cheat sheet for the final
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cheat sheet for the final

Course Number: ENG 77, Spring 2008

College/University: Tufts

Word Count: 1151

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Final Exam Current and the Motion of Charges -the rate of flow of electrical charge through a cross-sectional area I = Q/t (1A = 1 C/s) I = qnAvd, n = # of free charges = (Navog.)/Mmolar, vd = drift velocity Resistance and Ohm's Law potential drop V = Va Vb = EL Resistance : ratio of potential drop to the current R = V/I (1 = 1V/A) R = L/A -resistance of a conductivity wire Energy in Electrical Circuits P = IV =...

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Exam Final Current and the Motion of Charges -the rate of flow of electrical charge through a cross-sectional area I = Q/t (1A = 1 C/s) I = qnAvd, n = # of free charges = (Navog.)/Mmolar, vd = drift velocity Resistance and Ohm's Law potential drop V = Va Vb = EL Resistance : ratio of potential drop to the current R = V/I (1 = 1V/A) R = L/A -resistance of a conductivity wire Energy in Electrical Circuits P = IV = I2R = V2/R (1 watt = 1 J/s) -power dissipated (potential energy loss per unit time) in the conducting segment -electromotive force (emf) supplies energy to a circuit (volts), has the same equations as V -charge through a source of emf (battery), PE increased by Q P = I -rate at which energy is supplied by the source Va,+ Vb,- = Ir, r = internal resistance of battery I = /(R + r) W = Q, Q = total charge the battery can deliver -total energy stored in a battery Combinations of Resistors Series: Req = R1 + R2 + R3 +... V = IR1 + IR2 = I(R1 + R2) Parallel: Req = (1/R1 + 1/R2 + 1/R3 +...)-1 V = I1R1 = I2R2 -the current divides into each branch proportionally Kirchhoff's Rules 1. when any closed loop surface is traversed, the sum of the changes in potential must = 0 2. at any junction (branch point), the sum of the currents into the junction must equal the sum of the currents out of the junction Single Loop Circuits: ex.) IR1 2 Ir2 IR2 + 1 Ir1 = 0 Multi-Loop Circuits: -the V must still sum to 0 for each loop, do the inner loop then th outer loop RC Circuits -contain a resistor and a capacitor Discharging a capacitor: no battery, switch is closed and a current flows, current = the rate of decrease of the charge on the + plate of the capacitor I0 = V0/R = Q0/RC Q/C IR = 0, Q/C RdQ/dt I = I0e-t/, Q = Q0e-t/ time constant = RC = time it takes the charge to decrease by a factor of e-1 Charging a capacitor: add a battery IR Q/C = 0 I = dQ/dt = (/R)(e-t/RC) = I0e-t/, I0 = /R Q = Qmax(1 e-t/) Energy conservation in charging a capacitor : W = Qf = C2 -work done by the battery -half the work is accounted for by the energy stored in the capacitor, the rest goes as thermal energy WR = 2C = Qf -total amount of joule heating The Force Exerted by a Magnetic Field F = qv x B (1T = 1 N/(Cm/s) = 1 N/Am) -right hand rule: fingers in direction of v, thumb in v x B, fingers curl towards B -F is perpendicular to v and B I = nqvA, A=cross-sectional area -current in a wire F = IL x B Motion of a Point Charge in a Magnetic Field F = ma qvB = mv2/r r = mv/qB T = 2m/qB = 2f = qB/m Torques on Current Loops and Magnets Fnet on loop = 0 F1 = F2 = IaB (ab = A = area of the loop) = F2Bsin = IABsin = NIA (Am2) =xB -magnetic dipole moment PE of a Magnetic Dipole in a Magnetic Field rotated through d: dW = - d = -Bsin d dU = -dW U = -Bcos = -B if U0 = 0 Magnetic Field of Moving Point Charges -a moving charge produces a field B = (0/4)(qv x r /r2) 0 = 4 x 10-7Tm/A = 4 x 10 -7N/A2 The Magnetic Field of Currents dB = (0/4)(Idl x r /r2), dB=0 if dl and r are parallel -Biot-Savart Law B = (0/4)(I /R2)2R = 0I /2R -field due to a current in a loop Bx = (0/4)(2R2I /(x2 + R2)3/2 at x >> R, Bx = (0/4)(2 / |x|3), = IR2 -field the on axis of a current loop Bx = 0nI /2 [x2 /(x22 + R2) x1 /(x12 + R2)] n = turns/unit length x1x2 =dist. ends-origin -field on the axis of a solenoid @ x = 0 Bx = 0nI -field inside a long solenoid B = (0/4)(I/R)(sin2 sin1) -field due to a straight wire segment B = (0/4)(2I /R) -field due to an infinitely long straight wire Magnetic Force between Parallel Wires dF2 = |I2dl2 x B1| -segment I2dl2 dist. R away from a || wire, mag. of the force on the segment dF2/dl2 = 0I1I2 / 2R = (20/4)(I1I2 /R) -when R<<wire length Gauss's Law for Magnetism -net flux inside a closed magnetic surface is = 0 m,net = Bn dA = 0 Ampere's Law -used when high degree of symmetry c Bt dt = B dl = 0Ic B = (0NI /2R) B = (0NI/2r) r = mid-radius btwn outer/inner N = # of turns -field inside a tightly wound toroid Magnetization and Susceptibility M = d /dV = A di/A dl = di/dl d = A di -di is the Amperian surface current Bm = 0M -field inside the cylinder, far from its ends Atomic Magnetic Moments L = mvr -angular momentum = IA = Ir2 = qvr -magnetic moment I = q/T = qv/2r -current = -e/2me (L/) = -BL/ -due to the motion of an electron B = e/2me = 9.27 x 10-24 Am2 or J/T -Bohr magneton s = -2e/2me (s/) = -2Bs/ -due to electron spin Magnetic Flux = s Bn dA = s Bn dA 1 Wb = 1 Tm2 m = BAcos = BnA -flat surface with area A and uniform B m = NBAcos -surface bound by coils of wire Induced EMF and Faraday's Law = -dm/dt -induced emf in a stationary circuit in a changing magnetic field Lenz's Law: induced emf tends to oppose the change that produces it Motional EMF = c (v x B) dl = -dm/dt -induced by motion of conductor in a magnetic field = vBl -magnitude of emf for a rod moving _|_ to both the rod and B V = vBl Ir Inductance m = LI 1 H = 1 Wb/A = 1 Tm2/A -self-inductance L = m/I = 0n2Al -self-inductance of a solenoid V = Ir = -LdI/dt Ir -potential difference across an inductor M1,2 = M2,1 I1 -mutual inductance M B1 = 0n1I1 (from the inner) m = 0n2n1l(r12)I1 M2.1 = 0n2n1 lr12 -inductance of concentric solenoids Magnetic Energy Energy stored in an inductor: Um = LI2 Magnetic energy density: um = B2/20 RL Circuits Ifinal = 0/R 0 = emf of battery I = 0/R (1 e-(R/L)t) = L/R Flux Quantization = nh /2e n = 1, 2, 3... -total flux through the surface of a superconductor 0 = h/2e = 2.0678 x 10-15 Tm2 -fluxon, smallest unit of flux Alternating Current Generators m = NBAcos peak = NBA = t + = peaksin(t + ) Alternating Current in a Resistor VR = = peaksin(t + ) = VR,peak cos(t) -voltage drop across a resistor VR,peak = cost = IR I = (VR,peak/R)cos(t) = Ipeakcos(t) P = I2R = (Ipeakcos(t))2R = Ipeak2Rcos2(t) -power dissipated varies with time Pavg = (I2R)avg = Ipeak2R -average power dissipated Root Mean Square Values Irms = (I2avg) by definition (I2)avg = Ipeak2 Irms = Ipeak/(2) Pavg = Irms2R rms = peak/(2) Pavg = rmsIrms Alternating Current Circuits XL = L = 2fL Irms = Vrms/XL -inductive resistance Vc = = peakcos(t) = Vc,peakcos(t) -voltage drop across a capacitor Ipeak = CVc = Vc,peak/(1/C) = Vc,peak/Xc Irms = Vc,rms/Xc XL = 1/C capacitive reactance V2 = (N2/N1)V1 N1= primary, fewer coils V2I2 = V1I1 -transformer voltage drop

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