Chapter 25 & 28 Solutions

Transcription

1 Chapter 25 & 28 Solutions 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. Q 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. Q 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. Q 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. If the diamond-shaped loop were rotated 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.

2 Problems P25.2. Prepare: The motional emf induced may be determined by Solve: The motional emf is 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. Solve: (a) Using Equation 25.6, (b) Using Equation 25.6 again, Assess: This is a reasonable field for the circumstances given. 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. Solve: The field is constant in each region so we will use Equation Take 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 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. Solve: The field is constant inside the solenoid so we will use Equation Take the field. The magnetic flux is to be in the same direction as When the loop is tilted, the component of in the direction of 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

3 P Prepare: Equation gives the induced emf for a loop of wire in a changing magnetic field. Since the loop is not changing size, and the orientation stays perpendicular, this boils down to Since is (positive and) constant, the ranking of will be the same as the ranking for Solve: Plug in the respective values of the data and compute for each case. The ranking from greatest to least is 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. P Prepare: The emf produced by a changing magnetic flux is related to the changing magnetic flux by. Magnetic flux is related to the strength of the magnetic field, the area and the angle between the area and magnetic field vectors by In order to calculate the maximum emf generated, let s use the angle Solve: Since the cross-sectional area of the eyeball does not change, the emf generated is 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. P Prepare: The induced emf is determined by The induced emf is in the opposite direction of the applied voltage (the 9.0 volts from the battery, so the net emf is The net emf and current are related by Combining these expressions and solving for the current we obtain: I Solve: The current in the circuit and hence the resistor is

4 Assess: The current in the circuit is decreased slightly due to the fact that the induced emf opposes the applied voltage. P Prepare: Electromagnetic waves are sinusoidal and travel with the speed of light. The electric field is E y = E 0 cos (kx ωt), where E 0 = 20 V/m and k = m 1. Solve: (a) The wavelength is (b) The frequency is (c) The magnetic field amplitude is Assess: This field is approximately 500 times smaller than the earth s magnetic field, and is thus reasonable. P Prepare: We will use Malus s law for the polarized light. From Equation the relationship between the incident and the transmitted polarized light is I transmitted = I incident cos 2 θ, 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 the intensity is (a) I transmitted = I incident cos 2 θ = (10 W/m 2 ) cos 2 = 10 W/m 2 (b) I transmitted = I incident cos 2 θ = (10 W/m 2 ) cos 2 = 7.5 W/m 2 (c) I transmitted = I incident cos 2 θ = (10 W/m 2 ) cos 2 = 5 W/m 2 (d) I transmitted = I incident cos 2 θ = (10 W/m 2 ) cos 2 = 2.5 W/m 2 (e) I transmitted = I incident cos 2 θ = (10 W/m 2 ) cos 2 = 0 W/m 2 Assess: The final intensity must be between 0 W/m 2 and 10 W/m 2 and all values are in this range. P Prepare: We will use Malus s law for the polarized light. From Equation the relationship between the incident and the transmitted polarized light is I transmitted = I 0 cos 2 θ, where θ is the angle between the electric field and the axis of the filter. Solve: From Equation 25.20, 0.25 I 0 = I 0 cos 2 θ θ = Assess: Note that θ is the angle between the electric field and the axis of the filter. P Prepare: The energy of a photon is given by Equation 25.21: We first compute from Equation 25.16: Solve: (a)

5 We want to know the answer in ev as well. (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: 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). P Prepare: Wavelength and energy of photons are related by Solve: The maximum wavelength that can cause the desired transition is Assess: As expected, this wavelength is in the visible range. P Prepare: We follow somewhat the strategy outlined in Example 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 where. We are given and the intensity Solve: The energy per photon is Now we estimate the number of photons by applying the relationships. Assess: Don t confuse the usage of 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 P Prepare: The intensity, power, and area are related by Power, energy, and time are related by The total amount of energy is related to the energy of each photon by The energy of a photon is related to its wavelength by Solve: Combining these results we have Solving this expression for the number of photons per second, obtain

6 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. P Prepare: For both cases we will use to find the frequency, then to arrive at the wavelength. In the second case the energy of the photons is Solve: For the 25 KeV photons: For the 62.5 kev photons: Assess: We see that the more energetic photons have a shorter wavelength, as we expected. These wavelengths are in the x-ray range. P Prepare: The power is the energy/time. We ll plug in our answer from part (a) for the power into Equation for part (b). Solve: (a) (b) Solve Equation for The area of the focus circle is Assess: This is a very strong field, but that is what is needed for surgery to ablate the thin layer of the cornea. P Prepare: Use the photon model of light. Solve: (a) The wavelength is calculated as follows:

7 (b) The energy of a visible-light photon of wavelength 500 nm is The number of photons n such that is 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. P Prepare: We will use Equation 25.21, to find the energy of the photons. We are given 590 nm, and Solve: (a) (b) Think through this in words before putting in numbers. Remember that power = energy/time. 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. P 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. P 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. P 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

8 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. P 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. Chapter 28 Problems P Prepare: The relationship between the energy and wavelength of a photon is Note that the shorter the wavelength, the greater the energy. We have been given a photon threshold wavelength for damage. Photons with energy greater than these photons will do damage, hence photons with a wavelength less than these photons will do damage. We will call this energy the photon threshold energy for DNA damage. Solve: The photon threshold energy for DNA damage is Assess: This is a reasonable photon energy. P Prepare: We are given We also recall that for photons Solve: (a) The minimum chemical energy required to generate a photon of wavelength is (b) The number of ATP molecules needed to produce that much energy is However, the number of ATP molecules must be an integer, so the minimum number is 8. Assess: The firefly must be using quite a few molecules of ATP since it will emit many photons of light. P Prepare: The energy and wavelength of a photon are related by Before starting to solve the problem we might want to note that the arithmetic will be easier if we first calculate the quantity hc (hc = ev m). Solve: Photon energies corresponding to the wavelengths given are In like manner for =533 nm obtain E = 2.33 ev, and for = 564 nm obtain E = 2.20 ev. Assess: These are reasonable photon energies. Note that as the wavelength increases the energy decreases. This is consistent with the fact that the photon energy is inversely proportional to the wavelength.

9 P Prepare: The energy of a single photon is related to the wavelength by energy is related the energy of a single photon by is related to the total amount of energy by Solve: The number of photons reaching the retina is determined by The total The amount of energy that reaches the retina Assess: We expect a small number of photons since the total amount of energy reaching the retina is small. P Prepare: Power is related to the total amount of energy and time by The total amount of energy is related to the energy of a single photon by The energy of a photon is related to the wavelength by Solve: The power or the flash is Assess: This is a small power rating, but considering that the dinoflagellates might flash numerous times per second, it is a reasonable number.