p2F09_Exam1_Solutions - 15B B01 9.42x10-5 T 4 B i R = B02...

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F 2009 Exam 1A Solutions A) 1B 2A 2B 4D 5C 6C 7A 8E 9B 10C 11A 12D 13B 14E 15A B01) 6.28x10 -5 T 0 / 4 B i R μ = B02) 500Tm 2 B A Φ = r r g B03) 3.14x10 14 V/m 2 2 cos y kQ E r θ = B04) 30uF 2 0 D C L ε = Ca) 1A Cb) 20V Cc) 20V Cd) 1A Ce) 2x10 -3 J Cf) Exponential decay / -4 0 with RC=10 t RC V V e s - = F2009 Exam 1B Solutions A) 1A 2E 3D 4B 5A 6C 7B 8E 9B 10D 11E 12C 13B 14D
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Unformatted text preview: 15B B01) 9.42x10-5 T / 4 B i R = B02) 75Tm 2 B A Φ = r r g B03) 4.6x10 14 V/m 2 2 cos y kQ E r = B04) 20uF 2 D C L = Ca) 1A Cb) 10V Cc) 10V Cd) 1A Ce) 5x10-4 J Cf) /-5 with RC=5x10 t RC V V e s-=...
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This note was uploaded on 10/17/2011 for the course PHYS 1010 taught by Professor Schroeder during the Spring '07 term at Rensselaer Polytechnic Institute.

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