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1999 E&M C2 Solution

1999 E&M C2 Solution - E& M SOLUTIONS#2 1999...

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Unformatted text preview: E & M SOLUTIONS #2 1999 Physics C Solutions Distribution of Points E & M 2 (15 points) (a) 5 points For using Faraday's law for a loop 1 point an M or =-—— dt A! , For relating magnetic flux to magnetic field and area 1 point flzAiB. or fl=A£ d: d: A! A: For using the proper expression for the area of a loop I point A=rrr2 8=Jrr2£ or 8:127:52 d: A: For using the correct radius, i.e. the radius of the field I point a = rr(0.6 m)1 (0.40 T/s) For cornputing the correct answer 1 point 5 = 0.45 V (b) 3 pom For any statement of Ohm's law 1 point V = [R Solving for the current: I = VI}? = 457R I =(0.45 v)/(5.0 n) For computing the correct answer 1 point I = 0.090 A For indicating a clockwise direction for the current 1 point (e) 3 points For relating the energy dissipated to the power in the resistor 1 point E=IP d! or E=Pr For an expression for electric power 1 point . V3 P = PR or --- or IV Example using P = 12R: E = 13R! E = (0.090 ii)2 (5.0 r2)(15 s) For computing the correct answer 1 point E = 0.61 .l E 8: M SOLUTION f2 1999 Physics C Solutions Distribution of Points E & M 2 (continued) (d) 3 points For stating that the brightness of thebe will be less 1 point Pct indicating that the reduction in brightnms is due to a decrease in cuncntoradecreaseinthecmf lpoint For indicating that the deans: in current or emf. our the reduction in brightness, isduetoadeczeaseinthearcaofthelooporadecreaseinthcchangtngflux lpo'mt For using correct units with three numerical answers 1 point ...
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