value for the induced emf is reasonable.
Since the resistance of the ring is small, the current will be large—too large for comfort.
The resistance of the ring is R = ρ
P25.61. Prepare: Please refer to Figure P25.61. The moving wire will have a motional emf that produces a
current in the loop. We will assume there is no resistance in the rails, and if there is any resistance, it is accounted
for by the resistor.
Solve: (a) At constant velocity the external pushing force is balanced by the magnetic force, so using Equation 25.5
Fpush = Fmag = IlB =
E
B 2l 2v (0.50 T) 2 (0.10 m) 2 (0.50 m/s)
⎛ Blv ⎞
=
= 6.25 × 10−4 N = 6.3 × 10−4 N
lB = ⎜
lB =
⎟
2.0 Ω
R
R
⎝ R ⎠
(b) The power is P = Fv = (6.25 × 10−4 N)(0.50 m/s) = 3.1 × 10−4 W.
(c) The flux is out of the page and decreasing and the induced current/field will oppose the change. The induced
field must have a flux that is out of the page so the current will be counterclockwise.
The magnitude of the current using Equation 25.4 is
⎛ Blv ⎞ (0.50 T)(0.10 m)(0.50 m/s)
= 1.25 × 10−2 A = 1.3 × 10−2 A
I =⎜
⎟=
2.0 Ω
⎝ R ⎠
(d) The power is P = I 2 R = (1.25 × 10−2 A) 2 (2.0 Ω) = 3.1 × 10−4 W.
Assess: From energy conservation we see that the mechanical energy put in by the pushing force shows up as
electrical energy in the resistor.
P25.62. Prepare: Please refer to Figure P25.62 and assume that the magnetic field is uniform over the loop.
We are just interested in the emf due to the motion of the eye so we can ignore the details of the time dependence
of the change. The motion of the eye will change the orientation of the loop relative to the fixed field direction,
resulting in an induced emf.
Solve: From Faraday’s law,
Ecoil = N
=
ΔΦ
ΔΦ
Δ cosθ
Nπ r 2 B|cosθ f − cosθi |
=N
= NAB
=
Δt
Δt
Δt
Δt
20π (3 × 10−3 m) 2 (1.0 T) | cos85 ° − cos90 ° |
= 2.5 × 10−4 V
0.20 s
Assess: This is a reasonable emf to measure, although you might need some amplification.
P25.63. Prepare: Light is an electromagnetic wave. A quick measurement of a light bulb shows that its radius
is r ≈ 2.8 cm. Since 4.0 W is given off as visible light, 4.0 W is the energy transported per second by the
electromagnetic light wave. This energy is carried in all directions.
25-26 Chapter 25
Solve:
The light intensity is given by Equation 25.17:
P
P
cE
=
= 0 E02
2
A 4π r 2
4.0 W
(3.0 × 108 m/s)(8.85 × 10−12 C 2 /(N ⋅ m 2 )) 2
V
⇒
=
E0 ⇒ E0 ≈ 550
2
−2
4π (2.8 × 10 m)
2
m
I=
⇒ B0 =
E0 ⎛ 553 V/m ⎞
−6
=⎜
⎟ ≈ 1.8 × 10 T
c ⎝ 3.0 × 108 m/s ⎠
Assess: This is on the order of earth’s magnetic field and is reasonable.
P25.64. Prepare: The power is the energy/time. We’ll plug in our answer from part (a) for the power into
Equation 25.17 for part (b).
I=
Solve:
P 1
= cE0 E02
A 2
(a)
P=
E 2 . 5 × 10−3 J
=
= 250000 W = 250 kW
Δ t 10 × 10−9 s
(b) Solve Equation 25.17 for E0. The area of the focus circle is A = π r 2 = π (0. 425 × 10−3 m) 2 = 5. 67 × 10−7 m 2.
E0 =
P 2
250000 W
2
=
= 1 . 8 × 107 N/C = 1 .8 × 107 V/m
−7
2
8
A c E0
5 . 67 × 10 m (3 . 0 × 10 m/s)(8 .85 × 10−12 C 2 /(N ⋅ m 2 ))
Assess: This is a very strong field, but that is what is needed for surgery to ablate the thin layer of the cornea.
P25.65. Prepare: Radio waves are electromagnetic waves. The radio transmitter is radiating energy in all
directions at the rate of 21 J per second.
Solve: (a) The signal intensity received at the earth is given by Equation 25.17:
I=
P
P
21 W
=
=
= 8.252 × 10−26 W/m2 = 8.3 × 10−26 W/m2
2
A 4π r
4π (4.5 × 1012 m) 2
(b) Using Equation 25.17 again,
I=
cE0 2
2I
2(8.252 × 10−26 W/m 2 )
V
E0 ⇒ E0 =
=
= 7.9 × 10−12
8
2
cE0
(3.0 × 10 m/s)(8.85 × 10 −12 C 2 /(N ⋅ m 2 ))
m
Assess: Given such a small signal and the large distance, we expect a small signal intensity and electric field.
This problem helps one appreciate the technology associated with the space program.
P25.66. Prepare: Intensity is related to power and area by I = P / A. Intensity is also related to the amplitude
of the oscillating electric field by I = cEo Eo2 / 2.
Solve: The area of a sphere with a 100 ft radius is A = 4π r 2 = 4π [(100 ft )(m / (3.26 ft))]2 = 1.17 × 104 m 2
Combine these expressions for the intensity and solve for the amplitude of the electric field.
1/ 2
⎛ 2P ⎞
Eo = ⎜
⎟
⎝ cEo A ⎠
1/ 2
⎛
⎞
2(4.0 × 10−3 W)
=⎜
−12 2
8
2 ⎟
⎝ (3.0 × 10 m/s)(8.85 × 10 C / (N ⋅ m ) ⎠
= 2.6 × 10−4
V
m
Assess: Since the power output of the phone signal is small, we expect the amplitude of the oscillating electric
field to be small.
P25.67. Prepare: The intensity of the beam is related to the magnitude of the electric field by I = cEo Eo2 / 2.
The intensity of the beam before and after the polarizing filter are related by I after = I before cos 2 θ .
EM Induction and EM Waves 25-27
Solve: The intensity of the beam after the polarizing filter is I after = I before cos 2 θ = I before cos 2 45o = I before / 2.
Inserting the relationship between intensity and the magnitude of the electric field obtain.
2
cε o Eafter
cε E 2
E
= (1 / 2) o before ⇒ before = 2
Eafter
2
2
Assess: Since the intensity of the beam depends on the square of the amplitude oscillating electric field, if
I before
E
= 2 then before = 2.
I after
Eafter
P25.68. Prepare: Radio waves are electromagnetic waves. Assume that the transmitter unit radiates in all
directions. The transmitting unit radiates energy in all directions at the rate of 250 mJ per second.
Solve: From Equation 25.17, the signal intensity at a distance of 42 m is
I=
P
P
250 × 10−3 W
=
=
= 1.13 × 10−5 W/m 2
4π (42 m) 2
A 4π r 2
Using Equation 25.17 again,
I=
cE0 2
2I
2(1.13 × 10−5 W/m 2 )
V
=
= 0.092
E0 ⇒ E0 =
8
2
2
− 12
2
(3.0 × 10 m/s)(8.85 × 10 C /(N ⋅ m ))
m
cE0
A few steps before 42 m, the field strength was 0.100 V/m and the door opened. The manufacturer’s claims are correct.
Assess: Find the instruction manual for a garage door and/or your car opener and check it out. Don’t be
surprised if you are the only one who gets a little excited by this project.
P25.69. Prepare: Use Malus’s law for the polarized light.
Solve: For unpolarized light, the electric field vector varies randomly through all possible values of θ. Because
the average value of cos2θ is 12 , the intensity transmitted by a polarizing filter when the incident light is
unpolarized is I1 =
1
2
I0. For polarized light, Itransmitted = I0 cos2θ. Therefore,
I2 = I1cos2 45° I3 = I2cos2 45° ⇒ I3 = (I1 cos2 45°)cos2 45° =
1
1
I0(cos4 45°) = I 0
2
8
Assess: For this situation, the intensity is reduced by a factor of two each time it encounters a polarizing filter.
P25.70. Prepare: Use the photon model of light.
Solve:
(a) The wavelength is calculated as follows:
(6.63 × 10−34 J ⋅ s)(3.0 × 108 m/s)
⎛c⎞
= 2.0 × 10−12 m
Egamma = hf = h ⎜ ⎟ ⇒ λ =
1.0 × 10−13 J
⎝λ⎠
(b) The energy of a visible-light photon of wavelength 500 nm is
−34
8
⎛ c ⎞ (6.63 × 10 J ⋅ s)(3.0 × 10 m/s)
Evisible = h ⎜ ⎟ =
= 3.978 × 10−19 J
500 × 10−9 m
⎝λ⎠
25-28 Chapter 25
The number of photons n such that Egamma = nEvisible is
n=
Egamma
Evisible
=
1.0 × 10−13 J
= 2.51 × 105
3.978 × 10−19 J
Assess: Since the energy of photons in the visible-light range is small compared to the energy of the gamma-ray
photon, we expect a large number of photons.
P25.71. Prepare: We will use Ephoton = hf to find the frequency, then c = λ f to arrive at the wavelength.
Solve:
2 . 0 × 1013 eV ⎛ 1 . 6 × 10−19 J ⎞
E
27
=
⎜
⎟ = 4 . 8 × 10 Hz
1 eV
h 6 . 63 × 10−34 J ⋅ s ⎝
⎠
c 3 . 0 × 108 m/s
λ= =
= 6 .2 × 10−20 m
f 4 . 8 × 1027 Hz
f =
Assess: These are extremely high-energy photons, with correspondingly large frequencies and small
wavelengths. This wavelength is much smaller than the diameter of a proton.
P25.72. Prepare: We will use the photon model of light, that is, Equation 25.21.
Solve:
The energy of a 1000 kHz photon is Ephoton = hf = (6.63 × 10−34 J ⋅ s)(1000 × 103 Hz) = 6.63 × 10−28 J.
The energy transmitted each second is 20 × 103 J. The number of photons transmitted each second is 20 × 103 J/
(6.63 × 10−28 J) = 3.0 × 1031.
Assess: This number is large because of the significant power emitted by the antenna.
P25.73. Prepare: We will use Equation 25.21, Ephoton = hf , to find the energy of the photons. We are given
λ = 590 nm, P = 40 × 10−6 W, and Δ t = 0.15 s.
Solve:
(a)
⎛ 3.0 × 108 m/s ⎞
⎛c⎞
−19
−19
Ephoton = hf = h ⎜ ⎟ = (6.63 × 10−34 J ⋅ s) ⎜
⎟ = 3.37 × 10 J ≈ 3.4 × 10 J
−9
λ
590
10
m
×
⎝ ⎠
⎝
⎠
(b) Think through this in words before putting in numbers. Remember that power = energy/time.
#photons
=
flash
energy
time
energy
photon
⋅
time
flash
#photons
40 × 10−6 W
=
⋅ 0.15 s/flash = 1.8 × 1013 photons/flash
flash
3.37 × 10−19 J/photon
Assess: This is a lot of photons, but it isn’t a very big fraction of a mole of photons.
We used an extra significant figure from part (a) as a guard digit. The units cancel properly.
P25.74. Prepare: The energy emitted per unit of time by a surface is related to the area and temperature of the
surface by Q / Δ t = eσ AT 4 . The energy of a photon is related to its wavelength by Ephoton = hc / λ .
Solve: The energy emitted by the surface in a time Δ t is
Q = eσ AT 4 Δt = (0.97)(5.67 ×10 −8 W/(m 2 ⋅ K 4 ))(1.8m 2 )(3.03 ×10 2 K) 4 (1.0 s) = 835 J
The energy of each photon is
Ephoton = hc / λ = (6.63 × 10−34 J ⋅ s)(3.0 × 108 m/s) / 10−4 m = 19.9 × 10 −22 J
The number of photons with energy Ephoton need to provide the amount of energy Q is determined by
Q = nEphoton ⇒ n = Q / Ephoton = (835 J) / (19.9 × 10−22 J / photon) = 4.2 × 1023 photons
EM Induction and EM Waves 25-29
Assess: Since the energy of each photon is small compared to the amount of energy needed, we expect a large
number of photons. To help you keep things in perspective, this is about a mole of photons.
P25.75. Prepare: The depth of penetration is related to the wavelength by d ∝ λ 1/ 2 . If we introduce a
proportionality constant , say b, this can be written as d = bλ 1/ 2 . Frequency and wavelength are related by λ = c / f .
Solve: For the first case we may write d1 = bλ11/ 2 = b(c / f1 )1/ 2 . In like manner for the second case we have
d 2 = bλ21/ 2 = b(c / f 2 )1/ 2 . Dividing the second case by the first case and solving for d 2 obtain
d 2 = d1 ( f1 / f 2 )1/ 2 = (14 cm)(27 × 106 H z / (2.4 × 1012 Hz)1/ 2 = 4.5 × 10−2 cm
Assess: Since the wavelength frequency for the second case is much larger, we expect the penetration to be
much less.
P25.76. Prepare: Metal detectors work by inducing eddy currents in the material between the coils of the
detector. Insulators do not allow the movement of charged particles.
Solve: In order for a material to sustain eddy currents, charged particles must be able to move freely through the
material. Charged particles do not move freely through insulators, hence the metal detector will not detect
insulators. The correct response is B.
Assess: To create eddy currents in a material, charged particles must be able to move freely in that material.
This is not the case for an insulator.
P25.77. Prepare: Metal detectors produce induced eddy currents to sense the presence of any metal—not just
magnetic material. Eddy currents can only be established in materials that allow free movement of charged particles.
Solve: Since metal detectors sense all metals not just magnetic materials, the answer does not have to be A. The
detector does not require any special alignment of the metal, so the answer is not D. Body tissue has high
resistivity and is not a good conductor, so the answer is not B. In order for the detector to sense the metal screws,
the magnetic field would have to penetrate the tissues of the body. The correct choice is C.
Assess: The detector works by establishing eddy currents in the material to be detected. If the oscillating magnetic field
cannot penetrate body tissue, it cannot reach the metal screw to set up eddy currents and allow detection.
P25.78. Prepare: Induced currents always oppose the act that created them. Since two different currents are
induced in the receiver, assume that we are addressing only the current induced by the metal.
Solve: First the easy part. In order to oppose the act that created it, the induced current at the receiver due to the
changing magnetic field from the transmitter is in the opposite direction of the current in the transmitter. Now the
hard part, the induced eddy current in the metal is in the opposite direction of the current in transmitter—this in
turn sets up an opposite direction induced current in the receiver. Since this second induced current in the
receiver is in the opposite direction of the current in the detected metal (which is in the opposite direction of the
current in the transmitter), it must be in the same direction as the current in the transmitter. Finally, since we are
told that the initial field points toward the receiver and is increasing with time we can conclude that the current in
the transmitter is clockwise, the induced eddy current in the metal is counterclockwise and the current induced in
the receiver due to the induced current in the metal is clockwise. The correct choice is C.
Assess: Induced currents always oppose the act that created them. In this case we started with the current in the
transmitter and worked our way through successive elements of the detection system.
P25.79. Prepare: The magnetic field of a coil will increase if the current in the coil increases. Induced
currents depend on changing magnetic flux. Changing magnetic flux depends on both the magnitude of the
change in the magnetic field and how rapidly it is changing.
Solve: Increasing the frequency of the oscillating current will increase the rate of change of magnetic flux, and
hence a larger eddy current. This rules out choice A. Increasing the magnitude of the current in the transmitter
will increase the magnitude of the oscillating field, hence the change in the magnetic field, the magnetic flux, and
the induced eddy current. This rules out choice B. A metal with a higher resistivity will have more resistance and
this will decrease the magnitude of the induced current. This indicates that C would be a good choice. Given the
detection system, repositioning the metal between the transmitter and the receiver will not affect the eddy
currents. This rules out choice D. The correct choice is C.
Assess: Anything that increases the rate of change of the magnetic flux experienced by the metal will create a
larger eddy current in the metal.