formula1 - Magnetic ²eld from an in²nitely long current...

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Physics 2054 Midterm-I, Sept 30, 2008 Useful informations Coulomb’s Law: F = k q 1 q 2 r 12 2 ˆ r = 1 4 πǫ 0 q 1 q 2 r 12 2 ˆ r Electric Field due to a point charge: E = k Q r 2 ˆ r Electric potential due to a point charge: V = k Q r Potential energy between two charges: U = k q 1 q 2 r 12 Parallel combination of resistances 1 R = 1 R 1 + 1 R 2 Series combination of capacitances 1 C = 1 C 1 + 1 C 2 Series combination of resistances R = R 1 + R 2 R = ρ L A Parallel combination of capacitances C = C 1 + C 2 Capacitance for a parallel plate capacitor = C = ǫ 0 A d Q = VC Electrostatic energy U = 1 2 CV 2 Ohm’s Law: V = IR , Electric Power P = IV Kirchho±’s rule: loop V = 0 At any node I = 0 Magnetic force F = qvB sin θ
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Unformatted text preview: Magnetic ²eld from an in²nitely long current carrying wire B = μ o 2 π I R Magnetic force on a current carrying wire of length L: F = ILB sin θ The directions of the magnetic ²eld due to a straight wire and a circular loop B going in X B coming out B coming out Fundamental constants e = 1 . 6 × 10-19 C m e = 9 . 1 × 10-31 Kg m p = 1 . 67 × 10-27 Kg h = 6 . 62 × 10-34 J · s c = 1 √ ǫ μ = 3 . × 10 8 m / s μ = 4 π × 10-7 T · m / A ǫ = 8 . 85 × 10-12 C 2 · N · m 2...
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This note was uploaded on 09/21/2009 for the course PHY 2054 taught by Professor Zmudskyy during the Fall '08 term at University of Central Florida.

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