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MASSACHUSETTS INSTITUTE OF TECHNOLOGY Department of Physics 8.02 Exam 3 Equation Sheet Force Law: Fq = q(Eext + v q ´ Bext ) Force on Current Carrying Wire: F= ò Id s¢ ´ Bext wire Magnetic Dipole Faraday’s law ò E × ds = - closed fixed path d B × n̂ da dt openòò surface S e = I ind R Gauss’s Law for Magnetism: òò B × n̂ da = 0 closed surface Torque on a Magnetic Dipole Gauss’s Law: òò E× dA = qenc closed surface eo Force on a Magnetic Dipole Fz = m z Electric Potential Difference Electrostatics: ¶Bz b DV = Vb - Va º - ò E × d s ¶z a Source Equations: dq dq¢(r - r¢) E(r ) = ke ò 2 rˆ = ke ò 3 r r - r¢ source source m Ids ´ r̂ B(r) = o ò 4p source r 2 B(r) = mo Id s¢ ´ (r - r¢ ) ò 3 4p source r - r¢ r̂ points from source to field point Ampere’s Law ò B × d s = m0 òò J × n̂ da closed path S E = -ÑV Potential Energy: DU = qDV Capacitance: 1 Q2 1 Q C= UE = = CDV 2 2 C 2 DV Inductance: e back L= = - LdI / dt F B,Total I = NF B I U M = 12 LI 2 Energy Density Stored in Fields: uE = 12 e 0 E 2 ; uB = 12 B 2 / m0 1 Current Density and Current: I = òò J × n̂ da Power Dissipated in Resistor: PJoule = I 2 R = DV 2 / R open surface Ohm’s Law: DV = I R J = s c E where s c is the conductivity E = rr J where rr is the resistivity Power: P = F × v Constants: m0 º 4p ´ 10-7 T × m × A-1 e0 º 1 / m0c2 8.85 ´ 10-12 C2 × N -1 × m-2 ke = 1/ 4pe 0 9.0 ´ 109 N × m2 × C-2 Power from Voltage Source: Psouce = IDV 2