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# ch28 - 1(a Eq 28-3 leads to 6.50 10-17 N FB v= = = 4.00 105...

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1. (a) Eq. 28-3 leads to v F eB B == × ×× ° −− sin . .. s i n . φ 650 10 160 10 2 60 10 230 400 10 17 19 3 5 N CT ms c hc h (b) The kinetic energy of the proton is Km v × × = × 1 2 1 2 167 10 4 00 10 134 10 22 7 5 2 16 . kg m s J. c h This is (1.34 × 10 – 16 J) / (1.60 × 10 – 19 J/eV) = 835 eV.

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2. (a) We use Eq. 28-3: F B = |q| vB sin φ = (+ 3.2 × 10 –19 C) (550 m/s) (0.045 T) (sin 52°) = 6.2 × 10 –18 N. (b) a = F B / m = (6.2 × 10 – 18 N) / (6.6 × 10 – 27 kg) = 9.5 × 10 8 m/s 2 . (c) Since it is perpendicular to G G vF B , does not do any work on the particle. Thus from the work-energy theorem both the kinetic energy and the speed of the particle remain unchanged.
3. (a) The force on the electron is () ( ) 19 6 6 14 ˆˆ ˆ ˆ ij i k = 1.6 10 C 2.0 10 m s 0.15 T 3.0 10 m s 0.030 T ˆ 6.2 10 N k. Bx y x y x y y x F q v B qv v B Bj qvB vB = + × + = ªº −× × × ¬¼ GG G G Thus, the magnitude of G F B is 6.2 × 10 14 N, and G F B points in the positive z direction. (b) This amounts to repeating the above computation with a change in the sign in the charge. Thus, G F B has the same magnitude but points in the negative z direction, namely, 14 ˆ 6.2 10 N k. B F =− × G

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4. The magnetic force on the proton is F = q v × B where q = + e . Using Eq. 3-30 this becomes (4 × 10 17 )i ^ + (2 × 10 17 )j ^ = e [(0.03 v y + 40)i ^ + (20 – 0.03 v x )j ^ – (0.02 v x + 0.01 v y )k ^ ] with SI units understood. Equating corresponding components, we find (a) v x = 3.5 ×1 0 3 m/s, and (b) v y = 7.0 ×1 0 3 m/s.
5 . Using Eq. 28-2 and Eq. 3-30, we obtain G Fq v B v B q v B v B xy yx x x =− = di b g ## kk 3 where we use the fact that B y = 3 B x . Since the force (at the instant considered) is F z # k where F z = 6.4 × 10 –19 N, then we are led to the condition qv vB F B F qv v x z x z 3 3 −= ¡ = . Substituting V x = 2.0 m/s, v y = 4.0 m/s and q = –1.6 × 10 –19 C, we obtain B x = –2.0 T.

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6. Letting GG G G Fq EvB =+ × = di 0 , we get vB sin φ = E . We note that (for given values of the fields) this gives a minimum value for speed whenever the sin factor is at its maximum value (which is 1, corresponding to = 90°). So v min = E / B = (1.50 × 10 3 V / m) / (0.400 T) = 3.75 × 10 3 m / s.
7. Straight line motion will result from zero net force acting on the system; we ignore gravity. Thus, GG G G Fq EvB =+ × = di 0 . Note that G G v B so G G vBv B ×= . Thus, obtaining the speed from the formula for kinetic energy, we obtain () ( ) ( ) 3 4 31 9 3 1 100V 20 10 m 2.67 10 T. 2/ 2 1.0 10 V 1.60 10 C / 9.11 10 kg e EE B v Km −− × == = = × ×× × In unit-vector notation, 4 ˆ (2.67 10 T)k B =− × G .

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8. We apply GG G G G Fq EvB m a e =+ × = di to solve for G E : G G G G E ma q Bv 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 ij k V m c h b g b g b g e j µ
9. Since the total force given by G G G G Fe EvB =+ × di vanishes, the electric field G E must be perpendicular to both the particle velocity G v and the magnetic field G B . The magnetic field is perpendicular to the velocity, so G G v B × has magnitude vB and the magnitude of the electric field is given by E = vB . Since the particle has charge e and is accelerated through a potential difference V, 2 1 2 mv eV = and 2. ve V m = Thus, () ( ) 19 3 5 27 2 1.60 10 C 10 10 V 2 1.2 T 6.8 10 V m.

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ch28 - 1(a Eq 28-3 leads to 6.50 10-17 N FB v= = = 4.00 105...

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