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Download 3. (a) The force on the electron is Thus, the magnitude of FB is 6.2
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3. (a) The force on the electron is FB qv B q vx ˆi v y ˆj Bx ˆi By j q vx By v y Bx kˆ = 1.6 1019 C 2.0 106 m s 0.15 T 3.0 106 m s 0.030 T ˆ 6.2 1014 N k. Thus, the magnitude of FB is 6.2 1014 N, and FB points in the positive z direction. (b) This amounts to repeating the above computation with a change in the sign in the charge. Thus, FB has the same magnitude but points in the negative z direction, namely, ˆ F 6.2 1014 N k. B 4. (a) We use Eq. 28-3: FB = |q| vB sin = (+ 3.2 10–19 C) (550 m/s) (0.045 T) (sin 52°) = 6.2 10–18 N. (b) The acceleration is a = FB/m = (6.2 10– 18 N) / (6.6 10– 27 kg) = 9.5 108 m/s2. (c) Since it is perpendicular to v , FB 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. 11. Since the total force given by F e E v B vanishes, the electric field E must be perpendicular to both the particle velocity v and the magnetic field B . The magnetic field is perpendicular to the velocity, so 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, mv 2 / 2 eV and v 2eV m . Thus, d i 2 1.60 1019 C 10 103 V 2eV 1.2 T 6.8 105 V m. EB m 9.99 1027 kg