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Electromagnetic Theory > Formulaes > Flashcards

Formulaes Flashcards

1
Q

Coloumb’s Law

A

F = 1/4πε0 qQ/ℛ^2 ℛ(hat)

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2
Q

Principle of superposition

A

F = F1 + F2 + F3 + …

E = E1 + E2 + E3 + …

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3
Q

Density of field lines

A

n/4πr^2 where n is the number of field lines

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4
Q

What are Maxwell’s equations?

A

∇ . E = p/ϵ Gauss’s law

∇ . B = 0 Gauss’s law

∇ x E = - ∂B/∂t Faraday’s Law

∇ x B = μ0 (J + ϵ ∂E/∂t) Ampere-Maxwell Law

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5
Q

What are Maxwell’s equations in Integral Form?

A

∫ E . da = qenc/ϵ Gauss’s law

∫ B . da = 0 Gauss’s law

∫ E . dl = -∂/∂t ∫ B .da Faraday’s Law

∫ B .dl = μ0 (Ienc + ϵ ∂/∂t ∫ E .da) Ampere-Maxwell Law

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6
Q

The Electric Field Strength

A

F = QE

where E = 1/4πε0 ∫ qi/ℛ^2 ℛ(hat) dr

and ℛ(hat) = ℛ(vector)/ℛ

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7
Q

General work done or energy

A

W = Q ΔV

where ΔV = - ∫ E .dl

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8
Q

The separation vector

A

where ℛ = r-r’

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9
Q

Electric field for a line charge

A

E = 1/4πε0 ∫ λ(r)/ℛ^2 ℛ(hat) dl

where λ(r) is the charge-per-unit-length

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10
Q

Electric field for a surface charge

A

E = 1/4πε0 ∫ σ(r)/ℛ^2 ℛ(hat) da

where σ(r) is the charge-per-unit-area

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11
Q

The electric dipole moment

A

p = qd

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12
Q

The electric dipole potential

A

V(r) ~ 1/4πϵ qdcosθ/r^2

and p = qd

V(r) ~ p/4πϵ cosθ/r^2

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13
Q

Electric field for a volume charge

A

E = 1/4πε0 ∫ p(r)/ℛ^2 ℛ(hat) dτ

where p(r) is the charge-per-unit volume

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14
Q

the flux of E through a surface

A

Φ = ∫ E.da

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15
Q

Starting from Gauss’s law in integral form derive Gauss’s law in differential form.

A

∫ E.da = Q(enc)/ε0

∫ E.da = ∫ ∇ . E dτ

Qenc = ∫ p dτ

∫ ∇ . E dτ = ∫ p/ε0 dτ

∇ . E = p/ε0

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16
Q

Electric potential derivation

A

V(r) = - (r ∫ O) E.dl

where O is some standard reference point.

V(b) - V(a) = (b ∫ a) ∇V . dl

The potential difference is therefore

E = -∇V

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17
Q

Show that V(r) = q/4πε0r

A

V(b) - V(a) = - (b ∫ a) E . dl

E = 1/4πε0 q/r^2

integrate

V = q/4πε0r

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18
Q

Derive poisson’s equation and therefore Laplace’s equation

A

∇. E = ∇. (-∇V)

∇. E = -∇^2V

from Gauss’s law

∇^2V = -p/ε0

Laplace’s equation when regions lacking charge p = 0

∇^2V = 0

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19
Q

Derive the energy density associated with the electric field.

A

W = (b ∫ a) F. dl and F = -QE

W = Q[V(b) - V(a)]

W = 1/2 (N Σ i=1) (N Σ j =1) qi Vij on formula sheet

W = 1/2 ∫ pV dτ

W = ε0/2 ∫ (∇ .E) V dτ

use vector identitiy

W = ε/2 ∫ E^2 dτ

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20
Q

The potential for a volume charge

A

V(r) = 1/4πε0 ∫ p(r)/ℛ dτ

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21
Q

The energy of a continuous charge distribution

A

W = ε/2 ∫ E^2 dτ

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22
Q

Area of : Circle, Sphere and Cylinder

A

Circle : πr^2
Sphere : 4πr^2
Cylinder : 2πrl

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23
Q

ratio of conductors

A

R = distance from the conductor surface/ size of the conductor

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24
Q

Integral limits for outer and inner

A

(r ∫ R) , where r is inner and R is outer.

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25
Q

The torque in an electric field

A

N = p x E

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26
Q

The work done during rotation by a dipole in an electric field

A

U = -p . E

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27
Q

Surface charge density

A

σ = Pcosθ = P . n

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28
Q

Bound volume charge

A

pb = - ∇. P

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29
Q

Electric Displacement.

A

D = ϵ0E + P

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30
Q

Derive Gauss’s law in terms of electric displacement.

A

∇ . E = pf+pb/ϵ0

∇ . (ϵ0E + P) = pf

pb = - ∇. P

and σb = P . n

thus D = ϵ0E + P

hence ∇. D = pf in differential form

and ∫ D . da = qf,enc in integral form.

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31
Q

Polarisation

A

P = ϵ0 Χe E

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32
Q

Show that D =ϵ0ϵr E

A

D = ϵ0E + P
D = ϵ0E + ϵ0ΧeE
D = ϵ0(1+ Χe)E
D = ϵ0ϵrE

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33
Q

Derive the normal and parallel components of electric field.

A

∫ E .da = Qenc/ε = σA/ε

E⊥above -E⊥below = σ/ε

∫ E . dl = 0

E∥above = E∥below

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34
Q

Derive the normal and parallel components of electric displacement and polarisation.

A

∇ x D = ε (∇ x E) + ∇ x P
∇ x D = ∇ x P

such that D⊥above - D⊥below = σf = 0 if no free charge

D∥above - D∥below = P∥above - P∥below

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35
Q

Derive the relationship of capacitance

A

V = V+ - V-
V = (+plate ∫ -plate) E.dl

C = Q/V

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36
Q

Derive the Energy stored in Capacitance.

A

dW = Vdq = q/c dq
since C = Q/V
W = 1/C (Q ∫ 0) qdq
W = Q^2/2C = 1/2 CV^2 = 1/2 QV.

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37
Q

What is the Energy Stored in a Capacitor with Dielectric

A

W = ε/2 ∫ εr E^2 dτ

W = 1/2 ∫ D . E dτ

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38
Q

Derive the Lorentz Force Law

A

dFmag = I(dl x B)

dFmag = I dt(v x B)

Fmag = Q(v x B)

F = Q(E + v x B)

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39
Q

Work done by magnetic force

A

dW = Fmag . dl
dW = Q(v x B). vdt
dW = 0

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40
Q

Derive the Continuity Equation

A

J = I/πa^2

I = ∫ J .da

∫ J . da = ∫ ∇ . J dτ

∫ ∇ . J dτ = -d/dt ∫ p dτ

Giving the continuity equation

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41
Q

Ampere’s Law

A

∫ B . dl = µ0 I

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42
Q

Derive the Time-Independent Ampere-Maxwell Law.

A

Ienc = ∫ J .da

1/μ ∫ B .dl = ∫ J .da

1/μ ∫ ∇ x B . da = ∫ J.da

∇ x B = μJ

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43
Q

Amperian Loops

A

∫ (∇ x B) . da = ∫ B . dl = µ ∫ J .da

where ∫ J .da = Ienc

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44
Q

Derive Magnetic flux

A

∇ . B = 0

apply divergence theorem

∫ ∇ . B dτ = ∫ B . ds = 0

hence Φ = ∫ B . dS

where Φ1 + Φ2 = ∫ B . dS + ∫ B . dS = 0

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45
Q

Magnetic Vector Potential

A

B = ∇ x A

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46
Q

Vector Magnetic Potential for a dipole

A

A(r) = µ0 (m x r)/4πr^3

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47
Q

Show that ∇^2 A = -µ0J

A

∇ x B = ∇ x ( ∇ x A)
∇ x B = ∇ ( ∇ . A) - ∇^2 A = µ0J

such that ∇^2A = -µ0J

48
Q

Induced dipole moment and dipole moment

A

p = αE : induced dipole moment

where α is the atomic polarisability

P = Np : dipole moment

N atoms per unit volume and P polarisation

49
Q

In a vacuum

A

free space such that

Xe = 0 and Xm = 0

ϵ = ϵ0 and μ = μ0

50
Q

In matter

A

ϵ = ϵ0ϵr and μ = μ0μr

51
Q

Torque in the Magnetic Field

A

N = m x B

52
Q

The work done by a rotating dipole in a magnetic field

A

U = -m . B

53
Q

Magnetic moment

A

m = nIA

54
Q

Total current density

A

J = Jf + Jb

55
Q

Bound surface current

A

Kb = M x n

56
Q

Bound volume current

A

Jb = ∇ x M

for steady currents ∇ . Jb = 0

57
Q

Derive the auxiliary field H

A

1/µ0 ( ∇ x B) = Jf + Jb

Jb = ∇ x M

1/µ0 ( ∇ x B) = Jf + ∇ x M

∇ x (B/µ0 - M) = Jf

hence H = B/µ0 - M

and in integral form, ∫ H . dl = If,enc

58
Q

Magnetic Susceptibility

A

M = χm H

59
Q

Show that B = µ0µr H

A

B = µ0(H + M)
B = µ0(1+ χm) H

hence B = µ0µrH

60
Q

Relative permeability

A

µr = 1 + χm

61
Q

Perpendicular component of magnetic field.

A

B⊥above = B⊥below

62
Q

Derive the parallel component of magnetic field and hence the auxiliary parallel component.

A

H⊥above - H⊥below = -(M⊥above - M⊥below)

B∥above - B∥below = µ0(Kf + Kb)

=> H∥above - H∥below = Kf

63
Q

Motional electromotive force (EMF)

A

ε = -d/dt ∫ (B .dS)

the RHS of the integral form of faradays law.

64
Q

Magnetic induction

A

∫ E .dl = - ∫ ∂B/∂t . dS

∫ E .dl = - dΦm/dt

Φm = LI

65
Q

Derive Ohm’s Law

A

J = σf

f = F/q

J = σ(E + v x B)
J = σE

66
Q

Derive V = IR

A

I = JA = (σE)A

= (σ V/L)A = (σA/L)V

I = V/R

67
Q

Show that emf = electric potential

A

ε = ∫ f . dl = ∫ fs . dl

V = - (b ∫ a) E. dl
V = ( b ∫ a) fs .dl = ∫ fs . dl

V = ε

68
Q

Magnetic flux

A

Φ = ∫ B . dS

69
Q

flux rule for EMF

A

ε = - dΦ/dt

70
Q

Derive the differential form of Faraday’s law starting from the definition of EMF.

A

ε = -dΦ/dt = ∫ E . dl

  • ∫ dB/dt . dS = ∫ E . dl apply stokes theorem to RHS

∫ - dB/dt . dS = ∫ ∇ x E . dS

=> -dB/dt = ∇ x E

71
Q

derive the emf induced.

A

ε = - dΦ/dt

where Φ = LI

ε = -L dI/dt

72
Q

Derive the work done (induced) per unit time

A

dW/dt = -εI = LI dI/dt

W = 1/2 LI^2

73
Q

energy stored in the magnetic field

A

W = 1/2µ0 ∫ B^2 dτ

such that

Umag = 1/2 ∫ B. H dτ

74
Q

Derive the Ampere-Maxwell Law

A

Start from the continuity equation

∇ . J = -∂p/∂t = -∂/∂t (ε0 ∇. E)

∇ . J = - ∇ . (ε0 ∂E/∂t)

such that

∇ x B = µ0J + µ0ε0 ∂E/∂t

75
Q

Displacement current

A

jd = ε0 ∂E/∂t

such that ∂D/∂t = jd

76
Q

Derive the polarisation current

A

pb = - ∇ . P

∂pb/∂t = - ∂/∂t ∇ . P

∂pb/∂t = - ∇ . ∂P/∂t

  • ∇ . Jp = -∇ . ∂P/∂t

=> Jp = ∂P/∂t

77
Q

The free current density

A

Jf

78
Q

The bound current density, in magnetic materials

A

Jb

79
Q

Maxwell’s fix for ∂E/∂t ≠ 0

A

jd

80
Q

The polarisation current in dielectrics when ∂E/∂t ≠ 0

A

Jp

81
Q

Completed Ampere-Maxwell law in matter

A

∇ × B = μ0 [Jf +Jb+jd+Jp]

82
Q

Derive the auxiliary field H in terms of Jf, Jb, jd and Jp

A

∇ × B = μ0 [Jf +∇×M+jd+Jp]
∇ × ( B − μ0M ) = μ0 [Jf +jd +Jp]
μ0 ∇ × H = μ0 [Jf + jd + Jp]
∇ × H = Jf + jd + Jp.

where substitutions can be made for jd and Jp

such that

∇ × H = Jf + ∂D/∂t

83
Q

Maxwell’s equations in static fields

A

no time-dependence

∇ · E = p/ε0
∇ · B = 0
∇ × E = 0
∇ × B = μ0J

84
Q

Maxwell’s equations in quasi-static conditions

A

a good conductor

∇ · E = p/ε0
∇ · B = 0
∇ × E = −∂B/∂t
∇ × B = μ0J

85
Q

Maxwell’s equations in free space.

A

No sources or sinks

∇ · E = 0
∇ · B = 0
∇ × E = −∂B/∂t
∇ × B = μ ε ∂E/∂t

86
Q

speed in wave equation

A

c = 1/√µε

87
Q

Derive the Wave equation.

A

Start from faraday’s law in free space

Take cross product of both sides

and use the vector identity.

88
Q

To determine if solutions are of the wave equation.

A

Substitute solution in the plane wave solutions.

where they are related through c = ω/k

89
Q

Concentric

A

Spherical - point charge

90
Q

Coaxial

A

Cylindrical (line charge)

91
Q

Pillbox

A

Plane (surface charge)

92
Q

Out of the page and into the page

A

𝇇 out of the page
⨂ into the page

93
Q

the total energy stored in electromagnetic fields

A

u = 1/2(ε0E^2 +1/μ0B^2) energy density

U = 1/2 ∫ (ε0E^2 +1/μ0B^2) dτ

94
Q

Poynting vector and energy stored in fields Uem.

A

Energy crossing per area per unit time

S = 1/μ0 (E x B)

dW/dt = ∫ S . da

95
Q

Power through a surface area.

A

P = ∫ S . da

96
Q

Derive the displacement current Id

A

Start from the integral form of ampere-maxwell’s equation.

Replace dE/dt with Jd.

Id = ∫ Jd . da

97
Q

Derive the time-averaged poynting vector. Given the plane wave solution E = E0 exp[i(kz-ωt)] and B = B0 exp[i(kz-ωt)].

A

take the plane wave solutions and substitute them into the Poynting vector.

looking at the real fields Re

B0 = E0/c where cos^2 -> 1/2 over one cycle.

giving <S> = E0^2/2cµ0</S>

98
Q

refractive index

A

v = 1/√µε

c = 1/√µ0ε0

v = c/n

n = √(εµ/ε0µ0) = √εrµr

in general n = √εr

99
Q

complex permittivity

A

εr = ε1 + iε2

100
Q

relationship between v, k and

A

v ̃= ω/k ̃= 1/√ε0εrµ0µr

101
Q

Skin depth

A

d = c/ωη = 2c√ε1 /ωε2

102
Q

Reflection

A

R = (E0,r/E0,i)^2 which gives R on the formula sheet

103
Q

Transmission

A

T = ε2v2/ε1v1(E0,t/E0,i)^2 which gives transmission on the formula sheet

104
Q

What is the relation between transmission and reflection

A

R + T = 1

105
Q

Current

A

I = dq/dt

106
Q

Derive the relationship between pb and σb from polarisation

A

Use divergence theorem

∫ ∇ · P dτ + ∫ P . da = 0

P . da = P. n(hat) dS

and relationships on the formula sheet

107
Q

r(hat) x dl

A

|r(hat)||dl|sin theta

where r(hat) = r(vector)/r

108
Q

For a non magnetic media

A

μr = 1

109
Q

Diamagnetism

A

I = e/T = ev/2πR

110
Q

derive the change in magnetic moment

Δm = -e^2R^2/4me B

A

coloumb’s attraction = centripetal acceleration

1/4πϵ e^2/R^2 = me v^2/R

speed changes in presence of magnetic field

1/4πϵ e^2/R^2 + eṽB = me ṽ^2/R

eṽB = me/R (ṽ^2 -v^2)

Δv = ṽ -v = eRB/2me use binomial expansion

Δm = -1/2 e(Δv)Rz(hat)

Δm = -e^2R^2/4me B

111
Q

Derive Poynting’s Theorem

A

dW = F.dl

v = dl/dt
dW = q(E+vxB).vdt
q = pdV and pv = J
dW = q(E.v)dt

dW/dt = ∫ E.J dV
E. J = 1/μ E. (∇ x B) – ϵE . ∂E/ ∂t

Curl identity
Then substitute for faradays law
B . ∂B/∂t = 1/2 ∂/∂t (B^2)
E . ∂E/∂t = 1/2 ∂/∂t (E^2)

Gather E^2 and B^2 terms

Arriving at pointing theorem

112
Q

β

A

β = μ1v1/μ2v2 = μ1n2/μ2n1

113
Q

cross product and dot product

A

cross product => sin(theta)
dot product => cos(theta)

114
Q

parallel and perpendicular

A

parallel => sin(theta)
perpendicular => cos(theta)

115
Q

Conduction current

A

Jc = σE

Electromagnetic Theory (2 decks)

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