c m/s v 343 m/s sound 116 Hz 2.96 m I I 2 m 50 m I50 m 2 m W/m W/m

Transcription

1 0.. Model: The wave is a traveling wave on a stretched string. Solve: The wave speed on a stretched string with linear density is v string T S /. The wave speed if the tension is doubled will be TS v string vstring 00 m/s 83 m/s 0.4. Model: Radio waves are electromagnetic waves that travel with speed c. Solve: (a) The wavelength is 8 c 3.00 m/s.96 m f 0.3 MHz (b) The speed of sound in air at 0 C is 343 m/s. The frequency is f v 343 m/s.96 m sound 6 Hz Solve: If a source of spherical waves radiates uniformly in all directions, the ratio of the intensities at distances r and r is I I r r I50 m m I m 50 m.6 0 I I m m W/m W/m 3 Assess: The power generated by the sound source is P I m [4π( m) ] (.0 W/m )(50.7) 0 W. This is a significant amount of power Model: Sound frequency is altered by the Doppler effect. The frequency increases for an observer approaching the source and decreases for an observer receding from a source. Solve: You need to ride your bicycle away from your friend to lower the frequency of the whistle. The minimum speed you need to travel is calculated as follows: f v0 v0 f0 0 khz khz v 343 m/s v0 6.3 m/s 6 m/s Assess: A speed of 6.3 m/s corresponds to approximately 35 mph. This is a possible but very fast speed on a bicycle..3. Model: The laser light forms a standing wave inside the cavity. Solve: The wavelength of the laser beam is L m m 00, m m 00,000 The frequency is 8 c m/s 00, , m f Hz.4. Solve: (a) For the open-open tube, the two open ends exhibit antinodes of a standing wave. The possible wavelengths for this case are

2 L m m,, 3, m The three longest wavelengths are. m. m. m.4 m. m m 3 (b) In the case of an open-closed tube, The three longest wavelengths are 4L m m, 3, 5, m 4. m 4. m 4. m 4.84 m.6 m m Model: Reflections at the string boundaries cause a standing wave on a stretched string. Solve: Because the vibrating section of the string is.9 m long, the two ends of this vibrating wire are fixed, and the string is vibrating in the fundamental harmonic. The wavelength is L m L.90 m 3.80 m m The wave speed along the string is v f (7.5 Hz)(3.80 m) 04.5 m/s. The tension in the wire can be found as follows: TS mass kg v T S v v 04.5 m/s 80 N length.00 m.9. Solve: The beat frequency is f f f 3 Hz f 00 Hz f 03 Hz beat f is larger than f because the increased tension increases the wave speed and hence the frequency... Model: Two closely spaced slits produce a double-slit interference pattern. The interference pattern looks like the photograph of Figure.3(b). It is symmetrical, with the m fringes on both sides of and equally distant from the central maximum. Solve: The two paths from the two slits to the m bright fringe differ by r r r, where r m 500 nm 000 nm Thus, the position of the m bright fringe is 000 nm farther away from the more distant slit than from the nearer slit..8. Model: Two closely spaced slits produce a double-slit interference pattern. The interference pattern looks like the photograph of Figure.3(b). Solve: In a span of fringes, there are gaps between them. The formula for the fringe spacing is 3 9 L 50 m (6330 m)(3.0 m) y d 0.40 mm d d 4 Assess: This is a reasonable distance between the slits, ensuring d L.340.

3 .0. Model: A diffraction grating produces a series of constructive-interference fringes at values of m determined by Equation.5. Solve: We have dsin m m m0,,, 3, d sin 0.0 and d sin Dividing these two equations, sin sin Model: Light passing through a circular aperture leads to a diffraction pattern that has a circular central maximum surrounded by a series of secondary bright fringes. Solve: The width of the central maximum for a circular aperture of diameter D is.44l w D 9 (.44)(5000 m)(.0 m) m 4.9 mm 3.4. Model: Use the ray model of light. The sun is a point source of light. A ray that arrives at the diver 50 above horizontal refracted into the water at water 40. Solve: Using Snell s law at the water-air boundary n sin nwater.33 n sin sinair sinwater sin 40 n.0 air air water water air 58.7 Thus the height above the horizon is = 90 air = Because the sun is far away from the fisherman (and the diver), the fisherman will see the sun at the same angle of 3 above the horizon. air 3.6. Model: Use the ray model of light. For an angle of incidence greater than the critical angle, the ray of light undergoes total internal reflection.

4 Solve: The critical angle of incidence is given by Equation 3.9: c n.48 ncore.60 cladding sin sin 67.7 Thus, the maximum angle a light ray can make with the wall of the core to remain inside the fiber is Assess: We can have total internal reflection because n core n cladding Model: Assume the lens is thin. Solve: fs s s s f s f fs (0 cm)(60 cm) s 30 cm s f 60 cm 0 cm The magnification is ms s30 cm 60 cm. This means the image is inverted and has a height of 0.50 cm. Assess: Ray tracing will confirm these results Model: Use the ray model of light and the law of refraction. Assume the sun is a point source of light. When the bottom of the pool becomes completely shaded, a ray of light that is incident at the top edge of the swimming pool does not reach the bottom of the pool after refraction.

5 Solve: The depth of the swimming pool is d 4.0 m tan water. We will find the angle by using Snell s law. We have sin m nwater sinwater nair sin 70 water sin d 4.0 m.33 tan Model: Ignore the small space between the lens and the eye. Refer to Example 4.5, but we want to solve for s, the far point. Solve: (a) The power of the lens is negative which means the focal length is negative, so Ellen wears diverging lenses. This is the remedy for myopia. (b) We want to know where the image should be for an object s m given f.0 m. When s m, f.0 m P s s f s f.0 m m s f So the far point is 00 cm. Assess: The negative sign on s is expected because we need the image to be virtual. 4.. Model: Assume the eyepiece is a simple magnifier with M eye 5 cm / feye. feye 5 cm/0.5 cm. Solve: (a) The magnification of a telescope is (b) M f 00 cm 40 obj feye.5 cm f.00 m f-number 5.0 D 0.0 m Assess: These results are in reasonable ranges for magnification and f-number Equation 4.5 gives the smallest resolvable distance: d 0.6 / NA. We are given 600 nm and dmin 0.75m 750 nm. Solve: Assess: This is in the normal range for NA. 0.6λ (0.6)(600 nm) NA 0.49 d 750 nm min min Model: Two objects are marginally resolved if the angular separation between the objects is. / D.

6 Solve: (a) The angular separation between the sun and Jupiter is m 7800 m 5.90 rad light years m m D m 3.8 cm D D (b) The sun is vastly brighter than Jupiter, which is much smaller and seen only dimly by reflected light. In theory it may be possible to resolve Jupiter and the sun, but in practice the extremely bright light from the sun will overwhelm the very dim light from Jupiter Model: The 5 Xe nucleus and the proton are point charges. That is, all the charge on the Xe nucleus is assumed to be at its center. Solve: (a) The magnitude of the force between the nucleus and the proton is given by Coulomb s law: F Kq q N m /C C.600 C m nucleus proton nucleus on proton (b) Applying Newton s second law to the proton, r 5.00 N on proton proton proton proton 7 F m a a 3.00 m/s.670 kg N Model: The charges are point charges. Please refer to the figure in the problem. Solve: The electric force on charge q is the vector sum of the forces F on and F, 3 on charge, q is the.0 nc charge, and q 3 is the.0 nc charges. We have where q is the.0 nc

7 F Kq q r, away from q on N m /C.00 C.00 C, away from q.0 0 m N, away from q.80 0 Ncos60 iˆ sin 60 ˆj Kq q 4 4 F3 on, toward q N, toward q3.800 Ncos60iˆsin 60 ˆj. r N cos60 ˆ F F F i iˆ N on on 3 on So, the force on the.0 nc charge is N and it is directed to the right Model: The charged particles are point charges. Solve: The two.0 nc charges exert an upward force on the.0 nc charge. Since the net force on the nc charge is zero, the unknown charge must exert a downward force of equal magnitude. This implies that q is a positive charge. The force of charges on charge is F Kq q, away from q on r (0.00 m) (0.030 m) (9.0 0 N m / C )(.0 0 C)(.0 0 C) From the figure, tan (/3) Thus 5 ˆ 5 F (.50 i ˆj ) N on (cos iˆsin ˆj) From symmetry, F 3 on is the same except the x-component is reversed. When we add F and F, on 3 on the x- components cancel and the y-components add to give 5 F on F3 on.5360 ˆj N F q on must have the same magnitude, pointing in the ĵ direction, so F Kqq (.5360 N)(0.00 m).5360 N 0.68 nc r (9.00 N m /C )(.0 0 C) 5 5 q on q 9 9 A positive charge q 0.68 nc will cause the net force on the.0 nc charge to be zero Model: The charged particles are point charges.

8 Solve: (a) The force on q is the vector sum of the force from Q and Q. We have F K Q q KQq, away from Q i ˆ ax ax Q on q F K Q q KQq, toward Q i ˆ ax ax Q on q KQq a x Fnet KQq x ax ax a x To arrive at the final expression we used ax ax axax a x (b) There are two cases when x a. For x a, F K Q q KQq, away from Q i ˆ xa xa Q on q F For x a (that is, for negative values of x), F K Q q KQq, toward Q i ˆ xa xa Q on q 4KQqax Fnet KQq x KQq iˆ x a Q on q xa xa x a F 4KQqax x a net x F KQq iˆ a x Q on q That is, the net force is to the right when x a and to the right when x a. We can combine these two cases into a single equation for x a :.

9 Here, the force is always to the right when x a. F 4KQqa x x a net x 7.9. Model: The electric field in a region of space between the plates of a parallel-plate capacitor is uniform. Solve: The electric field inside a capacitor is E Q 0 A. Thus, the charge needed to produce a field of strength E is 6 Q0AE C /N m 0.00 m 3.00 N/C 33.4 nc The number of electrons transferred from one plate to the other is C C 7.. Model: A uniform electric field causes a charge to undergo constant acceleration. Solve: Kinematics yields the acceleration of the electron. v v (4.00 m/s) (.00 m/s) x 0.0 m m/s v v a x a The magnitude of the electric field required to obtain this acceleration is E 3 6 Fnet mea (9.0 kg)(5.00 m/s ) 5.80 N/C. 9 e e.60 C Model: The electron orbiting the proton experiences a force given by Coulomb s law. Solve: The force that causes the circular motion is where we used v r T rf. The frequency is qp qe mv m e e 4 r f F 4 r r r N m /C.600 C 3 qp qe 5 f Hz mr e kg m Model: The rod is thin and is assumed to be a line of charge of length L.

10 Solve: (a) The -versus-y graph over the length of the rod is shown in the figure. (b) Consider a segment of charge q of length y at a distance y from the center of the rod. The amount of charge in this segment is qy a y y Converting q to dq, y to dy, and integrating from y L/ to y L/, the total charge is Thus the constant a is L/ L/ L / y al Qdq a y dy aydy a 4 L / 0 0 4Q a L (c) With the origin of the coordinate system at the center of the rod, consider two symmetrically located charge segments q with length y. The electric field at P from the top charge segment makes an angle below the x- axis and the electric field of the bottom charge segment makes an angle above the x-axis. Because the charge density is symmetric about the origin (i.e., (at y) (at y)), the y-components of the two contributions cancel out. Thus, we have to calculate only the x-component of the electric field at P. Because the electric field strength of the lower half of the rod is the same as that of the upper half, we only need to obtain the electric field strength of half the rod, then multiply by two. The electric field along the x-direction due to a charge segment q is q y x Ex cos x y x y x y x y 3/ xa yy Changing E to de, y dy, integrating y from y 0 m to y L/, and multiplying by to take into account the entire rod, the electric field is L/ L/ ax y dy ax y dy ax Ex 3/ 3/ x y x y x y ax 8Q x 4 0 x x L 4 4 0L x L 4 The last step used the expression for a from part (b). L /

11 For any closed surface that encloses a total charge Q in, the net electric flux through the closed surface is Q. in 0 e Model: The electric field over the five surfaces is uniform. Please refer to Figure in the problem. Solve: The electric flux through a surface area A is e EA EAcos where is the angle between the electric field and a line perpendicular to the plane of the surface. The electric field is perpendicular to side and is parallel to sides, 3, and 5. Also the angle between E and A 4 is 60. The electric fluxes through these five surfaces are EA cos 400 N/C m 4 mcos N m /C EA cos N m /C EAcos 400 N/C m sin30 4 m cos N m /C Assess: Because the flux into these five faces is equal to the flux out of the five faces, the net flux is zero, as we found Solve: For any closed surface that encloses a total charge Q in, the net electric flux through the closed surface is e Q. in 0 The total flux through the cube is Q in e 600 N m /C Q in 600 N m /C8.850 C /N m C 5.3 nc Model: The excess charge on a conductor resides on the outer surface. Solve: (a) Consider a Gaussian surface surrounding the cavity just inside the conductor. The electric field inside a conductor in electrostatic equilibrium is zero, so E is zero at all points on the Gaussian surface. Thus

12 e 0. Gauss s law tells us that e Q in / 0, so the net charge enclosed by this Gaussian surface is Q Q Q in point wall 0. We know that Q point 00 nc, so Q wall 00 nc. The positive charge in the cavity attracts an equal negative charge to the inside surface. (b) The conductor started out neutral. If there is 00 nc on the wall of the cavity, then the exterior surface of the conductor was initially 00 nc. Transferring 50 nc to the conductor reduces the exterior surface charge by 50 nc, leaving it at 50 nc. Assess: The electric field inside the conductor stays zero Model: The charges are point charges. Please refer to the figure in the problem. Solve: The electric potential energy of the electron is U U electron 3 3 U 9.00 N m /C C.60 0 C.60 0 C.60 0 C.00 m m.00 m m J.0 J.40 J Model: The electric potential difference between the plates is determined by the uniform electric field in the parallel-plate capacitor. Solve: (a) The potential of an ordinary AA or AAA battery is.5 V. Actually, this is the potential difference between the two terminals of the battery. If the electric potential of the negative terminal is taken to be zero, then the positive terminal is at a potential of.5 V. (b) If a battery with a potential difference of.5 V is connected to a parallel-plate capacitor, the potential difference between the two capacitor plates is also.5 V. Thus, V C.5 V V V Ed where d is the separation between the two plates. The electric field inside a parallel-plate capacitor is Q.5 V A.5 V.00 m C /Nm d 0 Q Q E.5 V d 0 A0 A m C Thus, the battery moves C of electron charge from the positive to the negative plate of the capacitor.

13 9.36. Model: While the net potential is the sum of the scalar potentials due to each charge, the net electric field is the vector sum of the electric fields. Solve: (a) Let the two charges be labeled as Q.0 nc and Q.0 nc. As pictorially shown in the figure, the net electric field will not become zero anywhere except at infinity. We will now show this result mathematically. Let the point of zero electric field be at a distance x from the origin. Then, C.00 C Enet E E 0 N/C E E 4 x.0 cm 4 x.0 cm 0 0 (x.0 cm) (x.0 cm) Obviously, this equation can only be satisfied if x. That is, the electric field is zero at and at. (b) Because potential is a scalar quantity, the potential of charge Q is always negative and the potential of charge Q is always positive. These two scalar numbers will add to give zero potential at x 0 m and at. Mathematically this can be shown as follows. When.0 cm x.0 cm, let the point of zero potential be at x. Then, the condition V V 0 V/m is When x >.0 cm, C 0 C 4.0 cm x 4.0 cm x C 0 C 4 x.0 cm 4 x.0 cm This equation can only be satisfied at x. (c) The graphs are shown in the figure above. 0 (.0 cm x) (.0 cm x) 0 x 0 cm. 0 (x.0 cm) (x.0 cm) Model: Energy is conserved.

14 Solve: (a) The electric field inside a parallel-plate capacitor is constant with strength 3 50 V V E d 0.0 m (b) Assuming the initial velocity is zero, energy conservation yields U K U i f f 0 mv e Ed f e 6. 0 V/m C. 0 V/m 0.0 m m/s f 3 v 9.0 kg Assess: This speed is about 3% the speed of light. At that speed, relativity must be taken into account Model: The charged metal spheres are isolated and far from each other and anything else. Solve: (a) The charge on a sphere of radius r, charged to a potential V 0 is given by equation 9.33: The ratio of charge densities is Q 4 0rV 0 Q 4 0rV 0 Q A Q A r 4r r R Q Q A r 4r r R A (b) The electric field strengths at the surface of the conductors is E E and E, thus. 0 0 E Assess: Halving the radius doubles the electric field strength as well as the charge density. Sharp corners (i.e., parts of conductors with a small radius of curvature, like a small sphere) have strong electric fields. This characteristic is why lightning rods work Model: Capacitance is a geometric property of two electrodes. Solve: The ratio of the charge to the potential difference is called the capacitance: C Q V. C The potential difference across the capacitor is

15 Using R R.0 mm, Q Q Q VC 4 0 R 4 0 R 4 0 R R RR RR C R R R R.00 m 0 F RR 000 F.00 m 9.00 N m /C 9000 m R R.00 m 9000 m R.00 m R 9000 m m.00 m m R m.95 cm The outer radius is R R 0.00 m m 3.05 cm. So, the diameters are 5.9 cm and 6. cm Model: Assume the battery is ideal. The circuit in the Figure (see problem) has been redrawn to show that the six capacitors are arranged in three parallel combinations, each combination being a series combination of two capacitors. Solve: (a) The equivalent capacitance of the two capacitors in series is C. The equivalent capacitance of the six capacitors is 3 C. (b) As points a and b are midpoints of identical capacitors, V a V b 6.0 V. Therefore, the potential difference between points a and b is zero Model: Assume the battery is ideal. Solve: When the capacitors are individually charged, their charges are Q CV 0 F 0 V 00 C Q CV 0 F 0 V 00 C These two capacitors are then connected with the positive plate of C connected with the negative plate of C as shown in the figure. Let the new charges on C and C be Q and Q. Then, Q Q 00 C 00 C 00 C The voltages across parallel capacitors are the same, so Q Q C VV Q Q C C C

16 Substituting this expression for Q into the previous equation, C 0 F Q Q 00 C Q 00 C C 0 F Solving these equations, we get Q 33 C and Q 67 C. Finally, Q 33 C V 3.3 V V C 0 F Solve: (a) From Table 3., the resistivity of aluminum is m. From Equation 3.3, the length L of a wire with a cross-sectional area A and having a resistance R is 6 00 m 000 AR L 3.57 m 3.6 m 8.80 m (b) The number of turns is the length of the wire divided by the circumference of one turn. Thus, 3.57 m m Solve: (a) A current of.8 pa for the potassium ions means that a charge of.8 pc flows through the potassium ion channel per second. The number of potassium ions that pass through the ion channel per second is.80 C/s 9.60 C.50 s Since the channel opens only for.0 ms, the total number of potassium ions that pass through the channel is.5 0 s.0 0 s atoms. (b) The current density in the ion channel is.8 pa.80 A 7 I J.50 A/m A nm 0.50 m Solve: Equation 3.3 defines the current density as J I A. This means Assess: Fuse wires are usually thin A A D I D I cm 0.50 mm 4 J J 500 A/cm Solve: (a) The charge delivered is A 500 s.5 C. (b) The current in the lightning rod and the potential drop across it are related by Equation 3.. Using for iron from Table 3., m5.0 m500 3 A A LI 4 I V A.430 m L V 00 V This is the area required for a maximum voltage drop of 00 V. The corresponding diameter of the lightning rod is

17 A r m m 8.8 mm Please refer to the figure in the problem. Solve: (a) Only bulb A is in the circuit when the switch is open. The bulb s resistance R is in series with the internal resistance r, giving a total resistance R eq R r. The current is.50 V I E bat 0.3 A Rr 6.50 This is the current leaving the battery. But all of this current flows through bulb A, so I A I bat 0.3 A. (b) With the switch closed, bulbs A and B are in parallel with an equivalent resistance R eq R Their equivalent resistance is in series with the battery s internal resistance, so the current flowing from the battery is.50 V I E bat 0.48 A R r 3.50 eq But only half this current goes through bulb A, with the other half through bulb B, so IA I bat 0.4 A. (c) The change in I A when the switch is closed is 0.07 A. This is a decrease of 7.4%. (d) If r 0, the current when the switch is open would be I A I bat 0.50 A. With the switch closed, the current would be I bat A and the current through bulb A would be IA I bat 0.50 A. The current through A would not change when the switch is closed Model: The batteries and the connecting wires are ideal. The figure shows how to simplify the circuit in the figure in the problem using the laws of series and parallel resistances. Having reduced the circuit to a single equivalent resistance, we will reverse the procedure and build up the circuit using the loop law and the junction law to find the current and potential difference of each resistor. Solve: From the last circuit in the figure and from Kirchhoff s loop law, V 3 V I.0 A 9 Thus, the current through the batteries is.0 A. As we rebuild the circuit, we note that series resistors must have the same current I and that parallel resistors must have the same potential difference.

18 In Step of the above figure, the 9 resistor is returned to the 6 and 3 resistors in series. Both resistors must have the same.0 A current as the 6 resistor. We use Ohm s law to find V 3 (.0 A)(3 ) 3.0 V V V As a check, 3.0 V 6.0 V 9 V, which was V ( V 3 V) 9 V of the 9 resistor. In Step, the 6 resistor is returned to the 4 and 8 resistors in parallel. The two resistors must have the same potential difference V 6.0 V. From Ohm s law, 6.0 V 3 I7 A V I4 A 4 4 As a check, 0.75 A 0.5 A.0 A which was the current I of the 6 resistor. In Step 3, the 8 resistor is returned to the 3 and 5 (right) resistors in series, so the two resistors must have the same current of 0.88 A. We use Ohm s law to find V 3 (3/4 A)(3 ) 9/4 V V 4 (3/4 A)(5 ) 5/4 V As a check, 9/4 V 5/4 V 4/4 V = 6.0 V, which was V of the 8 resistor. In Step 4, the 3 resistor is returned to 4 (left) and resistors in parallel, so the two must have the same potential difference V 9/4 V. From Ohm s law, 9/4 V I4 9/6 A = 0.56 A 4 9/4 V 9 I A 0.9 A 48 As a check, 0.56 A 0.9 A 0.75 A, which was the same as the current through the 3 resistor. Resistor Potential difference (V) Current (A) Model: The wires are ideal, but the batteries are not.

19 Solve: (a) The good battery alone can drive a current through the starter motor V I 00 A (b) Alone, the dead battery drives a current 8.0 V I A (c) Let I, I, I 3 be defined as shown in the figure above. Kirchhoff s laws applied to the good battery and dead battery loop, good battery and starter motor loop, and the top middle junction yield three equations in the three unknown currents: V I0.0 I V I0.0 I V 0 I I I 3 Substituting for I from the third equation into the first and second equations gives V I0.0 I V I0.5 I Solving for I from the first equation, V I I 0.0 Substituting into the second equation and solving for I 3 yields the current through the starter motor is 99 A. (d) Substituting the value for I 3 into the expression for I yields the current through the dead battery is 3.9 A. Assess: The good battery is charging the dead battery as well as running the started motor. A total of 03 A flows through the good battery Model: The battery and the connecting wires are ideal. Please refer to the figure in the problem. Solve: (a) A very long time after the switch has closed, the potential difference V C across the capacitor is This is because the capacitor charges until V C E, while the charging current approaches zero. (b) The full charge of the capacitor is Q max C(V C ) max C E. (c) In this circuit, I dq dt because the capacitor is charging, that is, because the charge on the capacitor is increasing. (d) From Equation 3.36, capacitor charge at time t is Q Q max ( e t/ ). So, A graph of I as a function of t is shown below. dq d t t t E I CE e CE e CE e e dt dt RC R t E.

20 Model: Two parallel wires carrying currents in the same direction exert attractive magnetic forces on each other. Please refer to the figure in the problem. The current in the circuit on the left is I and has a clockwise direction. The current in the circuit on the right is I and has a counterclockwise direction. Solve: Since I 9 V 4.5 A, the force between the two wires is F N 0 7 T m/a0.0 m4.5 A LI I d m 9 V I 3.0 A R A I Model: Charged particles moving perpendicular to a uniform magnetic field undergo circular motion at constant speed. Please refer to the figure in the problem. Solve: The potential difference causes an ion of mass m to accelerate from rest to a speed v. Upon entering the magnetic field, the ion follows a circular trajectory with cyclotron radius r mv/eb. To be detected, an ion s trajectory must have radius d r 8 cm. This means the ion needs the speed ebr ebd v m m This speed was acquired by accelerating from potential V to potential 0. We can use the conservation of energy equation to find the voltage that will accelerate the ion: mv K U K U 0 J ev mv 0 J V e Using the above expression for v, the voltage that causes an ion to be detected is mv m ebd eb d V e e m 8m An ion s mass is the sum of the masses of the two atoms minus the mass of the missing electron. For example, the mass of is N m m N m N m elec (4.003 u)( kg/u) kg kg Note that we re given the atomic masses very accurately in Exercise 8. We need to retain this accuracy to tell the difference between and CO. The voltage for N is N C0.00 T 0.08 m kg V 0.07 V Ion Mass (kg) Accelerating voltage (V) N O CO Assess: The difference between N and CO is not large but is easily detectable Model: The loop will not rotate about the axle if the torque due to the magnetic force on the loop balances the torque of the weight. Please refer to Figure in the problem.

21 Solve: The rotational equilibrium condition net 0 N m is about the axle and means that the torque from the weight is equal and opposite to the torque from the magnetic force. We have kg g 0.05 m Bsin90 NIA kg 9.8 m/s 0.05 m B 0.3 T 0.0 A0.050 m0.00 m Assess: The current in the loop must be clockwise for the two torques to be equal. B Model: The bar is a current-carrying wire in a perpendicular uniform magnetic field. The current is constant. Please refer to figure in the problem. Solve: (a) The right-hand rule as described in section 33.8 requires the current to be into the page. (b) The net force on the bar is F IlB, and is constant throughout the motion. The acceleration of the bar is thus F IlB. Using constant acceleration kinematics, m m IlB vf v0 as 0 m/s d vf m IlBd m Model: Assume the wire is infinitely long. See Figure in the problem. The current through the wire produces a steady magnetic field, but the magnetic field strength depends on the distance from the wire. The loop moves away from the wire at constant speed. The flux through the loop varies due to its motion. Faraday s law gives the magnitude of the induced emf, and Ohm s law will yield the current strength. The direction of the magnetic field is into the paper at the location of the loop. The magnetic field is perpendicular to the plane of the loop, so B da BdA. Solve: The flux through the loop is decreasing as it moves away from the wire. Lenz s law implies that the induced current is clockwise in order to increase the flux. Using the results of Example 34.5, which treats the rectangular loop as a series of tall infinitesimally thin strips, 0I(4.0 cm) x.0 cm m ln x where x is the distance from the wire to the closer edge of the loop, and changes with time. Note dm dm dx dm dx v, where v is the velocity of the loop. dt dx dt dx dt The induced emf is thus At x =.0 cm, dm dx 0 I(4.0 cm).0 cm E v dx dt x x.0 cm I.0 cm 0 6 E 4.0 cm 0 m/s.30 V.0 cm 3.0 cm Since the loop resistance is R = 0.00, the induced current is thus E Iloop R Assess: The induced emf is proportional to the current in the wire. For reasonable currents (~ A) the induced emf is reasonable. 5 I Model: Assume the magnetic field is uniform over the plane of the loop. The oscillating magnetic field strength produces a changing flux through the loop and an induced emf in the loop.

22 Solve: (a) The normal to the surface of the loop is in the same direction as the magnetic field so that ABBA. The induced emf is d db db E A r = r B0 cost dt dt dt The cosine will oscillate between and so the maximum emf is E r B r f B Hz 00 T 0.93 V 6 9 max 0 0 (b) If the loop is rotated so that the plane is perpendicular to the electric field, then the normal to the surface will be parallel to the magnetic field. There is no magnetic flux through the loop and no induced emf Model: Assume negligible resistance in the LC part of the circuit. With the switch in position for a long time the capacitor is fully charged. After moving the switch to position, there will be oscillations in the LC part of the circuit. Solve: (a) After a long time the potential across the capacitor will be that of the battery and Q0 CVb att. When the switch is moved, the capacitor will discharge through the inductor and LC oscillations will begin. The maximum current is H CVbatt C.0 0 F 0 0 batt batt 3 V A 76 ma I Q C V V LC L (b) The current will be a maximum one-quarter cycle after the maximum charge. The period is 3 6 T LC 500 H.00 F.0 ms So the current is first maximum at tmax T 0.50 ms Model: Assume an ideal inductor and an ideal (resistanceless) battery. Please refer to the figure in the problem. Solve: (a) Because the switch has been open a long time, no current is flowing the instant before the switch is closed. A basic property of an ideal inductor is that the current through it cannot change instantaneously. This is because the potential difference V L L(dI/dt) would become infinite for an instantaneous change of current, and that is not physically possible. Because the current through the inductor was zero before the switch was closed, it must still be zero (or very close to it) immediately after the switch is closed. Consequently, the inductor has no effect on the circuit. It is simply a 0 resistor and 0 resistor in series with the battery. The equivalent resistance is 30, so the current through the circuit (including through the 0 resistor) is I V bat /R eq (30 V)/(30 ).0 A. (b) After a long time, the currents in the circuit will reach steady values and no longer change. With steady currents, the potential difference across the inductor is V L L(dI/dt) 0 V. An ideal inductor has no resistance (R 0 ), so the inductor simply acts like a wire. In this case, the inductor shorts out the 0 resistor. All current from the 0 resistor flows through the resistanceless inductor, so the current through the 0 resistor is 0 A. (c) When the switch has been closed a long time, and the inductor is shorting out the 0 resistor, the current passing through the 0 resistor and through the inductor is I (30 V)/(0 ) 3.0 A. Because the current through an inductor cannot change instantaneously, the current must remain 3.0 A immediately after the switch reopens. This current must go somewhere (conservation of current), but now the open switch prevents the current from going back to the battery. Instead, it must flow upward through the 0 resistor. That is, the current flows around the LR circuit consisting of the 0 resistor and the inductor. This current will decay with time, with time constant L/R, but immediately after the switch reopens the current is 3.0 A Model: Use the Galilean transformation of fields.

23 A current of.5 A flows to the right through the wire, and the plastic insulation has a charge of linear density.5 n C/cm. Solve: The magnetic field B at a distance r from the wire is On the other hand, E is radially out along 0 I B, clockwise seen from left r ˆr, that is, E rˆ r As the mosquito is.0 cm from the center of the wire at the top of the wire, T m/a.5 A 0.00 m B T.5 0 C/m N m /C 5 V E m m where the direction of B is out of the page and the direction of E is radially outward. In the mosquito s frame (let us call it S), we want B 0 T. Thus, B B V E 0 B V E c c Because V E must be in the direction of B and E is radially outward, according to the right-hand rule V must be along the direction of the current. The magnitude of the velocity is cb m/s T V 5 E 4.50 V/m m/s The mosquito must fly at m/s parallel to the current. This is highly unlikely to happen unless the mosquito is from Planet Krypton, like Superman Model: The microwave beam is an electromagnetic wave. The water does not lose heat during the process. Solve: The rate of energy transfer from the beam to the cube is C 0 P IA E0 A 8 30 m/s C /Nm V/m 0.0 m.9 kw The amount of energy required to raise the temperature by 50 C is E mct 0.0 m 000 kg/m 486 J/kg/ C 50 C.09 0 J The time required for the water to absorb this much energy from the microwave beam is

24 5 E.090 J t 6 s 3 P.90 W Assess: Raising kg of water by 50 C in a microwave oven takes around 3 minutes, so this is reasonable Model: Use the particle model for the astronaut. Solve: According to Newton s third law, the force of the radiation on the astronaut is equal to the momentum delivered by the radiation. For this force we have P 000 W F prad A c m/s Using Newton s second law, the acceleration of the astronaut is N N m/s a 80 kg Using v f v i a(t f t i ) and a time equal to the lifetime of the batteries, v f 0 m/s ( m/s )(3600 s) m/s The distance traveled in the first hour is calculated as follows: v v a s f i first hour ( m/s) (0 m/s) ( m/s )(s) first hour (s) first hour 0.70 m This means the astronaut must cover a distance of 5.0 m 0.7 m 4.73 m in a time of 9 hours. The acceleration is zero during this time. The time it will take the astronaut to reach the space capsule is 4.73 m t 3,533 s 8.76 hours m/s Because this time is less than 9 hours, the astronaut is able to make it safely to the space capsule Model: Use Malus s law for the polarized light. Solve: For unpolarized light, the electric field vector varies randomly through all possible values of. Because the average value of cos is, the intensity transmitted by a polarizing filter when the incident light is unpolarized is I I 0. For polarized light, I transmitted I 0 cos. Therefore, I I cos 45 I 3 I cos 45 I 3 (I cos 45 )cos 45 I 0(cos 4 45 ) I Model: Assume the spacecraft is an inertial reference frame. Solve: Light travels at speed c in all inertial reference frames, regardless of how the reference frames are moving with respect to the light source. Relative to the spacecraft, the starlight is approaching at the speed of light c m/s.

25 37.9. Model: The clocks are in the same reference frame. Solve: The speed of light is c 300 m/ s 0.30 m/ns. The distance from the origin to the point (x, y, z) (30 m, 40 m, 0 m) is 30 m 40 m 50 m. The clock should be preset to 67 ns. So, the time taken by the light to travel 50 m is 50 m 67 ns 0.30 m/ns Model: Bjorn and firecrackers and are in the same reference frame. Light from both firecrackers travels towards Bjorn at 300 m/s. Solve: Bjorn is 600 m from the origin. Light with a speed of 300 m/s takes.0 s to reach Bjorn. Since this flash reaches Bjorn at t 3.0 s, it left firecracker at t.0 s. The flash from firecracker takes.0 s to reach Bjorn. So, the light left firecracker at t.0 s. Note that the two events are not simultaneous although Bjorn sees the events as occurring at the same time Model: Let S be the earth s reference frame and S the rocket s reference frame. Solve: (a) The astronauts measure proper time i t. Thus (b) In frame S, the distance of the distant star is 0 y t 0 y v0.9965c vc vc xvt c 60 y ly y 60 y 59.8 ly