Lecture 1 Vector algebra: A brief revie... · Lecture 1 Vector algebra: A brief review Outline 1....
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Transcript of Lecture 1 Vector algebra: A brief revie... · Lecture 1 Vector algebra: A brief review Outline 1....
Lecture 1 Vector algebra: A brief review
Outline
1. Scalars and Vectors
2. Unit vectors
3. Position and distance vectors
4. Algebraic operations with vectors: addition, pair and triple products.
5. Vector projections
1
• Scalar quantities have only magnitudes (mass, pressure, temperature, speed,distance, charge, potential)
• Vector quantities have both magnitudes and directions (velocity, force, electric�eld).
Any vector in terms of coordinates, A = (Ax, Ay, Az)
a =A
|A| ,⇐⇒, unit vector;
the magnitude by Pythagorus' theorem
|A| =√
A2x + A2
y + A2z.
Alternative representation,
A = (Ax, Ay, Az) ⇐⇒ A = Axax + Ayay + Azaz.
a =Axax + Ayay + Azaz√
A2x + A2
y + A2z
.
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Vector addition
C = A±B ⇐⇒ C = (Ax ±Bx)ax + (Ay ±By)ay + Az ±Bz)az.
Position vector
rP = xax + yay + zaz
Distance vector
rPQ = rQ − rP = (xQ − xP )ax + (yQ − yP )ay + (zQ − zP )az
3
Dot Product
A ·B = |A||B| cos θAB = AxBx + AyBy + AzBz.
Properties:
• A ·B = B ·A, commutativity,
• A · (B + C) = A ·B + A ·C, distributivity
• A ·A = |A|2
Corollaries:|A| =
√A ·A.
ai · aj =
0, i 6= j = x, y, z
1, i = j = x, y, z,
4
Cross Product
A×B = |A||B| sin θABan.
A×B =
∣∣∣∣∣∣∣∣
ax ay az
Ax Ay Az
Bx By Bz
∣∣∣∣∣∣∣∣Properties:
• A ·B = −B ·A, anti-commutativity,
• A× (B + C) = A×B + A×C, distributivity
• A×A = 0
Corollaries:ai · aj = ak, i, j, k = x, y, z, cyclic permutations
For example, ax × ay = az , az × ax = ay.
5
Example 1. Let A = αax + 3ay − 2az and B = 4ax + βay + 8az . Find α and β
such that (i)A ‖ B, (ii)A⊥B.Example 2 Show that (A ·B)2 + (A×B)2 = (AB)2.
Triple Products
Scalar:A · (B×C) = B · (A×C) = C · (A×B).
A · (B×C) =
∣∣∣∣∣∣∣∣
Ax Ay Az
Bx By Bz
Cx Cy Cz
∣∣∣∣∣∣∣∣.
Vector:A× (B×C) = B(A ·C)−C(A ·B).
Example 3 Show that
ax =ay × az
ax · ay × az
.
Example 4 Simplify the expressions (a)A× (A×B), (b)A× [A× (A×B)].
6
Scalar component of A in direction of a:
A‖ = |A| cos θ = |A||a| cos θ = A · a.
Vector components of A, parallel and perpendicular to a:
A‖ = (A · a)a,
A⊥ = A− (A · a)a
Example 5 Show that A⊥⊥a.Example 6 Given H = 2xyax − (x + z)ay + z2az , �nd a unit vector parallel to H atP (1, 3,−2).
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Lecture 2 Vector calculus: Part I
Outline
1. Del Operator.
2. Gradient of a scalar �eld.
3. Flux of a vector �eld.
4. Divergence of a vector �eld and divergence theorem
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The del operator, denoted∇ is given by the expressions
∇ = ax∂
∂x+ ay
∂
∂y+ az
∂
∂z, (Cartesian);
∇ = aρ∂
∂ρ+ aφ
1
ρ
∂
∂φ+ az
∂
∂z, (cylindrical);
∇ = ar∂
∂r+ aθ
1
r
∂
∂θ+ aφ
1
r sin θ
∂
∂φ, (spherical).
1. The gradient of a scalar, grad V ≡ ∇V .
2. The divergence of a vector, div E ≡ ∇ · E, (Cartesian coordinates only).
3. The curl of a vector, curl A ≡ ∇×A, (Cartesian coordinates only).
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Gradient
The vector separation of the points M and N :
dr = axdx + aydy + azdz. (1)
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The scalar �eld V change in going from M to N :
dV =∂V
∂xdx +
∂V
∂ydy +
∂V
∂zdz. (2)
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Introduce the gradient of the scalar �eld V to be a vector �eld such that
grad V ≡ ∂V
∂xax +
∂V
∂yay +
∂V
∂zaz. (3)
dV =
(∂V
∂xax +
∂V
∂yay +
∂V
∂zaz
)·
·(axdx + aydy + azdz) = grad V · dr. (4)
gradV lies in the direction of maximum increase of V !
It follows from the de�nition of the Del operator that
grad V = ∇V.
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Alternative interpretation of the gradient:
• dV = 0 =⇒ ∇V⊥dr.
• dr is a tangent to the surface V (x, y, z) = const =⇒ ∇V ‖an, (an · dr = 0).
The gradient of V is a vector �eld that is everywhere normal to the surfaceV (x, y, z) = const!
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Gradient in curvilinear coordinate systems:
∇V =∂V
∂xax +
∂V
∂yay +
∂V
∂zaz; (Cartesian),
∇V =∂V
∂ρaρ +
1
ρ
∂V
∂φaφ +
∂V
∂zaz; (cylindrical),
∇V =∂V
∂rar +
1
r
∂V
∂θaθ +
1
r sin θ
∂V
∂φaφ. (spherical).
Example 1 Find the gradient of the �eld V = E0r cos θ.
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Outward �ux of a vector �eld
The outward �ux Ψ of a vector �eld D through a closed surface is de�ned as
Ψ ≡∮
S
D · dS =
∮
S
D · andS.
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Divergence of vector �elds
The divergence of A at a point P is the outward �ux per unit volume in the limit∆v → 0:
divA ≡ lim∆v→0
∮SA · dS∆v
= lim∆v→0
∮SA · andS
∆v
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Physical interpretation of divergence:
• measure of the �eld divergence at a point,
• measure of the �eld source/sink strength per unit volume (source/sink density).
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In curvilinear coordinate systems:
∇ ·A =∂Ax
∂x+
∂Ay
∂y+
∂Az
∂z; (Cartesian),
divA =1
ρ
∂(ρAρ)
∂ρ+
1
ρ
∂Aφ
∂φ+
∂Az
∂z; (cylindrical),
divA =1
r2
∂(r2Ar)
∂r+
1
r sin θ
∂(Aθ sin θ)
∂θ+
1
r sin θ
∂Aφ
∂φ; (spherical).
Example 2 Find the divergence of the position vector (a) in Cartesian coordinates and(b) in spherical coordinates.Example 3 Determine the divergence of the �eld B = (k/ρ)aφ.
∇ · F = 0 ⇐⇒ F is solenoidal.
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Note: divA 6= ∇ ·A in cylindrical and spherical coordinates!!
For, example, in cylindrical coordinates
divA =1
ρ
∂(ρAρ)
∂ρ+
1
ρ
∂Aφ
∂φ+
∂Az
∂z;
by the same token,
∇ ·A =
(aρ
∂
∂ρ+ aφ
1
ρ
∂
∂φ+ az
∂
∂z
)· (Aρaρ + Aφaφ + Azaz)
=∂Aρ
∂ρ+
1
ρ
∂Aφ
∂φ+
∂Az
∂z6= divA! (5)
Note: even though divA 6= ∇ ·A in curvilinear coordinates, we use∇ ·A instead ofdivA for notational simplicity!
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Divergence theorem (Gauss-Ostrogradsky theorem)
Divergence theorem: The total outward �ux of a vector �eld A through a closedsurface = the volume integral of divergence of A.
Example 4 Verify Gauss's theorem for the �eld F = krar and the spherical shellsurface, R1 ≤ r ≤ R2.
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Lecture 3 Vector calculus: Part II
Outline
1. Laplacian of scalar and vector �elds
2. Circulation of a vector �eld
3. Curl of a vector �eld
4. Stokes's theorem
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Laplacian of scalar and vector �elds
Laplacian of a scalar �eld:
∇2V ≡ div (gradV ) .
In Cartesian coordinates,
∇2V = ∇ · ∇V =
(ax
∂
∂x+ ay
∂
∂y+ az
∂
∂z
)·
·(
∂V
∂xax +
∂V
∂yay +
∂V
∂zaz
). (6)
Hence,
∇2V =∂2V
∂x2+
∂2V
∂y2+
∂2V
∂z2.
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In cylindrical coordinates:
∇2V =1
ρ
∂
∂ρ
(ρ∂V
∂ρ
)+
1
ρ2
∂2V
∂φ2+
∂2V
∂z2;
In spherical coordinates:
∇2V =1
r2
∂
∂r
(r2∂V
∂r
)+
1
r2 sin θ
∂
∂θ
(sin θ
∂V
∂θ
)+
1
r2 sin2 θ
∂2V
∂φ2.
Laplacian of a vector �eld (rational: div(gradA) doesn't make sense!):
∇2A ≡ grad (div A)− curl curlA.
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Circulation of a vector �eld
The circulation of a vector �eld A along a closed path:
circulation of A along closed path L ≡∮
L
A · dl.
Note: if the path is not closed,∫
LA · dl is the circulation.
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Curl of a vector �eldCurl of A: a vector �eld whose
• magnitude = maximum circulation of A per unit area as the area tends to zero;
• direction coincides with the unit normal to the area oriented so as to maximize thecirculation of A.
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Mathematically,
curlA ≡(
lim∆S→0
∮LA · dl∆S
)
maxan.
Physically, curl
• (a) provides the maximum value of the circulation of a vector �eld per unit area and
• (b) indicates the direction along which the maximum is attained.
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Curl in Cartesian coordinates
curlA = ∇×A =
∣∣∣∣∣∣∣∣
ax ay az
∂∂x
∂∂y
∂∂z
Ax Ay Az
∣∣∣∣∣∣∣∣or,
∇×A =
[∂Az
∂y− ∂Ay
∂z
]ax +
[∂Ax
∂z− ∂Az
∂x
]ay
+
[∂Ay
∂x− ∂Ax
∂y
]az. (7)
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Curl in curvilinear coordinates
curlA =1
ρ
∣∣∣∣∣∣∣∣
aρ ρaφ az
∂∂ρ
∂∂φ
∂∂z
Aρ ρAφ Az
∣∣∣∣∣∣∣∣; (cylindrical).
curlA =1
r2 sin θ
∣∣∣∣∣∣∣∣
ar raθ r sin θaφ
∂∂r
∂∂θ
∂∂φ
Ar rAθ r sin θAφ
∣∣∣∣∣∣∣∣; (spherical).
Note: even though curlA 6= ∇×A in curvilinear coordinates, we use∇×A insteadof curlA for notational simplicity!Example 1 Calculate curl of (a) A = (k/ρ)aφ; (b) A = f(r)ar.
∇×A = 0 ⇐⇒A is irrotational/conservative.
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Stokes's theorem
Stokes's theorem: The circulation of a vector �eld A around a path L = the surfaceintegral of curl of A over the surface bounded by L.
Example 2 Verify Stokes's theorem for A = sin(φ/2)aφ over the hemisphere surfaceand its circular contour lying in the plane z = 0.Example 3 Given F = (3y − c1z)ax + (c2x− 2z)ay + (c3y + z)az . Find c1, c2
and c3 such that F is irrotational.
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Lecture 4 Coulomb's Law and Field Intensity
Outline
1. Electric charge, Coulomb's Law and superposition principle.
2. Electrostatic �eld intensity.
3. Electric �elds of continuous charge distributions
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Coulomb's law and superposition principle
The force F between two point charges Q1 and Q2 is:
1. along the line joining the charges;
2. directly porportional to Q1Q2;
3. inversely proportional to the distance R12 = r1 − r2 between the charges.
Hence, force on charge Q1 due to Q2:
F12 = k︸︷︷︸constant
×Q1Q2
|R12|2 × aR12︸︷︷︸unit vector
=Q1Q2
4πε0|R12|2aR12 .
where the permittivity of free space
ε0 = 8.854× 10−12 [F/m ] (farads per meter).
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F12 =Q1Q2
4πε0|R12|2aR12 =Q1Q2
4πε0|R12|2R12
|R12|︸ ︷︷ ︸aR12
=Q1Q2
4πε0|R12|3R12.
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F12 =Q1Q2
4πε0|R12|2aR12 .
Note
1. aR21 = −aR12 =⇒ F21 = −F12
2.
sign(Q1) =
−sign(Q2) attraction,
sign(Q2) repulsion.
3. Q1 and Q2 are at rest.
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Consider charges Q1, Q2 . . . QN at positions r1, r2, . . . rN .
Superposition principle. The force on a probe charge Q at r is
F =Q
4πε0
N∑
k=1
QkRk
|Rk|3 , Rk = r− rk.
Units:
• Charge=Coulombs, [C], 1C = 6× 1018 electronic charges (large unit!).
• Force=Newtons, [N].
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Electric �eld intensity
The electric �eld intensity(or simply electric �eld)=force per unit charge:
E ≡ F
Q. (8)
Example. Force acting on a probe charge q at r due to Q at r′:
F =qQ
4πε0|R|2aR, R = r− r′. (9)
It follows from equations (8) and (9) that the electric �eld intensity at the position of q is
E = F/q =Q
4πε0|R|2aR.
Field at r due to Q1, Q2 . . . QN at r1, r2, . . . rN :
E =1
4πε0
N∑
k=1
QkRk
|Rk|3 , Rk = r− rk.
Units: Newton/Coulomb=Volt/meter.
35
Example 1 Calculate the electrostatic �eld of a dipole, consisting of two equal andopposite charges +Q and−Q, separated by a distance d, at a large distance from thedipole, r À d.
36
Electric �elds of continuous distributions of charges
Elementary (in�nitesimal) charge
dQ =
ρvdv volume charge
ρSdS surface charge
ρLdl line charge
37
Summing (integrating) contributions of elementary charges using the superpositionprinciple:
E =
∫ρvdv
4πε0|R|2aR, (volume charge).
E =
∫ρSdS
4πε0|R|2aR, (surface charge).
E =
∫ρLdl
4πε0|R|2aR, (line charge).
38
Lecture 5 Electric Flux Density and Gauss's Law
Introduce the electric �ux density �eld as
D ≡ ε0E,
and de�ne the electric �ux Ψ by the expression
Ψ =
∫D · dS.
Gauss's law. The total �ux Ψ through a closed surface S = total enclosed charge Qin:
Ψ =
∮
S
D · dS = Qenc =
∫
v
ρvdv (10)
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Thus∮
S
D · dS =
∫
v
ρvdv, (integral form of the �rst Maxwell's equation). (11)
Divergence theorem ∮
S
D · dS =
∫
v
∇ ·Ddv. (12)
It follows by comparing Eqs. (11) and (12):
∇ ·D = ρv, (differential form of the �rst Maxwell's equation). (13)
40
Applications of Gauss's law to calculating electric �elds
Use∮
SD · dS = Qenc for charge distributions with
• spherical symmetry (example: sphere),
• axial symmetry (example: in�nite cylinder),
• re�ectional symmetry (examples: in�nite straight line, in�nite plane).
41
Choose a special Gaussian surface such that
1. The surface is closed.
2. At each point of the surface D is either normal or tangential to the surface.
3. D is constant over the parts of the surface where D‖an.
=⇒ ∮SD · dS = D × total area where D‖an!!
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Example: a point charge
∮D · dS = D
∫ π
0
r2 sin θdθ
∫ 2π
0
dφ
︸ ︷︷ ︸4πR2
= Q;
E =D
ε0
=Q
4πε0r2ar.
43
Example Use Gauss's law to determine the electric �eld due to the following chargecon�gurations
• spherical shell of radius R carrying a charge Q;
• uniformly charged, with the density ρv, sphere of radius R;
• uniformly charged in�nitely long cylinder of radius a with the charge density ρv;
• in�nite straight line, carrying a charge ρl per unit length;
• in�nite plane charged with ρs per unit surface.
44
Lecture 6 Electrostatic potential
Outline
1. Work done to move a charge in an electric �eld. Electric potential.
2. Electric potentials of continuous charge distributions.
3. Relation between electric �eld and potential.
4. Example: potential of a dipole.
45
Motivation: why need potential?
Argument:
• It is easier to deal with a scalar potential then with a vector electric �eld.
• We will show that the latter can be obtained from the former.
46
Work done to move a charge in an electric �eld. Electric potential.
To keep a charge in equilibrium in the �eld E, a force Fa must be applied such that
Fa = −F = −QE.
47
The work done in displacing Q by dl is (work=force× displacement)
dW = Fa · dl = −QE · dl.
dl = dxax + dyay + dzaz, (Cartesian),
dl = dρaρ + ρdφaφ + dzaz, (cylindrical),
dl = drar + rdθaθ + r sin θdφaφ, (spherical).
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Hence to move Q from A to B:
W = −Q
∫ B
AE · dl.
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De�ne a potential difference as
VAB =W
Q= −
∫ B
AE · dl.
• VA and VB are referred to as the initial and �nal points.
• Units of VAB : [Joules/Coulomb=Volts].
• the work done by an external agent = change in potential energy of the charge in the�eld.
• VAB < 0 =⇒ work is done by the �eld; VAB > 0 =⇒ work is done by the externalagent.
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Example: potential of a point charge
From Coulomb's law for the �eld due to a point charge at the origin:
E(r) =Q
4πε0r2ar
Hence, by de�nition
VAB = VB − VA = −∫ rB
rA
Q
4πε0r2ar
︸ ︷︷ ︸E
· drar︸︷︷︸dl‖
=Q
4πε0
(1
rB
− 1
rA
)(14)
Pick a reference point rA → const, and choose rB → |r| ≡ r:
V (r) =Q
4πε0r+ const.
51
Choose the convenient reference point at in�nity, rA →∞, const ∝ 1/rA → 0:
V (r) =Q
4πε0r, (charge at the origin).
Generalization. Potential at r due to a charge at r′: shifting the origin to the point r′;r → r− r′, so that
V (r) =Q
4πε0|r− r′| , (charge at r′).
Superposition of point charges. Potential at r due to charges Q1, . . . Qn at pointswith r1, . . . rn:
V (r) =1
4πε0
n∑
k=1
Qk
|r− rk| , (by superposition principle).
52
Potential due to continuous distributions of charge
V (r) =1
4πε0
∫
L
ρL(r′)dl′
|r− r′| , (line charge);
V (r) =1
4πε0
∫
S
ρS(r′)dS ′
|r− r′| , (surface charge);
V (r) =1
4πε0
∫
v
ρv(r′)dv′
|r− r′| , (volume charge).
Example 1 Obtain an expression for the electric �eld potential and intensity on the axisof a uniformly charged disk of radius b with the charge density ρl.
53
Relation between electric �eld and potential.
The work done in moving a charge in time-independent (electrostatic) �eld isindependent of the path:
∫
1
E · dl = −∫
2
E · dl, or∮
1−2
E · dl = 0. (15)
54
Stokes's theorem: ∮
1−2
E · dl =
∫
S
∇× E · dS = 0 (16)
∫
S
∇× E · dS = 0 =⇒ ∇× E = 0, (potentiality of electrostatic �elds)!
Two forms to represent the electrostatic potential:
V = −∫
E · dl, (integral form) ⇐⇒ dV = −E · dl (differential form).
55
For an in�nitesimally small length element dl = dr (straight line approximation).
dV = −E · dr.
Recall the de�nition of the gradient:
dV = ∇V · dr.
56
Thus,dV = −E · dr; dV = ∇V · dr. (17)
It follows from equation (17) that
E = −∇V, (connection between potential and �eld).
57
Example: potential due to a dipole
Dipole: two equal, opposite charges, separated by a small distance d.
Potential due to a dipole by superposition:
V =Q
4πε0
(1
r1
− 1
r2
)=
Q
4πε0
(r2 − r1)
r1r2
.
58
Consider the potential at a remote point, r À d:
r2 − r1 ' d cos θ; r1r2 ' r2, (in denomenator).
59
It follows that
V =Q
4πε0
d cos θ
r2,
d = daz =⇒ d · ar = d cos θ.
De�ne the dipole moment: p = Qd.
V =p · ar
4πε0r2.
The corresponding electric �eld:
E = −∇V =p
4πε0r3(2ar cos θ + aθ sin θ).
60
Lecture 7 Energy and Energy Density of Electrostatic Fields
Energy of assembly of charges = work done to assemble them.
• Move Q1 ⇒ P1, Q2 ⇒ P2, Q3 ⇒ P3 in this order.
61
• Vij : potential at Pi due to charge at Pj .
WE = W1 + W2 + W3 = 0︸︷︷︸no �eld
+Q2V21 + Q3(V31 + V32). (18)
Reverse the order of charge movement
WE = W1 + W2 + W3 = 0 + Q2V23 + Q1(V12 + V13). (19)
Adding (1) and (2) and grouping terms
2WE = Q1(V12 + V13) + Q2(V21 + V23)
+ Q3(V13 + V32) = Q1V1 + Q2V2 + Q3V3. (20)
WE =1
2(Q1V1 + Q2V2 + Q3V3). (21)
62
Generalizations
• N point chrages
WE =1
2
N∑
k=1
QkVk .
• Continuous distributions of charge
WE =1
2
∫
L
ρLV dl, (line charge);
WE =1
2
∫
S
ρSV dS, (surface charge);
WE =1
2
∫
v
ρvV dv, (volume charge).
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Energy density of the electrostatic �eld
WE =1
2
∫
v
ρvV dv =1
2
∫
vsphere
∇ ·D︸ ︷︷ ︸=ρv
V dv (22)
64
Use the vector identity
∇ · (V A) = A · ∇V + V∇ ·A ⇐⇒ (fg)′ = f ′g + fg′.
scalar �eld lhs=scalar �eld rhs.
WE =1
2
∫
vsphere
∇ ·DV dv
=1
2
∫
vsphere
∇ · (V D)dv − 1
2
∫
vsphere
D · ∇V dv. (23)
Using divergence theorem,
1
2
∫
vsphere
∇ · (V D)dv =1
2
∮
Ssphere
V D · dS. (24)
65
As R →∞, (�nite charged volume = point charge) =⇒|D| ∝ 1/R2 V ∝ 1/R V D ∝ 1/R3 and surface area∝ R2.
Hence,
limR→∞
1
2
∮
Ssphere
V D · dS = 0.
So that in the limit R →∞
WE = −1
2
∫D · ∇V︸︷︷︸
=−E
dv =1
2
∫(D · E)dv.
66
Since D = ε0E,
WE =1
2
∫(D · E)dv =
1
2
∫ε0E
2dv
The density of the �eld can be inferred from the de�nition
WE =
∫wE︸︷︷︸
density
dv
It follows that
wE =1
2(D · E) =
1
2ε0E
2 .
Example Determine the electrostatic energy needed to assemble a uniformly chargedsphere of radius R with the charge density ρv.
67
Lecture 8 Conductors and Dielectrics
• abundance of free charge carriers in conductors;
• no volume charge or electric �eld inside a conductor, ρv = 0, Ein = 0 ;
• E = −∇V = 0 =⇒ V = const , conductor=equipotential body.
Example 1 Two spherical conductors of radii R1 and R2, respectively carry a totalcharge of Q. The conductors are connected by a very long conducting wire. Find thecharges on the two spheres.
68
Polarization in Dielectrics
Dielectrics: positive nuclear charge (+Q) +negative electron charge, (−Q)=neutralatom.
Response of a dielectric to an applied electric �eld, E; p=dipole moment.
69
Classi�cation of dielectric materials
• nonpolar; p = 0 at E = 0 =⇒, dipole alignment at E 6= 0.
• polar; p 6= 0, even at E = 0, random orientation of permanent dipoles=⇒ dipoles rotate to align along E 6= 0.
70
Dipole moment of an individual atom (molecule):
p = Qd.
N dipoles in a small volume ∆v,
p =N∑
k=1
Qkdk.
De�ne the polarization as a dipole moment per unit volume,
P = lim∆v→0
∑Nk=1 Qkdk
∆v
71
Field due to a polarized dielectric
Starting with the potential dV due to an in�nitesimal polarized volume dv:
dV =
dp︷ ︸︸ ︷(Pdv′) ·aR
4πε0R2
where R =√
(x− x′)2 + (y − y′)2 + (z − z′)2,
72
we can demonstrate � see Appendix A � that the expression for the potential due topolarized charge,
V =
∫dv′(P · aR)
4πε0R2,
can be transformed into a volume and surface contributions as
V =
∫
S′
ρps︷ ︸︸ ︷P · a′n4πε0R
dS ′ +∫
v′
ρpv︷ ︸︸ ︷−∇′ ·P4πε0R
dv′
ρps = P · an, polarization surface charge,
ρpv = −∇ ·P, polarization volume charge.
73
• surface polarization charge:
Q =
∫ρpsdS =
∮P · dS,
• volume polarization charge:∫
v
ρpvdv = −∫
v
∇ ·Pdvdivergence
= −∮
P · dS.
It follows ∫ρpsdS +
∫
v
ρpvdv = 0, neutrality of dielectric!
74
Dielectric susceptibility and permittivity
Consider a dielectric region containing free charge of density ρv.
The total charge density ρt:
ρt = ρv + ρpvGauss= ∇ · (ε0E).
Hence,ρv = ∇ · (ε0E)− ρpv = ∇ · (ε0E + P)︸ ︷︷ ︸
D
.
De�ne the �ux density �eld in dielectric D:
D ≡ ε0E + P
∇ ·D = ρv , Gauss's law in dielectrics
75
De�ne the electric susceptibility χ by the expression
P = χε0E, if χ = scalar, D‖E.
De�ne the dielectric permittivity ε such that
D = ε0E + P = ε0(1 + χ)E = εE .
De�ne the relative permittivity as
εr = 1 + χ =ε
ε0
.
• εr and χ are dimensionless, while ε and ε0 are measured in [farads/meter].
Example 2 A positive charge Q is at the center of a spherical dielectric shell (ε = εrε0)of the inner and outer radii R1 and R2, R1 < R2. Determine E, D, P, ρps and ρpv.
76
ε is independent of the applied �eld E, of the position and direction withindielectric =⇒ linear, homogeneous, isotropic dielectric.
Energy density of the �eld in dielectric
w =1
2(D · E) =
1
2ε0εrE
2
Example 3 Let V (x, y, z) = x2y2z in a region de�ned by−1 < x, y, z < 1. Theregion is �lled with the dielectric ε = 2ε0. Determine the volume charge density ρv
within the region.Example 4 A solid sphere of radius a and dielectric constant εr has a uniform volumecharge density ρ0. Show that the potential at the center of the sphere is given by
V =ρ0a
2
6εrε0
(2εr + 1).
77
Lecture 9 Boundary conditions in electrostatics
Outline
1. Dielectric-dielectric interface boundary conditions. Law of refraction.
2. Conductors: a brief overview
3. Dielectric-conductor interface boundary conditions
78
Introductory comments
Boundary conditions take place at the interface separating two different media.
• Decompose the �eld E into normal (to the interface) and tangential (to theinterface) components on both sides of the interface
Ej = Ejn + Ejt, j = 1, 2.
• Use Maxwell's equations:∮
E · dl = 0; (potentiality of electrostatic �eld),
∮D · dS = Qencl; (Gauss's law).
79
Dielectric-dielectric boundary conditions
Apply∮
E · dl = 0 to the closed path abcda:
E1t∆w − E1n∆h
2− E2n
∆h
2
−E2t∆w + E2n∆h
2+ E1n
∆h
2= 0. (25)
80
It follows from Eq. (40) that as ∆h → 0,
E1t = E2t . (26)
D = εE =⇒ Dt = εEt.
Thus,D1t
ε1
=D2t
ε2
.
• Tangential component of the �eld is continuous across the interface.
• Tangential component of the �ux density is discontinuous across the interface.
81
Use Gaussian surface in �gure (b) to get
∆Qencl = ρS∆S = D1n∆S −D2n∆S.
ThusD1n −D2n = ρS . (27)
82
In the absence of the free charge on the surface, ρS = 0,
D1n = D2n . (28)
Since D = εE,ε1E1n = ε2E2n.
• Normal component of the �eld is discontinuous across the interface.
• Normal component of the �ux density is continuous across the interface in theabsence of free surface charge.
Example 1 A dielectric sheet with εr is introduced into a uniform �eld E = E0ax in freespace. Find E, D and P inside the dielectric sheet.
83
Refraction of the electric �eld at the interface of two dielectic media
E1t = E2t =⇒ E1 sin θ1 = E2 sin θ2,
84
D = εE =⇒ Dn = εEn
D1nno free charge
= D2n =⇒ ε1E1 cos θ1 = ε2E2 cos θ2.
85
E1 sin θ1 = E2 sin θ2, (29)
ε1E1 cos θ1 = ε2E2 cos θ2. (30)
Dividing Eq. (38) by eq. (37) term by term,
tan θ1
ε1
=tan θ2
ε2
⇐⇒ tan θ1
tan θ2
=ε1
ε2
. (31)
• law of refraction of the �eld at a boundary free of surface charge.
Example 2 Work out the magnitude of E2 in terms of E1 as well.
86
Brief overview of conductors
• abundance of free charge carriers;
• no volume charge or electric �eld inside a conductor, ρv = 0, Ein = 0 ;
• E = −∇V = 0 =⇒ V = const , conductor=equipotential body.
87
Dielectric-conductor boundary conditions
Consider a circulation of E along the contour abcda (E = 0 inside conductor)
Ein=0︷︸︸︷0 ×∆w − 0× ∆h
2+ En
∆h
2
−Et∆w − En∆h
2+ 0× ∆h
2= 0. (32)
88
As ∆h → 0,Et = 0 . (33)
Applying Gauss's law to the �ux across the interface,
∆Qencl = ρS∆S = Dn∆S − 0×∆S,
Dn = ρS . (34)
• The electric �eld outside of a conductor is normal to its surface,
Et = 0, En = ρS/ε
89
Lecture 10 Poisson and Laplace equations
Outline
1. Poisson equation.
2. Laplace equation
3. Uniqueness of the solution to Laplace equation
90
Poisson and Laplace equations
Recall Gauss's law in differential form,
∇ ·D = ∇ · ( εE︸︷︷︸D
) = ρv. (35)
The potential V is de�ned asE = −∇V. (36)
On substituting from Eq. (44) into Eq. (43), we obtain a general form of the Poissonequation
∇ · (−ε∇V ) = ρv . (37)
Only If ε = const (homogeneous, isotropic, linear dielectric media),
∇2V = −ρv
ε. (38)
91
In the absence of charge, ρv = 0, Poisson's equation reduces to Laplace's equation:
∇ · (ε∇V ) = 0 . (39)
Only If ε = const (homogeneous, isotropic, linear dielectric media),
∇2V = 0 . (40)
92
Laplace's equation in curvilinear coordinates
∇2V = 0 :
∂2V
∂x2+
∂2V
∂y2+
∂2V
∂z2= 0
1
ρ
∂
∂ρ
(ρ∂V
∂ρ
)+
1
ρ2
∂2V
∂φ2+
∂2V
∂z2= 0 ,
1
r2
∂
∂r
(r2∂V
∂r
)+
1
r2 sin θ
∂
∂θ
(sin θ
∂V
∂θ
)+
1
r2 sin2 θ
∂2V
∂φ2= 0 .
Uniqueness theorem: A solution to Poisson's equation satisfying given boundarycondition(s) is unique.
93
Example 1 Find the electrostatic potential inside an in�nite strip, 0 ≤ z ≤ a of aninhomogeneous dielectric medium with the dielectric constant εr(z) = εr0/(1 + z/a).The potential satis�es the boundary conditions V (z = 0) = 0 and V (z = a) = V0.Example 2 Determine the electrostatic potential in the space between two very longconcentric cylinders of radii a and b, b > a. The internal cylinder is grounded, whereasthe external one is held at the �xed potential V0.Example 3 What is a potential distribution in between the two planes, φ = 0 andφ = α, if the �rst plane is grounded, and the second has the potential V0?Example 4 Determine the potential distribution in the region of space between twocones, θ = θ1 and θ = θ2, θ1 < θ2. The surfaces of the cones are held at the potentialsV = 0 and V = Vex, respectively. The common vertex of the cones is insulated.Example 5 Calculate E inside and outside a spherical cloud of a uniform volumecharge, ρv. The radius of the cloud is R.
94
Lecture 11 Capacitance and capacitors
Outline
1. De�nition and calculation of the capacitance.
2. Capacitance of parallel-plate, cylindrical and spherical capacitors.
3. Muliti-dielectric capacitors.
95
Consider two conductors, referred to as plates, carrying equal but opposite charges,±Q=capacitor.
Potential difference between the conductors:
V = V1 − V2 = −∫ 2
1
E · dl.The charge on each capacitor is related to the �eld by Gauss's law
Q =
∮D · dS = ε
∮E · dS; assuming ε = const.
96
Capacitance of such a capacitor is de�ned as
C =Q
V=
ε∮
E · dS∫ 2
1E · dl
Methods of calculating capacitance
• Assume Q is given and determine V in terms of Q using Coulomb or Gauss's law.(Method I)
• Assume V0 ≡ V is given and determine the corresponding Q by solving theappropriate Laplace's equation (Method II).
97
Method I
• Choose a suitable coordinate system.
• Let the two conductors carry charges +Q and−Q, respectively.
• Determine E using Coulomb's or Gauss's law and �nd the potential fromV = − ∫
E · dS.
• Obtain the capacitance from C = Q/V .
Illustrating Method I by examples.
98
Parallel-plate capacitor
99
Neglecting fringing, the uniform surface charge density is
ρS = Q/A.
Use Gauss's law to calculate the �eld between the plates:
Qenc =
∫
top
D · dS︸ ︷︷ ︸
=0
+
∫
bottom
D · dS +
∫
side
D · dS︸ ︷︷ ︸
=0
.
100
Consequently,
Qenc = ρSA = D
∫dS = DA =⇒ D = ρSan.
Hence
E =ρS
εan =
Q
εAan
Calculate the potential difference between the plates, an = −az
V = −∫ 2
1
E · dl = −∫ d
0
Q
εAaz · (−az)dz
=Q
εA
∫ d
0
dz (az · az)︸ ︷︷ ︸=1
=Q
εAd. (41)
101
Thus, the capacitance of a parallel-plate capacitor:
C =Q
V=
εA
d.
Application: Measuring C with the dielectric and with the air, we infer the relativedielectric constant εr viz.,
εr =C
C0
.
102
Cylindrical (coaxial) capacitor
Choose Gaussian surface to be another coaxial cylinder such that a < ρ < b.
Q =
∮D · dS = ε
∮E · dS = εE(2πρ)L,
103
It follows that
E =Q
2περLaρ
Neglecting fringing, the potential difference between the plates is
V = −∫ 2
1
E · dl = −∫ a
b
(Q
2περLaρ
)· aρdρ
=Q
2πεL
∫ a
b
dρ
ρ(aρ · aρ)︸ ︷︷ ︸
=1
=Q
2πεLln
b
a. (42)
Thus,
C =Q
V=
2πεL
ln b/a.
104
Spherical capacitor
Choose Gaussian surface to be another coaxial sphere such that a < r < b.
Q =
∮D · dS = ε
∮E · dS = εE4πr2.
105
Hence,
E =Q
4πεr2ar
And, the potential difference is calculated as follows
V = −∫ 2
1
E · dl = −∫ a
b
(Q
4πεr2ar
)· ardr
=Q
4πε
∫ a
b
dr
r2(ar · ar)︸ ︷︷ ︸
=1
=Q
4πε
(1
a− 1
b
). (43)
Thus, the capacitance of such a capacitor is
C =Q
V=
4πε1a− 1
b
.
• b →∞, C = 4πεa, useful to estimate capacitance of a single conductor (theother plate in in�nity) of a typical size a.
106
Multiple-dielectric capacitors
• Compound capacitor with the interface parallel E and D⇐⇒ arrangement ofcapacitors in parallel.
Example. parallel-plate capacitors:
Ceq = C1 + C2 =
ε0εr1︷︸︸︷ε1 A1
d+
ε0εr2︷︸︸︷ε2 A2
d.
107
• Compound capacitor with the interface normal to E and D⇐⇒ arrangement ofcapacitors in a series.
Example. parallel-plate capacitors:
1
Ceq
=1
C1
+1
C2
=ε1d1 + ε2d2
ε1ε2A.
108
Energy stored in a capacitor
Recall the expression for the energy to assemble a system of surface charges
WE =1
2
∫ρSV dS.
Applying to a capacitor with∫
ρ1SdS = − ∫ρ2SdS = Q:
WE =1
2
∑j=1,2
∫ρjSVjdS =
1
2(V1
∫ρ1SdS + V2
∫ρ2SdS)
= Q(V1 − V2)/2 = QV/2 (44)
WE =1
2QV =︸︷︷︸
Q=CV
1
2CV 2 =︸︷︷︸
V =Q/C
Q2
2C(45)
Example 1 In the arrangement displayed in �gure (a), determine the voltage dropacross each dielectric.
109
Lecture 12 Method of images
Motivation
1. hard to solve Poisson's equation for complicated systems of point charges andconductors
2. uniqueness theorem: any solution to Poisson's equation is unique =⇒ a cleverguess will work!
Areas of application: useful to �nd potential and electric �eld strength for systems ofpoint/line charges in the vicinity of conducting bodies.
110
A point charge +Q, located a distance d above a grounded conducting plane
The solution must satisfy the conditions:
1. V (ρ, 0) = 0, the plane z = 0 is grounded;
2. V (ρ, z) → 0 as ρ, z →∞;
3. by symmetry, V (ρ, z) = V (−ρ, z).
Solution satisfying 1-3:
V (ρ, z) =
Q4πε0
(1
R+− 1
R−
)z > 0,
0 z < 0.
HereR+ = [ρ2 + (z − d)2]1/2,
R− = [ρ2 + (z + d)2]1/2.
111
Interpretation: the effect of the plane⇐⇒ an image charge−Q, located at z = −d.
The electric �eld:
E = −∇V =
Q4πε0
[ρaρ+(z−d)az
R3+
− ρaρ+(z+d)az
R3−
]z > 0,
0 z < 0.
The force on the charge +Q due to the image charge−Q
F = − Q2
4πε0(2d)2az.
112
Surface density of induced charge
Boundary condition at the plane surface, (ε1 = ε2 = ε0):
(E− − E+) · an = ρs/ε0.
an = az ; E− = 0 ( the �eld in the half-space z < 0) and E+ = −∇V .
ρs = − ε0∂V
∂z
∣∣∣∣z=0
.
Thus
ρs = − Qd
2π(ρ2 + d2)3/2.
113
Further Examples to be discussed in class
Example 1. A point charge Q is placed a distance d above a grounded conductingplane. Find the work done to remove the charge far away from the plane.Example 2. Determine the electrostatic potential distribution and the induced surfacecharge due to an in�nitely long, charged � with the charge κ per unit length � �lament,located a distance h above a grounded conducting plane.
114
Lecture 13 Electric current and current density. Continuity equation
Outline
1. De�nition of the current and current density.
2. Convection and conduction currents. Point form of Ohm's law.
3. Continuity equation.
4. Volume charge relaxation in materials.
115
Electric current and current density
Electric current, I : rate of charge transport past a point or across a speci�ed surface.
I =dQ
dt, [Coulomb/second]
116
Current density, J : current per unit cross-section [Amp�eres/meter2]:
dI = J · dS I =
∫
S
J · dS .
117
Convection versus conduction currents
• Convection current⇐⇒ simply �ow of charged particles.
• Conduction current⇐⇒ �ow of particles in a conductor.
118
Derive an expression for the density of any type of current.
∆I =∆Q
∆t= ρv∆S
∆l
∆t︸︷︷︸U
= ρvU ·∆S
Since J = ∆I/∆S, we obtainJ = ρvU .
119
Point form of Ohm's law
Drift velocity of electrons in a conductor (liquid, crystalline, gas) = average velocity ofsuch electrons due to applied �eld, subject to collisions.
120
Drift velocity is related to mobility µ [meter2/(Volt·second)] as follows
U = µE.
Using the derived expression for the current density in terms of the drift velocity,
J = ρvU,
we obtain the point form of Ohm's law:
J = ρvµE = σE .
• σ = conductivity (Siemens/meter).
• σ decreases with the temperature =⇒ increase of resistivity of the material tocurrent �ow.
Recall the integral form of Ohm's law:
V = IR R = resistance .
121
Resistance calculations
Ohm's law:
R =V
I=− ∫
LE · dl∮
SJ · dS =︸︷︷︸
σ=const
− ∫LE · dl
σ∮
SE · dS .
Recall the expression for the capacitance
C =Q
V=
∮SD · dS
− ∫LE · dl =︸︷︷︸
ε=const
ε∮
SE · dS
− ∫LE · dl .
It follows
RC =ε
σ
Example 1 Calculate the resistance per unit length of a coaxial cable of inner and outerradii a and b, a < b.Example 2 A conducting material of uniform thickness h and conductivity σ has theshape of the quarter of an annulus in the xy-plane: 0 ≤ φ ≤ π/2, a ≤ ρ ≤ b.Determine the resistance between the end faces.
122
Continuity equation
Law of charge conservation: time rate of decrease of charge inside a closed volume =outward current �ux through the surface of the volume.
−dQin
dt=
∮J · dS. (46)
Using the divergence theorem,∮
J · dS =
∫
v
∇ · Jdv (47)
The time rate of decrease of the charge in terms of volume charge density:
−dQin
dt= − d
dt
∫
v
ρvdv = −∫
v
∂ρv
∂tdv. (48)
123
It follows from equations (54), (55) and (56) that∫
v
∇ · Jdv = −∫
v
∂ρv
∂tdv. (49)
In other words, ∫
v
dv
(∂ρv
∂t+∇ · J
)= 0. (50)
Implying that∂ρv
∂t+∇ · J = 0 .
• current continuity equation
• stationary case, ∂ρv/∂t = 0 =⇒ ∇ · J = 0 .
124
Relaxation of introduced charge inside volume
Introduce volume charge ρv at the interior of a given conducting material.
Ohm's + Gauss's laws, assuming ε = const and σ = const:
J = σE Ohm,
∇ · E = ρv/ε.
It follows∇ · J = ∇ · (σE) = σ∇ · E = σρv/ε.
125
On substituting to the continuity equation,
∂ρv
∂t+∇ · J = 0,
one obtains∂ρv
∂t+
σ
ε︸︷︷︸const
ρv = 0. (51)
The solution to this homogeneous, linear differential equation is
ρv = ρv0︸︷︷︸ρv(t=0)
exp
(−σt
ε
)= ρv0e
−t/τ .
• τ = ε/σ ⇐⇒ relaxation time
126
Lecture 14 Biot-Savart's and Amp�ere's Laws
Biot-Savart's law: The magnetic �eld intensity dH due to a current element Idl at apoint P is proportional to Idl, to sine of the angle α between the element and the linejoining the element and P and inversely proportional to the square of the distance fromthe element to P
127
Mathematical formulation of Bio-Savart law
dH =Idl sin α
4πR2=
Idl×R/R︷︸︸︷aR
4πR2=
Idl×R
4πR3
128
Direction of H due to the current element:
129
Volume, surface and line current con�gurations
De�ne J (volume current in Amp�eres/meter square); K (surface current inAmp�eres/meter) such that
Idl = KdS = Jdv
130
Applying the principle of superposition yields:
H =
∫
L
Idl×R
4πR3, (line current);
H =
∫
S
KdS ×R
4πR3, (surface current);
H =
∫
v
Jdv ×R
4πR3, (volume current).
Example 1 A current �lament of length 2L, carrying the current I , is placed along thez-axis. Calculate the magnetic �eld H generated by the current at the point P (ρ, 0, 0)
Example 2 Calculate the magnetic �eld on the axis of a ring of radius R carrying thecurrent I .
131
Amp�ere's Law
Circulation of H around a closed path = net current enclosed by the path∮
H · dl = Ienc , (integral form of Amp�ere's law).
Applying Stokes's theorem,
Ienc =
∮
L
H · dlStokes︷︸︸︷
=
∫
S
(∇×H) · dS.
132
By de�nition of the current density,
Ienc =
∫
S
J · dS.
Hence, ∫
S
J · dS =
∫
S
(∇×H) · dS,
Implying that
∇×H = J (differential form of Amp�ere's law= 3rd Maxwell's equation).
133
Application of Amp�ere's law
1. Current distribution possesses axial, spherical or re�ectional symmetry.
2. At each point of the closed path H is either tangential or normal to the path.
3. H has the same magnitude at all points of the path where H is tangential.
Illustrative examples of the Amp�ere's law application to be considered:
• in�nite �lamentary current
• in�nitely long solenoid
• in�nitely long coaxial transmission line.
134
Lecture 15 Magnetic �ux density and magnetic �ux
Introduce the magnetic �ux density in free space by analogy with the correspondingelectric �ux density (D = ε0E):
B = µ0H.
where
µ0 = 4π × 10−7, (Henries/meter)=magnetic permiability of free space.
135
The magnetic �ux through a �nite surface is de�ned as
Φ =
∫
S
B · dS.
Example 1 A coaxial cylindrical conductor with an inner conductor of radius a and anouter one of radius b carries current I inside the inner conductor. Find the magnetic �uxper unit length crossing the surface φ = const between the conductors.
136
Comparison between electric and magnetic �uxes
• Electrostatic �eld: electric �ux lines are not necessarily closed.
• Magnetostatic �eld: magnetic �ux lines must be either closed or go to in�nity⇐⇒ no isolated magnetic charges in nature!!
Can one isolate a pole of a permanent magnet then?
137
No isolated magnetic charge of any form in nature ⇐⇒∮
B · dS = 0.
Note
• This is a law of conservation of magnetic �ux (Gauss's law for magnetic �elds).
138
0 =
∮B · dS
divergence︷︸︸︷=
∫
v
∇ ·Bdv.
It follows that
∇ ·B = 0, (differential form of magnetic �ux conservation).
139
Scalar magnetic potential
Recall two identities from the vector analysis that hold for any scalar �eld f and vector�eld F:
∇× (∇f) = 0; (curl of a gradient =0)
∇ · (∇× F) = 0; (divergence of a curl=0).
In the absence of a volume current, J = 0, introduce a scalar magnetic potential Vm
(in Amp�eres) such thatH = −∇Vm if J = 0.
140
Justi�cation
0 = JMaxwell′s eq︷︸︸︷
= ∇×H = ∇× (−∇Vm)
curl(grad(...))=0︷︸︸︷= 0.
In free space, J = 0,
∇ ·H = 0, H = −∇Vm =⇒ ∇2Vm = 0.
• Vm satis�es Laplace's equation in free space.
141
Vector magnetic potential
Introduce a vector magnetic potential A (in Weber/meter) such that
B = ∇×A.
Justi�cation:
0no magnetic charge︷︸︸︷
= ∇ ·B = ∇ · (∇×A)
div(curl(...))=0︷︸︸︷= 0.
• Scalar �eld Vm can be used only if J = 0.
• Vector �eld A can always be employed.
Example 2 Given A = −µ0Iρ2/(4πa2)az , what is B?Example 3 Determine A of the magnetic �eld produced by an in�nitely long, straight�lamentary current I .
142
Superposition principle yields
A =
∫
L
µ0Idl
4πR, (line current);
A =
∫
S
µ0KdS
4πR, (surface current);
A =
∫
v
µ0Jdv
4πR, (volume current).
143
The �ux of a magnetic �eld in terms of the vector potential
Φ =
∫
S
B · dS =
∫
S
(∇×A) · dS =︸︷︷︸Stokes
∮A · dl.
Thus
Φ =
∮A · dl.
144
Summary of Maxwell's equations for static �elds
Gauss's law for electrostatic (left) and magnetostatic (right) �elds
Differential (point) form:
∇ ·D = ρv, ∇ ·B = 0.
Integral form:∮
S
D · dS = Qenc,
∮
S
B · dS = 0.
145
Amp�ere's law for electrostatic (left) and magnetostatic (right) �elds
Differential (point) form:
∇× E = 0, ∇×B = J.
Integral form:∮
L
E · dl = 0,
∮
L
H · dl = Ienc.
146
Lecture 16 Forces and torques due to magnetic �elds
Outline
1. Lorentz force on a charged particle
2. Force on a current element
3. Force between two current elements
4. Torque on a current loop.
147
Force on a charged particle
Force on a stationary or moving charge due to an electric �eld (Coulomb's law):
Fe = QE.
Force on a moving charge due to a magnetic �eld (Lorentz force):
Fm = QU×B.
In presence of both electric and magnetic �eld =⇒ Lorentz force equation:
F = Fe + Fm = Q(E + U×B).
148
Newton's equation of motion for a charged particle:
MdU
dt= Q(E + U×B).
Properties of the electric and magnetic forces
• Fe performs work.
• Fm does not perform work, (Fm · dl = 0!) =⇒ Fm changes direction of motionof a charged particle.
149
Force on a current element
Recall the connection between the current density J and the drift velocity U:
J = ρvU;
as well as the relation between the different current elements
Idl = KdS = Jdv.
Hence,Idl = Jdv = ρvdv︸︷︷︸
dQ
U = dQU.
150
Thus,Idl = dQU.
• charge element dQ moving with velocity U⇐⇒ a conduction current element Idl.
Consequently, the force (Amp�ere force) on a current element:
dF = Idl×B.
151
For a closed circuit,
F =
∮
L
Idl×B.
Similarly, for surface and volume current elements,
dF = KdS ×B, dF = Jdv ×B;
or, in the integral form,
F =
∫
S
KdS ×B, F =
∫
v
Jdv ×B.
152
Force between two current elements
The force between elements I1dl1 and I2dl2 ⇐⇒ force on I1dl1 due to the magnetic�eld dB2 generated by I2dl2:
dF12 = I1dl1 × dB2.
Example 1 Determine the force per unit length between two in�nitely long parallelstraight wires with the currents I1 and I2, respectively, placed a distance d apart.
153
Biot-Savart law implies
dB2 =µ0I2dl2 ×R12
4πR312
Consequently,
dF12 =µ0I1dl1 × (I2dl2 ×R12)
4πR312
The force between �nite current elements is found by the superposition principle:
F12 =µ0I1I2
4π
∮
L1
∮
L2
dl1 × (dl2 ×R12)
R312
.
• Analog of Coulomb's law for interaction between two charges.
• F21 = −F12 due to the third law of Newton (action=reaction).
154
Magnetic torque and moment
Torque on a magnetic loop is a vector product of the force and the moment arm:
T = r× F, (Newtons · meter).
155
Calculate a total force on a rectangular loop placed in a uniform magnetic �eld:
• Along the sides 12 and 34: dl‖B;
• dF ∝ dl×B =⇒ dF12 = dF34 = 0.
156
The elementary forces on the other sides,
dF23 = Idz(az × ax)B; dF14 = Idz(−az × ax)B.
The total force,
F =
∫ 3
2
dF23 +
∫ 4
1
dF14 =︸︷︷︸uniform B
IBlay − IBlay = 0.
157
However, the forces form a couple, whose torque has the magnitude
|T| = |F|w sin α = IB lw︸︷︷︸S
sin α = IBS sin α.
158
De�ne a magnetic dipole moment m as
m = ISan ; (Amp�ere·meter2). (52)
The torque on a small loopT = m×B . (53)
• The de�nition of magnetic moment, Eq. (60), holds for a loop of any geometricalshape.
• The expression for the torque, Eq. (61), holds only for a small loop such that B isuniform across the loop.
159
Lecture 17 Magnetization in materials
Outline
1. A small current loop as a magnetic dipole: Comparison with electric dipoles.
2. Magnetization of a material. Magnetization volume and surface currents.
3. Brief classi�cation of magnetic materials.
160
Small current loop as magnetic dipole
The magnetic vector potential A and magnetic �ux density B far away from a small loop,r À a:
A =µ0(m× ar)
4πr2
and
161
B =µ0m
4πr3(2 cos θar + sin θaθ)
• small loop⇐⇒ magnetic dipole!
162
Comparison with electric dipole
163
Magnetization
Orbital motion of electrons around atoms or internal degree of freedom (spin)⇐⇒magnetic moment m = ISan.
164
Magnetization M, (Amp�eres/meter) is a magnetic dipole moment per unit volume:
M = lim∆v→0
∑Nk=1 mk
∆v
165
By de�nition, the dipole moment of elementary volume, dv′ is dm = Mdv′, and
dA(r) =µ0
dm︷ ︸︸ ︷Mdv′×
R/R︷︸︸︷aR
4πR2=
µ0M×R
4πR3dv′. (54)
R = |r− r′| =√
(x− x′)2 + (y − y′)2 + (z − z′)2.
166
Hence,
A =
∫µ0M×R
4πR3dv′.
It can be shown � see Appendix A � in strict analogy with the similar derivation fordielectrics that the last expression can be transformed into
A =µ0
4π
∫
v′
Jmdv′
R+
µ0
4π
∮
S′
KmsdS ′
R, (55)
where we have introduced
Jm = ∇×M , (magnetization volume current);
Kms = M× an , (magnetization surface current).
167
Recall Amp�ere's law in free space, M = 0, Jf being a free current,
∇×H = Jf , ⇐⇒ ∇×(
B
µ0
)= Jf .
In a medium, M 6= 0, and Amp�ere's law is modi�ed as follows
∇×(
B
µ0
)= Jf + Jm = ∇×H +∇×M = ∇× (H + M).
Thus in the medium,B = µ0(H + M).
• This relation holds for any medium!
168
In linear media, one can introduce magnetic susceptibility such that
M = χmH
It follows that
B = µ0(H + M) = µ0(H + χmH) = µ0(1 + χm)︸ ︷︷ ︸µ
H;
so thatB = µH, µ ⇐⇒ magnetic permeabilty, Henries/meter.
Relative permeability:
µr = 1 + χm =µ
µ0
169
Classi�cation of magnetic materials
1. Diamagnetic materials: B = 0, zero magnetic moments of atoms; µr ≤ 1.
2. Paramagnetic materials: B = 0, zero magnetic moments of atoms; µr ≥ 1.
3. Ferromagnetic materials: permanent magnetic moments of atoms; µr À 1.
• Paramagnetics and diamagnetics µr ' 1 =⇒ weak magnetic materials.
• Ferromagnetics, µr À 1 =⇒ strong magnetic materials (permanent magnets).
Example A ferromagnetic sphere of radius R is magnetized uniformly such thatM = M0az . Determine the equivalent magnetization current densities, Jm and Jms.
170
Lecture 18 Magnetic boundary conditions
To derive the boundary conditions at the interface of two magnetic media, we use
• Gauss's law for magnetic �elds∮
B · dS = 0.
• Amp�ere's circuit law∮
H · dl = Ienc.
Express B and H in terms of normal (to the surface) and tangential (to the surface)components:
B = Bn + Bt, H = Hn + Ht.
171
Consider a Gaussian surface indicated in the �gure,
B1n∆S −B2n∆S = 0
B1n = B2n
B=µH︷︸︸︷=⇒ µ1H1n = µ2H2n.
172
Consider a closed abcda path indicated in the �gure, K⊥ path
K∆w = H1t∆w + H1n∆h
2+ H2n
∆h
2
−H2t∆w −H2n∆h
2−H1n
∆h
2(56)
173
K∆w = H1t∆w + H1n∆h
2+ H2n
∆h
2
−H2t∆w −H2n∆h
2−H1n
∆h
2(57)
As ∆h → 0,H1t −H2t = K.
Generalizing,
(H1 −H2)× an12 = K ; an12 ⇐⇒ unit normal from medium 1 to medium 2.
174
In the absence of the free surface current K = 0,
H1t = H2t =⇒︸︷︷︸B=µH
B1t/µ1 = B2t/µ2.
• The normal component of B is continuous and the tangential one isdiscontinuous across the interface.
• The tangential component of H is continuous and the normal one isdiscontinuous across the interface.
175
Refraction of magnetic �ux lines at the interface
B1n = B2n =⇒ B1 cos θ1 = B2 cos θ2;
H1t = H2t =⇒ B1
µ1
sin θ1 =B2
µ2
sin θ2.
176
B1 cos θ1 = B2 cos θ2;
B1
µ1
sin θ1 =B2
µ2
sin θ2.
Dividing these equations term by term, we obtain the law of refraction of the magnetic�ux density, similar to the one for the electric �ux density:
tan θ1
µ1
=tan θ2
µ2
=⇒ tan θ1
tan θ2
=µ1
µ2
Example 1 Work out the magnitude of H2 in terms of H1 as well.Example 2 A long circular rod of magnetic material with permeability µ is insertedcoaxially into a long solenoid. The radius a of the rod is less than the radius b of thesolenoid. The solenoid winding has n turns per unit length and it carries the current I .Find B, H and M everywhere. Determine also Jm and Jms.
177
Lecture 19 Inductance and inductors
Flux linkage λ is a �ux Φ =∫
B · dS through a circuit with N identical turns:
λ = NΦ = LI ; L ⇐⇒ inductance.
178
• Inductor⇐⇒ circuit with inductance.
Inductance as a property of the geometry of the inductor (independent of the current)
L =λ
I=
NΦ
I, (Henry=Weber/Amp�ere).
179
Mutual inductance and self-inductance
Generalization to two circuits
Φij is a magnetic �ux through ith circuit due to the magnetic �eld of the current in thejth circuit, (i, j = 1, 2).
Φij =
∫
Si
dSi ·Bj i, j = 1, 2.
180
De�ne mutual inductance L12 as
L12 =λ12
I2
=N1Φ12
I2
.
By the same token,
L21 =λ21
I1
=N2Φ21
I1
.
Reciprocity theorem: L12 = L21.
Self-inductances of circuits 1 and 2 are de�ned as
L1 =λ11
I1
=N1Φ1
I1
;
L2 =λ22
I2
=N2Φ2
I2
;
Here, the total �ux through ith circuit is
Φi =∑j=1,2
Φij, i = 1, 2.
181
Calculation of self-inductance and mutual inductance
Method I
1. Choose a suitable coordinate system.
2. Let the conductor carry a current I .
3. Determine B from either Biot-Savart law or from Amp�ere's law ( if there is symmetryof current distribution).
4. Calculate the �ux viz., Φ =∫
B · dS.
5. Find the self-inductance, L = NΦ/I .
182
Example 1 Calculate the self-inductance of a toroid consisting of N turns of wire, tightlywound on a frame of a rectangular cross-section with the inner and outer radii a and b,respectively, and the height h.Example 2 Find the inductance per unit length of a very long solenoid with the air corehaving n turns per unit length.Example 3 Find the mutual inductance between a very long straight wire and a coplanarcircular loop of radius R. The distance between the center of the loop and the wire is d.
183
Lecture 20 Electromagnetic Induction
Faraday's observations: The current is induced in a circuit whenever
• a steady current in an adjacent circuit is turned on/off;
• there is a relative motion of the primary and secondary circuits
• a permanent magnet is thrust into/out of the circuit
Qualitatively: the induced current is caused by an induced electromotive force (emf)brought about by the changing magnetic �ux!
184
Quantitative Description
The magnetic �ux though a closed circuit C with the cross-section S
Φ =
∫
S
dS ·B,
The electromotive force by de�nition,
E =
∮
C
dl · E.
Faraday's law:
E = −dΦ
dt.
Either a magnetic �eld changes or a circuit moves, or else the circuit moves in atime-dependent and/or inhomogeneous magnetic �eld.
185
Consider a static circuit C linked to a time-varying magnetic �eld B
E =
∮
C
dl · E = − d
dt
∫
S
dS ·B = −∫
S
dS · ∂B
∂t∮
dl · E =
∫
S
dS · (∇× E) = −∫
S
dS · ∂B
∂t.
The Maxwell equation for time-varying EM �elds:
∇× E = −∂B
∂t
Example 1 A metal bar slides over a pair of conducting rails in a uniform magnetic �eldB = Boaz , with a constant velocity u = uax. Determine the voltage generated acrossthe bar.
186
Example 2 An h× w rectangular conducting loop is placed in a magnetic �eldB(t) = ayB0 sin(ωt). The normal to the loop makes the angle α with y-axis. Find aninduced emf in the loop when: (a) the loop is at rest and (b) the loop rotates with theangular velocity ω around the x−axis.Example 3 Determine the emf induced in a triangular loop to be drawn in class. Thevertical bar moves with a constant velocity u = uax in the inhomogeneous magnetic�eld, B = Axaz , where A is a constant. Assume the bar starts at the origin at t = 0.
Application: Transformers
The voltage in a primary coil by Faraday's law: V1 = −N1dΦ/dt.The voltage in a secondary coil by Faraday's law: V2 = −N2dΦ/dt.
V1
V2
=N1
N2
.
Energy conservation yields for the powers:
P1 = P2 =⇒ I1V1 = I2V2 =⇒ I1
I2
=N2
N1
.
187
Magnetic Energy
In complete analogy with the electrostatic case, the magnetic energy stored in the �eld isgiven by
Wm =1
2
∫(B ·H)dv =
1
2
∫µH2dv
The derivation is less straightforward, though, as even static magnetic �elds aregenerated by steady currents which would have to be turned on at some instant. Thus, atransient regime must be considered � which is done in Appendix B � to determine thework done to switch on the currents generating the magnetic �eld. The self-inductancecan then be determined as follows
Wm =1
2
∫(B ·H)dv =
1
2LI2, =⇒ L =
∫(B ·H)
I2dv
188
Displacement current
Motivation: Incompatibility of Ampere's law and the charge conservation law
∇×H = J,
It follows that∇ · J = ∇ · (∇×H) = 0.
Charge conservation law implies contradiction
∇ · J = −∂ρv
∂t6= 0!
189
Resolution: introducing Maxwell's displacement current Jd as
∇×H = J + Jd.
∇ · (∇×H) = 0 = ∇ · J +∇ · Jd.
∇ · Jd = −∇ · J =∂ρv
∂t=
∂
∂t(∇ ·D) = ∇ · ∂D
∂tThus,
Jd =∂D
∂t.
The Maxwell equation for time-varying EM �elds:
∇×H = J +∂D
∂t.
Displacement current manifestations: electromagnetic waves!!
190
Appendix A: Derivation of Surface and Volume Polarized Charges
Starting with the potential dV due to an in�nitesimal polarized volume dv:
dV =
dp︷ ︸︸ ︷(Pdv′) ·aR
4πε0R2(58)
Noting, R =√
(x− x′)2 + (y − y′)2 + (z − z′)2, we can show that
∇′(
1
R
)=
aR
R2.
Exercise: Check the above expression explicitly calculating the gradient of 1/R.
191
∇′(
1
R
)=
aR
R2.
It follows thatP · aR
R2= P · ∇′
(1
R
). (59)
Using the vector identity,
∇′ · (fA) = f∇′ ·A + A · ∇′f (60)
It follows from Eq. (37) and (38) with the identi�cation, f → 1/R′ and A → P that
192
P · aR
R2= ∇′ ·
(P
R
)− ∇′ ·P
R(61)
Substituting back into equation (25) and integrating, we obtain
V =1
4πε0
∫
v′
[∇′ ·
(P
R
)− ∇′ ·P
R
]dv′. (62)
Applying the divergence theorem to the �rst term on the rhs of Eq. (29), we obtain
V =
∫
S′
ρps︷ ︸︸ ︷P · a′n4πε0R
dS ′ +∫
v′
ρpv︷ ︸︸ ︷−∇′ ·P4πε0R
dv′ (63)
ρps = P · an, polarization surface charge,
ρpv = −∇ ·P, polarization volume charge.
193
Appendix B: Derivation of Magnetization Currents
By de�nition, the dipole moment of elementary volume, dv′ is dm = Mdv′, and
dA(r) =µ0
dm︷ ︸︸ ︷Mdv′×
R/R︷︸︸︷aR
4πR2=
µ0M×R
4πR3dv′. (64)
R = |r− r′| =√
(x− x′)2 + (y − y′)2 + (z − z′)2.
194
Hence,
∇′(
1
R
)=
(x− x′)ax + (y − y′)ay + (z − z′)az
[(x− x′)2 + (y − y′)2 + (z − z′)2]3/2=
R
R3
Using the last expression, one can express R/R3 in equation (66) in terms of∇1/R
and integrating over the volume, obtain the expression for the vector potential:
A =µ0
4π
∫M×∇′
(1
R
)dv′. (65)
195
A =µ0
4π
∫M×∇′
(1
R
)dv′. (66)
Use the vector identity,
∇× (fF) = f∇× F− F× (∇f),
with f = 1/R and F = M, that is,
M×∇′(
1
R
)=
1
R∇′ ×M−∇′ ×
(M
R
).
Substituting the lhs back into equation (68), we obtain
A =µ0
4π
∫
v′
∇′ ×M
Rdv′ − µ0
4π
∫
v′∇′ ×
(M
R
)dv′. (67)
196
A =µ0
4π
∫
v′
∇′ ×M
Rdv′ − µ0
4π
∫
v′∇′ ×
(M
R
)dv′. (68)
Using yet another vector identity,∫
v′∇′ × Fdv′ = −
∮
S′F× dS,
we cast Eq. (70) into the form
A =µ0
4π
∫
v′
∇′ ×M
Rdv′ +
µ0
4π
∮
S′
M× an
RdS ′
=µ0
4π
∫
v′
Jmdv′
R+
µ0
4π
∮
S′
KmsdS ′
R. (69)
197
A =µ0
4π
∫
v′
∇′ ×M
Rdv′ +
µ0
4π
∮
S′
M× an
RdS ′
=µ0
4π
∫
v′
Jmdv′
R+
µ0
4π
∮
S′
KmsdS ′
R. (70)
Introduce,Jm = ∇×M , (magnetization volume current);
Kms = M× an , (magnetization surface current).
198
Appendix C: Magnetic energy derivation
Power associated with the Faraday emf inducing a current I :
P =dWm
dt= IE = −I
dΦ
dt.
Incremental change in the energy = incremental work against emf:
δWm = IδΦ
∆(δWm) = Jδσ︸︷︷︸I
∫
S
dS · δB︸ ︷︷ ︸
δΦ
= Jδσ
∫
S
dS · (∇× δA) = Jδσ
∮
C
dl · δA.
The total incremental work:
δWm =∑
l
Jδσ
∮
C
dl · δA.
using Jδσdl = Jδv, we arrive at
δWm =
∫dvδA · J.
199
Using the Maxwell equation:∇×H = J,
δWm =
∫dvδA · (∇×H).
Using the vector identity,
∇ · (P×Q) = Q · (∇×P)−P · (∇×Q),
we obtain
δWm =
∫dv[H · (∇× δA) +∇ · (H× δA)]
The second term on the r.h.s. is zero for any localized �eld distribution (why?). Hence,
δWm =
∫dv(H · δB).
200
For para- or diamagnetic media, H ∝ B so that � using the same rules as forderivatives � we have
H · δB =1
2δ(H ·B),
and adding all incremental contributions, we obtain for the total energy stored inmagnetic �eld,
Wm =1
2
∫dv(B ·H) =
1
2
∫dvµH2
201
