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Homework 6 Ch20: Q 10, 18; P 3, 5, 9, 11, 15, 17 Questions: 10. Three particles, a, b, and c, enter a magnetic field as shown in Fig. 20–46. What can you say about the charge on each particle? Solution Charge a experiences an initial upward force. By the right hand rule, charge a must be positive. Charge b experiences no force, and so charge b must be uncharged. Charge c experiences an initial downward force. By the right hand rule, charge c must be negative. 18. A beam of electrons is directed toward a horizontal wire carrying a current from left to right (Fig. 20–49). In what direction is the beam deflected? Solution The beam of electrons is deflected to the right. The magnetic field created by the current in the wire, I, is directed into the page on the side of the wire where the electron beam is located. The right hand rule tells us that the negative electrons will feel a force to the right as they move toward the wire in a downward magnetic field. Problems: 3. (I) How much current is flowing in a wire 4.80 m long if the maximum force on it is 0.750 N when placed in a uniform 0.0800-T field? Solution Use Eq. 20-2. Fmax IlB I Fmax lB 0.750 N 4.80 m 8.00 102 T 1.95 A 5. (II) The force on a wire carrying 8.75 A is a maximum of 1.28 N when placed between the pole faces of a magnet. If the pole faces are 55.5 cm in diameter, what is the approximate strength of the magnetic field? Solution Use Eq. 20-2. The length of wire in the B-field is the same as the diameter of the pole faces. Fmax IlB B Fmax Il 1.28 N 8.75 A 0.555 m 0.264 T 9. (I) Alpha particles of charge q 2e and mass m 6.6 10 27 kg are emitted from a radioactive source at a speed of 1.6 107 m s . What magnetic field strength would be required to bend them into a circular path of radius r 0.25 m? Solution The magnetic force is perpendicular to the velocity. In this scenario, the magnetic force is causing centripetal motion, and so must have the form of a centripetal force. Fmax qvB m v2 r B mv qr 6.6 10 kg 1.6 10 m s 1.3T 2 1.60 10 C 0.25 m 27 7 19 11. (I) Find the direction of the force on a negative charge for each diagram shown in Fig. 20–51, where v (green) is the velocity of the charge and B (blue) is the direction of the magnetic field. ( means the vector points inward. means it points outward, toward you.) Solution Since the charge is negative, the answer is the OPPOSITE of the result given from the right hand rule applied to the velocity and magnetic field. (a) left (b) left (c) upward (d) inward into the paper (e) no force (f) downward 15. (II) An electron experiences the greatest force as it travels 2.9 106 m s in a magnetic field when it is moving northward. The force is upward and of magnitude 7.2 10 13 N. What are the magnitude and direction of the magnetic field? Solution The magnetic field can be found from Eq. 20-4, and the direction is found from the right hand rule. Remember that the charge is negative. Fmax qvB B Fmax qv 7.2 1013 N 1.60 10 19 C 2.9 106 m s 1.6 T The direction would have to be East for the right hand rule, applied to the velocity and the magnetic field, to give the proper direction of force. 17. (II) A doubly charged helium atom whose mass is 6.6 1027 kg is accelerated by a voltage of 2100 V. (a) What will be its radius of curvature if it moves in a plane perpendicular to a uniform 0.340-T field? (b) What is its period of revolution? Solution (a) The velocity of the ion can be found using energy conservation. The electrical potential energy of the ion becomes kinetic energy as it is accelerated. Then, since the ion is moving perpendicular to the magnetic field, the magnetic force will be a maximum. That force will cause the ion to move in a circular path. 2qV Einitial Efinal qV 12 mv 2 v Fmax qvB m r mv qB m v m 2 r 2qV 1 m qB B 2mV q 2 6.6 10 27 kg 2100 V 1 19 2 1.60 10 C 0.340 T 2.7 10 2 m (b) The period can be found from the speed and the radius. Use the expressions for the radius and the speed from above. v 2r 2r 2m 2 6.6 10 27 kg T 3.8 10 7 s 19 T v qB 2 1.60 10 C 0.340T