Download Exam3_Equations - Massachusetts Institute of Technology

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