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Unformatted text preview: Figure 21.21 Solve: (a) (b) The magnetic force on a positive charge in the bar is eastward. Positive charge accumulates at end a of the bar and that end (the east end) is at higher potential. (c) If the bar is moving east to west the force is to the north. The force is perpendicular to the bar and there is no charge accumulation at its ends; the potential difference between its ends is zero. Reflect: If the bar moves parallel to the magnetic field there is no magnetic force and no potential difference is produced. 21.22. Set Up: The rod moves perpendicular to the magnetic field, so Eq. (21.6) applies. To determine which end is at higher potential, consider the direction of the magnetic force on a positive charge in the moving rod. The end where positive charge accumulates is the higher potential end. (a) so (b) The rod is sketched in Figure 21.22a. The magnetic force on a positive charge in the moving rod is shown. The righthand rule applied to and says this force is toward end b and this end is at higher potential. (c) The rod is sketched in Figure 21.22b. The magnetic force now is perpendicular to the rod. There is no charge accumulation at its ends and no induced emf. Figure 21.22 21.23. Set Up: Solve: This is a large speed and not practical. It is also difficult to produce a 5.00 cm wide region of 0.650 T magnetic field. 21.24. Set Up: Solve: (a) This is much too small to be noticeable. (b) This is too small to be noticeable. E 5 1 565 mph 2 1 0.4470 m / s 1 mph 2 1 0.50 3 10 2 4 T 21 64.4 m 2 5 0.813 mV. E 5 1 180 mph 2 1 0.4470 m / s 1 mph 2 1 0.50 3 10 2 4 T 21 1.5 m 2 5 6.0 mV. 1 G 5 10 2 4 T. 1 mph 5 0.4470 m / s. E 5 v BL . v 5 E BL 5 1.50 V 1 0.650 T 21 5.00 3 10 2 2 m 2 5 46.2 m / s 5 103 mph. E 5 v BL B b a v F v B a b F ( a ) ( b ) B S v S B 5 V ab v L 5 0.450 V 1 4.50 m / s 21 0.120 m 2 5 0.833 T. V ab 5 v BL V ab 5 v BL 5 1 11.5 m / s 21 1.22 T 21 0.150 m 2 5 2.10 V. v B a b F N S E W V ab 5 v BL . Electromagnetic Induction 217 21.25. Set Up: Use Lenz’s law to determine the direction of the induced current. The force required to maintain constant speed is equal and opposite to the force that the magnetic field exerts on the rod because of the current in the rod. Solve: (a) (b) is into the page. The flux increases as the bar moves to the right, so the magnetic field of the induced current is out of the page inside the circuit. To produce magnetic field in this direction the induced current must be counter clockwise, so from b to a in the rod. (c) is to the left. To keep the bar moving to the right at constant speed an external force with magnitude and directed to the right must be applied to the bar. Reflect: The force on the rod due to the induced current is directed to oppose the motion of the rod. This agrees with Lenz’s law....
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 Spring '09
 RODRIGUEZ
 Physics, Flux, Magnetic Field

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