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Unformatted text preview: PHY 184 Final Practice exam 1 Solutions 1. Resistors in series first, then resistors in parallel: 34 3 4 12 1 2 12 34 63.8 , 33.0 1 1 1 21.8 ab ab R R R R R R R R R R = + = Ω = + = Ω = + → = Ω 2. Use Gauss’ law: 2 6.35 5 / q E Nm C ε Φ = = + 3. Potentials add, but the two charges are opposite: 1 2 1 2 1 2 488.893 kQ kQ V V V V r r = + = + =  4. Capacitance with a dielectric: 6.228 A D C C D A ε κ κ ε = → = = 5. The area of the loop changes from 2 2 0.957 r m π = to 0 in 0.061s, thus cos 8.63 B emf d A V B V dt t θ Φ ∆ =  =  = ∆ , and V=RI, so 2.29 I A = . 6. Time behavior of the current: / / t RC t RC q q e V V e = → = because V=q/C. The initial voltage V is simply the full potential difference of the battery. Thus at time t=0.203s, the voltage will then be 137 V V = . 7. The force is given by : sin 1.815 14 F qvB E N θ = = 8. The velocity of the particle is given by: / 5.622 7 / qBr mv r qB v E m s m = → = = . At this velocity, it takes 3.64E8s to go around once, which is 27.45 MHz. this velocity, it takes 3....
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This note was uploaded on 10/03/2008 for the course PHY 184 taught by Professor Gade during the Spring '07 term at Michigan State University.
 Spring '07
 Gade
 Physics

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