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Unformatted text preview: Physics 212 Classical and Modern Physics Spring 2008 Solutions for Hand-In Set #3 Problem A13 — Resistance and Dimmer Switches (a) I find a length of 22 cm, give or take a little. (b) From the assignment sheet we get the nichrome re- sistivity ρ = 1 . × 10- 6 Ω · m. Therefore the resistance is R = ρL A = (1 . × 10- 6 Ω · m)(0 . 22 m) π (1 . 25 × 10- 4 m) 2 = 4 . 5 Ω (c) Each battery provides 1 . 5 V, so in series they pro- vide a potential difference of 3 V. Using Ohm’s Law: I = V R = 3 V 4 . 5 Ω = . 67 A Problem A17 — Magnetic force on a moving particle The magnetic force is ~ F = q~v × ~ B . The magnitude of this force is F = | q | vB sin θ , where θ is the angle be- tween ~v and ~ B if I line them up tail to tail. In this problem θ = 60 ◦ , so F = (5 × 10- 6 C)(1 . 3 × 10 5 m)(0 . 8 T) sin(60 ◦ ) = . 450 N As for direction, ~v × ~ B points into the page, by the cross- product RHR. But since this is a negative charge, the direction is flipped: the force points out of the page. Wolfson Chapter 24-32 Power due to current flow is given by P = V I , always. Since we’re given a resistance, we want to use Ohm’s Law: V = IR . Solve for I = V/R and plug this into the power expression to get P = V ( V/R ) = V 2 /R . This is exactly what we need, since we’re given P and R , and we want to find V : V = √...
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This note was uploaded on 04/11/2008 for the course PHYS 212 taught by Professor Ladd during the Spring '08 term at Bucknell.
- Spring '08