Gen Phys Final 2

(a) If two protons are traveling parallel to each other in the same direction and at the same speed, is the magnetic force between them (i) attractive or (ii) repulsive? (b) Is the net force between them (i) attractive, (ii) repulsive, or (iii) zero? (Assume that the protons' speed is much slower than the speed of light.) Young, Hugh D.; Freedman, Roger A.. University Physics with Modern Physics Plus MasteringPhysics with eText -- Access Card Package, 14/e (Page 924). Addison-Wesley. Kindle Edition.

28.1 (a) (i), (b) (ii) The situation is the same as shown in Fig. 28.2 except that the upper proton has velocity v S rather than -v S . The magnetic field due to the lower proton is the same as shown in Fig. 28.2, but the direction of the magnetic force F S � qv S : B S on the upper proton is reversed. Hence the magnetic force is attractive. Since the speed v is small compared to c, the magnetic force is much smaller in magnitude than the repulsive electric force and the net force is still repulsive. Young, Hugh D.; Freedman, Roger A.. University Physics with Modern Physics Plus MasteringPhysics with eText -- Access Card Package, 14/e (Page 954). Addison-Wesley. Kindle Edition.

The accompanying figure shows a top view of two conducting rails on which a conducting bar can slide. A uniform magnetic field is directed perpendicular to the plane of the figure as shown. A battery is to be connected to the two rails so that when the switch is closed, current will flow through the bar and cause a magnetic force to push the bar to the right. In which orientation, A or B, should the battery be placed in the circuit? ❙ Young, Hugh D.; Freedman, Roger A.. University Physics with Modern Physics Plus MasteringPhysics with eText -- Access Card Package, 14/e (Page 899). Addison-Wesley. Kindle Edition.

A This orientation will cause current to flow clockwise around the circuit and hence through the conducting bar in the direction from the top to the bottom of the figure. From the righthand rule, the magnetic force F S � Il S : B S on the bar will then point to the right. Young, Hugh D.; Freedman, Roger A.. University Physics with Modern Physics Plus MasteringPhysics with eText -- Access Card Package, 14/e (Page 920). Addison-Wesley. Kindle Edition.

Figure 15.22 shows two wave pulses with different shapes traveling in different directions along a string. Make a series of sketches like Fig. 15.21 showing the shape of the string as the two pulses approach, overlap, and then pass each other.

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You have two lightweight metal spheres, each hanging from an insulating nylon thread. One of the spheres has a net negative charge, while the other sphere has no net charge. (a) If the spheres are close together but do not touch, will they (i) attract each other, (ii) repel each other, or (iii) exert no force on each other? (b) You now allow the two spheres to touch. Once they have touched, will the two spheres (i) attract each other, (ii) repel each other, or (iii) exert no force on each other?

(a) (i), (b) (ii) Before the two spheres touch, the negatively charged sphere exerts a repulsive force on the electrons in the other sphere, causing zones of positive and negative induced charge (see Fig. 21.7b). The positive zone is closer to the negatively charged sphere than the negative zone, so there is a net force of attraction that pulls the spheres together, like the comb and insulator in Fig. 21.8b. Once the two metal spheres touch, some of the excess electrons on the negatively charged sphere will flow onto the other sphere (because metals are conductors). Then both spheres will have a net negative charge and will repel each other. Young, Hugh D.; Freedman, Roger A.. University Physics with Modern Physics Plus MasteringPhysics with eText -- Access Card Package, 14/e (Page 721). Addison-Wesley. Kindle Edition.

Consider the system of three point charges in Example 21.4 (Section 21.3) and shown in Fig. 21.14. (a) What is the sign of the total potential energy of this system? (i) Positive; (ii) negative; (iii) zero. (b) What is the sign of the total amount of work you would have to do to move these charges infinitely far from each other? (i) Positive; (ii) negative; (iii) zero. Young, Hugh D.; Freedman, Roger A.. University Physics with Modern Physics Plus MasteringPhysics with eText -- Access Card Package, 14/e (Page 758). Addison-Wesley. Kindle Edition.

(a) (i), (b) (ii) The three charges q1 , q2 , and q3 are all pos itive, so all three of the terms in the sum in Eq. (23.11)—q1 q2 > r 12 , q1 q3 > r 13 , and q2 q3 > r 23 —are positive. Hence the total electric potential energy U is positive. This means that it would take posi tive work to bring the three charges from infinity to the positions shown in Fig. 21.14, and hence negative work to move the three charges from these positions back to infinity. Young, Hugh D.; Freedman, Roger A.. University Physics with Modern Physics Plus MasteringPhysics with eText -- Access Card Package, 14/e (Page 784). Addison-Wesley. Kindle Edition.

(a) Suppose the magnet in Fig. 29.14a were stationary and the loop of wire moved upward. Would the induced current in the loop be (i) in the same direction as shown in Fig. 29.14a, (ii) in the direction opposite to that shown in Fig. 29.14a, or (iii) zero? (b) Suppose the magnet and loop of wire in Fig. 29.14a both moved downward at the same velocity. Would the induced current in the loop be (i) in the same direction as shown in Fig. 29.14a, (ii) in the direction opposite to that shown in Fig. 29.14a, or (iii) zero? ❙ Young, Hugh D.; Freedman, Roger A.. University Physics with Modern Physics Plus MasteringPhysics with eText -- Access Card Package, 14/e (Page 966). Addison-Wesley. Kindle Edition.

(a) (i), (b) (iii) In (a), as in the original situation, the magnet and loop are approaching each other and the downward flux through the loop is increasing. Hence the induced emf and induced current are the same. In (b), since the magnet and loop are moving together, the flux through the loop is not changing and no emf is induced. Young, Hugh D.; Freedman, Roger A.. University Physics with Modern Physics Plus MasteringPhysics with eText -- Access Card Package, 14/e (Page 989). Addison-Wesley. Kindle Edition.

The accompanying figure shows a wire coil being squeezed in a uniform magnetic field. (a) While the coil is being squeezed, is the induced emf in the coil (i) clockwise, (ii) counterclockwise, or (iii) zero? (b) Once the coil has reached its final squeezed shape, is the induced emf in the coil (i) clockwise, (ii) counterclockwise, or (iii) zero? ❙ Young, Hugh D.; Freedman, Roger A.. University Physics with Modern Physics Plus MasteringPhysics with eText -- Access Card Package, 14/e (Page 964). Addison-Wesley. Kindle Edition.

(a) (i), (b) (iii) In (a), initially there is magnetic flux into the plane of the page, which we call positive. While the loop is being squeezed, the flux is becoming less positive 1dΦB > dt 6 02 and so the induced emf is positive as in Fig. 29.6b 1E = -dΦB > dt 7 02. If you point the thumb of your right hand into the page, your fingers curl clockwise, so this is the direction of positive induced emf. In (b), since the coil's shape is no longer changing, the magnetic flux is not changing and there is no induced emf. Young, Hugh D.; Freedman, Roger A.. University Physics with Modern Physics Plus MasteringPhysics with eText -- Access Card Package, 14/e (Page 989). Addison-Wesley. Kindle Edition.

You want to connect a 4@mF capacitor and an 8@mF capacitor. (a) With which type of connection will the 4@mF capacitor have a greater potential difference across it than the 8@mF capacitor? (i) Series; (ii) parallel; (iii) either series or parallel; (iv) neither series nor parallel. (b) With which type of connection will the 4@mF capacitor have a greater charge than the 8@mF capacitor? (i) Series; (ii) parallel; (iii) either series or parallel; (iv) neither series nor parallel. Young, Hugh D.; Freedman, Roger A.. University Physics with Modern Physics Plus MasteringPhysics with eText -- Access Card Package, 14/e (Page 793). Addison-Wesley. Kindle Edition.

(a) (i), (b) (iv) In a series connection the two capacitors carry the same charge Q but have different potential differences Vab = Q>C; the capacitor with the smaller capacitance C has the greater potential difference. In a parallel connection the two capacitors have the same potential difference Vab but carry different charges Q = CVab ; the capacitor with the larger capacitance C has the greater charge. Hence a 4@mF capacitor will have a greater potential difference than an 8@mF capacitor if the two are connected in series. The 4@mF capacitor cannot carry more charge than the 8@mF capacitor no matter how they are connected: In a series connection they will carry the same charge, and in a parallel connection the 8@mF capacitor will carry more charge. Young, Hugh D.; Freedman, Roger A.. University Physics with Modern Physics Plus MasteringPhysics with eText -- Access Card Package, 14/e (Page 815). Addison-Wesley. Kindle Edition.

(a) A negative point charge moves along a straight-line path directly toward a stationary positive point charge. Which aspect(s) of the electric force on the negative point charge will remain constant as it moves? (i) Magnitude; (ii) direction; (iii) both magnitude and direction; (iv) neither magnitude nor direction. (b) A negative point charge moves along a circular orbit around a positive point charge. Which aspect(s) of the electric force on the negative point charge will remain constant as it moves? (i) Magnitude; (ii) direction; (iii) both magnitude and direction; (iv) neither magnitude nor direction.

(a) (ii), (b) (i) The electric field E S produced by a positive point charge points directly away from the charge (see Fig. 21.18a) and has a magnitude that depends on the distance r from the charge to the field point. Hence a second, negative point charge q 6 0 will feel a force F S � qE S that points directly toward the positive charge and has a magnitude that depends on the distance r between the two charges. If the negative charge moves directly toward the positive charge, the direction of the force remains the same but the force magnitude increases as the distance r decreases. If the negative charge moves in a circle around the positive charge, the force magnitude stays the same (because the distance r is constant) but the force direction changes. Young, Hugh D.; Freedman, Roger A.. University Physics with Modern Physics Plus MasteringPhysics with eText -- Access Card Package, 14/e (Page 721). Addison-Wesley. Kindle Edition.

(a) If you double the speed of the charged particle in Fig. 27.17a while keeping the magnetic field the same (as well as the charge and the mass), how does this affect the radius of the trajectory? (i) The radius is unchanged; (ii) the radius is twice as large; (iii) the radius is four times as large; (iv) the radius is 1 2 as large; (v) the radius is 1 4 as large. (b) How does this affect the time required for one complete circular orbit? (i) The time is unchanged; (ii) the time is twice as long; (iii) the time is four times as long; (iv) the time is 1 2 as long; (v) the time is 1 4 as long. ❙ Young, Hugh D.; Freedman, Roger A.. University Physics with Modern Physics Plus MasteringPhysics with eText -- Access Card Package, 14/e (Page 893). Addison-Wesley. Kindle Edition.

(a) (ii), (b) (i) The radius of the orbit as given by Eq. (27.11) is directly proportional to the speed, so doubling the particle speed causes the radius to double as well. The particle has twice as far to travel to complete one orbit but is traveling at double the speed, so the time for one orbit is unchanged. This result also follows from Eq. (27.12), which states that the angular speed v is independent of the linear speed v. Hence the time per orbit, T = 2p>v, likewise does not depend on v. Young, Hugh D.; Freedman, Roger A.. University Physics with Modern Physics Plus MasteringPhysics with eText -- Access Card Package, 14/e (Page 920). Addison-Wesley. Kindle Edition.

You want to measure the current through and the potential difference across the 2@Ω resistor shown in Fig. 26.12 (Example 26.6 in Section 26.2). (a) How should you connect an ammeter and a voltmeter to do this? (i) Both ammeter and voltmeter in series with the 2@Ω resistor; (ii) ammeter in series with the 2@Ω resistor and voltmeter connected between points b and d; (iii) ammeter connected between points b and d and voltmeter in series with the 2@Ω resistor; (iv) both ammeter and voltmeter connected between points b and d. (b) What resistances should these meters have? (i) Both ammeter and voltmeter resistances should be much greater than 2 Ω; (ii) ammeter resistance should be much greater than 2 Ω and voltmeter resistance should be much less than 2 Ω; (iii) ammeter resistance should be much less than 2 Ω and voltmeter resistance should be much greater than 2 Ω; (iv) both ammeter and voltmeter resistances should be much less than 2 Ω. Young, Hugh D.; Freedman, Roger A.. University Physics with Modern Physics Plus MasteringPhysics with eText -- Access Card Package, 14/e (Page 862). Addison-Wesley. Kindle Edition.

(a) (ii), (b) (iii) An ammeter must always be placed in series with the circuit element of interest, and a voltmeter must always be placed in parallel. Ideally the ammeter would have zero resistance and the voltmeter would have infinite resistance so that their presence would have no effect on either the resistor current or the voltage. Neither of these idealizations is possible, but the ammeter resistance should be much less than 2 Ω and the voltmeter resistance should be much greater than 2 Ω. Young, Hugh D.; Freedman, Roger A.. University Physics with Modern Physics Plus MasteringPhysics with eText -- Access Card Package, 14/e (Page 880). Addison-Wesley. Kindle Edition.

Imagine moving along the axis of the current-carrying loop in Fig. 27.13c, starting at a point well to the left of the loop and ending at a point well to the right of the loop. (a) How would the magnetic field strength vary as you moved along this path? (i) It would be the same at all points along the path; (ii) it would increase and then decrease; (iii) it would decrease and then increase. (b) Would the magnetic field direction vary as you moved along the path? Young, Hugh D.; Freedman, Roger A.. University Physics with Modern Physics Plus MasteringPhysics with eText -- Access Card Package, 14/e (Page 890). Addison-Wesley. Kindle Edition.

(a) (ii), (b) no The magnitude of B S would increase as you moved to the right, reaching a maximum as you passed through the plane of the loop. As you moved beyond the plane of the loop, the field magnitude would decrease. You can tell this from the spacing of the field lines: The closer the field lines, the stronger the field. The direction of the field would be to the right at all points along the path, since the path is along a field line and the direction of B S at any point is tangent to the field line through that point. Young, Hugh D.; Freedman, Roger A.. University Physics with Modern Physics Plus MasteringPhysics with eText -- Access Card Package, 14/e (Page 920). Addison-Wesley. Kindle Edition.

The earth's magnetic field points toward (magnetic) north. For simplicity, assume that the field has no vertical component (as is the case near the earth's equator). (a) If you hold a metal rod in your hand and walk toward the east, how should you orient the rod to get the maximum motional emf between its ends? (i) East-west; (ii) north-south; (iii) up-down; (iv) you get the same motional emf with all of these orientations. (b) How should you hold it to get zero emf as you walk toward the east? (i) East-west; (ii) north-south; (iii) up-down; (iv) none of these. (c) In which direction should you travel so that the motional emf across the rod is zero no matter how the rod is oriented? (i) West; (ii) north; (iii) south; (iv) straight up; (v) straight down. ❙ Young, Hugh D.; Freedman, Roger A.. University Physics with Modern Physics Plus MasteringPhysics with eText -- Access Card Package, 14/e (Page 969). Addison-Wesley. Kindle Edition.

(a) (iii); (b) (i) or (ii); (c) (ii) or (iii) You will get the maximum motional emf if you hold the rod vertically, so that its length is perpendicular to both the magnetic field and the direction of motion. With this orientation, L S is parallel to v S : B S . If you hold the rod in any horizontal orientation, L S will be perpendicular to v S : B S and no emf will be induced. If you walk due north or south, v S : B S � 0 and no emf will be induced for any orientation of the rod. Young, Hugh D.; Freedman, Roger A.. University Physics with Modern Physics Plus MasteringPhysics with eText -- Access Card Package, 14/e (Page 989). Addison-Wesley. Kindle Edition.

Figure 15.8 shows a sinusoidal wave of period T on a string at times 0, 1 8 T, 2 8 T, 3 8 T, 4 8 T, 5 8 T, 6 8 T, 7 8 T, and T. (a) At which time is point A on the string moving upward with maximum speed? (b) At which time does point B on the string have the greatest upward acceleration? (c) At which time does point C on the string have a downward acceleration and a downward velocity?

(a) 2 8 T, (b) 4 8 T, (c) 5 8 T Since the wave is sinusoidal, each point on the string oscillates in simple harmonic motion (SHM). Hence we can apply all of the ideas from Chapter 14 about SHM to the wave depicted in Fig. 15.8. (a) A particle in SHM has its maximum speed when it is passing through the equilibrium position ( y = 0 in Fig. 15.8). The particle at point A is moving upward through this position at t = 2 8 T. (b) In vertical SHM the greatest upward acceleration occurs when a particle is at its maximum downward displacement. This occurs for the particle at point B at t = 4 8 T. (c) A particle in vertical SHM has a downward acceleration when its displacement is upward. The particle at C Young, Hugh D.; Freedman, Roger A.. University Physics with Modern Physics Plus MasteringPhysics with eText -- Access Card Package, 14/e (Page 504). Addison-Wesley. Kindle Edition.

(a) Which of Maxwell's equations explains how a credit card reader works? (b) Which one describes how a wire carrying a steady current generates a magnetic field? ❙ Young, Hugh D.; Freedman, Roger A.. University Physics with Modern Physics Plus MasteringPhysics with eText -- Access Card Package, 14/e (Page 977). Addison-Wesley. Kindle Edition.

(a) Faraday's law, (b) Ampere's law A credit card reader works by inducing currents in the reader's coils as the card's magnetized stripe is swiped (see the answer to the chapter opening question). Ampere's law describes how currents of all kinds (both conduction currents and displacement currents) give rise to magnetic fields. Young, Hugh D.; Freedman, Roger A.. University Physics with Modern Physics Plus MasteringPhysics with eText -- Access Card Package, 14/e (Page 989). Addison-Wesley. Kindle Edition.

(a) Is it possible to have a purely electric wave propagate through empty space—that is, a wave made up of an electric field but no magnetic field? (b) What about a purely magnetic wave, with a magnetic field but no electric field? ❙ Young, Hugh D.; Freedman, Roger A.. University Physics with Modern Physics Plus MasteringPhysics with eText -- Access Card Package, 14/e (Page 1054). Addison-Wesley. Kindle Edition.

(a) no, (b) no A purely electric wave would have a varying electric field. Such a field necessarily generates a magnetic field through Ampere's law, Eq. (29.21), so a purely electric wave is impossible. In the same way, a purely magnetic wave is impossible: The varying magnetic field in such a wave would automatically give rise to an electric field through Faraday's law, Eq. (29.20). Young, Hugh D.; Freedman, Roger A.. University Physics with Modern Physics Plus MasteringPhysics with eText -- Access Card Package, 14/e (Page 1077). Addison-Wesley. Kindle Edition.

Figure 27.13c depicts the magnetic field lines due to a circular current-carrying loop. (a) What is the direction of the magnetic moment of this loop? (b) Which side of the loop is equivalent to the north pole of a magnet, and which side is equivalent to the south pole? ❙ Young, Hugh D.; Freedman, Roger A.. University Physics with Modern Physics Plus MasteringPhysics with eText -- Access Card Package, 14/e (Page 905). Addison-Wesley. Kindle Edition.

(a) to the right; (b) north pole on the right, south pole on the left If you wrap the fingers of your right hand around the coil in the direction of the current, your right thumb points to the right (perpendicular to the plane of the coil). This is the direction of the magnetic moment M S . The magnetic moment points from the south pole to the north pole, so the right side of the loop is equivalent to a north pole and the left side is equivalent to a south pole. Young, Hugh D.; Freedman, Roger A.. University Physics with Modern Physics Plus MasteringPhysics with eText -- Access Card Package, 14/e (Page 920). Addison-Wesley. Kindle Edition.

Suppose all three of the resistors shown in Fig. 26.1 have the same resistance, so R1 = R2 = R3 = R. Rank the four arrangements shown in parts (a)-(d) of Fig. 26.1 in order of their equivalent resistance, from highest to lowest. Young, Hugh D.; Freedman, Roger A.. University Physics with Modern Physics Plus MasteringPhysics with eText -- Access Card Package, 14/e (Page 853). Addison-Wesley. Kindle Edition.

(a), (c), (d), (b) Here's why: The three resistors in Fig. 26.1a are in series, so Req = R + R + R = 3R. In Fig. 26.1b the three resistors are in parallel, so 1>Req = 1>R + 1>R + 1>R = 3>R and Req = R> 3. In Fig. 26.1c the second and third resistors are in parallel, so their equivalent resistance R23 is given by 1>R23 = 1>R + 1>R = 2>R; hence R23 = R> 2. This combination is in series with the first resistor, so the three resistors together have equivalent resistance Req = R + R> 2 = 3R> 2. In Fig. 26.1d the second and third resistors are in series, so their equivalent resistance is R23 = R + R = 2R. This combination is in parallel with the first resistor, so the equivalent resistance of the threeresistor combination is given by 1>Req = 1>R + 1> 2R = 3> 2R. Hence Req = 2R> 3. Young, Hugh D.; Freedman, Roger A.. University Physics with Modern Physics Plus MasteringPhysics with eText -- Access Card Package, 14/e (Page 880). Addison-Wesley. Kindle Edition.

You want to connect a 4@mF capacitor and an 8@mF capacitor. With which type of connection will the 4@mF capacitor have a greater amount of stored energy than the 8@mF capacitor? (i) Series; (ii) parallel; (iii) either series or parallel; (iv) neither series nor parallel. Young, Hugh D.; Freedman, Roger A.. University Physics with Modern Physics Plus MasteringPhysics with eText -- Access Card Package, 14/e (Page 797). Addison-Wesley. Kindle Edition.

(i) Capacitors connected in series carry the same charge Q. To compare the amount of energy stored, we use the expression U = Q 2 > 2C from Eq. (24.9); it shows that the capacitor with the smaller capacitance 1C = 4 mF2 has more stored energy in a series combination. By contrast, capacitors in parallel have the same potential difference V, so to compare them we use U = 1 2 CV 2 from Eq. (24.9). It shows that in a parallel combination, the capacitor with the larger capacitance 1C = 8 mF2 has more stored energy. (If we had instead used U = 1 2 CV 2 to analyze the series combination, we would have to account for the different potential differences across the two capacitors. Likewise, using U = Q 2 > 2C to study the parallel combination would require us to account for the different charges on the capacitors.) Young, Hugh D.; Freedman, Roger A.. University Physics with Modern Physics Plus MasteringPhysics with eText -- Access Card Package, 14/e (Page 815). Addison-Wesley. Kindle Edition.

The space between the plates of an isolated parallel-plate capacitor is filled by a slab of dielectric with dielectric constant K. The two plates of the capacitor have charges Q and -Q. You pull out the dielectric slab. If the charges do not change, how does the energy in the capacitor change when you remove the slab? (i) It increases; (ii) it decreases; (iii) it remains the same. Young, Hugh D.; Freedman, Roger A.. University Physics with Modern Physics Plus MasteringPhysics with eText -- Access Card Package, 14/e (Page 803). Addison-Wesley. Kindle Edition.

(i) Here Q remains the same, so we use U = Q 2 > 2C from Eq. (24.9) for the stored energy. Removing the dielectric lowers the capacitance by a factor of 1>K; since U is inversely proportional to C, the stored energy increases by a factor of K. It takes work to pull the dielectric slab out of the capacitor because the fringing field tries to pull the slab back in (Fig. 24.16). The work that you do goes into the energy stored in the capacitor. Young, Hugh D.; Freedman, Roger A.. University Physics with Modern Physics Plus MasteringPhysics with eText -- Access Card Package, 14/e (Page 815). Addison-Wesley. Kindle Edition.

Metal objects reflect not only visible light but also radio waves. This is because at the surface of a metal, (i) the electric-field component parallel to the surface must be zero; (ii) the electric-field component perpendicular to the surface must be zero; (iii) the magneticfield component parallel to the surface must be zero; (iv) the magnetic-field component perpendicular to the surface must be zero; (v) more than one of these. Young, Hugh D.; Freedman, Roger A.. University Physics with Modern Physics Plus MasteringPhysics with eText -- Access Card Package, 14/e (Page 1050). Addison-Wesley. Kindle Edition.

(i) Metals are reflective because they are good conductors of electricity. When an electromagnetic wave strikes a conductor, the electric field of the wave sets up currents on the conductor surface that generate a reflected wave. For a perfect conductor, the requirement that the electric-field component parallel to the surface must be zero implies that this reflected wave is just as intense as the incident wave. Tarnished metals are less shiny because their surface is oxidized and less conductive; polishing the metal removes the oxide and exposes the conducting metal. Young, Hugh D.; Freedman, Roger A.. University Physics with Modern Physics Plus MasteringPhysics with eText -- Access Card Package, 14/e (Page 1077). Addison-Wesley. Kindle Edition.

In a complex circuit like the one on this circuit board, is it possible to connect several resistors with different resistances so that all of them have the same potential difference? (i) Yes, and the current will be the same through all of the resistors; (ii) yes, but the current may be different through different resistors; (iii) no; (iv) the answer depends on the value of the potential difference. Young, Hugh D.; Freedman, Roger A.. University Physics with Modern Physics Plus MasteringPhysics with eText -- Access Card Package, 14/e (Page 848). Addison-Wesley. Kindle Edition.

(ii) The potential difference V is the same across resistors connected in parallel. However, there is a different current I through each resistor if the resistances R are different: I = V>R. Young, Hugh D.; Freedman, Roger A.. University Physics with Modern Physics Plus MasteringPhysics with eText -- Access Card Package, 14/e (Page 880). Addison-Wesley. Kindle Edition.

Which of the following factors will, if increased, make it more difficult to produce a certain amount of current in a conductor? (There may be more than one correct answer.) (i) The mass of the moving charged particles in the conductor; (ii) the number of moving charged particles per cubic meter; (iii) the amount of charge on each moving particle; (iv) the average time between collisions for a typical moving charged particle. ❙ Young, Hugh D.; Freedman, Roger A.. University Physics with Modern Physics Plus MasteringPhysics with eText -- Access Card Package, 14/e (Page 838). Addison-Wesley. Kindle Edition.

(i) The difficulty of producing a certain amount of current increases as the resistivity r increases. From Eq. (25.24), r = m> ne 2 t, so increasing the mass m will increase the resistivity. That's because a more massive charged particle will respond more sluggishly to an applied electric field and hence drift more slowly. To produce the same current, a greater electric field would be needed. (Increasing n, e, or t would decrease the resistivity and make it easier to produce a given current.) Young, Hugh D.; Freedman, Roger A.. University Physics with Modern Physics Plus MasteringPhysics with eText -- Access Card Package, 14/e (Page 847). Addison-Wesley. Kindle Edition.

A parallel-plate capacitor has charges Q and -Q on its two plates. A dielectric slab with K = 3 is then inserted into the space between the plates as shown in Fig. 24.20. Rank the following electric-field magnitudes in order from largest to smallest. (i) The field before the slab is inserted; (ii) the resultant field after the slab is inserted; (iii) the field due to the bound charges. Young, Hugh D.; Freedman, Roger A.. University Physics with Modern Physics Plus MasteringPhysics with eText -- Access Card Package, 14/e (Page 804). Addison-Wesley. Kindle Edition.

(i), (iii), (ii) Equation (24.14) says that if E0 is the initial electric-field magnitude (before the dielectric slab is inserted), then the resultant field magnitude after the slab is inserted is E0 >K = E0 > 3. The magnitude of the resultant field equals the difference between the initial field magnitude and the magnitude Ei of the field due to the bound charges (see Fig. 24.20). Hence E0 - Ei = E0 > 3 and Ei = 2E0 > 3. Young, Hugh D.; Freedman, Roger A.. University Physics with Modern Physics Plus MasteringPhysics with eText -- Access Card Package, 14/e (Page 815). Addison-Wesley. Kindle Edition.

The needle of a magnetic compass points north. This alignment is due to (i) a magnetic force on the needle; (ii) a magnetic torque on the needle; (iii) the magnetic field that the needle itself produces; (iv) both (i) and (ii); (v) all of (i), (ii), and (iii). Young, Hugh D.; Freedman, Roger A.. University Physics with Modern Physics Plus MasteringPhysics with eText -- Access Card Package, 14/e (Page 881). Addison-Wesley. Kindle Edition.

(ii) A magnetized compass needle has a magnetic dipole moment along its length, and the earth's magnetic field (which points generally northward) exerts a torque that tends to align that dipole moment with the field. See Section 27.7 for details. Young, Hugh D.; Freedman, Roger A.. University Physics with Modern Physics Plus MasteringPhysics with eText -- Access Card Package, 14/e (Page 920). Addison-Wesley. Kindle Edition.

The energy stored in a capacitor is equal to q 2 > 2C. When a capacitor is discharged, what fraction of the initial energy remains after an elapsed time of one time constant? (i) 1> e; (ii) 1> e 2 ; (iii) 1 - 1> e; (iv) 11 - 1> e2 2 ; (v) answer depends on how much energy was stored initially. Young, Hugh D.; Freedman, Roger A.. University Physics with Modern Physics Plus MasteringPhysics with eText -- Access Card Package, 14/e (Page 866). Addison-Wesley. Kindle Edition.

(ii) After one time constant, t = RC and the initial charge Q0 has decreased to Q0 e -t/RC = Q0 e -RC/RC = Q0 e -1 = Q0 > e. Hence the stored energy has decreased from Q0 2 > 2C to 1Q0 > e2 2 > 2C =Q0 2 > 2Ce 2 , a fraction 1> e 2 = 0.135 of its initial value. This result doesn't depend on the initial value of the energy. Young, Hugh D.; Freedman, Roger A.. University Physics with Modern Physics Plus MasteringPhysics with eText -- Access Card Package, 14/e (Page 880). Addison-Wesley. Kindle Edition.

An electric dipole is placed in a region of uniform electric field E S , with the electric dipole moment p S pointing in the direction opposite to E S . Is the dipole (i) in stable equilibrium, (ii) in unstable equilibrium, or (iii) neither? (Hint: You many want to review Section 7.5.)

(ii) Equations (21.17) and (21.18) tell us that the potential energy for a dipole in an electric field is U = -p S ~ E S = -pEcosf, where f is the angle between the directions of p S and E S . If p S and E S point in opposite directions, so that f = 180°, we have cosf = -1 and U = +pE. This is the maximum value that U can have. From our discussion of energy diagrams in Section 7.5, it follows that this is a situation of unstable equilibrium. Young, Hugh D.; Freedman, Roger A.. University Physics with Modern Physics Plus MasteringPhysics with eText -- Access Card Package, 14/e (Page 721). Addison-Wesley. Kindle Edition.

You maintain a constant electric field inside a piece of semiconductor while lowering the semiconductor's temperature. What happens to the current density in the semiconductor? (i) It increases; (ii) it decreases; (iii) it remains the same. ❙ Young, Hugh D.; Freedman, Roger A.. University Physics with Modern Physics Plus MasteringPhysics with eText -- Access Card Package, 14/e (Page 823). Addison-Wesley. Kindle Edition.

(ii) Figure 25.6b shows that the resistivity r of a semiconductor increases as the temperature decreases. From Eq. (25.5), the magnitude of the current density is J = E> r, so the current density decreases as the temperature drops and the resistivity increases. Young, Hugh D.; Freedman, Roger A.. University Physics with Modern Physics Plus MasteringPhysics with eText -- Access Card Package, 14/e (Page 847). Addison-Wesley. Kindle Edition.

A copper wire of square cross section is oriented vertically. The four sides of the wire face north, south, east, and west. There is a uniform magnetic field directed from east to west, and the wire carries current downward. Which side of the wire is at the highest electric potential? (i) North side; (ii) south side; (iii) east side; (iv) west side. ❙ Young, Hugh D.; Freedman, Roger A.. University Physics with Modern Physics Plus MasteringPhysics with eText -- Access Card Package, 14/e (Page 909). Addison-Wesley. Kindle Edition.

(ii) The mobile charge carriers in copper are negatively charged electrons, which move upward through the wire to give a downward current. From the right-hand rule, the force on a positively charged particle moving upward in a westward-pointing magnetic field would be to the south; hence the force on a negatively charged particle is to the north. The result is an excess of negative charge on the north side of the wire, leaving an excess of positive charge—and hence a higher electric potential—on the south side. Young, Hugh D.; Freedman, Roger A.. University Physics with Modern Physics Plus MasteringPhysics with eText -- Access Card Package, 14/e (Page 920). Addison-Wesley. Kindle Edition.

The six strings of a guitar are the same length and under nearly the same tension, but they have different thicknesses. On which string do waves travel the fastest? (i) The thickest string; (ii) the thinnest string; (iii) the wave speed is the same on all strings.

(ii) The relationship v = 2F>m [Eq. (15.14)] says that the wave speed is greatest on the string with the smallest linear mass density. This is the thinnest string, which has the smallest amount of mass m and hence the smallest linear mass density m = m> L (all strings are the same length). Young, Hugh D.; Freedman, Roger A.. University Physics with Modern Physics Plus MasteringPhysics with eText -- Access Card Package, 14/e (Page 504). Addison-Wesley. Kindle Edition.

Water makes life possible: The cells of your body could not function without water in which to dissolve essential biological molecules. Water is such a good solvent because its molecules (i) have zero net charge; (ii) have zero net charge, but the positive and negative charges are separated; (iii) have nonzero net charge; (iv) do not respond to electric forces; (v) exert repulsive electric forces on each other.

(ii) Water molecules have a permanent electric dipole moment: One end of the molecule has a positive charge and the other end has a negative charge. These ends attract negative and positive ions, respectively, holding the ions apart in solution. Water is less effective as a solvent for materials whose molecules do not ionize (called nonionic substances), such as oils. Young, Hugh D.; Freedman, Roger A.. University Physics with Modern Physics Plus MasteringPhysics with eText -- Access Card Package, 14/e (Page 721). Addison-Wesley. Kindle Edition.

In one type of welding, electric charge flows between the welding tool and the metal pieces that are to be joined. This produces a glowing arc whose high temperature fuses the pieces together. Why must the tool be held close to the metal pieces? (i) To maximize the potential difference between tool and pieces; (ii) to minimize this potential difference; (iii) to maximize the electric field between tool and pieces; (iv) to minimize this electric field; (v) more than one of these. Young, Hugh D.; Freedman, Roger A.. University Physics with Modern Physics Plus MasteringPhysics with eText -- Access Card Package, 14/e (Page 752). Addison-Wesley. Kindle Edition.

(iii) A large, constant potential difference Vab is maintained between the welding tool 1a2 and the metal pieces to be welded 1b2. For a given potential difference between two conductors a and b, the smaller the distance d separating the conductors, the greater is the magnitude E of the field between them. Hence d must be small in order for E to be large enough to ionize the gas between the conductors (see Section 23.3) and produce an arc through this gas. Young, Hugh D.; Freedman, Roger A.. University Physics with Modern Physics Plus MasteringPhysics with eText -- Access Card Package, 14/e (Page 784). Addison-Wesley. Kindle Edition.

Suppose that the magnetic field in Fig. 29.20 were directed out of the plane of the figure and the disk were rotating counterclockwise. Compared to the directions of the force F S and the eddy currents shown in Fig. 29.20b, what would the new directions be? (i) The force F S and the eddy currents would both be in the same direction; (ii) the force F S would be in the same direction, but the eddy currents would be in the opposite direction; (iii) the force F S would be in the opposite direction, but the eddy currents would be in the same direction; (iv) the force F S and the eddy currents would be in the opposite directions. ❙ Young, Hugh D.; Freedman, Roger A.. University Physics with Modern Physics Plus MasteringPhysics with eText -- Access Card Package, 14/e (Page 973). Addison-Wesley. Kindle Edition.

(iii) By Lenz's law, the force must oppose the motion of the disk through the magnetic field. Since the disk material is now moving to the right through the field region, the force F S is to the left—that is, in the opposite direction to that shown in Fig. 29.20b. Young, Hugh D.; Freedman, Roger A.. University Physics with Modern Physics Plus MasteringPhysics with eText -- Access Card Package, 14/e (Page 989). Addison-Wesley. Kindle Edition.

If all of the dimensions of the box in Fig. 22.2a are increased by a factor of 3, how will the electric flux through the box change? (i) The flux will be 3 2 = 9 times greater; (ii) the flux will be 3 times greater; (iii) the flux will be unchanged; (iv) the flux will be 1 3 as great; (v) the flux will be 1 1 3 2 2 = 1 9 as great; (vi) not enough information is given to decide. Young, Hugh D.; Freedman, Roger A.. University Physics with Modern Physics Plus MasteringPhysics with eText -- Access Card Package, 14/e (Page 725). Addison-Wesley. Kindle Edition.

(iii) Each part of the surface of the box will be three times farther from the charge +q, so the electric field will be 1 1 3 2 2 = 1 9 as strong. But the area of the box will increase by a factor of 3 2 = 9. Hence the electric flux will be multiplied by a factor of 1 1 9 2 192 = 1. In other words, the flux will be unchanged. Young, Hugh D.; Freedman, Roger A.. University Physics with Modern Physics Plus MasteringPhysics with eText -- Access Card Package, 14/e (Page 751). Addison-Wesley. Kindle Edition.

A single point charge q is embedded in a very large block of dielectric of dielectric constant K. At a point inside the dielectric a distance r from the point charge, what is the magnitude of the electric field? (i) q> 4pP 0 r 2 ; (ii) Kq> 4pP 0 r 2 ; (iii) q> 4pKP 0 r 2 ; (iv) none of these. Young, Hugh D.; Freedman, Roger A.. University Physics with Modern Physics Plus MasteringPhysics with eText -- Access Card Package, 14/e (Page 806). Addison-Wesley. Kindle Edition.

(iii) Equation (24.23) shows that this situation is the same as an isolated point charge in vacuum but with E S replaced by KE S . Hence KE at the point of interest is equal to q> 4pP 0 r 2 , and so E = q> 4pKP 0 r 2 . As in Example 24.12, filling the space with a dielectric reduces the electric field by a factor of 1>K. Young, Hugh D.; Freedman, Roger A.. University Physics with Modern Physics Plus MasteringPhysics with eText -- Access Card Package, 14/e (Page 815). Addison-Wesley. Kindle Edition.

In Example 27.6 He + ions with charge +e move at 1.00 * 10 5 m> s in a straight line through a velocity selector. Suppose the He + ions were replaced with He 2+ ions, in which both electrons have been removed from the helium atom and the ion charge is +2e. At what speed must the He 2+ ions travel through the same velocity selector in order to move in a straight line? (i) 4.00 * 10 5 m> s; (ii) 2.00 * 10 5 m> s; (iii) 1.00 * 10 5 m> s; (iv) 0.50 * 10 5 m> s; (v) 0.25 * 10 5 m> s. ❙ Young, Hugh D.; Freedman, Roger A.. University Physics with Modern Physics Plus MasteringPhysics with eText -- Access Card Package, 14/e (Page 896). Addison-Wesley. Kindle Edition.

(iii) From Eq. (27.13), the speed v = E>B at which particles travel straight through the velocity selector does not depend on the magnitude or sign of the charge or the mass of the particle. All that is required is that the particles (in this case, ions) have a nonzero charge. Young, Hugh D.; Freedman, Roger A.. University Physics with Modern Physics Plus MasteringPhysics with eText -- Access Card Package, 14/e (Page 920). Addison-Wesley. Kindle Edition.

In a certain region of space the potential is given by V = A + Bx + Cy 3 + Dxy, where A, B, C, and D are positive constants. Which of these statements about the electric field E S in this region of space is correct? (There may be more than one correct answer.) (i) Increasing the value of A will increase the value of E S at all points; (ii) increasing the value of A will decrease the value of E S at all points; (iii) E S has no z-component; (iv) the electric field is zero at the origin 1x = 0, y = 0, z = 02. Young, Hugh D.; Freedman, Roger A.. University Physics with Modern Physics Plus MasteringPhysics with eText -- Access Card Package, 14/e (Page 774). Addison-Wesley. Kindle Edition.

(iii) From Eqs. (23.19), the components of the electric field are Ex = - 0V>0x = -1B + Dy2, Ey = - 0V>0y = - (3Cy 2 + Dx2, and Ez = - 0V>0z = 0. The value of A has no effect, which means that we can add a constant to the electric potential at all points without changing E S or the potential difference between two points. The potential does not depend on z, so the zcomponent of E S is zero. Note that at the origin the electric field is not zero because it has a nonzero xcomponent: Ex = -B, Ey = 0, Ez = 0. Young, Hugh D.; Freedman, Roger A.. University Physics with Modern Physics Plus MasteringPhysics with eText -- Access Card Package, 14/e (Page 784). Addison-Wesley. Kindle Edition.

Suppose you increase the voltage across the copper wire in Examples 25.2 and 25.3. The increased voltage causes more current to flow, which makes the temperature of the wire increase. (The same thing happens to the coils of an electric oven or a toaster when a voltage is applied to them. We'll explore this issue in more depth in Section 25.5.) If you double the voltage across the wire, the current in the wire increases. By what factor does it increase? (i) 2; (ii) greater than 2; (iii) less than 2. Young, Hugh D.; Freedman, Roger A.. University Physics with Modern Physics Plus MasteringPhysics with eText -- Access Card Package, 14/e (Page 826). Addison-Wesley. Kindle Edition.

(iii) Solving Eq. (25.11) for the current shows that I = V>R. If the resistance R of the wire remained the same, doubling the voltage V would make the current I double as well. However, we saw in Example 25.3 that the resistance is not constant: As the current increases and the temperature increases, R increases as well. Thus doubling the voltage produces a current that is less than double the original current. An ohmic conductor is one for which R = V>I has the same value no matter what the voltage, so the wire is nonohmic. (In many practical problems the temperature change of the wire is so small that it can be ignored, so we can safely regard the wire as being ohmic. We do so in almost all examples in this book.) Young, Hugh D.; Freedman, Roger A.. University Physics with Modern Physics Plus MasteringPhysics with eText -- Access Card Package, 14/e (Page 847). Addison-Wesley. Kindle Edition.

A capacitor has vacuum in the space between the conductors. If you double the amount of charge on each conductor, what happens to the capacitance? (i) It increases; (ii) it decreases; (iii) it remains the same; (iv) the answer depends on the size or shape of the conductors. Young, Hugh D.; Freedman, Roger A.. University Physics with Modern Physics Plus MasteringPhysics with eText -- Access Card Package, 14/e (Page 790). Addison-Wesley. Kindle Edition.

(iii) The capacitance does not depend on the value of the charge Q. Doubling Q causes the potential difference Vab to double, so the capacitance C = Q>Vab remains the same. These statements are true no matter what the geometry of the capacitor. Young, Hugh D.; Freedman, Roger A.. University Physics with Modern Physics Plus MasteringPhysics with eText -- Access Card Package, 14/e (Page 815). Addison-Wesley. Kindle Edition.

In a flashlight, how does the amount of current that flows out of the bulb compare to the amount that flows into the bulb? (i) Current out is less than current in; (ii) current out is greater than current in; (iii) current out equals current in; (iv) the answer depends on the brightness of the bulb. Young, Hugh D.; Freedman, Roger A.. University Physics with Modern Physics Plus MasteringPhysics with eText -- Access Card Package, 14/e (Page 816). Addison-Wesley. Kindle Edition.

(iii) The current out equals the current in. In other words, charge must enter the bulb at the same rate as it exits the bulb. It is not "used up" or consumed as it flows through the bulb. Young, Hugh D.; Freedman, Roger A.. University Physics with Modern Physics Plus MasteringPhysics with eText -- Access Card Package, 14/e (Page 847). Addison-Wesley. Kindle Edition.

This child acquires an electric charge by touching the charged metal shell. The charged hairs on the child's head repel and stand out. What would happen if the child stood inside a large, charged metal shell? She would acquire (i) the same sign of charge as on the shell, and her hairs would stand out; (ii) the opposite sign of charge as on the shell, and her hairs would stand out; (iii) no charge, and her hairs would be relaxed; (iv) any of these, depending on the amount of charge on the shell. Young, Hugh D.; Freedman, Roger A.. University Physics with Modern Physics Plus MasteringPhysics with eText -- Access Card Package, 14/e (Page 722). Addison-Wesley. Kindle Edition.

(iii) The electric field inside a cavity within a conductor is zero, so there would be no electric effect on the child. (See Section 22.5.) Young, Hugh D.; Freedman, Roger A.. University Physics with Modern Physics Plus MasteringPhysics with eText -- Access Card Package, 14/e (Page 751). Addison-Wesley. Kindle Edition.

Rank the following circuits in order from highest to lowest current: (i) A 1.4@Ω resistor connected to a 1.5@V battery that has an internal resistance of 0.10 Ω; (ii) a 1.8@Ω resistor connected to a 4.0@V battery that has a terminal voltage of 3.6 V but an unknown internal resistance; (iii) an unknown resistor connected to a 12.0@V battery that has an internal resistance of 0.20 Ω and a terminal voltage of 11.0 V. Young, Hugh D.; Freedman, Roger A.. University Physics with Modern Physics Plus MasteringPhysics with eText -- Access Card Package, 14/e (Page 832). Addison-Wesley. Kindle Edition.

(iii), (ii), (i) For circuit (i), we find the current from Eq. (25.16): I = E> 1R + r2 = 11.5 V2> 11.4 Ω+ 0.10 Ω2 = 1.0 A. For circuit (ii), we note that the terminal voltage vab = 3.6 V equals the voltage IR across the 1.8@Ω resistor: Vab = IR, so I = Vab >R = 13.6 V2> 11.8 Ω2 = 2.0 A. For circuit (iii), we use Eq. (25.15) for the terminal voltage: Vab = E - Ir, so I = 1E - Vab 2> r = 112.0 V - 11.0 V2> 10.20 Ω2 = 5.0 A. Young, Hugh D.; Freedman, Roger A.. University Physics with Modern Physics Plus MasteringPhysics with eText -- Access Card Package, 14/e (Page 847). Addison-Wesley. Kindle Edition.

Rank the following circuits in order from highest to lowest values of the net power output of the battery. (i) A 1.4@Ω resistor connected to a 1.5@V battery that has an internal resistance of 0.10 Ω; (ii) a 1.8@Ω resistor connected to a 4.0@V battery that has a terminal voltage of 3.6 V but an unknown internal resistance; (iii) an unknown resistor connected to a 12.0@V battery that has an internal resistance of 0.20 Ω and a terminal voltage of 11.0 V. ❙ Young, Hugh D.; Freedman, Roger A.. University Physics with Modern Physics Plus MasteringPhysics with eText -- Access Card Package, 14/e (Page 836). Addison-Wesley. Kindle Edition.

(iii), (ii), (i) These are the same circuits that we analyzed in Test Your Understanding of Section 25.4. In each case the net power output of the battery is P = Vab I, where Vab is the battery terminal voltage. For circuit (i), we found that I = 1.0 A, so Vab = E - Ir = 1.5 V - 11.0 A210.10 Ω2 = 1.4 V, so P = 11.4 V211.0 A2 = 1.4 W. For circuit (ii), we have Vab = 3.6 V and found that I = 2.0 A, so P = 13.6 V212.0 A2 = 7.2 W. For circuit (iii), we have Vab = 11.0 V and found that I = 5.0 A, so P = 111.0 V215.0 A2 = 55 W. Young, Hugh D.; Freedman, Roger A.. University Physics with Modern Physics Plus MasteringPhysics with eText -- Access Card Package, 14/e (Page 847). Addison-Wesley. Kindle Edition.

Four identical strings each carry a sinusoidal wave of frequency 10 Hz. The string tension and wave amplitude are different for different strings. Rank the following strings in order from highest to lowest value of the average wave power: (i) tension 10 N, amplitude 1.0 mm; (ii) tension 40 N, amplitude 1.0 mm; (iii) tension 10 N, amplitude 4.0 mm; (iv) tension 20 N, amplitude 2.0 mm.

(iii), (iv), (ii), (i) Equation (15.25) says that the average power in a sinusoidal wave on a string is Pav = 1 2 2mFv 2 A 2 . All four strings are identical, so all have the same mass, length, and linear mass density m. The frequency f is the same for each wave, as is the angular frequency v = 2pf. Hence the average wave power for each string is proportional to the square root of the string tension F and the square of the amplitude A. Compared to string (i), the average power in each string is (ii) 14 = 2 times greater; (iii) 4 2 = Young, Hugh D.; Freedman, Roger A.. University Physics with Modern Physics Plus MasteringPhysics with eText -- Access Card Package, 14/e (Page 504). Addison-Wesley. Kindle Edition.

In flash photography, the energy used to make the flash is stored in a capacitor, which consists of two closely spaced conductors that carry opposite charges. If the amount of charge on the conductors is doubled, by what factor does the stored energy increase? (i) 22; (ii) 2; (iii) 222; (iv) 4; (v) 8. Young, Hugh D.; Freedman, Roger A.. University Physics with Modern Physics Plus MasteringPhysics with eText -- Access Card Package, 14/e (Page 785). Addison-Wesley. Kindle Edition.

(iv) Equation (24.9) shows that the energy stored in a capacitor with capacitance C and charge Q is U = Q 2 > 2C. If Q is doubled, the stored energy increases by a factor of 2 2 = 4. Note that if the value of Q is too great, the electric-field magnitude inside the capacitor will exceed the dielectric strength of the material between the plates and dielectric breakdown will occur (see Section 24.4). This puts a practical limit on the amount of energy that can be stored. Young, Hugh D.; Freedman, Roger A.. University Physics with Modern Physics Plus MasteringPhysics with eText -- Access Card Package, 14/e (Page 815). Addison-Wesley. Kindle Edition.

When a cut diamond is illuminated with white light, it sparkles brilliantly with a spectrum of vivid colors. These distinctive visual features are a result of (i) light traveling much slower in diamond than in air; (ii) light of different colors traveling at different speeds in diamond; (iii) diamond absorbing light of certain colors; (iv) both (i) and (ii); (v) all of (i), (ii), and (iii). Young, Hugh D.; Freedman, Roger A.. University Physics with Modern Physics Plus MasteringPhysics with eText -- Access Card Package, 14/e (Page 1078). Addison-Wesley. Kindle Edition.

(iv) The brilliance and color of a diamond are due to total internal reflection from its surfaces (Section 33.3) and to dispersion, which spreads this light into a spectrum (Section 33.4). Young, Hugh D.; Freedman, Roger A.. University Physics with Modern Physics Plus MasteringPhysics with eText -- Access Card Package, 14/e (Page 1110). Addison-Wesley. Kindle Edition.

Suppose that charge q2 in Example 21.4 were -2.0 mC. In this case, the total electric force on Q would be (i) in the positive x-direction; (ii) in the negative x-direction; (iii) in the positive y-direction; (iv) in the negative y-direction; (v) zero; (vi) none of these.

(iv) The force exerted by q1 on Q is still as in Example 21.4. The magnitude of the force exerted by q2 on Q is still equal to F1 on Q, but the direction of the force is now toward q2 at an angle a below the x-axis. Hence the x-components of the two forces cancel while the (negative) y-components add together, and the total electric force is in the negative y-direction. Young, Hugh D.; Freedman, Roger A.. University Physics with Modern Physics Plus MasteringPhysics with eText -- Access Card Package, 14/e (Page 721). Addison-Wesley. Kindle Edition.

The immense cylinder in this photograph is a current-carrying coil, or solenoid, that generates a uniform magnetic field in its interior as part of an experiment at CERN, the European Organization for Nuclear Research. If two such solenoids were joined end to end, the magnetic field along their common axis would (i) become four times stronger; (ii) double in strength; (iii) become 12 times stronger; (iv) not change; (v) weaken. Young, Hugh D.; Freedman, Roger A.. University Physics with Modern Physics Plus MasteringPhysics with eText -- Access Card Package, 14/e (Page 921). Addison-Wesley. Kindle Edition.

(iv) There would be no change in the magnetic field strength. From Example 28.9 (Section 28.7), the field inside a solenoid has magnitude B = m0 nI, where n is the number of turns of wire per unit length. Joining two solenoids end to end doubles both the number of turns and the length, so the number of turns per unit length is unchanged. Young, Hugh D.; Freedman, Roger A.. University Physics with Modern Physics Plus MasteringPhysics with eText -- Access Card Package, 14/e (Page 954). Addison-Wesley. Kindle Edition.

Suppose that the line of charge in Fig. 21.24 (Example 21.10) had charge +Q distributed uniformly between y = 0 and y = +a and had charge -Q distributed uniformly between y = 0 and y = -a. In this situation, the electric field at P would be (i) in the positive x-direction; (ii) in the negative x-direction; (iii) in the positive y-direction; (iv) in the negative y-direction; (v) zero; (vi) none of these.

(iv) Think of a pair of segments of length dy, one at coordinate y 7 0 and the other at coordinate -y 6 0. The upper segment has a positive charge and produces an electric field dE S at P that points away from the segment, so this dE S has a positive x-component and a negative y-component, like the vector dE S in Fig. 21.24. The lower segment has the same amount of negative charge. It produces a dE S that has the same magnitude but points toward the lower segment, so it has a negative x-component and a negative y-component. By symmetry, the two x-components are equal but opposite, so they cancel. Thus the total electric field has only a negative y-component. Young, Hugh D.; Freedman, Roger A.. University Physics with Modern Physics Plus MasteringPhysics with eText -- Access Card Package, 14/e (Page 721). Addison-Wesley. Kindle Edition.

Two charged objects repel each other through the electric force. The charges on the objects are (i) one positive and one negative; (ii) both positive; (iii) both negative; (iv) either (ii) or (iii); (v) any of (i), (ii), or (iii).

(iv) Two charged objects repel if their charges are of the same sign (either both positive or both negative). Young, Hugh D.; Freedman, Roger A.. University Physics with Modern Physics Plus MasteringPhysics with eText -- Access Card Package, 14/e (Page 721). Addison-Wesley. Kindle Edition.

Rank the following surfaces in order from most positive to most negative electric flux: (i) a flat rectangular surface with vector area A S � 16.0 m 2 2nd in a uniform electric field E S � 14.0 N>C2ne; (ii) a flat circular surface with vector area A S � 13.0 m 2 2ne in a uniform electric field E S � 14.0 N>C2nd � 12.0 N>C2ne; (iii) a flat square surface with vector area A S � 13.0 m 2 2nd � 17.0 m 2 2ne in a uniform electric field E S � 14.0 N>C2nd � 12.0 N>C2ne; (iv) a flat oval surface with vector area A S � 13.0 m 2 2nd � 17.0 m 2 2ne in a uniform electric field E S � 14.0 N>C2nd � 12.0 N>C2ne. Young, Hugh D.; Freedman, Roger A.. University Physics with Modern Physics Plus MasteringPhysics with eText -- Access Card Package, 14/e (Page 729). Addison-Wesley. Kindle Edition.

(iv), (ii), (i), (iii) In each case the electric field is uniform, so the flux is ΦE = E S ~ A S . We use the relationships for the scalar products of unit vectors: nd ~ nd = ne ~ ne = 1, nd ~ ne = 0. In case (i) we have ΦE = 14.0 N>C216.0 m 2 2nd ~ ne = 0 (the electric field and vector area are perpendicular, so there is zero flux). In case (ii) we have ΦE = 314.0 N>C2nd � 12.0 N>C2ne4 ~ 13.0 m 2 2ne = 12.0 N>C2 # 13.0 m 2 2 = 6.0 N # m 2 >C. Similarly, in case (iii) we have ΦE = 314.0 N>C2nd �12.0 N>C2ne4 ~ 313.0 m 2 2nd �17.0 m 2 2ne4 = 14.0 N>C213.0 m 2 2 - 12.0 N>C217.0 m 2 2 = -2 N # m 2 >C, and in case (iv) we have ΦE = 314.0 N>C2nd � 12.0 N>C2ne4 ~ 313.0 m 2 2nd � 17.0 m 2 2ne4 = 14.0 N>C213.0 m 2 2 + 12.0 N>C2 # 17.0 m 2 2 = 26 N # m 2 >C. Young, Hugh D.; Freedman, Roger A.. University Physics with Modern Physics Plus MasteringPhysics with eText -- Access Card Package, 14/e (Page 751). Addison-Wesley. Kindle Edition.

Suppose we replaced the wire in Example 25.1 with 12-gauge copper wire, which has twice the diameter of 18-gauge wire. If the current remains the same, what effect would this have on the magnitude of the drift velocity vd ? (i) None—vd would be unchanged; (ii) vd would be twice as great; (iii) vd would be four times greater; (iv) vd would be half as great; (v) vd would be one-fourth as great. Young, Hugh D.; Freedman, Roger A.. University Physics with Modern Physics Plus MasteringPhysics with eText -- Access Card Package, 14/e (Page 820). Addison-Wesley. Kindle Edition.

(v) Doubling the diameter increases the cross-sectional area A by a factor of 4. Hence the current-density magnitude J = I >A is reduced to 1 4 of the value in Example 25.1, and the magnitude of the drift velocity vd = J> n 0 q 0 is reduced by the same factor. The new magnitude is vd = 10.15 mm> s2> 4 = 0.038 mm> s. This behavior is the same as that of an incompressible fluid, which slows down when it moves from a narrow pipe to a broader one (see Section 12.4). Young, Hugh D.; Freedman, Roger A.. University Physics with Modern Physics Plus MasteringPhysics with eText -- Access Card Package, 14/e (Page 847). Addison-Wesley. Kindle Edition.

An infinitesimal current element located at the origin 1x = y = z = 02 carries current I in the positive y-direction. Rank the following locations in order of the strength of the magnetic field that the current element produces at that location, from largest to smallest value. (i) x = L, y = 0, z = 0; (ii) x = 0, y = L, z = 0; (iii) x = 0, y = 0, z = L; (iv) x = L>12 , y = L>12 , z = 0. Young, Hugh D.; Freedman, Roger A.. University Physics with Modern Physics Plus MasteringPhysics with eText -- Access Card Package, 14/e (Page 926). Addison-Wesley. Kindle Edition.

28.2 (i) and (iii) (tie), (iv), (ii) From Eq. (28.5), the magnitude of the field dB due to a current element of length dl carrying current I is dB = 1m0 > 4p21I dlsinf> r 2 2. In this expression r is the distance from the element to the field point, and f is the angle between the direction of the current and a vector from the current element to the field point. All four points are the same distance r = L from the current element, so the value of dB is proportional to the value of sinf. For the four points the angle is (i) f = 90°, (ii) f = 0, (iii) f = 90°, and (iv) f = 45°, so the values of sinf are (i) 1, (ii) 0, (iii) 1, and (iv) 1>12 . Young, Hugh D.; Freedman, Roger A.. University Physics with Modern Physics Plus MasteringPhysics with eText -- Access Card Package, 14/e (Page 954). Addison-Wesley. Kindle Edition.

The accompanying figure shows a circuit that lies on a horizontal table. A compass is placed on top of the circuit as shown. A battery is to be connected to the circuit so that when the switch is closed, the compass needle deflects counterclockwise. In which orientation, A or B, should the battery be placed in the circuit? Young, Hugh D.; Freedman, Roger A.. University Physics with Modern Physics Plus MasteringPhysics with eText -- Access Card Package, 14/e (Page 929). Addison-Wesley. Kindle Edition.

28.3 A This orientation will cause current to flow clockwise around the circuit. Hence current will flow south through the wire that lies under the compass. From the right-hand rule for the magnetic field produced by a long, straight, current-carrying conductor, this will produce a magnetic field that points to the left at the position of the compass (which lies atop the wire). The combination of the northward magnetic field of the earth and the westward field produced by the current gives a net magnetic field to the northwest, so the compass needle will swing counterclockwise to align with this field. Young, Hugh D.; Freedman, Roger A.. University Physics with Modern Physics Plus MasteringPhysics with eText -- Access Card Package, 14/e (Page 954). Addison-Wesley. Kindle Edition.

A solenoid is a wire wound into a helical coil. The accompanying figure shows a solenoid that carries a current I. (a) Is the magnetic force that one turn of the coil exerts on an adjacent turn (i) attractive, (ii) repulsive, or (iii) zero? (b) Is the electric force that one turn of the coil exerts on an adjacent turn (i) attractive, (ii) repulsive, or (iii) zero? (c) Is the magnetic force between opposite sides of the same turn of the coil (i) attractive, (ii) repulsive, or (iii) zero? (d) Is the electric force between opposite sides of the same turn of the coil (i) attractive, (ii) repulsive, or (iii) zero? ❙ Young, Hugh D.; Freedman, Roger A.. University Physics with Modern Physics Plus MasteringPhysics with eText -- Access Card Package, 14/e (Page 930). Addison-Wesley. Kindle Edition.

28.4 (a) (i), (b) (iii), (c) (ii), (d) (iii) Current flows in the same direction in adjacent turns of the coil, so the magnetic forces between these turns are attractive. Current flows in opposite directions on opposite sides of the same turn, so the magnetic forces between these sides are repulsive. Thus the magnetic forces on the solenoid turns squeeze them together in the direction along its axis but push them apart radially. The electric forces are zero because the wire is electrically neutral, with as much positive charge as there is negative charge. Young, Hugh D.; Freedman, Roger A.. University Physics with Modern Physics Plus MasteringPhysics with eText -- Access Card Package, 14/e (Page 954). Addison-Wesley. Kindle Edition.

Figure 28.12 shows the magnetic field dB S produced at point P by a segment dl S that lies on the positive y-axis (at the top of the loop). This field has components dBx 7 0, dBy 7 0, dBz = 0. (a) What are the signs of the components of the field dB S produced at P by a segment dl S on the negative y@axis (at the bottom of the loop)? (i) dBx 7 0, dBy 7 0, dBz = 0; (ii) dBx 7 0, dBy 6 0, dBz = 0; (iii) dBx 6 0, dBy 7 0, dBz = 0; (iv) dBx 6 0, dBy 6 0, dBz = 0; (v) none of these. (b) What are the signs of the components of the field dB S produced at P by a segment dl S on the negative z@axis (at the right-hand side of the loop)? (i) dBx 7 0, dBy 7 0, dBz = 0; (ii) dBx 7 0, dBy 6 0, dBz = 0; (iii) dBx 6 0, dBy 7 0, dBz = 0; (iv) dBx 6 0, dBy 6 0, dBz = 0; (v) none of these. Young, Hugh D.; Freedman, Roger A.. University Physics with Modern Physics Plus MasteringPhysics with eText -- Access Card Package, 14/e (Page 933). Addison-Wesley. Kindle Edition.

28.5 (a) (ii), (b) (v) The vector dB S is in the direction of dl S : r S . For a segment on the negative y@axis, dl S � �k n dl points in the negative z@direction and r S � xnd � ane. Hence dl S : r S � 1a dl2nd � 1x dl2ne, which has a positive x@component, a negative y@component, and zero z@component. For a segment on the negative z@axis, dl S � ne dl points in the positive y@direction and r S � xnd � ak n . Hence dl S : r S � 11a dl2nd � 11x dl2k n , which has a positive x@component, zero y@component, and a negative z@component. Young, Hugh D.; Freedman, Roger A.. University Physics with Modern Physics Plus MasteringPhysics with eText -- Access Card Package, 14/e (Page 954). Addison-Wesley. Kindle Edition.

The accompanying figure shows magnetic field lines through the center of a permanent magnet. The magnet is not connected to a source of emf. One of the field lines is colored red. What can you conclude about the currents inside the permanent magnet within the region enclosed by this field line? (i) There are no currents inside the magnet; (ii) there are currents directed out of the plane of the page; (iii) there are currents directed into the plane of the page; (iv) not enough information is given to decide. ❙ Young, Hugh D.; Freedman, Roger A.. University Physics with Modern Physics Plus MasteringPhysics with eText -- Access Card Package, 14/e (Page 936). Addison-Wesley. Kindle Edition.

28.6 (ii) Imagine carrying out the integral A B S ~ dl S along an integration path that goes counterclockwise around the red magnetic field line. At each point along the path the magnetic field B S and the infinitesimal segment dl S are both tangent to the path, so B S ~ dl S is positive at each point and the integral A B S ~ dl S is likewise positive. It follows from Ampere's law A B S ~ dl S = m0 I encl and the right-hand rule that the integration path encloses a current directed out of the plane of the page. There are no currents in the empty space outside the magnet, so there must be currents inside the magnet (see Section 28.8). Young, Hugh D.; Freedman, Roger A.. University Physics with Modern Physics Plus MasteringPhysics with eText -- Access Card Package, 14/e (Page 954). Addison-Wesley. Kindle Edition.

Consider a conducting wire that runs along the central axis of a hollow conducting cylinder. Such an arrangement, called a coaxial cable, has many applications in telecommunications. (The cable that connects a television set to a local cable provider is an example of a coaxial cable.) In such a cable a current I runs in one direction along the hollow conducting cylinder and is spread uniformly over the cylinder's cross-sectional area. An equal current runs in the opposite direction along the central wire. How does the magnitude B of the magnetic field outside such a cable depend on the distance r from the central axis of the cable? (i) B is proportional to 1> r; (ii) B is proportional to 1> r 2 ; (iii) B is zero at all points outside the cable. ❙ Young, Hugh D.; Freedman, Roger A.. University Physics with Modern Physics Plus MasteringPhysics with eText -- Access Card Package, 14/e (Page 939). Addison-Wesley. Kindle Edition.

28.7 (iii) By symmetry, any B S field outside the cable must circulate around the cable, with circular field lines like those surrounding the solid cylindrical conductor in Fig. 28.20. Choose an integration path like the one shown in Fig. 28.20 with radius r 7 R, so that the path completely encloses the cable. As in Example 28.8, the integral A B S ~ dl S for this path has magnitude B12pr2. From Ampere's law this is equal to m0 I encl . The net enclosed current I encl is zero because it includes two currents of equal magnitude but opposite direction: one in the central wire and one in the hollow cylinder. Hence B12pr2 = 0, and so B = 0 for any value of r outside the cable. (The field is nonzero inside the cable; see Exercise 28.43.) Young, Hugh D.; Freedman, Roger A.. University Physics with Modern Physics Plus MasteringPhysics with eText -- Access Card Package, 14/e (Page 954). Addison-Wesley. Kindle Edition.

Which of the following materials are attracted to a magnet? (i) Sodium; (ii) bismuth; (iii) lead; (iv) uranium. ❙ Young, Hugh D.; Freedman, Roger A.. University Physics with Modern Physics Plus MasteringPhysics with eText -- Access Card Package, 14/e (Page 944). Addison-Wesley. Kindle Edition.

28.8 (i), (iv) Sodium and uranium are paramagnetic materials and hence are attracted to a magnet, while bismuth and lead are diamagnetic materials that are repelled by a magnet. (See Table 28.1.) Young, Hugh D.; Freedman, Roger A.. University Physics with Modern Physics Plus MasteringPhysics with eText -- Access Card Package, 14/e (Page 954). Addison-Wesley. Kindle Edition.

For each of the following electromagnetic waves, state the direction of the magnetic field. (a) The wave is propagating in the positive z-direction, and E S is in the positive x-direction; (b) the wave is propagating in the positive y-direction, and E S is in the negative z-direction; (c) the wave is propagating in the negative x-direction, and E S is in the positive z-direction. Young, Hugh D.; Freedman, Roger A.. University Physics with Modern Physics Plus MasteringPhysics with eText -- Access Card Package, 14/e (Page 1059). Addison-Wesley. Kindle Edition.

32.2 (a) positive y-direction, (b) negative x-direction, (c) positive y-direction You can verify these answers by using the right-hand rule to show that E S : B S in each case is in the direction of propagation, or by using the rule shown in Fig. 32.9. Young, Hugh D.; Freedman, Roger A.. University Physics with Modern Physics Plus MasteringPhysics with eText -- Access Card Package, 14/e (Page 1077). Addison-Wesley. Kindle Edition.

The first of Eqs. (32.17) gives the electric field for a plane wave as measured at points along the x-axis. For this plane wave, how does the electric field at points off the x-axis differ from the expression in Eqs. (32.17)? (i) The amplitude is different; (ii) the phase is different; (iii) both the amplitude and phase are different; (iv) none of these. Young, Hugh D.; Freedman, Roger A.. University Physics with Modern Physics Plus MasteringPhysics with eText -- Access Card Package, 14/e (Page 1063). Addison-Wesley. Kindle Edition.

32.3 (iv) In an ideal electromagnetic plane wave, at any instant the fields are the same anywhere in a plane perpendicular to the direction of propagation. The plane wave described by Eqs. (32.17) is propagating in the x-direction, so the fields depend on the coordinate x and time t but do not depend on the coordinates y and z. Young, Hugh D.; Freedman, Roger A.. University Physics with Modern Physics Plus MasteringPhysics with eText -- Access Card Package, 14/e (Page 1077). Addison-Wesley. Kindle Edition.

Figure 32.13 shows one wavelength of a sinusoidal electromagnetic wave at time t = 0. For which of the following four values of x is (a) the energy density a maximum; (b) the energy density a minimum; (c) the magnitude of the instantaneous (not average) Poynting vector a maximum; (d) the magnitude of the instantaneous (not average) Poynting vector a minimum? (i) x = 0; (ii) x = l> 4; (iii) x = l> 2; (iv) x = 3l> 4. Young, Hugh D.; Freedman, Roger A.. University Physics with Modern Physics Plus MasteringPhysics with eText -- Access Card Package, 14/e (Page 1068). Addison-Wesley. Kindle Edition.

32.4 (a) (i) and (iii), (b) (ii) and (iv), (c) (i) and (iii), (d) (ii) and (iv) Both the energy density u and the Poynting vector magnitude S are maximum where the E S and B S fields have their maximum magnitudes. (The directions of the fields don't matter.) From Fig. 32.13, this occurs at x = 0 and x = l> 2. Both u and S have a minimum value of zero; that occurs where E S and B S are both zero. From Fig. 32.13, this occurs at x = l> 4 and x = 3l> 4. Young, Hugh D.; Freedman, Roger A.. University Physics with Modern Physics Plus MasteringPhysics with eText -- Access Card Package, 14/e (Page 1077). Addison-Wesley. Kindle Edition.

In the standing wave described in Example 32.7, is there any point in the cavity where the energy density is zero at all times? If so, where? If not, why not? Young, Hugh D.; Freedman, Roger A.. University Physics with Modern Physics Plus MasteringPhysics with eText -- Access Card Package, 14/e (Page 1071). Addison-Wesley. Kindle Edition.

32.5 no There are places where E S � 0 at all times (at the walls) and the electric energy density 1 2 P 0 E 2 is always zero. There are also places where B S � 0 at all times (on the plane midway between the walls) and the magnetic energy density B 2 > 2m0 is always zero. However, there are no locations where both E S and B S are always zero. Hence the energy density at any point in the standing wave is always nonzero. Young, Hugh D.; Freedman, Roger A.. University Physics with Modern Physics Plus MasteringPhysics with eText -- Access Card Package, 14/e (Page 1077). Addison-Wesley. Kindle Edition.

Some crystals are not isotropic: Light travels through the crystal at a higher speed in some directions than in others. In a crystal in which light travels at the same speed in the x@ and z@directions but faster in the y@direction, what would be the shape of the wave fronts produced by a light source at the origin? (i) Spherical, like those shown in Fig. 33.3; (ii) ellipsoidal, flattened along the y@axis; (iii) ellipsoidal, stretched out along the y@axis. ❙ Young, Hugh D.; Freedman, Roger A.. University Physics with Modern Physics Plus MasteringPhysics with eText -- Access Card Package, 14/e (Page 1080). Addison-Wesley. Kindle Edition.

33.1 (iii) The waves go farther in the y@direction in a given amount of time than in the other directions, so the wave fronts are elongated in the y@direction. Young, Hugh D.; Freedman, Roger A.. University Physics with Modern Physics Plus MasteringPhysics with eText -- Access Card Package, 14/e (Page 1110). Addison-Wesley. Kindle Edition.

You are standing on the shore of a lake. You spot a tasty fish swimming some distance below the lake surface. (a) If you want to spear the fish, should you aim the spear (i) above, (ii) below, or (iii) directly at the apparent position of the fish? (b) If instead you use a high-power laser to simultaneously kill and cook the fish, should you aim the laser (i) above, (ii) below, or (iii) directly at the apparent position of the fish? Young, Hugh D.; Freedman, Roger A.. University Physics with Modern Physics Plus MasteringPhysics with eText -- Access Card Package, 14/e (Page 1086). Addison-Wesley. Kindle Edition.

33.2 (a) (ii), (b) (iii) As shown in the figure, light rays coming from the fish bend away from the normal when they pass from the water 1n = 1.332 into the air 1n = 1.002. As a result, the fish appears to be higher in the water than it actually is. Hence you should aim a spear below the apparent position of the fish. If you use a laser beam, you should aim at the apparent position of the fish: The beam of laser light takes the same path from you to the fish as ordinary light takes from the fish to you (though in the opposite direction). Young, Hugh D.; Freedman, Roger A.. University Physics with Modern Physics Plus MasteringPhysics with eText -- Access Card Package, 14/e (Page 1110). Addison-Wesley. Kindle Edition.

In which of the following situations is there total internal reflection? (i) Light propagating in water 1n = 1.332 strikes a water-air interface at an incident angle of 70°; (ii) light propagating in glass 1n = 1.522 strikes a glass-water interface at an incident angle of 70°; (iii) light propagating in water strikes a water-glass interface at an incident angle of 70°. Young, Hugh D.; Freedman, Roger A.. University Physics with Modern Physics Plus MasteringPhysics with eText -- Access Card Package, 14/e (Page 1089). Addison-Wesley. Kindle Edition.

33.3 (i), (ii) Total internal reflection can occur only if two conditions are met: nb must be less than na , and the critical angle ucrit (where sin ucrit = nb > na ) must be smaller than the angle of incidence u a . In the first two cases both conditions are met: The critical angles are (i) ucrit = sin -1 11> 1.332 = 48.8° and (ii) ucrit = sin -1 11.33> 1.522 = 61.0°, both of which are smaller than u a = 70°. In the third case nb = 1.52 is greater than na = 1.33, so total internal reflection cannot occur for any incident angle. Young, Hugh D.; Freedman, Roger A.. University Physics with Modern Physics Plus MasteringPhysics with eText -- Access Card Package, 14/e (Page 1110). Addison-Wesley. Kindle Edition.

33.5 You are taking a photograph of a sunlit office building at sunrise, so the plane of incidence is nearly horizontal. In order to minimize the reflections from the building's windows, you place a polarizing filter on the camera lens. How should you orient the filter? (i) With the polarizing axis vertical; (ii) with the polarizing axis horizontal; (iii) either orientation will minimize the reflections just as well; (iv) neither orientation will have any effect. ❙ Young, Hugh D.; Freedman, Roger A.. University Physics with Modern Physics Plus MasteringPhysics with eText -- Access Card Package, 14/e (Page 1098). Addison-Wesley. Kindle Edition.

33.5 (ii) The sunlight reflected from the windows of the highrise building is partially polarized in the vertical direction, perpendicular to the horizontal plane of incidence. The Polaroid filter in front of the lens is oriented with its polarizing axis perpendicular to the dominant direction of polarization of the reflected light. Young, Hugh D.; Freedman, Roger A.. University Physics with Modern Physics Plus MasteringPhysics with eText -- Access Card Package, 14/e (Page 1110). Addison-Wesley. Kindle Edition.

Sound travels faster in warm air than in cold air. Imagine a weather front that runs north-south, with warm air to the west of the front and cold air to the east. A sound wave traveling in a northeast direction in the warm air encounters this front. How will the direction of this sound wave change when it passes into the cold air? (i) The wave direction will deflect toward the north; (ii) the wave direction will deflect toward the east; (iii) the wave direction will be unchanged. ❙ Young, Hugh D.; Freedman, Roger A.. University Physics with Modern Physics Plus MasteringPhysics with eText -- Access Card Package, 14/e (Page 1102). Addison-Wesley. Kindle Edition.

33.7 (ii) Huygens's principle applies to waves of all kinds, including sound waves. Hence this situation is exactly like that shown in Fig. 33.35, with material a representing the warm air, material b representing the cold air in which the waves travel more slowly, and the interface between the materials representing the weather front. North is toward the top of the figure and east is toward the right, so Fig. 33.35 shows that the rays (which indicate the direction of propagation) deflect toward the east. Young, Hugh D.; Freedman, Roger A.. University Physics with Modern Physics Plus MasteringPhysics with eText -- Access Card Package, 14/e (Page 1110). Addison-Wesley. Kindle Edition.

Figure 22.16 shows six point charges that all lie in the same plane. Five Gaussian surfaces—S1 , S2 , S3 , S4 , and S5 — each enclose part of this plane, and Fig. 22.16 shows the intersection of each surface with the plane. Rank these five surfaces in order of the electric flux through them, from most positive to most negative. Young, Hugh D.; Freedman, Roger A.. University Physics with Modern Physics Plus MasteringPhysics with eText -- Access Card Package, 14/e (Page 733). Addison-Wesley. Kindle Edition.

S2 , S5 , S4 , S1 and S3 (tie) Gauss's law tells us that the flux through a closed surface is proportional to the amount of charge enclosed within that surface. So an ordering of these surfaces by their fluxes is the same as an ordering by the amount of enclosed charge. Surface S1 encloses no charge, surface S2 encloses 9.0 mC + 5.0 mC + 1-7.0 mC2 = 7.0 mC, surface S3 encloses 9.0 mC + 1.0 mC + 1-10.0 mC2 = 0, surface S4 encloses 8.0 mC + 1-7.0 mC2 = 1.0 mC, and surface S5 encloses 8.0 mC + 1-7.0 mC2 + 1-10.0 mC2 + 11.0 mC2 + 19.0 mC2 + 15.0 mC2 = 6.0 mC. Young, Hugh D.; Freedman, Roger A.. University Physics with Modern Physics Plus MasteringPhysics with eText -- Access Card Package, 14/e (Page 751). Addison-Wesley. Kindle Edition.

What type of wave is "the wave" shown in Fig. 15.2? (i) Transverse; (ii) longitudinal; (iii) a combination of transverse and longitudinal.

The "wave" travels horizontally from one spectator to the next along each row of the stadium, but the displacement of each spectator is vertically upward. Since the displacement is perpendicular to the direction in which the wave travels, the wave is transverse. Young, Hugh D.; Freedman, Roger A.. University Physics with Modern Physics Plus MasteringPhysics with eText -- Access Card Package, 14/e (Page 504). Addison-Wesley. Kindle Edition.

When an earthquake strikes, the news of the event travels through the body of the earth in the form of seismic waves. Which aspects of a seismic wave determine how much power is carried by the wave: (i) the amplitude; (ii) the frequency; (iii) both the amplitude and the frequency; or (iv) neither the amplitude nor the frequency?

The power of a mechanical wave depends on both its amplitude and its frequency [see Eq. (15.25)].

If you double the wavelength of a wave on a particular string, what happens to the wave speed v and the frequency f ? (i) v doubles and f is unchanged; (ii) v is unchanged and f doubles; (iii) v becomes onehalf as great and f is unchanged; (iv) v is unchanged and f becomes one-half as great; (v) none of these.

The speed of waves on a string, v, does not depend on the wavelength. We can rewrite the relationship v = lf as f = v> l, which tells us that if the wavelength l doubles, the frequency f becomes one-half as great. Young, Hugh D.; Freedman, Roger A.. University Physics with Modern Physics Plus MasteringPhysics with eText -- Access Card Package, 14/e (Page 504). Addison-Wesley. Kindle Edition.

Suppose the frequency of the standing wave in Example 15.6 were doubled from 440 Hz to 880 Hz. Would all of the nodes for f = 440 Hz also be nodes for f = 880 Hz? If so, would there be additional nodes for f = 880 Hz? If not, which nodes are absent for f = 880 Hz?

Yes, yes Doubling the frequency makes the wavelength half as large. Hence the spacing between nodes (equal to l> 2) is also half as large. There are nodes at all of the previous positions, but there is also a new node between every pair of old nodes. Young, Hugh D.; Freedman, Roger A.. University Physics with Modern Physics Plus MasteringPhysics with eText -- Access Card Package, 14/e (Page 504). Addison-Wesley. Kindle Edition.

The card reader at your bank's cash machine scans the information that is coded in a magnetic pattern on the back of your card. Why must you remove the card quickly rather than hold it motionless in the card reader's slot? (i) To maximize the magnetic force on the card; (ii) to maximize the magnetic force on the mobile charges in the card reader; (iii) to generate an electric force on the card; (iv) to generate an electric force on the mobile charges in the card reader. Young, Hugh D.; Freedman, Roger A.. University Physics with Modern Physics Plus MasteringPhysics with eText -- Access Card Package, 14/e (Page 955). Addison-Wesley. Kindle Edition.

iv As the magnetic stripe moves through the card reader, the coded pattern of magnetization in the stripe causes a varying magnetic flux. An electric field is induced, which causes a current in the reader's circuits. If the card does not move, there is no induced current and none of the credit card's information is read. Young, Hugh D.; Freedman, Roger A.. University Physics with Modern Physics Plus MasteringPhysics with eText -- Access Card Package, 14/e (Page 989). Addison-Wesley. Kindle Edition.

Subtract Eq. (1) from Eq. (2) in Example 26.6. To which loop in Fig. 26.12 does this equation correspond? Would this equation have simplified the solution of Example 26.6? ❙ Young, Hugh D.; Freedman, Roger A.. University Physics with Modern Physics Plus MasteringPhysics with eText -- Access Card Package, 14/e (Page 858). Addison-Wesley. Kindle Edition.

loop cbdac, no Equation (2) minus Eq. (1) gives -I 2 11 Ω2 - 1I 2 + I 3 212 Ω2 + 1I 1 - I 3 211 Ω2 + I 1 11 Ω2 = 0. We can obtain this equation by applying the loop rule around the path from c to b to d to a to c in Fig. 26.12. This isn't an independent equation, so it would not have helped with the solution of Example 26.6. Young, Hugh D.; Freedman, Roger A.. University Physics with Modern Physics Plus MasteringPhysics with eText -- Access Card Package, 14/e (Page 880). Addison-Wesley. Kindle Edition.

While a guitar string is vibrating, you gently touch the midpoint of the string to ensure that the string does not vibrate at that point. Which normal modes cannot be present on the string while you are touching it in this way?

n = 1, 3, 5, NWhen you touch the string at its center, you are producing a node at the center. Hence only standing waves with a node at x = L> 2 are allowed. From Figure 15.26 you can see that the normal modes n = 1, 3, 5, ccannot be present. Young, Hugh D.; Freedman, Roger A.. University Physics with Modern Physics Plus MasteringPhysics with eText -- Access Card Package, 14/e (Page 504). Addison-Wesley. Kindle Edition.

A hollow conducting sphere has no net charge. There is a positive point charge q at the center of the spherical cavity within the sphere. You connect a conducting wire from the outside of the sphere to ground. Will you measure an electric field outside the sphere? Young, Hugh D.; Freedman, Roger A.. University Physics with Modern Physics Plus MasteringPhysics with eText -- Access Card Package, 14/e (Page 742). Addison-Wesley. Kindle Edition.

no Before you connect the wire to the sphere, the presence of the point charge will induce a charge -q on the inner surface of the hollow sphere and a charge q on the outer surface (the net charge on the sphere is zero). There will be an electric field outside the sphere due to the charge on the outer surface. Once you touch the conducting wire to the sphere, however, electrons will flow from ground to the outer surface of the sphere to neutralize the charge there (see Fig. 21.7c). As a result the sphere will have no charge on its outer surface and no electric field outside. Young, Hugh D.; Freedman, Roger A.. University Physics with Modern Physics Plus MasteringPhysics with eText -- Access Card Package, 14/e (Page 751). Addison-Wesley. Kindle Edition.

If the electric field at a certain point is zero, does the electric potential at that point have to be zero? (Hint: Consider the center of the ring in Example 23.11 and Example 21.9.) Young, Hugh D.; Freedman, Roger A.. University Physics with Modern Physics Plus MasteringPhysics with eText -- Access Card Package, 14/e (Page 769). Addison-Wesley. Kindle Edition.

no If E S � 0 at a certain point, V does not have to be zero at that point. An example is point O at the center of the charged ring in Figs. 21.23 and Fig. 23.21. From Example 21.9 (Section 21.5), the electric field is zero at O because the electricfield contribu tions from different parts of the ring completely cancel. From Example 23.11, however, the potential at O is not zero: This point corresponds to x = 0, so V = 11> 4pP 0 21Q> a2. This value of V corresponds to the work that would have to be done to move a unit positive test charge along a path from infinity to point O; it is nonzero because the charged ring repels the test charge, so posi tive work must be done to move the test charge toward the ring. Young, Hugh D.; Freedman, Roger A.. University Physics with Modern Physics Plus MasteringPhysics with eText -- Access Card Package, 14/e (Page 784). Addison-Wesley. Kindle Edition.

If the electric potential at a certain point is zero, does the electric field at that point have to be zero? (Hint: Consider point c in Example 23.4 and Example 21.8.) Young, Hugh D.; Freedman, Roger A.. University Physics with Modern Physics Plus MasteringPhysics with eText -- Access Card Package, 14/e (Page 765). Addison-Wesley. Kindle Edition.

no If V = 0 at a certain point, E S does not have to be zero at that point. An example is point c in Figs. 21.23 and 23.13, for which there is an electric field in the +xdirection (see Exam ple 21.9 in Section 21.5) even though V = 0 (see Example 23.4). This isn't a surprising result because V and E S are quite different quantities: V is the net amount of work required to bring a unit charge from infinity to the point in question, whereas E S is the electric force that acts on a unit charge when it arrives at that point. Young, Hugh D.; Freedman, Roger A.. University Physics with Modern Physics Plus MasteringPhysics with eText -- Access Card Package, 14/e (Page 784). Addison-Wesley. Kindle Edition.

Would the shapes of the equipotential surfaces in Fig. 23.23 change if the sign of each charge were reversed? Young, Hugh D.; Freedman, Roger A.. University Physics with Modern Physics Plus MasteringPhysics with eText -- Access Card Package, 14/e (Page 771). Addison-Wesley. Kindle Edition.

no If the positive charges in Fig. 23.23 were replaced by negative charges, and vice versa, the equipotential surfaces would be the same but the sign of the potential would be reversed. For example, the surfaces in Fig. 23.23b with potential V = +30 V and V = -50 V would have potential V = -30 V and V = +50 V, respectively. Young, Hugh D.; Freedman, Roger A.. University Physics with Modern Physics Plus MasteringPhysics with eText -- Access Card Package, 14/e (Page 784). Addison-Wesley. Kindle Edition.

In the circuit shown in Fig. 27.39, you add a switch in series with the source of emf so that the current can be turned on and off. When you close the switch and allow current to flow, will the rotor begin to turn no matter what its original orientation? ❙ Young, Hugh D.; Freedman, Roger A.. University Physics with Modern Physics Plus MasteringPhysics with eText -- Access Card Package, 14/e (Page 907). Addison-Wesley. Kindle Edition.

no The rotor will not begin to turn when the switch is closed if the rotor is initially oriented as shown in Fig. 27.39b. In this case there is no current through the rotor and hence no magnetic torque. This situation can be remedied by using multiple rotor coils oriented at different angles around the rotation axis. With this arrangement, there is always a magnetic torque no matter what the orientation. Young, Hugh D.; Freedman, Roger A.. University Physics with Modern Physics Plus MasteringPhysics with eText -- Access Card Package, 14/e (Page 920). Addison-Wesley. Kindle Edition.

To prevent the circuit breaker in Example 26.14 from blowing, a home electrician replaces the circuit breaker with one rated at 40 A. Is this a reasonable thing to do? Young, Hugh D.; Freedman, Roger A.. University Physics with Modern Physics Plus MasteringPhysics with eText -- Access Card Package, 14/e (Page 870). Addison-Wesley. Kindle Edition.

no This is a very dangerous thing to do. The circuit breaker will allow currents up to 40 A, double the rated value of the wiring. The amount of power P = I 2 R dissipated in a section of wire can therefore be up to four times the rated value, so the wires could get very warm and start a fire. (This assumes the resistance R remains unchanged. In fact, R increases with temperature, so the dissipated power can be even greater, and more dangerous, than we have estimated.) Young, Hugh D.; Freedman, Roger A.. University Physics with Modern Physics Plus MasteringPhysics with eText -- Access Card Package, 14/e (Page 880). Addison-Wesley. Kindle Edition.

You place a known amount of charge Q on the irregularly shaped conductor shown in Fig. 22.17. If you know the size and shape of the conductor, can you use Gauss's law to calculate the electric field at an arbitrary position outside the conductor? Young, Hugh D.; Freedman, Roger A.. University Physics with Modern Physics Plus MasteringPhysics with eText -- Access Card Package, 14/e (Page 738). Addison-Wesley. Kindle Edition.

no You might be tempted to draw a Gaussian surface that is an enlarged version of the conductor, with the same shape and placed so that it completely encloses the conductor. While you know the flux through this Gaussian surface (by Gauss's law, it's ΦE = Q> P 0 ), the direction of the electric field need not be perpendicular to the surface and the magnitude of the field need not be the same at all points on the surface. It's not possible to do the flux integral AE # dA, and we can't calculate the electric field. Gauss's law is useful for calculating the electric field only when the charge distribution is highly symmetric. Young, Hugh D.; Freedman, Roger A.. University Physics with Modern Physics Plus MasteringPhysics with eText -- Access Card Package, 14/e (Page 751). Addison-Wesley. Kindle Edition.

The accompanying figure shows a uniform magnetic field B S directed into the plane of the paper (shown by the blue * >s). A particle with a negative charge moves in the plane. Which path—1, 2, or 3— does the particle follow? ❙ Young, Hugh D.; Freedman, Roger A.. University Physics with Modern Physics Plus MasteringPhysics with eText -- Access Card Package, 14/e (Page 887). Addison-Wesley. Kindle Edition.

path 3 Applying the right-hand rule to the vectors v S (which points to the right) and B S (which points into the plane of the figure) says that the force F S � qv S : B S on a positive charge would point upward. Since the charge is negative, the force points downward and the particle follows a trajectory that curves downward. Young, Hugh D.; Freedman, Roger A.. University Physics with Modern Physics Plus MasteringPhysics with eText -- Access Card Package, 14/e (Page 920). Addison-Wesley. Kindle Edition.

Suppose the electric field lines in a region of space are straight lines. If a charged particle is released from rest in that region, will the trajectory of the particle be along a field line?

yes If the field lines are straight, E S must point in the same direction throughout the region. Hence the force F S � qE S on a particle of charge q is always in the same direction. A particle released from rest accelerates in a straight line in the direction of F S , and so its trajectory is a straight line along a field line. Young, Hugh D.; Freedman, Roger A.. University Physics with Modern Physics Plus MasteringPhysics with eText -- Access Card Package, 14/e (Page 721). Addison-Wesley. Kindle Edition.

Suppose you cut off the part of the compass needle shown in Fig. 27.5a that is painted gray. You discard this part, drill a hole in the remaining red part, and place the red part on the pivot at the center of the compass. Will the red part still swing when a current is applied as in Fig. 27.5b? Young, Hugh D.; Freedman, Roger A.. University Physics with Modern Physics Plus MasteringPhysics with eText -- Access Card Package, 14/e (Page 883). Addison-Wesley. Kindle Edition.

yes When a magnet is cut apart, each part has a north and south pole (see Fig. 27.4). Hence the small red part behaves much like the original, full-sized compass needle. Young, Hugh D.; Freedman, Roger A.. University Physics with Modern Physics Plus MasteringPhysics with eText -- Access Card Package, 14/e (Page 920). Addison-Wesley. Kindle Edition.

If you wiggle a magnet back and forth in your hand, are you generating an electric field? If so, is this field conservative? ❙ Young, Hugh D.; Freedman, Roger A.. University Physics with Modern Physics Plus MasteringPhysics with eText -- Access Card Package, 14/e (Page 972). Addison-Wesley. Kindle Edition.

yes, no The magnetic field at a fixed position changes as you move the magnet, which induces an electric field. Such induced electric fields are not conservative. Young, Hugh D.; Freedman, Roger A.. University Physics with Modern Physics Plus MasteringPhysics with eText -- Access Card Package, 14/e (Page 989). Addison-Wesley. Kindle Edition.