solutions_chapter21

Q 5 mc d t 5 1 0.250 kg 21 4190 j kg c° 21 80.0 c°

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Unformatted text preview: Q 5 mc D T 5 1 0.250 kg 21 4190 J / kg # C° 21 80.0 C° 2 5 8.38 3 10 4 J. U 5 1 2 Li 2 . Q 5 mc D T i 5 0.641 A. U max 5 1 2 Li max 2 5 1.03 mJ. i 5 0. v max 5 11.7 V. q 5 0. i max 5 v max Å C L 5 1 11.7 V 2 Å 15.0 3 10 2 6 F 5.00 3 10 2 3 H 5 0.641 A. 1 2 C v max 2 5 1 2 Li max 2 . v max 5 Q C 5 175 m C 15.0 m F 5 11.7 V. U L 5 1 2 Li 2 . U C 5 1 2 C v 2 . v S B S v S E 5 1 0.350 m / s 21 cos 45° 21 0.120 T 21 0.707 m 2 5 0.0210 V. v cos 45°. v S B S . v S E 5 1 0.350 m / s 21 cos 45° 21 0.120 T 21 0.500 m 2 5 0.0148 V. v cos 45°. v S B S . v S B S , v S E 5 v BL , " 2 1 0.500 m 2 2 5 0.707 m. 21-14 Chapter 21 (d) Let clockwise currents be positive. At the loop is entering the field. It is totally in the field at time and beginning to move out of the field at time The graph of the induced current as a function of time is sketched in Figure 21.64. Figure 21.64 21.65. Set Up: At the angle between the normal to the loop and the magnetic field is zero. From the discussion of generators in Section 21.3, the rate of change of is The induced emf is Solve: (a) The graph of versus t is sketched in Figure 21.65a. (b) The graph of versus t is sketched in Figure 21.65b. (c) The graph of versus t is sketched in Figure 21.65c. (d) Doubling halves the period of and doubles its amplitude. The graph of versus t is sketched in Figure 21.65d. Figure 21.65 F B t ( a ) t ( c ) E t ( b ) t ( d ) E E 5 2 v AB sin 1 2 v t 2 E v E 5 v AB sin v t 2v BA sin v t F B F B 5 BA cos v t . v AB sin v t . 2v BA sin v t . F B F B 5 BA cos f . f 5 v t . f t 5 I t t b t a t b . t a t 5 Electromagnetic Induction 21-15 21.66. Set Up: The magnetic field of the long wire has magnitude and is directed into the page at the location of the loop. when is perpendicular to the conductor and when is parallel to the conductor. Lenz’s law says the flux of the induced current opposes the change in flux. The direction of the magnetic force on the positive charges in each side of the loop is given by applying the right-hand rule to and Label the sides of the loop as shown in Figure 21.66. Figure 21.66 Solve: (a) The velocity is parallel to sides 1 and 3 so there is no emf in those sides. The emf in side 2 is and is directed to produce a counterclockwise current. The emf in side 1 is and is directed to produce a clockwise current. The emfs in these two sides are in opposite directions so the net emf is (b) (i) Lenz’s law: The magnetic field is into the page and the flux through the loop is decreasing as the loop moves toward the smaller field. Therefore, the magnetic field of the induced current is into the page inside the loop and to produce magnetic field in this direction the induced current is clockwise. (ii) Magnetic forces: The magnetic force on a positive charge in side 4 is toward the top of the page. The magnetic force on a positive charge in side 2 is toward the bottom of the page. The force in side 4 is larger since the magneitc field of the wire is larger there. The net forcethe bottom of the page....
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Q 5 mc D T 5 1 0.250 kg 21 4190 J kg C° 21 80.0 C° 2 5...

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