Course Hero has millions of student submitted documents similar to the one

below including study guides, practice problems, reference materials, practice exams, textbook help and tutor support.

Course Hero has millions of student submitted documents similar to the one below including study guides, practice problems, reference materials, practice exams, textbook help and tutor support.

The CHAPTER Magnetic Field
28
1* When a cathode-ray tube is placed horizontally in a magnetic field that is directed vertically upward, the electrons emitted from the cathode follow one of the dashed paths to the face of the tube in Figure 28-30. The correct path is ____. (a) 1 (b) 2 (c) 3 (d) 4 (e) 5 (b) 2 Why not define B to be in the direction of F, as we do for E? One cannot define the direction of the force by fiat. By experiment, F is perpendicular to B. 3 Find the magnetic force on a proton moving with velocity 4.46 Mm/s in the positive x direction in a magnetic field of 1.75 T in the positive z direction. Use Equ. 28-1 F = 1.25 pN i k = 1.25 pN j 4 A charge q = 3.64 nC moves with a velocity of 2.75106 m/s i. Find the force on the charge if the magnetic field is (a) B = 0.38 T j, (b) B = 0.75 T i + 0.75 T j, (c) B = 0.65 T i, (d) B = 0.75 T i + 0.75 T k . (a), (b), (c), (d) Use Equ. 28-1 (a) F = 3.8 mN ij = -3.8 mN k (b) F = 7.5 mN k (c) F = 0 (d) F = 7.5 mN j
5* A uniform magnetic field of magnitude 1.48 T is in the positive z direction. Find the force exerted by the field on a proton if the proton's velocity is (a) v = 2.7 Mm/s i, (b) v = 3.7 Mm/s j, (c) v = 6.8 Mm/s k , and (d) v = 4.0 Mm/s i + 3.0 Mm/s j. (a), (b), (c), (d) Use Equ. 28-1 (a) F = 0.639 pN ik = 0.639 pN j (b) F = 0.876 pN i (c) F = 0 (d) F = 0.71 pN i 0.947 pN j 6 An electron moves with a velocity of 2.75 Mm/s in the xy plane at an angle of 60 to the x axis and 30 to the y axis. A magnetic field of 0.85 T is in the positive y direction. Find the force on the electron. o o Use Equ. 28-1 F = 0.44(cos 60 i + cos 30 j)0.85 j pN = 0.187 pN k
7
A straight wire segment 2 m long makes an angle of 30 with a uniform magnetic field of 0.37 T. Find the magnitude of the force on the wire if it carries a current of 2.6 A. F = BI l sin ? = 0.962 N Use Equ. 28-4 A straight wire segment I l = (2.7 A)(3 cm i + 4 cm j) is in a uniform magnetic field B = 1.3 T i. Find the
8
Chapter 28 force on the wire. Use Equ. 28-4
The Magnetic Field
F = 0.140 N k
9* What is the force (magnitude and direction) on an electron with velocity v = (2i 3j)106 m/s in a magnetic field B = (0.8i + 0.6j 0.4k ) T? Use Equ. 28-4 F = 0.192 pN i 0.128 pN j 0.576 pN k ; F = 0.621 pN 10 The wire segment in Figure 28-31 carries a current of 1.8 A from a to b. There is a magnetic field B = 1.2 T k . Find the total force on the wire and show that it is the same as if the wire were a straight segment from a to b. Use Equ. 28-4 F = 0.0684 N j + 0.0864 N i; l = 3 cm i + 4 cm j If the wire is straight from a to b, F = I l B = 0.0864 N i 0.0684 N j 11 A straight, stiff, horizontal wire of length 25 cm and mass 50 g is connected to a source of emf by light, flexible leads. A magnetic field of 1.33 T is horizontal and perpendicular to the wire. Find the current necessary to float the wire, that is, the current such that the magnetic force balances the weight of the wire. I = mg/ l B = 1.48 A F = (I l B mg) k = 0 12 A simple gaussmeter for measuring horizontal magnetic fields consists of a stiff 50-cm wire that hangs from a conducting pivot so that its free end makes contact with a pool of mercury in a dish below. The mercury provides an electrical contact without constraining the movement of the wire. The wire has a mass of 5 g and conducts a current downward. (a) What is the equilibrium angular displacement of the wire from vertical if the horizontal magnetic field is 0.04 T and the current is 0.20 A? (b) If the current is 20 A and a displacement from vertical of 0.5 mm can be detected for the free end, what is the horizontal magnetic field sensitivity of this gaussmeter? (a) At equilibrium, mg sin ? = I l B sin ? = 0.08154; ? = 4.68 (b) ? = 0.001 rad = sin ?; solve for B B = 4.91 T 13* A current-carrying wire is bent into a semicircular loop of radius R that lies in the xy plane. There is a uniform magnetic field B = Bk perpendicular to the plane of the loop (Figure 28-32). Show that the force acting on the loop is F = 2IRBj. With the current in the direction indicated and the magnetic field in the z direction, pointing out of the plane of the paper, the force is in the radial direction. On an element of length d l the force is dF = BIR d? with x and y components dFx = BIR cos ? d ? and dFy = BIR sin ? d ?. By symmetry, the x component of the force is zero.
Fy = BIR sin d = 2 IBR .
0
14 A 10-cm length of wire carries a current of 4.0 A in the positive z direction. The force on this wire due to a magnetic field B is F = (0.2i + 0.2j) N. If this wire is rotated so that the current flows in the positive x direction, the force on the wire is F = 0.2k N. Find the magnetic field B.
Chapter 28 1. Write the F = (0.2 i + 0.2 j) N in terms of B 2. Solve for Bx, By, and Bz 3. Repeat for F = 0.2 k N 4. Write B
The Magnetic Field (0.4 k )(Bx i + By j + Bz k ) = 0.2 i + 0.2 j By = 0.5 T, Bx = 0.5 T, Bz is undetermined (0.4 i)(Bx i + By j + Bz k ) = 0.2 k ; Bz = 0 B = (0.5 i + 0.5 j) T
15 A 10-cm length of wire carries a current of 2.0 A in the positive x direction. The force on this wire due to the presence of a magnetic field B is F = (3.0j + 2.0k ) N. If this wire is now rotated so that the current flows in the positive y direction, the force on the wire is F = (3.0i 2.0k ) N. Determine the magnetic field B. 1. Write F = (3.0 j + 2.0 k ) N in terms of B (0.2 i)(Bx i + By j + Bz k ) = 3 j + 2 k 2. Solve for the components of B Bx is undetermined; By = 10 T, Bz = 15 T 3. Repeat for F = (3.0 i 2.0 k ) N (0.2 j)(Bx i + By j + Bz k ) = 3.0 i 2.0 k ; Bx = 10 T, Bz = 15 T 4. Write B B = (10 i + 10 j 15 k ) T 16 A wire bent in some arbitrary shape carries a current I in a uniform magnetic field B. Show explicitly that the total force on the part of the wire from some point a to some point b is F = I l B, where l is the vector from a to b. From Equ. 28-5 we have F = dF = Idl B . But I and B are constant.Thus, F = I l B , where l = dl
a a
a
b
b
b
is just the vector from a to b. 17* True or false: The magnetic force does not accelerate a particle because the force is perpendicular to the velocity of the particle. False 18 A moving charged particle enters a region in which it is suddenly deflected perpendic ular to its motion. How can you tell if the deflection was caused by a magnetic field or an electric field? 1. Determine if the deflecting force depends on the particle's speed. If so, it is due to a magnetic field. 2. Examine the path of the particle. If it is circular, the deflection is due to a magnetic field; if it is parabolic, it is due to an electric field. 19 A proton moves in a circular orbit of radius 65 cm perpendicular to a uniform magnetic field of magnitude 0.75 T. (a) What is the period for this motion? (b) Find the speed of the proton. (c) Find the kinetic energy of the proton. T = (2p1.671027/1.610190.75) s = 87.4 ns (a) Use Equ. 28-7 (b) v = 2pr/T v = 4.67107 m/s (c) K = 1/2mv2 K = 1.821012 J = 11.4 MeV 20 An electron of kinetic energy 45 keV moves in a circular orbit perpendicular to a magnetic field of 0.325 T. (a) Find the radius of the orbit. (b) Find the frequency and period of the motion. 2Km r = 2.20 mm (a) Use Equ. 28-6 and v = 2K/m ; r =
qB
(b) Use Equs. 28-7 and 28-8
T = 0.11 ns; f = 1/T = 9.08 GHz
Chapter 28
The Magnetic Field
21* An electron from the sun with a speed of 1107 m/s enters the earth's magnetic field high above the equator where the magnetic field is 4107 T. The electron moves nearly in a circle except for a small drift along the direction of the earth's magnetic field that will take it toward the north pole. (a) What is the radius of the circular motion? (b) What is the radius of the circular motion near the north pole where the magnetic field is 2105 T? (a), (b) Use Equ. 28-6 (a) r = 142 m (b) r = 2.85 m 22 Protons and deuterons (each with charge +e) and alpha particles (with charge +2e) of the same kinetic energy enter a uniform magnetic field B that is perpendicular to their velocities. Let rp, rd, and ra be the radii of their circular orbits. Find the ratios rd/rp and ra /rp. Assume that ma = 2md = 4mp. r = 2Km /qB (see Problem 20). With K and B the same for the three particles, r m1/2/q. Consequently, rd/rp = 2 and ra/rp = 1. 23 A proton and an alpha particle move in a uniform magnetic field in circles of the same radii. Compare (a) their velocities, (b) their kinetic energies, and (c) their angular momenta. (See Problem 22.) (a) From Equ. 28-6, v = qBr/m. With r and B constant, v q/m. Consequently, va = vp/2. (b) K mv2 = q 2/m. Consequently, Ka = Kp. (c) L = mvr mv for constant r. Consequently, La = 2Lp. 1 2 mv 24 A particle of charge q and mass m has momentum p = mv and kinetic energy K = 2 = p2/2m. If the particle moves in a circular orbit of radius r perpendicular to a uniform magnetic field B, show that (a) p = Bqr and (b) K = B2q 2r2/2m. (a) The centripetal force is qBv and this must equal mv2/r. So qB = p/r and p = qBr. (b) K = p 2/2m = q 2B2r2/2m. 25* A beam of particles with velocity v enters a region of uniform magnetic field B that makes a small angle ? with v. Show that after a particle moves a distance 2p (m/qB)v cos ? measured along the direction of B, the velocity of the particle is in the same direction as it was when it entered the field. The particle's velocity has a component v1 parallel to B and a component v2 normal to B. v1 = v cos ? and is constant. v2 = v sin ? ; the magnetic force due to this velocity component is qBv 2 and results in a circular motion perpendicular to B. The period of that circular motion is given by Equ. 28-7. At the end of one period, v2 is the same as at the start of the period. In that time, the particle has moved a distance v1T = (v cos ?)(2pm/qB) in the direction of B. 26 A proton with velocity v = 107 m/s enters a region of uniform magnetic field B = 0.8 T, which is into the page, as shown in Figure 28-33. The angle ? = 60. Find the angle and the distance d.
Chapter 28
The Magnetic Field
The trajectory of the particle is shown. From symmetry, it is evident that the angle in Figure 28-33 equals the angle ? = 60. From the adjoining figure we see that d/2 = r sin 30, or d = r. From Equ. 28-6, r = d = 0.1305 m.
27 Suppose that in Figure 28-33 B = 0.6 T, the distance d = 0.4 m, and ? = 24. Find the speed v and the angle f if the particles are (a) protons and (b) deuterons. (a) r = d/(2 cos ?); = ? (see Problem 26) r = 0.2/cos 24 m = 0.219 m = mpvp/q pB Solve for vp vp = 1.26107 m/s; = 24 (b) Solve for vd; md = 2mp, q d = q p vd = 6.29106 m/s; = 24 28 A beam of positively charged particles passes undeflected from left to right through a velocity selector in which the electric field is up. The beam is then reversed so that it travels from right to left. Will the beam now be deflected in the velocity selector? If so, in which direction? Yes; it will be deflected up. 29* A velocity selector has a magnetic field of magnitude 0.28 T perpendicular to an electric field of magnitude 0.46 MV/m. (a) What must the speed of a particle be for it to pass through undeflected? What energy must (b) protons and (c) electrons have to pass through undeflected? (a) Use Equ. 28-9 v = (0.46106/0.28) m/s = 1.64106 m/s (b) K = 1/2mpv2 K = 2.251015 J = 14.1 keV (c) K = 1/2mev2 K = 7.68 eV 30 A beam of protons moves along the x axis in the positive x direction with a speed of 12.4 km/s through a region of crossed fields balanced for zero deflection. (a) If there is a magnetic field of magnitude 0.85 T in the positive y direction, find the magnitude and direction of the electric field. (b) Would electrons of the same velocity be deflected by these fields? If so, in what direction? (a) Use Equ. 28-9 E = vB = 12.41030.85 V/m = 10.5 kV/m q v B is in the z direction E = 10.5 kV/m k (b) For electrons, both FB and FE are reversed Electrons are not deflected 31 The plates of a Thomson q/m apparatus are 6.0 cm long and are separated by 1.2 cm. The end of the plates is 30.0 cm from the tube screen. The kinetic energy of the electrons is 2.8 keV. (a) If a potential of 25.0 V is
Chapter 28
The Magnetic Field
applied across the deflection plates, by how much will the beam deflect? (b) Find the magnitude of the crossed magnetic field that will allow the beam to pass through undeflected. v = 2K/m (a) 1. Find the speed of the electrons; v = 3.14107 m/s 2. Find E = V/d E = 2083 V/m 2 qEx1 x 2 qE x1 vx = 3.14107 m/s; vy = eEt/me = 6.99105 m/s y = + 2 3. (see Equ. 28-10) 2m v mv ?y = 7.35 mm (b) B = E/v B = 25/(0.0123.14107) T = 6.63105 T
32 Chlorine has two stable isotopes, 35Cl and 37Cl, whose natural abundances are about 76% and 24%, respectively. Singly ionized chlorine gas is to be separated into its isotopic components using a mass spectrometer. The magnetic field in the spectrometer is 1.2 T. What is the minimum value of the potential through which these ions must be accelerated so that the separation between them is 1.4 cm? 2m V 0.007 m = 1.20 10-4 ( 37 - 35 ) V From Equ. 2812, r = ; write ?r = ?s/2 q B2 Solve for ?V ?V = 122 kV
33* A singly ionized 24Mg ion (mass 3.9831026 kg) is accelerated through a 2.5-kV potential difference and deflected in a magnetic field of 557 G in a mass spectrometer. (a) Find the radius of curvature of the orbit for the ion. (b) What is the difference in radius for 26Mg and 24Mg ions? (Assume that their mass ratio is 26/24.) (a) r =
2 m V ; evaluate for 24Mg q B2
r24 = 63.5 cm ?r = 63.5( 26/24 1) cm = 2.59 cm
(b) r26 = r24 26/24 ; evaluate ?r = r26 r24
34 A beam of 6Li and 7Li ions passes through a velocity selector and enters a magnetic spectrometer. If the diameter of the orbit of the 6Li ions is 15 cm, what is the diameter of that for 7Li ions? For constant v, r m D7 = D6(7/6) = 17.5 cm 35 In Example 28-6, determine the time required for a 58Ni ion and a 60Ni ion to complete the semicircular path. 1. v = 2q V/m ; find v58 and v60 v58 = 9.96104 m/s; v60 = 9.79104 m/s t58 = 0.501p/9.96104 s = 15.8 s; t60 = 16.4 s 2. t = pr/v; find t58 and t60 36 Before entering a mass spectrometer, ions pass through a velocity selector consisting of parallel plates separated by 2.0 mm and having a potential difference of 160 V. The magnetic field between the plates is 0.42 T. The magnetic field in the mass spectrometer is 1.2 T. Find (a) the speed of the ions entering the mass spectrometer and (b) the difference in the diameters of the orbits of singly ionized 238U and 235U. (The mass of a 235 U ion is 3.9031025 kg.) (a) Use Equ. 28-9 v = 8104/0.42 m/s = 1.90105 m/s (b) r235 = m235v/qB; find ?D = 2r235(238/235 1) ?D = 9.89 mm
Chapter 28
The Magnetic Field
37* A cyclotron for accelerating protons has a magnetic field of 1.4 T and a radius of 0.7 m. (a) What is the cyclotron frequency? (b) Find the maximum energy of the protons when they emerge. (c) How will your answers change if deuterons, which have the same charge but twice the mass, are used instead of protons? (a) Use Equ. 28-8 f = 21.3 MHz (b) Use Equ. 28-13 K = 7.361012 J = 46 MeV (c) From Equs. 28-8 and 28-13, K and f 1/m f d = 10.7 MHz; Kd = 23 MeV 38 A certain cyclotron with magnetic field of 1.8 T is designed to accelerate protons to 25 MeV. (a) What is the cyclotron frequency? (b) What must the minimum radius of the magnet be to achieve a 25-MeV emergence energy? (c) If the alternating potential applied to the dees has a maximum value of 50 kV, how many revolutions must the protons make before emerging with an energy of 25 MeV? (a) Use Equ. 28-8 f = 27.4 MHz (b) From Equ. 28-13, r = 2 Km /qB ; find r r = 0.401 m; this is the minimum radius Number of revolutions = 25106/105 = 250 (c) Energy gain/revolution = 100 keV 39 Show that the cyclotron frequencies of deuterons and alpha particles are the same and are half that of a proton in the same magnetic field. (See Problem 22.) From Equ. 28-8, f q/m. Since q d = e and q a = 2e, and md = 1/2ma, q d/md = q a/ma. Consequently, f d = f a. Similarly, f d = f a = 1/2f p. 40 Show that the radius of the orbit of a charged particle in a cyclotron is proportional to the square root of the number of orbits completed. Since the energy gain per revolution is constant, K N, where N is the number of complete revolutions. The radius of an orbit is proportional to K1/2, so r N1/2. 41* What orientation of a current loop gives maximum torque? The normal to the plane of the loop should be perpendicular to B. 42 A small circular coil of 20 turns of wire lies in a uniform magnetic field of 0.5 T such that the normal to the plane of the coil makes an angle of 60 with the direction of B. The radius of the coil is 4 cm, and it carries a current of 3 A. (a) What is the magnitude of the magnetic moment of the coil? (b) What is the magnitude of the torque exerted on the coil? 2 (a) = NIA = 0.302 A.m (b) t = B sin ? t = 0.131 N.m 43 What is the maximum torque on a 400-turn circular coil of radius 0.75 cm that carries a current of 1.6 mA and resides in a uniform magnetic field of 0.25 T? 5 tmax = B; = NIA; tmax = NIAB tmax = 2.8310 N.m 44 A current-carrying wire is bent into the shape of a square of sides L = 6 cm and is placed in the xy plane. It carries a current I = 2.5 A. What is the torque on the wire if there is a uniform magnetic field of 0.3 T (a) in the z direction, and (b) in the x direction? (a) t = B; = IA k ; B = B k t=0 3 (b) B = B I t = 2.710 N.m j
Chapter 28 45*
The Magnetic Field
Repeat Problem 44 if the wire is bent into an equilateral triangle of sides 8 cm. t=0 3 t = 2.0810 N.m j
(a) t = B; = IA k ; B = B k (b) B = B I
46 A rigid, circular loop of radius R and mass M carries a current I and lies in the xy plane on a rough, flat table. There is a horizontal magnetic field of magnitude B. What is the minimum value of B such that one edge of the loop will lift off the table? Bmin = mg/pRI (see Example 28-9). 47 A rectangular, 50-turn coil has sides 6.0 and 8.0 cm long and carries a current of 1.75 A. It is oriented as shown in Figure 28-34 and pivoted about the z axis. (a) If the wire in the xy plane makes an angle ? = 37 with the y axis as shown, what angle does the unit normal n make with the x axis? (b) Write an expression for n in terms of the unit vectors i and j. (c) What is the magnetic moment of the coil? (d) Find the torque on the coil when there is a uniform magnetic field B = 1.5 T j. (e) Find the potential energy of the coil in this field. (a) From the figure it is evident that n makes an angle of 37 with the x axis. (b) n = cos 37 i sin 37 j = 0.8 i 0.6 j 2 2 (c) = NIA n = 0.336 A.m i 0.252 A.m j (d) t = B t = 0.504 N.m k (e) U = .B U = 0.378 J 48 The coil in Problem 47 is pivoted about the z axis and held at various positions in a uniform magnetic field B = 2.0 T j. Sketch the position of the coil and find the torque exerted when the unit normal is (a) n = i, (b) n = j, (c) n = j, and (d) n = ( i + j) / 2 . (a) The configuration of the coil is shown. Since B = 2.0 T j, t = B = B k = 0.84 N.m k
(b) The configuration of the coil is shown. Since B = 2.0 T j and = j, t = B = 0
Chapter 28 (c) The configuration of the coil is shown. As in part (b), t = 0.
The Magnetic Field
(d) The configuration of the coil is shown. In this case, t = B = 0.594 N.m k
49* The SI unit for the magnetic moment of a current loop is Am2. Use this to show that 1 T = 1 N/Am. Since [t] = [][B], [B] = [t]/[] = N.m/A.m2 = N/A.m 50 A small magnet of length 6.8 cm is placed at an angle of 60 to the direction of a uniform magnetic field of magnitude 0.04 T. The observed torque has a magnitude of 0.10 Nm. Find the magnetic moment of the magnet. 2 t = B sin ?; = t/(B sin ?) = 2.89 A.m 51 A wire loop consists of two semicircles connected by straight segments (Figure 28-35). The inner and outer radii are 0.3 and 0.5 m, respectively. A current of 1.5 A flows in this loop with the current in the outer semicircle in the clockwise direction. What is the magnetic moment of this current loop? 2 2 2 2 = IA; use righthand rule to determine n = 1.5p(0.5 0.3 )/2 A.m = 0.377 A.m ; points into the paper. 52 A wire of length L is wound into a circular coil of N loops. Show that when this coil carries a current I, its magnetic moment has the magnitude IL2/4pN. The circumference of each loop is L/N = 2pR and the area of each loop is pR2 = L2/4pN2. The magnetic moment is = NIA = IL2/4pN. 53* A particle of charge q and mass m moves in a circle of radius r and with angular velocity ?. (a) Show that the average current is I = q ?/2p and that the magnetic moment has the magnitude = 1 q r 2 . (b) Show that 2 the angular momentum of this particle has the magnitude L = mr2? and that the magnetic moment and angular momentum vectors are related by = (q/2m)L. (a) I = ?q/?t = q/T = qf = q ?/2p. = IA = (q ?/2p)(p r2) = q ? r2/2 (b) The moment of inertia of the particle is mr2, and so L = mr2?. Both and L point in the direction ?; so = (q/2m)L. 54 A single loop of wire is placed around the circumference of a rectangular piece of cardboard whose length and width are 70 and 20 cm, respectively. The cardboard is now folded along a line perpendicular to its length
Chapter 28
The Magnetic Field
and midway between the two ends so that the two planes formed by the folded cardboard make an angle of 90. If the wire loop carries a current of 0.2 A, what is the magnitude of the magnetic moment of this system? The two parts can be considered as two loops; each loop carries a current I, and at the fold line the two loop currents cancel. The magnetic moments of the two loops make an angle of 90 with one another. 2 1. Find the magnetic moment of each loop = 0.014 A.m 2 2 2. Add the two moments vectorially tot = 0.014 2 A.m = 0.0198 A.m 55 Repeat 54 Problem if the line along which the cardboard is folded is 40 cm from one end. The two parts can be considered as two loops; each loop carries a current I, and at the fold line the two loop currents cancel. The magnetic moments of the two loops make an angle of 90 with one another. 2 2 1. Find the magnetic moment of each loop 1 = 0.016 A.m ; 2 = 0.012 A.m 2 2 2 2 2. Add the two moments vectorially tot = (0.016 + 0.012 )1/2 A.m = 0.020 A.m 56 A hollow cylinder has length L and inner and outer radii Ri and Ro, respectively (Figure 28-36). The cylinder carries a uniform charge density ?. Derive an expression for the magnetic moment as a function of ?, the angular velocity of rotation of the cylinder about its axis. Consider an element of charge dq in a cylinder of length L, radius r, and thickness dr. We have dq = 2pL? r dr. The element of current due to this rotating charge is dI = ? dq/2p = L?? r dr, and the corresponding element of magnetic moment is d = A dI, where A = pr2. We now integrate:
= L r 3 dr =
Ri Ro
L 4 Ro - Ri4 4
(
)
57* A nonconducting rod of mass M and length l has a uniform charge per unit length ? and rotates with angular velocity ? about an axis through one end and perpendicular to the rod. (a) Consider a small segment of the rod of length dx and charge dq = ? dx at a distance x from the pivot (Figure 28-37). Show that the magnetic moment of this segment is 1 ??x2 dx. (b) Integrate your result to show that the total magnetic moment of the 2 1 3 rod is = 6 ?? l . (c) Show that the magnetic moment and angular momentum L are related by = (Q/2M)L, where Q is the total charge on the rod. (a) The area enclosed by the rotating element of charge is px2. The time required for one revolution is 1/f = 2p/?. The average current element is then dI = ? dx?/2p and d = A dI = 1/2?? x2dx. l 1 1 2 3 (b) = x dx = l 2 6 0 (c) The angular momentum L = I?, where I is the moment of inertia of the rod, I = (1/3)M l 2 . The total charge carried by the rod is Q = ? l . Thus = (Q/2M)L. Moreover, since ? and L = I? point in the same direction, = (Q/2M)L. 58 A nonuniform, nonconducting disk of mass M, radius R, and total charge Q has a surface charge density s = s0r/R and a mass per unit area sm = (M/Q)s. The disk rotates with angular velocity ? about its axis. (a) Show 3 that the magnetic moment of the disk has a magnitude = 1 p?s0R 4 = 10 Q?R2. (b) Show that the magnetic 5 moment and angular momentum L are related by = (Q/2M)L. (a) We are given that s = s0r/R and sm = (M/Q)s. An element of magnetic moment is then given by d = A dI = pr2s (?/2p)(2pr) dr = (p?s0/R)r4dr. Integrating from r = 0 to r = R one obtains = (1/5) p?s0R4.
Chapter 28
The Magnetic Field
The total charge is the integral of 2prs dr from r = 0 to r = R, where s = s0r/R. Thus Q = 2pR2s0/3, and 2 2 s0 = 3Q/2pR . Substituting this expression for s0 into the result for one obtains = 3Q?R /10. (b) The moment of inertia of the disk with a surface mass density sm is given by R 2M 0 4 I = dI = (M 0 / Q )(r / R) r 2 ( 2r )dr = R = (3/5)MR2 and L = I? = (3/5)MR2?. Since also points 5Q 0 in the direction of ? we see that = (Q/2M)L. 59 A spherical shell of radius R carries a surface charge density s. The sphere rotates about its diameter with angular velocity ?. Find the magnetic moment of the rotating sphere. For the shell, Q = 4pR2s. The angular momentum of a shell of mass M for rotation about a diameter is given by 2 4 L= I? = (2/3)MR ?. Applying the general expression = (Q/2M)L one finds = (4p/3)sR ?. 60 A solid sphere of radius R carries a uniform volume charge density ?. The sphere rotates about its diameter with angular velocity ?. Find the magnetic moment of this rotating sphere. For the sphere, Q = (4p/3)? R3 and I = (2/5)MR2. Applying the general result (Q/2M)I? , = (4p/15) ? R5?. 61* A solid cylinder of radius R and length L carries a uniform charge density +? between r = 0 and r = Rs and an equal charge density of opposite sign, ?, between r = Rs and r = R. What must be the radius Rs so that on rotation of the cylinder about its axis the magnetic moment is zero? For the solid cylinder of radius Rs, Q+ = p? Rs2L and L = I? = 1/2MRs2?. Hence + = (Q+/2M)L = p?LRs4?/4. For the cylindrical shell, Q = p?L(R2 Rs2) and L = I? = 1/2M(Rs2 + R2)?. Hence = p?L(R4 Rs4)?/4. Setting + + = 0 and solving for Rs one obtains Rs = R/21/4 = 0.841R. 62 A solid cylinder of radius R and length L carries a uniform charge density ? = ?0 between r = 0 and r = 1 1 2 R and a positive charge density of equal magnitude, +?0, between r = 2 R and r = R (Figure 28-38). The cylinder rotates about its axis with angular velocity ?. Derive an expression for the magnetic moment of the cylinder. Here we can use the results of the previous problem. We now set Rs = R/2 in the expressions for + and , changing the signs of the magnetic moments since ? = ?0 for the inner cylinder and ? = +?0 for the cylindrical shell. The total magnetic moment is then given by = (p?0L?/4)[R4 R4/16 R4/16] = 7p?0LR4?/32. 63 A cylindrical shell of length L with inner radius Ri and outer radius Ro carries a uniform charge density, +?0, between Ri and radius Rs and an equal charge density of opposite sign, ?0, between Rs and Ro. The cylinder rotates about its axis with angular velocity ?. Derive an expression for the magnetic moment of this cylinder. For the inner cylindrical shell (see Problem 61), i = p?0L(Rs4 Ri4)?/4, while for the outer shell, 4 4 4 4 4 o = p?0L(Ro Rs )?/4. The total magnetic moment is i + o = (p?0L?/4)(Ro + Ri 2Rs ). 64 A solid sphere of radius R carries a uniform charge density, +?0, between r = 0 and r = Rs and an equal charge density of opposite sign, ?0, between r = Rs and r = R. The sphere rotates about its diameter with angular velocity ?. Find Rs such that magnetic moment of the sphere is zero. What is the net charge carried by the sphere? For the inner sphere of radius Rs, Qi = 4p?0Rs3/3 and L = 2MRs2?/5. Hence i = 4p?0Rs5?/15. For the outer spherical shell, the charge is Qo = (4p?0/3)(R3 Rs3). If ? is the mass density, then M o = (4p?/3)(R3 Rs3) and the moment of inertia of the spherical shell is I = (2/5)(4p?/3)(R5 Rs5). Using the general result = (Q/2M)L one obtains o = (4p?0/15)(R5 Rs5)?. Setting i + o = 0 and solving for Rs one finds Rs = R/21/5 = 0.871R. The net charge carried by the sphere is Q = Qi + Qo, where Qi and Qo are given above. Setting Rs = R/21/5 one
Chapter 28
The Magnetic Field
obtains Q = 22/5(4p?0/3)R3. 65* A solid sphere of radius R carries a uniform charge density, +?0, between r = 0 and r = charge density of opposite sign, ?0, between r =
1 2
1 2
R and an equal
0R and r = R. The sphere rotates about its diameter with
angular velocity ?. Derive an expression for the magnetic moment of this rotating sphere. For the inner sphere of radius Ri, Qi = 4p?0Ri3/3 and L = 2MRi2?/5. Hence i = 4p?0Ri5?/15. For the outer spherical shell, the charge is Qo = (4p?0/3)(R3 Ri3). If ? is the mass density, then M o = (4p?/3)(R3 Ri3) and the moment of inertia of the spherical shell is I = (2/5)(4p?/3)(R5 Ri5). Using the general result = (Q/2M)L one obtains o = (4p?0/15)(R5 Ri5)?. We now set Ri = R/2 and = i + o and obtain = p?0R5?/4. 66 A metal strip 2.0 cm wide and 0.1 cm thick carries a current of 20 A in a uniform magnetic field of 2.0 T, as shown in Figure 28-39. The Hall voltage is measured to be 4.27 V. (a) Calculate the drift velocity of the electrons in the strip. (b) Find the number density of the charge carriers in the strip.(c) Is point a or b at the higher potential? (a) From Equ. 28-17, vd = VH/Bw vd = 0.107 mm/s (b) Use Equ. 28-18 n = 5.851028 m3 (c) E = vdB; vd directed opposite to I E points from b to a; Va > Vb 67 The number density of free electrons in copper is 8.471022 electrons per cubic centimeter. If the metal strip in Figure 28-39 is copper and the current is 10 A, find (a) the drift velocity vd and (b) the Hall voltage. (Assume that the magnetic field is 2.0 T.) (a) vd = I/wten vd = 3.69105 m/s VH = 1.48 V (b) VH = vdBw 68 A copper strip (n = 8.471022 electrons per cubic centimeter) 2 cm wide and 0.1 cm thick is used to measure the magnitudes of unknown magnetic fields that are perpendicular to the strip. Find the magnitude of B when I = 20 A and the Hall voltage is (a) 2.00 V, (b) 5.25 V, and (c) 8.00 V. (a), (b), (c) B = VHnte/I (a) B = 1.36 T (b) B = 3.56 T (c) B = 5.42 T 69* Because blood contains charged ions, moving blood develops a Hall voltage across the diameter of an artery. A large artery with a diameter of 0.85 cm has a flow speed of 0.6 m/s. If a section of this artery is in a magnetic field of 0.2 T, what is the potential difference across the diameter of the artery? VH = vdBw VH = 1.02 mV 70 The Hall coefficient R is defined as R = E y / J x B z , where Jx is the current per unit area in the x direction in
the slab, Bz is the magnetic field in the z direction, and Ey is the resulting Hall field in the y direction. Show that the Hall coefficient is 1/nq, where q is the charge of the charge carriers, 1.61019 C if they are electrons. (The Hall coefficients of monovalent metals, such as copper, silver, and sodium, are therefore negative.) VH = EHw and J = I/wt. Using the defintion of R we have R = VHt/IB = (IB/ntq)t/IB = 1/nq. The direction of E is vdB, where vd is in the direction of I if q is positive. It follows that if J is in the x direction and B points in the z direction, then EH is in the y direction for q positive. If q is negative, vd is reversed, and so is EH. 71 Aluminum has a density of 2.7103 kg/m3 and a molar mass of 27 g/mol. The Hall coefficient of aluminum is R = 0.31010 m3/C. (See Problem 70 for the definition of R.) Find the number of conduction electrons per
Chapter 28 aluminum atom. n = 1/Re
The Magnetic Field
n = 2.081029 m3; number of electrons/atom = 3.46
72 Magnesium is a divalent metal. Its density is 1.74103 kg/m3 and its molar mass is 24.3 g/mol. Assuming that each magnesium atom contributes two conduction electrons, what should be the Hall coefficient of magnesium? How does your result compare to the measured value of 0.941010 m3/C? 1. Find n n = 26.021029 1.74/24.3 m3 = 8.621028 m3 2. R = 1/ne R = 0.7251010 m3/C; thus n exp 1.5 electrons/atom 73* True or false: (a) The magnetic force on a moving charged particle is always perpendicular to the velocity of the particle. (b) The torque on a magnet tends to align the magnetic moment in the direction of the magnetic field. (c) A current loop in a uniform magnetic field behaves like a small magnet. (d) The period of a particle moving in a circle in a magnetic field is proportional to the radius of the circle. (e) The drift velocity of electrons in a wire can be determined from the Hall effect. (a) True (b) True (c) True (d) False (e) True 74 Show that the force on a current element is the same in direction and magnitude regardless of whether positive charges, negative charges, or a mixture of positive and negative charges create the current. From Equ. 28-4, the direction of the force does not depend on the sign of the charges that carry the current. 75 A proton with a charge +e is moving with a speed v at 50 to the direction of a magnetic field B. The component of the resulting force on the proton in the direction of B is (a) evB sin 50 cos 50. (b) evB cos 50. (c) zero. (d) evB sin 50. (e) none of these. (c) 76 If the magnetic field vector is directed toward the north and a positively charged particle is moving toward the east, what is the direction of the magnetic force on the particle? From Equ. 28-1, F points up. 77* A positively charged particle is moving northward in a magnetic field. The magnetic force on the particle is toward the northeast. What is the direction of the magnetic field? (a) Up (b) West (c) South (d) Down (e) This situation cannot exist. (e) 78 A 7Li nucleus with a charge of +3e and a mass of 7 u (1 u = 1.661027 kg) and a proton with charge +e and mass 1 u are both moving in a plane perpendicular to a magnetic field B. The two particles have the same momentum. The ratio of the radius of curvature of the path of the proton Rp to that of the 7Li nucleus, RLi is (a) Rp/RLi = 3. (b) Rp/RLi = 1/3. (c) Rp/RLi = 1/7. (d) Rp/RLi = 3/7. (e) none of these. (a) 79 An electron moving with velocity v to the right enters a region of uniform magnetic field that points out of the paper. After the electron enters this region, it will be (a) deflected out of the plane of the paper. (b) deflected into the plane of the paper. (c) deflected upward. (d) deflected downward. (e) undeviated in its motion.
Chapter 28
The Magnetic Field
(c) 80 How are magnetic field lines similar to electric field lines? How are they different? Magnetic field lines are similar to electric field lines in that their density is a measure of the strength of the field; the lines point in the direction of the field; also, magnetic field lines do not cross. They differ from electric field lines in that magnetic field lines must close on themselves (there are no isolated magnetic poles), and the force on a charge depends on the velocity of the charge and is perpendicular to the magnetic field lines. 81* A long wire parallel to the x axis carries a current of 6.5 A in the positive x direction. There is a uniform magnetic field B = 1.35 T j. Find the force per unit length on the wire. Use Equ. 28-4 F = 8.775 N/m k 82 An alpha particle (charge +2e) travels in a circular path of radius 0.5 m in a magnetic field of 1.0 T. Find (a) the period, (b) the speed, and (c) the kinetic energy (in electron volts) of the alpha particle. Take m = 6.6510 27 kg for the mass of the alpha particle. (a) T = 2pm/qB; q = 2e, ma = 6.651027 kg T = 0.131 s (b) v = 2pr/T v = 2.40107 m/s (c) K = 1/2mav2/e eV K = 12.0 MeV 83 If a current I in a given wire and a magnetic field B are known, the force F on the current is uniquely determined. Show that knowing F and I does not provide complete knowledge of B. If only F and I are known, one can only conclude that the magnetic field B is in the plane perpendicular to F. The specific direction of B is undetermined. 84 The pole strength q m of a bar magnet is defined by q m = ||/L, where L is the length of the magnet. Show that the torque exerted on a bar magnet in a uniform magnetic field B is the same as if a force + q m B is exerted on the north pole and a force - q m B is exerted on the south pole. The configuration of the magnet and field are shown in the figure. Then t = (Bq mL/2) sin ? + (Bq mL/2) sin ? = B sin ? = B, where = q mL.
85* A particle of mass m and charge q enters a region where there is a uniform magnetic field B along the x axis. The initial velocity of the particle is v = v0x i + v0y j so the particle moves in a helix. (a) Show that the radius of the helix is r = mv0y /qB. (b) Show that the particle takes a time t = 2pm/qB to make one orbit around the helix. (a) Since B = B i, v0 B = v0yB k ; i.e., vx = v0x and motion on the direction of the magnetic field is not affected by the field. In the plane perpendicular to i the motion is as described in Section 282, and the radius of the circular path is given by Equ. 28-6 with v = v0y, i.e., r = mv0y /qB. (b) The time for one complete orbit is given by Equ. 28-7, i.e., t = 2pm/qB.
Chapter 28
The Magnetic Field
86 A metal crossbar of mass M rides on a pair of long, horizontal conducting rails separated by a distance l and connected to a device that supplies constant current I to the circuit, as shown in Figure 28-40. A uniform magnetic field B is established as shown. (a) If there is no friction and the bar starts from rest at t = 0, show that at time t the bar has velocity v = (BI l /M)t. (b) In which direction will the bar move? (c) If the coefficient of static friction is s, find the minimum field B necessary to start the bar moving. (a), (b) The force on the bar is F = BI l and its acceleration is F/M = BI l /M = a. In the absence of friction the velocity is v = at = BI l t/M and is directed to the right since I l B is directed to the right. (c) To start the bar moving, Fmin = sMg = BminI l . So Bmin = sMg/I l . 87 Assume that the rails in Figure 28-40 are frictionless but tilted upward so that they make an angle ? with the horizontal. (a) What vertical magnetic field B is needed to keep the bar from sliding down the rails? (b) What is the acceleration of the bar if B has twice the value found in part (a)? (a) Note that with the rails tilted, F still points horizontally to the right. The component of F along the rail is then BI l cos ? and the component of Mg along the rail is -Mg sin ?. To hold the bar in position Ftot = 0. Solving for B one obtains B = (Mg/I l ) tan ?. (b) If B has twice the value found in (a), then the net force along the rail is Mg sin ? directed upward. Consequently, a = g sin ?. 88 A long, narrow bar magnet that has magnetic moment parallel to its long axis is suspended at its center as a frictionless compass needle. When placed in a magnetic field B, the needle lines up with the field. If it is displaced by a small angle ?, show that the needle will oscillate about its equilibrium position with frequency f = 21 B / I , where I is the moment of inertia about the point of suspension.
2 2 2 2 t = I(d ?/dt ) = B sin ?. For ? << 1, sin ? ?. Thus (d ?/dt ) = (B/I)?. This is the differential equation B/I for the SHO (see Equ. 14-2), and on comparison with Equ. 14-12 one obtains f = . 2
89* A conducting wire is parallel to the y axis. It moves in the positive x direction with a speed of 20 m/s in a magnetic field B = 0.5 T k . (a) What are the magnitude and direction of the magnetic force on an electron in the conductor? (b) Because of this magnetic force, electrons move to one end of the wire leaving the other end positively charged, until the electric field due to this charge separation exerts a force on the electrons that balances the magnetic force. Find the magnitude and direction of this electric field in the steady state. (c) Suppose the moving wire is 2 m long. What is the potential difference between its two ends due to this electric field? 18 (a) Use Equ. 281; q = 1.61019 C F = 1.610 N j (b) At steady state, q E + F = 0 E = 10 V/m j (c) ?V = E ?x ?V = 20 V 90 The rectangular frame in Figure 28-41 is free to rotate about the axis A-A on the horizontal shaft. The frame is 10 cm long and 6 cm wide and the rods that make up the frame have a mass per unit length of 20 g/cm. A uniform magnetic field B = 0.2 T is directed as shown. A current may be sent around the frame by means of the wires attached at the top. (a) If no current passes through the frame, what is the period of this physical pendulum for small oscillations? (b) If a current of 8.0 A passes through the frame in the direction indicated by the arrow, what is then the period of this physical pendulum? (c) Suppose the direction of the current is opposite to that shown. The frame is displaced from the vertical by some angle ?. What must be the magnitude of the
Chapter 28
The Magnetic Field
current so that this frame will be in equilibrium? (a) 1. Find the moment of inertia of the frame 2. Find D, the distance from the pivot to the CM 3. Use Equ. 14-31 (b) 1. With B and I as shown, Fm is downward; total restoring torque = (MgD + BIA)? 2. Use Equ. 14-31 with MgD (MgD + BIA) (c) The total torque = 0; MgD sin ? = BIA sin ?
I = [2(1/3) 0.20.12 + 0.120.12] kg.m2 = 2.53 g.m2 D = (0.40.05 + 0.120.1)/0.52 m = 0.0615 m T = 0.565 s ttot = (0.529.810.0615 + 0.28.00.006) ? N.m = 0.323? N.m T = 0.556 s I = MgD/BA = 261 A
91 A stiff, straight, horizontal wire of length 25 cm and mass 20 g is supported by electrical contacts at its ends, but is otherwise free to move vertically upward. The wire is in a uniform, horizontal magnetic field of magnitude 0.4 T perpendicular to the wire. A switch connecting the wire to a battery is closed and the wire is shot upward, rising to a maximum height h. The battery delivers a total charge of 2 C during the short time it makes contact with the wire. Find the height h. 1. ?p = F?t = BI l ?t = BQ l = mv0; evaluate v0 v0 = 0.420.25/0.02 m/s = 10 m/s 2 2. h = v0 /2g h = 5.10 m 92 A solid sphere of radius R carries a charge density ?0 in the region r = 0 to r = Rs and an equal charge density of opposite sign, +?0, between r = Rs and r = R. The net charge carried by the sphere is zero. (a) What must be the ratio R/Rs? (b) If this sphere rotates with angular velocity ? about its diameter, what is its magnetic moment? (a) Q+ = (4p/3)Rs3?0 ; Q = (4p/3)(R3 Rs3)?0 ; 2Rs3 = R3; Rs = R/21/3 = 0.794R Q+ + Q = 0; solve for Rs. 5 2/3 5 (b) Use the result of Problem 64 with Rs = R/21/3 = (4p/15) ?0R (1 2 )? = 0.31?0R ? 93* A circular loop of wire with mass M carries a current I in a uniform magnetic field. It is initially in equilibrium with its magnetic moment vector aligned with the magnetic field. The loop is given a small twist about a diameter and then released. What is the period of the motion? (Assume that the only torque exerted on the loop is due to the magnetic field.) 2 2 2 2 t = B sin ? IAB? = pR IB? = MR (d ?/dt ). This is the differential equation for a SHO, and comparison with Equs. 14-2 and 14-12 shows that T = 2p M/( I B) . 94 A small bar magnet has a magnetic moment that makes an angle ? with the x axis and lies in a nonuniform magnetic field given by B = Bx (x) i + By (y) j. 0 Use Fx = dU/dx and Fy = dU/dy to show that there is a net force on the magnet that is given by By F x Bx i + y j x y Let = x i + y j + z k . U = .B = xBx yBy. Since is constant but B depends on x and y, Fx = dU/dx = x( Bx/ x), Fy = dU/dy = y( By/ y), and F = x( B/ x) i + y( By/ y) j.

**Find millions of documents on Course Hero - Study Guides, Lecture Notes, Reference Materials, Practice Exams and more. Course Hero has millions of course specific materials providing students with the best way to expand their education.**

Below is a small sample set of documents:

Dartmouth - PHYS - 13/14

CHAPTER27The Microscopic Theory of Electrical Conduction1* In the classical model of conduction, the electron loses energy on average in a collision because it loses the drift velocity it had picked up since the last collision. Where does this

Dartmouth - PHYS - 13/14

CHAPTER18Temperature and the Kinetic Theory of Gases1* True or false: (a) Two objects in thermal equilibrium with each other must be in thermal equilibrium with a third object. (b) The Fahrenheit and Celsius temperature scales differ only in th

Dartmouth - PHYS - 13/14

CHAPTER39Relativity1* You are standing on a corner and a friend is driving past in an automobile. Both of you note the times when the car passes two different intersections and determine from your watch readings the time that elapses between t

Dartmouth - PHYS - 13/14

CHAPTER Alternating-Current Circuits31Note: Unless otherwise indicated, the symbols I, V, E, and P denote the rms values of I, V, and E and the average power. 1* A 200-turn coil has an area of 4 cm2 and rotates in a magnetic field of 0.5 T. (a)

Dartmouth - CHEM - 51

127Acidity of Carboxylic AcidsO O OH O Cl O Cl Cl O Cl Cl Cl O OH Cl Cl O OH O Cl OH OH ClWeek 51. Match the carboxylic acids below with the list of pKa's at right. Explain your ranking.Cl Cl Cl O OH Cl O OH O0.70OH1.48OHClMore ele

Dartmouth - CHEM - 51

52ChiralityWeek 21. For each of the f ollowing species, indicate if it is chiral or achiral . For those molecules that are chiral, circle all stereocenters, and draw the structure of the enantiomer.H H3C F Cl F ClH CH3chiralHCH3H3C

Dartmouth - CHEM - 51

123More Anomeric EffectWeek 51. Each of the following observations can be explained by a frontier molecular orbital (FMO) argument. Provide such an explanation for each observation. Be sure to label the relevant donor-acceptor interactions and

Dartmouth - CHEM - 51

1Lewis StructuresWeek 11. Draw complete Lewis structures, showing all atoms and lone pairs, for the following molecules.HCH4OHH O HHC HHH BrCH3BrHC HCH3OHC HOHH C H O HHCH3CH2OHHC HCH3NH2HC HN H

Dartmouth - CHEM - 51

178Frontier Molecular Orbitals of PolyenesWeek 71. Draw a simple representation of both the HOMO and LUMO for the following:LUMO:1 node2 nodes3 nodesHOMO:0 nodes1 node2 nodes2. Draw resonance structures for the following ion:

Dartmouth - CHEM - 51

185More Mechanisms and FMO AnalysisWeek 71. Consider the following interesting transformation:O MgBr+OEtH+carbocation intermediate a) Provide a curved-arrow mechanism f or the f ormation of the carbocation intermediate.EtO OOOMg

Dartmouth - PHYS - 13

Lecture 19 Torque and Angular Momentum Principle1Review from Last Time Rotational Kinematics Angular velocity: magnitude and direction Angular acceleration: magnitude and direction2Torque What causes angular acceleration? What causes an o

Dartmouth - PHYS - 13

Lecture 4 Applications of Newton's 2nd Law1Review from Last Time Newton's First Law (Law of Inertia) Newton's Second Law (The Momentum Principle)2Problem Solving - ExamplesNewton's second law can be applied in many different situations an

Dartmouth - PHYS - 13

Lecture 21 Conservation of Angular Momentum1Review: Analogies between Linear and Rotational Motion TABLE of anaologies1.1Angular Momentum of a single particle For a single particle a distance r from the point about which it rotates (like o

Dartmouth - PHYS - 13

Lecture 23 Orbital Mechanics1Brief Review of Gravity Gravitational force In general F = where r = |r2 - r1 | is the distance between the centers of the two objects, and the minus sign indicates that the force is attractive. Near the surface of

Dartmouth - PHYS - 13/14

CHAPTER Optical Images341* Can a virtual image be photographed? Yes. Note that a virtual image is "seen" because the eye focuses the diverging rays to form a real image on the retina. Similarly, the camera lens can focus the diverging rays onto

Dartmouth - CHEM - 51

59CyclohexanesWeek 31. Look at a chair. Notice the axial and equatorial substituents. Pay close attention to the parallel lines!2. Draw each of the following dimethyl cyclohexanes (on the planar ring), then identify whether the substituents w

Dartmouth - CHEM - 51

71Making Alcohols into Leaving GroupsWeek 31. The OH from an alcohol is not a good leaving group. Why? OH is a very strong base and strong bases make poor leaving groups.One way of making it a good leaving group is to protonate it. This often

Dartmouth - CHEM - 51

1Lewis StructuresWeek 11. Draw complete Lewis structures, showing all atoms and lone pairs, for the following molecules.CH4OHCH3 BrCH 3OCH3 CH 2OHCH 3NH2H 2CONH 4+HCNH 3O +C 2H 2C 2H 42Formal ChargeWeek 11. Given

Dartmouth - CHEM - 51

141Acidity of Carbonyl CompoundsCompoundO OWeek 61. All of the following compounds have acdic C-H bonds. The pKa of each compound is indicated: pKa Conjugate Base11R H O RO H O H H O OR13OR20R O CH2R25RO N C CH2R CH 2R31With e

Dartmouth - CHEM - 51

Sixth Hourly Examination Chemistry S-20August 6, 2007 8:10 9:20 a.m.Last Name: First Name: Section TF: Section Time:Important Notes:1. This exam consists of 8 problems on 8 pages, plus this cover sheet, and one page for scratch work at the end o

Dartmouth - PHYS - 13

Dartmouth - PHYS - 13

Lecture 6 Mass-spring oscillation (Simple Harmonic Motion)1Review from Last Time Gravitational Force- Review depends on product of masses and one over distance squared (DRAW) direction of force - always attractive along line connecting objects

Dartmouth - CHEM - 51

165Beckmann, Curtius, and Hofmann RearrangementsWeek 61. Provide the intermediates and products for the following transformations, and write a curved-arrow mechanism for the crucial rearrangement step.OH O N H NNH2OHBeckmannOH+OH2 N N

Dartmouth - PHYS - 13

Physics 13 - Winter 2008Homework #1-Due 10:00 PM on Monday, Jan. 14, 2007PART I: Complete the "p13-Homework 1" assignment at www.webassign.net. This part is submitted electronically.PART II: Complete the following problems. You will be graded in

Dartmouth - CHEM - 5/6

1Chem 6 Sample Exam 1 brief answers1. a. Consider the first 2 columns. Doubling [HI] quadruples the rate, so it must be second order in HI. Note that you get the same result by considering columns 2 and 3. So the rate law must be rate = k[HI]2. b.

Dartmouth - CHEM - 51

Third Hourly Examination Chemistry S-20July 16, 2007 8:10 9:20 a.m.Last Name: First Name: Section TF: Section Time:Important Notes:1. This exam consists of 8 problems on 8 pages, plus this cover sheet, and one page for scratch work at the end o

Dartmouth - PHYS - 13

n d 6 WaYA`FWVTgABR hbAUcB 9aP)R` d Q2cU8 cWi QAb APeR B P g8EYp)G8)cXPDiA8WfaDrT6GB@GRTRhBW r`eRV`9eRW QW 9caPE`D ha`AUEWFBb6 ThRQfBP iD A8(i@ cicW8 9R WW R W P P P ` W B 8 gP R b ` b ` U W R R R b R` 8 i b `ADyAvhR9`oAPQR )G8T6GBGRTRhBhABhiG8i

Dartmouth - CHEM - 5/6

Chem 6 sample Exam 3 solutions1. (a) CO, (c) BrF3 , BBr3 , like BF3 , is trigonal planar and the B-Br bond dipoles cancel out. The same is true for trigonal bipyramidal PF5 . Linear CO has a dipole moment since C and O have different electronegativ

Dartmouth - CHEM - 5/6

Chem 6 sample exam 1 (100 points total)@ This is a closed book exam to which the Honor Principle applies. @ The last page contains several equations which may be useful; you can detach it for easy reference. @ Please write clearly and SHOW YOUR WORK

Dartmouth - PHYS - 13

Lecture 7 Introducing a Model of a Solid1Review from Last Time Hook's Law: F=-kx - this force depends on position (compression of the spring). Last time we derived the equations of motion for a mass on a spring using this expression for the for

Dartmouth - CHEM - 5/6

Sample Exam 1 solutions 1. (a) To find the reaction orders, look at the dependence of rate onconcentrations. From rows 1 and 3, cutting the [OCl] in half cuts the rate in half, so the reaction is 1st order in OCl . Similarly, from rows 2 and 3, cut

Dartmouth - CHEM - 5/6

Chem 6 Sample exam 2 (150 points total)@ This is a closed book exam to which the Honor Principle applies. @ The last page contains equations and physical constants; you can detach it for easy reference. @ Please write clearly and SHOW YOUR WORK. If

Dartmouth - CHEM - 5/6

C h e m 5, 9 Section Exam 2SolutionsWinter, 2005 February 22, 20051.(4 points each) Here are a few quick problems to get you going. (a) Imagine formic acid, HCOOH(g), decomposing into CO2(g) and H2(g) according to HCOOH(g) CO2(g) + H2(g). Th

Dartmouth - CHEM - 5/6

1Chem 6 Sample Exam 2 Brief Answers1a. Two possible approaches here to find the energy difference and hence the frequency. Either find the energy of both states and subtract, or use one of the formulas that includes the subtraction for you. E = h

Dartmouth - CHEM - 5/6

Chem 6 Sample exam 3 (final) (200 points total)@ This is a closed book exam to which the Honor Principle applies. @ The last three pages contain equations, physical constants, a periodic table and other good stuff; you can detach them for easy refer

Dartmouth - CHEM - 5/6

1Chem 6 Sample exam 3 (final) brief answers1. Because E = hc/, shortest wavelength means biggest energy. This must correspond to a transition from n = infinity to the unknown quantum number n. The energy of this transition is given by E = 2.18x101

Dartmouth - CHEM - 5/6

Sample final exam answers1. For both these compounds, the formal oxidation state is Mn(II), which is d5 . Since NCS is a weak-field, high-spin ligand (from the spectrochemical series), the electron configuration for the octahedral complex will be (

Dartmouth - CHEM - 5/6

Chem 6 sample Exam 2 solutions1. (a) (1/2)mv2 = h . Since 274 nm is the maximum wavelength that will eject electrons, E = hc/ = = (6.626x1034Js) (3x108ms1)/(274 nm) (1m/109 nm) = 7.25x1019J Convert to kJ/mol (7.25x1019J) (1kJ/1000J) (6.02x1023/mo

Dartmouth - CHEM - 5/6

C h e m 5, 9 Section Exam 1SolutionsWinter, 2005 January 27, 20051.(10 + 5 points) In preparation for a study of the equilibrium constant for the reaction 2 NO2(g) N2O4(g) you take an 8.6 g frozen sample of pure N2O4(s) and place it in an o

Dartmouth - CHEM - 5/6

Chem 6 sample final exam (200 points total)@ This is a closed book exam to which the Honor Principle applies. @ The last 3 pages contain equations, physical constants, and the periodic table. You can detach them for easy reference. @ Please write cl

Dartmouth - CHEM - 5/6

Chem 5Winter 2005Exam 2 Study GuideHere are the central topics for the second exam. If you understand this material well and understand the problems on the past several problem sets, you'll do well on the exam!Chapter 7: Acids and Bases Can yo

Dartmouth - CHEM - 5/6

Chem 5Practice Exam 1 (with solutions)Winter 2005These are typical problems from past Chem 5 exams I have given. Note that your exam will include a Periodic Table (with atomic numbers and masses), values of important constants, and a list of us

Dartmouth - CHEM - 5/6

Chem 5Winter, 2005Practice Second Exam (the actual second exam last year) 1. (5 points each) Here are a few quick problems to get you going. (a) What is the pH of a saturated solution of Ca(OH)2, for which Ksp = 1.3 106? Let the equilibrium conc

Dartmouth - PHYS - 13/14

CHAPTER16Superposition and Standing Waves1* True or false: (a) The waves from two coherent sources that are radiating in phase interfere constructively everywhere in space. (b) Two wave sources that are out of phase by 180o are incoherent. (c)

Dartmouth - CHEM - 51

177Frontier Molecular Orbitals of PolyenesWeek 71. Draw a simple representation of both the HOMO and LUMO for the following:LUMO:HOMO:2. Draw resonance structures for the following ion:Now construct the complete molecular orbitals. Ide

University of Louisville - CALC - 205

IntegralsGeneral Integration Rulesb a ba cdx = c(b - a )b b a ab [ f ( x) - g ( x)]dx = f ( x)dx - g ( x)dxa ca [ f ( x) + g ( x)]dx = f ( x)dx + g ( x)dxa abb f ( x) dx + f ( x)dx = f ( x) dxc abbComparison Propert

Villanova - PHY - 2400

Villanova - PHY - 2400

Villanova - PHY - 2400

Villanova - PHY - 2400

Washington - MATH - 125

1. (7 pts) Find the function F (x) such that F (x) = x 3x + 1 and F (0) = 0.2. (8 pts) Use the Midpoint Rule (taking y-values at the midpoints of the intervals) with n = 3 6 x2 + 5 dx. Give your answer to two decimal subdivisions to find the appro

Washington - MATH - 125

University of Washington Midterm Exam IDepartment of Mathematics Math 125 A January 29, 2002NameStudent number Quiz section:ProblemTotal PointsScore114212310414510Total60Instructions 1. Print your name, student I

Washington - MATH - 125

Math 125G - Spring 2002 First Mid-Term Exam April 23, 2002NameSection1 2 3 4 5 6 7 8 Total10 10 10 10 10 10 10 10 80 Complete all questions. You may use a scientific calculator during this examination. Other calculating devices are not all

Washington - MATH - 125

Math 125G - Spring 2002 First Mid-Term Exam Solutions 1. Is 1 1 2 x ln x - x2 an antiderivative of x ln x ? Explain. 2 41 2 1 x ln x - x2 is x ln x (details left to the reader, must be shown 2 4 1 2 1 for credit), so, yes, x ln x - x2 an antideriva

Christian Brothers - ECE - 172

public class Seat { private String name; private Double pricePerTicket; private int ticketsSold;public Seat(String ticketName) {name = ticketName; pricePerTicket =00.00; ticketsSold = 00000;} public void setName (String ticketName){ name = tick

Christian Brothers - ECE - 172

import java.util.*; import javax.swing.*; import java.text.*; class Sin { private Double sin; private Integer i=1; private Double btmSum, Sin;public void Sin() { sin = 0.0; } public Double getSin(){ return sin; } public void setSin(double X, int N,

Christian Brothers - ECE - 172

public class CoinTest{ public static void main(String[] args){ Coin coin1, coin2, coin3, coin4; coin1 = new Coin(); coin2 = new Coin(); coin3 = new Coin(); coin4 = new Coin(); coin1.Toss(); coin2.Toss(); coin3.Toss(); coin4.Toss();System.out.printl

Christian Brothers - ECE - 172

/* */ public class Die{ private static final int MAX_NUMBER = 6; private static final int MIN_NUMBER = 1; private static final int NO_NUMBER = 0; private int number;/-public Die(){ number = NO_NUMBER; } /--public void Roll(){ number = (int)(Math.fl

Christian Brothers - ECE - 172

class Chapter5_14 { public static void main(String[] args) { WageComputer controller = new WageComputer(); controller.start(); controller.getNumbers(); controller.wageCompute(); }}

Christian Brothers - ME - 101

NAME_ CIVL 7133-8133 Final Examination Spring Semester 2008 April 25, 2008 Due NLT Thursday May 1 at 1200 This is a take-home examination. Show all work and calculations. All equations must be shown. Indicate and show all steps of your analysis to in

Christian Brothers - ECE - 172

import javax.swing.*; import java.util.*; import java.text.*; public class SeatTest {public static void main(String[] args){ Seat A, B, C; Double priceA, priceB, priceC; int SoldA, SoldB, SoldC; Scanner scanner; A = new Seat("Type A"); B = new Seat