ch28 tủ tài liệu training

1 (a) Eq 28-3 leads to 6.50 × 10−17 N FB v= = = 4.00 × 105 m s −19 −3 eB sin φ 160 × 10 C 2.60 × 10 T sin 23.0° c hc h (b) The kinetic energy of the proton is K= 2 mv = 167 × 10−27 kg 4.00 × 105 m s = 134 × 10−16 J 2 c hc This is (1.34 × 10– 16 J) / (1.60 × 10– 19 J/eV) = 835 eV h (a) We use Eq 28-3: FB = |q| vB sin φ = (+ 3.2 × 10–19 C) (550 m/s) (0.045 T) (sin 52°) = 6.2 × 10–18 N (b) a = FB/m = (6.2 × 10– 18 N) / (6.6 × 10– 27 kg) = 9.5 × 108 m/s2 G G (c) Since it is perpendicular to v , FB does not any work on the particle Thus from the work-energy theorem both the kinetic energy and the speed of the particle remain unchanged (a) The force on the electron is G G G G FB = qv × B = q vx ˆi + v y ˆj × Bx ˆi + By j = q ( vx By − v y Bx ) kˆ ( ( ( )( ) ) ( ) ) ( ) = 1.6 ì 1019 C êơ 2.0 ì 106 m s ( −0.15 T ) − 3.0 ×106 m s ( 0.030 T ) ẳ = 6.2 ì1014 N k G G Thus, the magnitude of FB is 6.2 × 1014 N, and FB points in the positive z direction (b) This amounts G to repeating the above computation with a change in the sign in the charge Thus, FB has the same magnitude but points in the negative z direction, namely, G ˆ F = − 6.2 ×10−14 N k B ( ) The magnetic force on the proton is → → → F = qv×B where q = +e Using Eq 3-30 this becomes (4 × 10−17 )i + (2 × 10−17)j = e[(0.03vy + 40)i + (20 – 0.03vx)j – (0.02vx + 0.01vy)k] ^ ^ ^ ^ with SI units understood Equating corresponding components, we find (a) vx = −3.5×103 m/s, and (b) vy = 7.0×103 m/s ^ Using Eq 28-2 and Eq 3-30, we obtain G F = q v x By − v y Bx k = q v x 3Bx − v y Bx k d i d b g i where we use the fact that By = 3Bx Since the force (at the instant considered) is Fz k where Fz = 6.4 × 10–19 N, then we are led to the condition d i q 3v x − v y Bx = Fz Ÿ Bx = Fz q 3v x − v y d i Substituting Vx = 2.0 m/s, vy = 4.0 m/s and q = –1.6 × 10–19 C, we obtain Bx = –2.0 T G G G G Letting F = q E + v × B = , we get vB sin φ = E We note that (for given values of d i the fields) this gives a minimum value for speed whenever the sin φ factor is at its maximum value (which is 1, corresponding to φ = 90°) So vmin = E / B = (1.50 × 103 V / m) / (0.400 T) = 3.75 × 103 m / s Straight line motion will result from zero net force acting on the system; we ignore G G G G G G G G gravity Thus, F = q E + v × B = Note that v ⊥B so v × B = vB Thus, obtaining the d i speed from the formula for kinetic energy, we obtain 100 V ( 20 ×10−3 m ) E E B= = = = 2.67 ×10−4 T − 19 − 31 v K / me (1.0 ×10 V ) (1.60 ×10 C ) / ( 9.11×10 kg ) G In unit-vector notation, B = −(2.67 ×10−4 T)kˆ G G G G G G We apply F = q E + v × B = me a to solve for E : d i G me aG G G + B×v E= q c9.11 × 10 = −31 kg 2.00 × 1012 m s i hd −19 −160 × 10 C  i − 6.00j + 4.80k V m = −114 e j i + b400àTgi ì b12.0 km sgj + b15.0 km sgk G G G G G Since the total force given by F = e E + v × B vanishes, the electric field E must be G G perpendicular to both the particle velocity v Gand the magnetic field B The magnetic G field is perpendicular to the velocity, so v × B has magnitude vB and the magnitude of the electric field is given by E = vB Since the particle has charge e and is accelerated through a potential difference V, mv = eV and v = 2eV m Thus, d ( i )( ) 1.60 ×10−19 C 10 ×103 V 2eV E=B = (1.2 T ) = 6.8 ×105 V m −27 m 9.99 ×10 kg ( ) → → 10 (a) The force due to the electric field ( F = q E ) is distinguished from that associated → → → with the magnetic field ( F = q v × B ) in that the latter vanishes at the speed is zero and the former is independent of speed The graph (Fig.28-34) shows that the force (ycomponent) is negative at v = (specifically, its value is –2.0 × 10–19 N there) which (because q = –e) implies that the electric field points in the +y direction Its magnitude is E = (2.0 × 10–19)/(1.60 × 10–19) = 1.25 V/m (b) We are told that the x and z components of the force remain zero throughout the motion, implying that the electron continues to move along the x axis, even though magnetic forces generally cause the paths of charged particles to curve (Fig 28-11) The exception to this is discussed in section 28-3, where the forces due to the electric and magnetic fields cancel This implies (Eq 28-7) B = E/v = 2.50 × 10−2 T → → → → → → → For F = q v × B to be in the opposite direction of F = q E we must have v × B in the → opposite direction from E which points in the +y direction, as discussed in part (a) Since the velocity is in the +x direction, then (using the right-hand rule) we conclude that ^ ^ ^ the magnetic field must point in the +z direction ( i × k = −j ) In unit-vector notation, we G have B = (2.50 ×10−2 T)kˆ G 75 The current is in the + i direction Thus, the i component of B has no effect, and (with x in meters) we evaluate G F = ( 3.00A ) ³ ( −0.600 T m ) x dx ( ˆi × ˆj) = ă 1.80 13 A T m k = (0.600N)k. 2 Đ â ã 76 (a) The largest value of force occurs if the velocity vector is perpendicular to the field Using Eq 28-3, FB,max = |q| vB sin (90°) = ev B = (1.60 × 10– 19 C) (7.20 × 106 m/s) (83.0 × 10– T) = 9.56 × 10– 14 N (b) The smallest value occurs if they are parallel: FB,min = |q| vB sin (0) = G (c) G By Newton’s second law, a = FB/me = |q| vB sin θ /me, so the angle θ between v and B is θ = sin −1 F m a I = sin GH q vB JK e LM c9.11 × 10 kghd4.90 × 10 m s i × 10 Chc7.20 × 10 m shc83.0 × 10 MN c160 −31 −1 −16 14 −3 OP = 0.267° Th P Q G G G G 77 (a) We use = ì B, where points into the wall (since the current goes clockwise G around the clock) Since B points towards the one-hour (or “5-minute’’) mark, and (by G the properties of vector cross products) τ must be perpendicular to it, then (using the G right-hand rule) we find τ points at the 20-minute mark So the time interval is 20 (b) The torque is given by G G =| ì B |= µ B sin 90° = NiAB = πNir B = 6π ( 2.0A )( 0.15m ) ( 70 ×10−3 T ) = 5.9 ×10−2 N ⋅ m 78 From m = B2qx2/8V we have ∆m = (B2q/8V)(2x∆x) Here x = 8Vm B q , which we substitute into the expression for ∆m to obtain F B q IJ ∆m = G H 8V K mq 8mV ∆x = B ∆x Bq 2V Thus, the distance between the spots made on the photographic plate is ∆x = = ∆m 2V B mq ( 37 u − 35 u ) (1.66 ×10−27 kg = 8.2 × 10−3 m 0.50 T u ) ( 7.3 ×103 V ( 36 u ) (1.66 ×10−27 kg )( ) u 1.60 ×10−19 C ) 79 (a) Since K = qV we have K p = 12 Kα ( as qα = K p ) , or K p / Kα = 0.50 (b) Similarly, qα = K d , K d / Kα = 0.50 (c) Since r = 2mK qB ∝ mK q , we have rd = md K d q p rp = m p K p qd ( 2.00u ) K p r = 10 (1.00u ) K p p 2cm=14cm (d) Similarly, for the alpha particle, we have rα = mα Kα q p rp = m p K p qα ( 4.00u ) Kα erp = 10 (1.00u ) ( Kα ) 2e 2cm=14cm 80 (a) Equating the magnitude of the electric force (Fe = eE) with that of the magnetic force (Eq 28-3), we obtain B = E / v sin φ The field is smallest when the sin φ factor is at its largest value; that is, when φ = 90° Now, we use K = mv to find the speed: v= c hc h × 10−19 J eV 2.5 × 103 eV 160 2K = = 2.96 × 107 m s −31 me 9.11 × 10 kg Thus, B= E 10 × 103 V m = = 3.4 × 10−4 T v 2.96 × 107 m s The direction of the magnetic field must be perpendicular to both the electric field ( −ˆj ) G G and the velocity of the electron ( + ˆi ) Since the electric force Fe = (−e) E points in the + ˆj G G G direction, the magnetic force FB = (−e)v × B points in the −ˆj direction Hence, the G ˆ direction of the magnetic field is − kˆ In unit-vector notation, B = (−3.4 ×10−4 T)k 81 (a) In Chapter 27, the electric field (called EC in this problem) which “drives” the current through the resistive material is given by Eq 27-11, which (in magnitude) reads EC = ρJ Combining this with Eq 27-7, we obtain EC = ρnevd Now, regarding the Hall effect, we use Eq 28-10 to write E = vdB Dividing one equation by the other, we get E/Ec = B/neρ (b) Using the value of copper’s resistivity given in Chapter 27, we obtain E B 0.65 T = = = 2.84 ×10−3 −19 −8 28 Ec ne ρ 8.47 × 10 m 1.60 ×10 C 1.69 ×10 Ω ⋅ m ( )( )( ) 82 (a) For the magnetic field moving electrons, we need a nonG to have an effect on the G negligible component of B to be perpendicular to v (the electron velocity) It is most efficient, therefore, to orient the magnetic field so it is perpendicular to the plane of the page The magnetic force on an electron has magnitude FB = evB, and the acceleration of the electron has magnitude a = v2/r Newton’s second law yields evB = mev2/r, so the radius of the circle is given by r = mev/eB in agreement with Eq 28-16 The kinetic energy of the electron is K = 21 me v , so v = K me Thus, r= This must be less than d, so me K 2me K = eB me e2 B 2me K 2me K ≤ d , or B ≥ 2 e2d e B (b) If the electrons are to travel as shown in Fig 28-33, the magnetic field must be out of the page Then the magnetic force is toward the center of the circular path, as it must be (in order to make the circular motion possible) 83 The equation of motion for the proton is G G G F = qv × B = q v x i + v y j + vz k × B i = qB vz j − v y k e e j LF dv IJ i + FG dv IJ j + FG dv IJ k OP G = m a = m MG NH dt K H dt K H dt K Q y x p j z p Thus, dv y dvx dvz = 0, = ω vz , = −ω v y , dt dt dt where ω = eB/m The solution is vx = v0x, vy = v0y cos ωt and vz = –v0y sin ωt In summary, G we have v t = v0 x i + v0 y cos ωt j − v0 y sin ωt k bg b g b g 84 Referring to the solution of problem 19 part (b), we see that r = 2mK qB implies the proportionality: r ∞ mK qB Thus, (a) rd md K d q p 2.0u e = = = ≈ 1.4 , and 1.0u e rp m p K p qd (b) rα mα Kα q p 4.0u e = = = 1.0 1.0u 2e rp m p K p qα 85 (a) The textbook uses “geomagnetic north” to refer to Earth’s magnetic pole lying in the northern hemisphere Thus, the electrons are traveling northward The vertical G G component of the magnetic field is downward The right-hand rule indicates that v × B is G G G to the west, but since the electron is negatively charged (and F = qv × B ), the magnetic force on it is to the east We combine F = mea with F = evB sin φ Here, B sin φ represents the downward component of Earth’s field (given in the problem) Thus, a = evB / me Now, the electron speed can be found from its kinetic energy Since K = mv , c hc h × 10−19 J eV 12.0 × 103 eV 160 2K v= = = 6.49 × 107 m s −31 me 9.11 × 10 kg Therefore, −19 −6 evB (1.60 ×10 C ) ( 6.49 × 10 m s ) ( 55.0 ×10 T ) 2 a= = = 6.27 ×1014 m s ≈ 6.3 ×1014 m s −31 me 9.11×10 kg (b) We ignore any vertical deflection of the beam which might arise due to the horizontal component of Earth’s field Technically, then, the electron should follow a circular arc However, the deflection is so small that many of the technicalities of circular geometry may be ignored, and a calculation along the lines of projectile motion analysis (see Chapter 4) provides an adequate approximation: ∆x = vt Ÿ t = ∆x 0.200m = v 6.49 × 107 m s which yields a time of t = 3.08 × 10–9 s Then, with our y axis oriented eastward, ∆y = 2 at = ( 6.27 ×1014 ) ( 3.08 ×10−9 ) = 0.00298m ≈ 0.0030 m 2 86 We replace the current loop of arbitrary shape with an assembly of small adjacent rectangular loops filling the same area which was enclosed by the original loop (as nearly as possible) Each rectangular loop carries a current i flowing in the same sense as the original loop As the sizes of these rectangles shrink to infinitesimally small values, the assembly gives a current distribution G equivalent to that of the original loop The G magnitude of the torque ∆τ exerted by B on the nth rectangular loop of area ∆An is given by ∆τ n = NiB sin θ∆An Thus, for the whole assembly τ = ¦ ∆τ n = NiB ¦ ∆An = NiAB sin θ n n 87 The total magnetic force on the loop L is G G G G G FB = i v³ dL × B = i ( v³ dL ) × B = L We note that z L ( ) L G G dL = If B is not a constant, however, then the equality z L ddLG × BGi = (z L G is not necessarily valid, so FB is not always zero G G dL) × B G 88 (a) Since B is uniform, G FB = where we note that z wire z wire G G idL × B = i ez wire G G G G dL × B = iLab × B , j G G G dL = Lab , with Lab being the displacement vector from a to b G G G G (b) Now Lab = , so FB = iLab × B = 89 With Fz = vz = Bx = 0, Eq 28-2 (and Eq 3-30) gives ^ ^ ^ ^ ^ Fx i + Fy j = q ( vyBz i − vxBz j + vxBy k ) where q = −e for the electron The last term immediately implies By = 0, and either of the other two terms (along with the values stated in the problem, bearing in mind that “fN” means femtonewtons or 10−15 N) can be used to solve for Bz We therefore find that → the magnetic field is given by B = (0.75 T)k^