Bukhovtsev-et-al-Problems-in-Elementary-Physics

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346 ANSWERS AND SOLUTIONS II /"---- ....., / " ~ I " I \. I ~ \ I \ '.0 \ / , / " fill' '---_.--" H H Fig. 459 Fig. 460 lem 560). This force is constant in magnitude and perpendicular to the velocity v. For this reason the acceleration of the electron is also constant in magnitude and constantly remains perpendicular to the velocity. The velocity changes only in direction. With a constant acceleration perpendicular to the velocity, the motion at a velocity constant in magnitude is uniform motion along a circle. mv 2 e On the basis of Newton's second law, R=c·vH. Therefore, the electron will move along a circle with a radius R = n;:: · 563. Let us resolve the velocity of the electron into the components v II parallel to Hand V...L perpendicular to H (Fig. 460). The component vu does not change in magnitude or direction since the Lorentz force does not act on a particle whose velocity is directed along the field. The component V.1. changes in direction in the same way as in Problem 562. Thus, rotation along a circle in a plane perpendicular to H is superposed on the uniform translational motion along H. This produces motion along a helical line with a constant pitch h=v u 1:, where 1: is the duration of one revolution of the electron along a circle with a radius R= mc:~n IX • . 2nR 2nmc 2nmc Since l'=--= -H ,then h= --;:;- v cos ex v~ e e'4 564. The action of the Lorentz force (see Problem 560) will cause the electrons to move towards the edge of the band. For this reason one edge of the band will receive a negative charge and the other a positive one. An additional electric field will be generated inside the band with an intensity E directed perpendicular to the current. The electrons will continue to move until the Lorentz force is equalized by the force acting on the electron from the side of the electric field E: eE=kevH. Hence, E=kvH. The potential difference CPA-CPB=Ea=kvHa or, since I = nevA, then I A-CPB= kHa ne Ā· 565. q> A - CPB =:: 23JLV. 566. The Lorentz force (see Problem 560) acts on both the free electrons and the positive ions at the points of a crystal lattice, since both move in a -,
ELECTRICITY AND- MAGNETISM 347 a cl It 0H I~ + + + + + + + + Fig. 46/ magnetic field. In accordance with the left-hand rule, the force f that acts on the free electrons will be directed as shown in Fig. 461. The electrons are displaced with respect to the lattice, and one side of the -parallelepiped is charged negatively and the other positively. An electric field is produced in the block, and when the intensity of this field satisfies the ratio eE=kevH, the electrons will no longer move with respect to the lattice. The sought intensity E=kvH. The density of the charges (J can be found from the equation 41(
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B. B. 6yX08I4e8. B. JI. KpU8IleHIC0
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First published 1971 Second printin
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6 Chapter 5. Geometrical Optics •
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8 PROBLEMS engineer arrived at the
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10 PROBLEMS c rIII a Fig. 3 01 IIII
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12 PROBLEMS ....------1----.... str
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18 PROBLEM~ friction if the weight
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20 PROBLEMS Fig. 28 H 11/ r ~--I---
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22 PROBLEMS Fig. 25 Fig. 26 B m; is
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24 PROBLEMS ~ 1 2 • ~ Fig. 29 tha
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2(; PROBLEMS m m Fig. 32 Fig. 33 90
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28 PROBLEMS o ;...-. ..... 8 D Fig.
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a Fig. 41 Fig. 42 for the rotating
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32 PROBLEMS Fig. 48 122. In Guerick
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34 PROBLEMS (I) What work is done b
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36 PROBLEMS ~ city of the weights a
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38 PROBLEMS wedge. The block can sl
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40 PROBLEMS s II ne. 61 Fl,. 62 182
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44 PROBLEMS 178. A ball bearing sup
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46 PROBLEMS Fig. 73 Fig. 74 189. At
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a 0' Fig. 80 Fig- 8/ ces '1 and " f
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50 PROBLEMS Fig. 84 Fig. 85 205. A
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52 PROBLEMS surface. Find the angul
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54 PROBLEMS '-... ' ..........-....
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56 PROBLEMS ~==-==~~~~---------- --
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58 PROBLEMS "/I I 11 L..-__----...,
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60 PROBLEMS the chamber closes herm
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--- --r-- - Fig. 114 ~--l---I"" Fig
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66 PROBLEMS ~ ~ I A ~ • B Fig. 11
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CHAPTER 2 HEAT. MOLECULAR PHYSICS 2
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70 PROBLEMS What quantity of heat w
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72 PROBLEMS 315. A barometer gives
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74 PROBLEMS pressure Po = 760 mm Hg
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76 PROBLEMS of the air in the room
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78 PROBLEMS -- II -- - --- -- - - -
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80 PROBLEMS angle will be different
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82 PROBLEMS ~ Fig. 141 force F will
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88 PROBLEMS 400. If only one charge
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90 PROBLEMS 417. A metal leaf is at
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92 PROBLEMS Fig. 153 Fig. 154 431.
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94 PROBLEMS Prove that if these cap
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96 PROBLEMS of the vessel is placed
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A Fig. /66 Fig. /67 463. Two conduc
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102 . PROBLEMS . Fig. 174 Fig. /75
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104 PROBLEMS length of L = 5.6 km t
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106 PROBLEMS an energy of QU/2 is l
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108 ....------------- PROBLEJ\\S Fi
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110 PROBLEMS Fig. 187 Fig. 188 Cons
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112 PROBLEMS 3-4. Magnetic Field of
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114 PROBLEMS 553. Will the density
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116 PROBLEMS tic field in the direc
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118 PROBLEMS Fig. 206 Fig. 207 ving
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120 PROBLEMS is , and its length l
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122 PROBLEMS -~ t Fig. 218 597. An
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124 PROBLEMS 608. Determine the eff
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PROBLEMS 625. Find the period of os
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130 PROBLEMS Fig. 227 tions of a we
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132 PROBLEMS 1; ~ ----- v Fig. 230
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134 PROBLEMS 00 DC- Fig. 232 Fig. 2
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136 PROBLEMS the third one that was
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138 PROBLEMS 680. Find graphically
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140 PROBLEMS W 689. Two flat mirror
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142 PROBLEMS 702. A ray incident on
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144 PROBLEMS .9' Fig. 247 length of
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146 PROBLEMS a point source arrange
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150 PROBLEMS 757. Two men, a far-si
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152 PROBLEMS H 767. Two point sourc
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154 PROBLEMS 773. A convergent lens
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156 PROBLEMS 6.. 2. Diffraction of
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158 P~OBLEMS focal length of f = 40
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ANSWERS AND SOLUTIONS CHAPTER 1 MEC
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162 ANSWERS AND SOLUTIONS I 2 9 5 G
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164 ANSWERS AND SOLUTIONS depend on
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166 ANSWERS AND SOLUTIONS Hence and
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o s Fig. 278 .D A B B' Fig. 277 25.
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172 ANSWERS AND SOLUTIONS '0; U2 OI
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174 ANSWERS AND SOLUTIONS 35. The a
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176 ANSWERS AND SOLUTIONS ice dry s
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178 ANSWERS AND SOLUTIONS The addit
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180 ANSWERS AND SOLUTIONS --+- 0 9
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182 ANSWERS AND SOLUTIONS 65. Initi
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184 ANSWERS AND SOLUTIONS Let us as
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186 ANSWERS AND SOLUTIONS where s i
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188 ANSWERS AND SOLUTIONS 1-5. Stat
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190 ANSWERS AND SOLUTIONS Hence, F=
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o • • • ~. Fig. 306 Fig. 907
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194 ANSWERS AND SOLUTIONS A C~~7;7l
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198 ANSWERS AND SOLUTIONS .0 Fig. 3
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200 ANSWERS AND SOLUTIONS rnicircle
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202 ANSWERS AND SOLUTIONS The wagon
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208 ANSWERS AND SOLUTION The kineti
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212 ANSWERS AND SOLUTIONS Fig. 326
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216 ANSWERS AND SOLUTIONS Fig.3d1 F
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220 ANSWERS AND SOLUTIONS " ...-- ,
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222 ANSWERS AND SOLUTIONS Therefore
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224 ANSWERS AND SOLUTIONS 171. The
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226 ANS\VERS AND SOLUTIONS A 2u (6)
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230 ANSWERS AND SOLUTIONS T=(M +m)
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232 ANSWERS AND SOLUTIONS 9-"-,,~,
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234 . ANSWERS AND SOLUTIONS Newton'
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236 o ANSWERS AND SOLUTIONS a A H F
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240 ANSWERS AND SOLUTIONS The close
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244 ANSWERS AND SOLUTIONS Since the
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248 ANSWERS AND SOLUTIONS 1-9. The
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250 ANSWERS AND SOLUTIONS ",' ,~ I
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254 ANSWE~S AND SOLUTIONS entire th
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266 ANSWERS AND SOLUTIONS time to b
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CHAPTER 2 HEAT. MOLECULAR PHYSICS 2
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270 ANSWERS AND SOLUTIONS ! This sy
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276 ANSWERS AND SOLUTIONS 320. Sinc
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286 ANSWERS AND SOLUT IONS JI!!!I!I
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290 ANSWERS AND SOLUTIONS 378. It i
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292 ANSWERS AND SOLUTIONS the kettl
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CHAPTER 3 ELECTRICITY AND MAGNETISM
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396 ANSWERS AND SOLUTIONS ~-"":;:'-
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8' Fig. 519 s* IIIIIIII N P H' Fig.
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404 ANSWERS AND SOLUTIONS ~---d---~
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406 ANSWERS AND SOLUTIONS E a S s'
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410 ANSWERS AND SOLUTIONS The dista
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412 ANS\\'ERS AND SOLUTIONS the sys
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414 J- ------....,.c 8- ----~--- AN
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416 ANSWERS AND SOLUTIONS Therefore
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418 ANSWERS AND SOLUTIONS N Fig. 54
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420 ANSWERS AND SOLUTIONS from the
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422 S, IT ........... In --.. U' .,
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424 ANSWERS AND SOLUTIONS dering tr
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426 ANSWERS AND SOLUTIONS Fig 556 T
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428 ANSWERS AND SOLUTIONS 779. When
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434 ANSWERS AND SOLUTIONS cent slit
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436 ANSWERS AND SOLUTIONS 805. For
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