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. (1912b), and Eq. (202a). 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. (1821b) and (202a). 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. (203b), 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)180v. 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 RIlR, E is directed into the page at the loop. Since B 0C r_1 for a long straight 772 ...
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 '08
 RICHARDSON, B
 Physics

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