lecture 1 vector algebra: a brief revie... · lecture 1 vector algebra: a brief review outline 1....
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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
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• 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
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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,
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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.
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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)].
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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.
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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.
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Electric �elds of continuous distributions of charges
Elementary (in�nitesimal) charge
dQ =
ρvdv volume charge
ρSdS surface charge
ρLdl line charge
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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).
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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)
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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).
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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.
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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.
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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.
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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.
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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.
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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.
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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).
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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.
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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)
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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).
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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.
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Thus,dV = −E · dr; dV = ∇V · dr. (17)
It follows from equation (17) that
E = −∇V, (connection between potential and �eld).
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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
.
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Consider the potential at a remote point, r À d:
r2 − r1 ' d cos θ; r1r2 ' r2, (in denomenator).
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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).
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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.
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• 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)
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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)
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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)
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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.
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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.
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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.
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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.
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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.
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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
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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,
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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.
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• 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!
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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
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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.
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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).
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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
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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).
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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)
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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.
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Use Gaussian surface in �gure (b) to get
∆Qencl = ρS∆S = D1n∆S −D2n∆S.
ThusD1n −D2n = ρS . (27)
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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.
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Refraction of the electric �eld at the interface of two dielectic media
E1t = E2t =⇒ E1 sin θ1 = E2 sin θ2,
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D = εE =⇒ Dn = εEn
D1nno free charge
= D2n =⇒ ε1E1 cos θ1 = ε2E2 cos θ2.
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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.
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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.
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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)
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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
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Lecture 10 Poisson and Laplace equations
Outline
1. Poisson equation.
2. Laplace equation
3. Uniqueness of the solution to Laplace equation
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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)
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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)
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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.
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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.
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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.
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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.
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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).
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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.
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Parallel-plate capacitor
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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
.
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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)
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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
.
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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,
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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.
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Spherical capacitor
Choose Gaussian surface to be another coaxial sphere such that a < r < b.
Q =
∮D · dS = ε
∮E · dS = εE4πr2.
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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.
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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.
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• 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.
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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.
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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.
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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.
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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.
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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.
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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.
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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.
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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]
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Current density, J : current per unit cross-section [Amp�eres/meter2]:
dI = J · dS I =
∫
S
J · dS .
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Convection versus conduction currents
• Convection current⇐⇒ simply �ow of charged particles.
• Conduction current⇐⇒ �ow of particles in a conductor.
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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 .
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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.
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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 .
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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.
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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)
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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 .
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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.
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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
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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
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Mathematical formulation of Bio-Savart law
dH =Idl sin α
4πR2=
Idl×R/R︷︸︸︷aR
4πR2=
Idl×R
4πR3
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Direction of H due to the current element:
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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
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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 .
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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.
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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).
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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.
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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.
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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.
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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?
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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).
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0 =
∮B · dS
divergence︷︸︸︷=
∫
v
∇ ·Bdv.
It follows that
∇ ·B = 0, (differential form of magnetic �ux conservation).
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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.
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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.
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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 .
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Superposition principle yields
A =
∫
L
µ0Idl
4πR, (line current);
A =
∫
S
µ0KdS
4πR, (surface current);
A =
∫
v
µ0Jdv
4πR, (volume current).
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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.
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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.
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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.
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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.
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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).
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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.
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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.
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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.
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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.
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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.
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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).
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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).
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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.
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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.
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However, the forces form a couple, whose torque has the magnitude
|T| = |F|w sin α = IB lw︸︷︷︸S
sin α = IBS sin α.
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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.
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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.
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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
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B =µ0m
4πr3(2 cos θar + sin θaθ)
• small loop⇐⇒ magnetic dipole!
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Comparison with electric dipole
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Magnetization
Orbital motion of electrons around atoms or internal degree of freedom (spin)⇐⇒magnetic moment m = ISan.
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Magnetization M, (Amp�eres/meter) is a magnetic dipole moment per unit volume:
M = lim∆v→0
∑Nk=1 mk
∆v
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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.
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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).
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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!
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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
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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.
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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.
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Consider a Gaussian surface indicated in the �gure,
B1n∆S −B2n∆S = 0
B1n = B2n
B=µH︷︸︸︷=⇒ µ1H1n = µ2H2n.
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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)
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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.
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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.
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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.
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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.
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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.
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• 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).
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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.
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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.
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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 .
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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.
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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!
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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.
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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.
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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
.
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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
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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!
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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!!
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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.
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∇′(
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
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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.
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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.
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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)
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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)
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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)
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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).
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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.
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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).
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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
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