f01x1aan - C C S 2 1 1 1 = = so C = 2C S = 2 5.0 ×10-7 F C...

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NAME PHYS-1200 PHYSICS II QUIZ 1 OCTOBER 3, 2001 SOLUTIONS PART A. 1. B 6. D 2. D 7. E 3. B 8. E 4. D 5. B PART B . 1. 2. The resistors are in series so the current is the same in both resistors. i = 0.60 A 3. The figure is below. The reasons for the results are as follows: a) According to the right hand rule, both wires produce fields that point upward at O . b) According to the right hand rule, both wires produce fields that go around the individual wires in a clockwise direction at point P . The veritcal components of the two fields add to zero, and both horizontal components point to the right. c) The wire on the left produces a field that points downward, and the wire on the right produces a field of equal magnitude that points upward at point Q . They add to zero.
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NAME PART C . 1. a) V 10 C 10 0 . 5 6 - × = = = E q V q C S C S = 5.0 ×10 -7 F = 0.50 μ F b) They are in series, so
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Unformatted text preview: C C S 2 1 1 1 = + = , so C = 2C S = 2( 5.0 ×10-7 F) C = 1.0 ×10-6 F = 1.0 F c) The capacitors are in series, so the charge on each capacitor equals q = 5.0 ×10-6 C. The surface encloses the negatively charged plate of the capacitor, so according to Gauss’ law 6 C)/ 10 . 5 ( / ε-×-= ∫ = ⋅ q d A E 2. a) b) According to Lenz’s law the field is DECREASING WITH TIME c) A 10 . 8 V 10 . 5 3 6--× × = = i R E R = 6.25 × 10-4 Ω = 0.625 m Ω d) dt d B Φ-= E , so the magnitude of dt d B Φ is dt d B Φ = 5.0 ×10-6 V = 5.0 V e) ( 29 2 2 A D D I B--= Φ , so ( 29 2 2 A D D dt dI dt d B--= Φ , and ( 29 ( 29 2 2 7-6 2 2 m) 40 . ( m) 50 . ( m 0.50 H/m) 10 (4 V 10 . 5--× × =--Φ =-π A D D dt d dt dI B A/s 20 A/s 9 . 19 = = dt dI...
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