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# test2 - force acting on moving charge in magnetic ﬁeld ~...

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Formula Sheet - Physics 317L - Test2 Electric Charge: electric current: i = dq dt , electrical force: F = 1 4 π 0 | q 1 || q 2 | r 2 permittivity constant : 0 = 8 . 85 × 10 - 12 C 2 /Nm 2 coulomb constant : k = 1 4 π 0 = 8 . 99 × 10 9 Nm 2 /C 2 elementary charge: e 1 . 602 × 10 - 19 C Electric Field: electric field : ~ E = ~ F q 0 , electric field by point charge q: ~ E = 1 4 π 0 | q | r 2 ~ r Electric Potential: electric potential energy: Δ U = U f - U i = - W (work : W = F Δ x ) electric potential difference: Δ V = V f - V i = - W q , V = U q potential of point like charge: V = 1 4 π 0 q r electric field: E = - Δ V Δ s Capacitance: capacitance: C = 0 A d , charge on plates: q = CV capacitor in parallel: C eq = C 1 + C 2 , capacitor in series: 1 C eq = 1 C 1 + 1 C 2 potential energy: U = 1 2 CV 2 Current and Resistance electrical current: i = dq dt resistance of conductor: R = V i ; R = ρ L A power: P = iV = i 2 R = V 2 R closed circuits emf : ε = dW dq = iR resistance in series: R eq = R 1 + R 2 , resistance in parallel: 1 R eq = 1 R 1 + 1 R 2 charging capacitor: q = Q (1 - e - t/RC ) discharging capacitor: q = q 0 e - t/RC , q 0 = initial charge Magnetic Fields:
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Unformatted text preview: force acting on moving charge in magnetic ﬁeld: ~ F B = q~v × ~ B charged particle circulation in magnetic ﬁeld: | q | vB = mv 2 r magnetic force on wire: ~ F B = i ~ L × ~ B , ~ L = length vector magnetic ﬁeld of long straight wire: B = μ i 2 πR , μ = 4 π x 10-7 [T m/A] or [H/m] ﬁeld of Solenoid: B = μ in , n =number of turns per unit length Induction and Inductance: magnetic ﬂux: Φ B = BA , ( A = area transverse to ~ B ) Faraday’s law: ε =-N d Φ B dt inductance: L = N Φ B i , for solenoid L l = μ n 2 A self-induction: ε L =-L di dt series RL circuit (rise of current): i = ε R (1-e-t/τ L ) , τ L = L/R magnetic energy: U B = 1 2 Li 2...
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