chp29 - Problem 13. A region contains an electric field E...

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Unformatted text preview: Problem 13. A region contains an electric field E = 7.43 + 2.83 kN/C and a magnetic field B = 153 + 36f: mT. Find the electromagnetic force on (a) a stationary proton, (b) an electron moving with velocity v = 6.1i Mm/s. Solution The force on a moving charge is given by Equa- tion 29-2 (called the Lorentz force) F = q(E + va). (a) For a stationary proton, q = e and v a 0, so F = eE = (1.6x10'19 C)(7.4r+ 2.83) kN/C = (1.15s + 0.4483) fN. (b) For the electron, q = —e and v = 6.1i Mm/s, so the electric force is the negative of the force in part (a) and the magnetic force‘is —-eva = (—1.5x10-19 C)(6.1i‘Mm/s)x(15j+ 36k) rnT = (—14.61? + 35.13) EN . The total Lorentz force is the sum of these. or (—1.18i+ 34.73 — 14.611) fN. Problem 19. Electrons and protons with the same kinetic energy are moving at right angles to a uniform magnetic field. How do their orbital radii compare? Solution It is convenient to anticipate the result of Problem 22 for the orbital radius of a non-relativistic charged particle in a plane perpendicular to a uniform magnetic field. Rom Equation 29-3, 1- = mv/qB. For a non-relativistic particle, K 2: §mv 2, or v = “Hf/m, therefore 1‘ = VZKm/qB. (Note: All quantities can, of course, be expressed in standard SI units, but in many applications, atomic units are more convenient. The conversion factor for electron volts to joules is just the numerical magnitude of the electronic charge, so if K is expressed in MeV, min MeV/cz, q in multiples of e, and B in teslas, we obtain _ ,/2K(ex105)m(ex105/E§) r _ (selB _ 105x/2Km _ WKm “ (3x108)qB _ 30qu ' Rom this expression, it follows that protons and electrons with the same kinetic energy have radii in the ratio rp/re = «mpime = Vl836 m 43, in the same magnetic field. Heavier particles are more difficult to bend. Problem 22. Show that the orbital radius of a charged particle moving at right angles to a magnetic field B can be written 2Km qB ’ where K is the kinetic energy in joules, m the particle mass, and 9 its charge. T: Solution See solution to Problem 19. Problem 2?. Figure 29-38 shows a simple mass spectrometer, designed to analyze and separate atomic and molecular ions with different charge-to—mass ratios. In the design shown, ions are accelerated through a potential difi‘erence V, after which they enter a region containing a uniform magnetic field. They describe semicircular paths in the magnetic field, and land on a detector a lateral distance :r from where they entered the field region, as shown. Show that a is given by 2V 2 B (tr/m)’ where B is the magnetic field strength, Vthe accelerating potential, and q/m the chargeto—mass ratio of the ion. By counting the number of ions accumulated at different positions 2, one can determine the relative abundanccs of different atomic or molecular species in a sample. I: Solution The positive ions enter the field region with speed - (determined from the work~energy theorem) of FIGURE 29-38 Problem 27 Solution. 113111132 : qV, or u z ,/2V(q/m). They are bent into a semicircle with diameter :6 = 21' = 2mu/qB = 2(m/qB)1/2V(q/m] = °\/2(m/q]V/B, as shown in Fig. 29-38 (see Equation 29-3). Problem 36. A wire of negligible resistance is bent into a rectangle as shown in Fig. 29—40, and a battery and resistor are connected as shown. The right-hand side of the circuit extends into a region containing a uniform magnetic field of 38 mT pointing into the page. Find the magnitude and direction of the net force on the circuit. Solution The forces on the upper and lower horizontal parts of the circuit are equal in magnitude, but opposite in direction and thus cancel (see Fig. 29-40), leaving the force on the righthand wire1 MB 2 (S/R)€B = (12 V/S Q)(0.1 m)(38 InT) = 15.2 mN toward the right, as the net force on the circuit. FIGURE 29—40 Problem 36 Solution. Problem 38. A 20-cm—long conducting rod with mass 18 g is suspended by wires of negligible mass, as shown in Fig. 29-41. The rod is in a region containing a uniform magnetic field of 0.15 T pointing horizontally into the page, as shown. An external circuit supplia current between the support points A and B. (a) What is the minimum current necessary to move the bar to the upper position shown? (b) Which direction should the current flow? Solution An upward magnetic force on the rod equal (in magnitude} to its weight is the minimum force necassary. (3) Since the rod is perpendicular to B, MB = mg implies I = rag/EB = (0.018x9.8 N)+ (0.2x0.15 Tm) = 5.88 A. (b) The force is upward for current flowing from A to B, consistent with the right hand rule for the cross product. h—Nm—‘l FIGURE 29—41 Problem 38 Solutiou. Problem 53. Nuclear magnetic resonance (NMR) is a technique for analyzing chemical structures and is also the basis of magnetic resonance imaging used for medical diagnosis. The NMR technique relies on sensitive measurements of the energy needed to flip atomic nuclei upside-down in a given magnetic. field. In an NMR apparatus with a 7.0-T magnetic field, how much energy is needed to flip a proton (is = 1.41x1rr28 A-m“) from parallel to antiparallel to the field? Solution From Equation 29-12, the energy required to reverse the orientation-of a proton‘s magnetic moment from parallel to antiparallel to the applied magnetic field is AU : 2pB = 2(1.41x10‘36 A-m2)(7.0 T) : 1.97x 10*25 J z 1.23x 10—6 eV. (This amount of energy is characteristic of radio waves of frequency 298 MHZ, see Chapter 39.) ...
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This note was uploaded on 09/10/2011 for the course PHYSICS 101 taught by Professor Wormer during the Spring '08 term at NYU Poly.

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chp29 - Problem 13. A region contains an electric field E...

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