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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 292 (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 =
(—1.5x1019 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 nonrelativistic charged particle in a plane perpendicular to a uniform
magnetic ﬁeld. Rom Equation 293, 1 = mv/qB. For
a nonrelativistic 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 2938 shows a simple mass spectrometer,
designed to analyze and separate atomic and
molecular ions with different chargeto—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 2938 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. 2938 (see Equation 293). 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
righthand 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. 2940), 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 20cm—long conducting rod with mass 18 g is
suspended by wires of negligible mass, as shown in
Fig. 2941. 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 upsidedown in a given magnetic.
ﬁeld. In an NMR apparatus with a 7.0T magnetic
ﬁeld, how much energy is needed to ﬂip a proton
(is = 1.41x1rr28 Am“) from parallel to
antiparallel to the ﬁeld? Solution From Equation 2912, the energy required to reverse
the orientationof a proton‘s magnetic moment from
parallel to antiparallel to the applied magnetic ﬁeld
is AU : 2pB = 2(1.41x10‘36 Am2)(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.
 Spring '08
 WORMER
 Physics

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