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Unit 13 Solutions_Page_02

Unit 13 Solutions_Page_02 - (C 2010 McGraw—Hill(b v is...

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Unformatted text preview: (C) 2010 McGraw—Hill (b) v is parallel to B, so F: 0 and, thus, 12E. (c) {TX E is out of the page. The charges are separated across the width of the rod. There is no current along the rod, so I 2 IE. (d) I: vaL : v(B 81n6)L 1 VBL sin20. 00: 0.3421531, R R R R According to the RHR and F = qi’x E, the current flows down the rod, or -. 5. (a) Strategy Use Newton’s second law, Eq. (19-12b), and Eq. (20-2a). Solution EFy 2 F13 — mg I 0, SO FB = mg when the rod is falling with constant velocity. The magnitude 2 2 ofthe magnetic forceis Fa : EB :ELB : VBL LB : VL B R R R . Set this equal to mg and solve for the terminal speed. 1212282 ng (0.0150 kg)(9.80 m/s2)(8.00 g2) :m,sov: :—: 3.44ms. R g L382 (1.30 m)2(0.450 T)2 (b) Strategy Use the potential energy in a uniform gravitational field and Eqs. (18-21b) and (20-2a). Solution The change in gravitational energy per second is 2 2 2 2 2 A_U _mgAy __mgv:_m2g;R :_(0.0150 kg) (9.80 m/s ) (28.00 (2) 30.505 W, 11—: Ar LB (1.30 m) (0.450 T) so the magnitude ofthe change is 0.505 W. The power dissipated is 2 2 2 2 2 2 2212122221” 2’29? BL =m23§=0505w R R L B R L B The magnitude of the change in gravitational potential energy per second and the power dissipated in the resistor are the same, 0.505 W. 6. Strategy According to Eq. (20-3b), the amplitude of the induced emf is proportional to the angular speed of the armature. Use a proportion. Solution Find the induced ernf. 50cm, soizfl and6f2— i=—'(27.0=V)18-0v. i i i Probhms: 20.15, 20.17, [20.18, 20.19], 20.21, 20.24- 15. (a) Strategy According to Lenz’s law, the direction of an induced current in a loop always opposes the charge in magnetic flux that induces the current. Solution According to the RI-lR, E is directed into the page at the loop. Since B 0C r_1 for a long straight 772 ...
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