s01x1ans - b) A L R = , so R L A = , and L = 2 r . Then, =...

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PHYS-1200 PHYSICS I I QUIZ 1 FEBRUARY 7, 2001 SOLUTIONS PART A. 1. D 6. D 2. D 7. E 3. A 8. E 4. D 5. A PART B . 1. a) Since the Gaussian surface is in the conducting metal, the electric field must be equal to zero everywhere on the Gaussian surface. Therefore, ( A E d ) A = 0 ¯¯¯¯¯¯¯¯¯¯¯¯ b) In this case, the integral equals the net charge enclosed by the surface divided by ε 0 . Then, ( A E d ) B = q / 0 ¯¯¯¯¯¯¯¯¯¯¯¯¯¯ 2. According to Ampere’s Law, the integral equals the net current that flows through the path of integration, times μ 0 . Two of the currents cancel, and the one that remains produces a magnetic field that is opposite in direction to the path of integration. Then, s B d = – (2.0 A) 0 ¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯
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PART C . 1. a) A 0 . 5 V 020 . 0 = = i R E R = 4.0 ×10 -3 = 4.0 m
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Unformatted text preview: b) A L R = , so R L A = , and L = 2 r . Then, = =--10 . 4 m) 10 . ( 2 m) 10 7 . 1 ( 2 3 8 R r A A = 2.7 10-6 m c) THE MAGNETIC FIELD IS INCREASING d) dt dB A dt BA d dt d B = = = ) ( E . Since A = r , this becomes, dt dB r 2 = E . Then, 2 2 m) (0.10 V 020 . = = r dt dB E dt dB = 0.64 T/s 2. Since the two capacitors are in series, the charge on each capacitor is equal to the charge on the equivalent capacitor. The capacitance of the equivalent caacitor is given by, C C C C 2 1 1 ' 1 = + = . Then, 2 ' C C = . Since V q C = ' , V q C = 2 and V q C 2 = . V 100 C 10 . 5 2 5- = C C = 1.0 10-6 F = 1.0 F...
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This test prep was uploaded on 04/18/2008 for the course PHYS 1200 taught by Professor Stoler during the Spring '06 term at Rensselaer Polytechnic Institute.

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s01x1ans - b) A L R = , so R L A = , and L = 2 r . Then, =...

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