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# HW28 - 98 2 Fundamentals of Physics Halliday Resnic 0401...

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Fundamentals of Physics 東東東東東東東 98 東東東東 2 東東 系系系□0401 系系系 □0409 系系系系系 系系系 系系系 . 系CH28系Magnetic Fields 3, 5, 9, 11, 14, 15, 17, 21, 27, 29, 31, 35, 39, 43, 47, 49, 53, 55, 59,63, 65 系Problem 28-3系 A proton traveling at 0 23 with respect to the direction of a magnetic field of strength 2.60 mT experiences a magnetic force of 17 6.5 10 N - × . Calculate (a) the proton’s speed and (b) its kinetic energy in electron-volts. <系>系(a) Eq. 28-3 leads to v F eB B = = × × × ° = × - - - sin . . . sin . . . φ 650 10 160 10 2 60 10 230 4 00 10 17 19 3 5 N C T m s c hc h (b) The kinetic energy of the proton is ( 29 ( 29 2 2 27 5 16 1 1 1.67 10 kg 4.00 10 m s 1.34 10 J 2 2 K mv - - = = × × = × , which is equivalent to 16 19 1.34 10 835 1.6 10 / J K eV J eV - - × = = × 系Problem 28-5系 An electron moves through a uniform magnetic field given by \$ (3 ) x x B B i B j = + r \$ . At a particular instant, the electron as velocity \$ (2 4 ) / v i j m s = + r \$ and the magnetic force acting on it is \$ 19 (6.4 10 ) N k - × . Find x B . <系>系Using Eq. 28-2 and Eq. 3-30, we obtain r F q v B v B q v B v B x y y x x x y x = - = - d i b g d i \$ \$ k k 3 where we use the fact that 3 y x B B = . Since the force (at the instant considered) is F z \$ k where 19 6.4 10 z F N - = × , then we are led to the condition ( 29 ( 29 3 . 3 z x y x z x x y F q v v B F B q v v - = = - Substituting 2 / x v m s = , 4 / y v m s = and 19 1.6 10 q C - = - × , we obtain 19 19 6.4 10 N 2.0 T. (3 ) ( 1.6 10 C)[3(2.0 m/s) 4.0 m] z x x y F B q v v - - × = = = - - - × - 系Problem 28-9系 An electron has an initial velocity of \$ \$ (12 15 ) / j k km s + and a constant acceleration of 12 2 (2 10 / ) m s i × \$ in a region in which uniform electric and magnetic fields are present. If (400 ) B T i μ = r \$ , find the electric field E r . <系>系We apply r r r r r F q E v B m a e = + × = d i to solve for r E : 東 1 東東東 11 東

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Fundamentals of Physics 東東東東東東東 98 東東東東 2 東東 r r r r E m a q B v e = + × = × × - × + × + = - - + - - 911 10 2 00 10 160 10 400 12 0 150 114 6 00 4 80 31 12 2 19 . . \$ . \$ . \$ . \$ . \$ . \$ . \$ . kg m s i C T i km s j km s k i j k V m c hd i b g b gb g e j μ 系Problem 28-11系 An ion source is producing 6 Li ions, which have charge e + and mass 27 9.99 10 kg - × . The ions are accelerated by a potential difference of 10 kV and pass horizontally into a region in which there is a uniform vertical magnetic field of magnitude 1.2 B T = . Calculate the strength of the smallest electric field, to be set up over the same region, that will allow the 6 Li ions to pass through undeflected. < 系 > 系 Since the total force given by
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HW28 - 98 2 Fundamentals of Physics Halliday Resnic 0401...

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