Electrodynamics 3
Transcript of Electrodynamics 3
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Q.1. Application of Maxwellβs eqns. [Griffiths Problem 7.42]
In a perfect conductor the conductivity π is infinite, so from
βOhmβs lawβ π½ = ππΈ , πΈ = 0 . Any net charge must be on the surface (as in electrostatics for an imperfect conductor).
(a) Show that the magnetic field is constant inside a perfect conductor.
Solution: Faradayβs law: π» Γ πΈ = βππ΅ ππ‘ so if πΈ = 0 then
ππ΅ ππ‘ = 0 , i.e. π΅ is independent of π‘, i.e. constant.
(b) Show that the magnetic flux through a perfectly conducting loop is constant.
Solution: Faradayβs law in integral form is πΈ β ππ π
= βπΞ¦π
ππ‘
Worked Examples Set 2
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Q.1. Solution (b) [continued]
Inside the perfectly conducting loop πΈ = 0 , so πΞ¦π
ππ‘= 0
i.e. Magnetic flux Ξ¦π is independent of π‘, i.e. constant.
(c) A superconductor is a perfect conductor with the additional
property that the constant π΅-field inside is actually zero.
Show that the current in a superconductor is confined to the surface (there are no volume currents).
Solution: Maxwell-Ampere law:
π» Γ π΅ = π0π½ + π0ν0 ππΈ ππ‘
so if πΈ = 0 and π΅ = 0 (given) then π½ = 0
i.e. there are no volume currents and so
any currents must be surface currents (πΎ).
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Q.2. Potentials and Gauge Transformations
[Griffiths Problems 10.3 and 10.5]
(a) Find the fields, and the charge and current distributions,
corresponding to potentials π π , π‘ = 0 , π΄ π , π‘ = β1
4ππ0
ππ‘
π2 π
(b) Use the gauge function π = β1
4ππ0
ππ‘
π to transform the
potentials in (a), and comment on the results.
Solution: (a) πΈ = βπ»π βππ΄
ππ‘= 0 +
1
4ππ0
π
π2 π =π
4ππ0π2 π
π΅ = π» Γ π΄ = 0 [π΄ = π΄π (radial component only) which does not depend on π or π, i.e. ππ΄π ππ = 0 , ππ΄π ππ = 0]
Fields of a stationary point charge, strange potentials.
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Q.2. Solution [continued] Charge distribution: we have a point charge π , which can be
written as a (3D) delta function : π = ππΏ3 π ; current density π½ = 0 (since π΅ = 0)
(b) Gauge transformation: π = β1
4ππ0
ππ‘
π
πβ² = π βππ
ππ‘= 0 β β
1
4ππ0
π
π=
π
4ππ0π
π΄ β² = π΄ + π»π = β1
4ππ0
ππ‘
π2 π + β1
4ππ0ππ‘ β
1
π2 π
= β1
4ππ0
ππ‘
π2 π +1
4ππ0
ππ‘
π2 π = 0
So this transformation transforms the strange potentials of (a) into the more standard potentials of a static point charge.
Q.3. Retarded Potentials
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[Griffiths, Example 10.2] A long straight wire has πΌ π‘ = 0 , π‘ β€ 0 and πΌ π‘ = πΌ0 , π‘ > 0 (βswitch onβ at π‘ = 0).
Find the resulting πΈ and π΅ fields.
Solution: Assume wire is neutral, so π = 0.
Retarded vector potential at π (see figure) is
π΄ π , π‘ =π0
4ππ§
πΌ π‘π
π
+β
ββππ§ [in same direction (π§ ) as current πΌ]
For the field to reach π takes time π π , so for π‘ < π π , π΄ = 0
For π‘ > π π the βsignalβ takes time π π to travel along π , so
range of π§ contributing to π΄ π , π‘ at π is π§ β€ ππ‘ 2 β π 2
Outside this range π‘π = π‘ β π π < 0 so πΌ π‘π = 0
[Note the π§ contributions from above & below the point]
Q.3. Solutions [continued]
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π΄ π , π‘ =π0πΌ0
4ππ§ 2
ππ§
π 2+π§2
ππ‘ 2βπ 2
0
=π0πΌ0
2ππ§ ln π 2 + π§2 + π§
ππ‘ 2βπ 2
=π0πΌ0
2ππ§ ln
ππ‘+ ππ‘ 2βπ 2
π
The derivatives of π΄ (= π΄π§ ) give the fields:
πΈ π , π‘ = βππ΄
ππ‘= β
π0πΌ0π
2π ππ‘ 2βπ 2π§
π΅ π , π‘ = π» Γ π΄ = βππ΄π§
ππ π =
π0πΌ0
2ππ
ππ‘
ππ‘ 2βπ 2π
Note that as π‘ β β , πΈ β 0 , π΅ βπ0πΌ0
2ππ π (the static fields)
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Q.4. Poynting Vector
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[Feynman Lectures in Physics, Vol. II, 27.5] A capacitor with circular plates, radius π , separation π, is being charged. Find the rate at which its stored energy is increasing and relate this to the Poynting vector flux.
Solution: We assume the field between the plates is uniform. The energy stored is
π = 1
2ν0πΈ
2 ππ±π±
=1
2ν0πΈ
2 ππ 2π and this increases at the
rate ππ
ππ‘= ππ 2π
π
ππ‘
1
2ν0πΈ
2 = ππ 2πν0πΈππΈ
ππ‘
Now in the capacitor space π½ π = 0 and Maxwell-Ampere
π». ππ = ππ·
ππ‘. ππ i.e. π». 2ππ = ν0
ππΈ
ππ‘. ππ 2 so π» =
π
2ν0
ππΈ
ππ‘
[π» is azimuthal, field lines around the cylindrical air space.]
Q.4. Solution [continued]
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We have ππ
ππ‘= ππ 2πν0πΈ
ππΈ
ππ‘ , π» =
π
2ν0
ππΈ
ππ‘
Now the Poynting vector is π = πΈ Γ π» and we see that it points inwards; this is
the direction of energy flow! (note πΈ β₯ π» ) The Poynting vector flux, through the curved surface that defines the capacitor space , is
π . ππ = βπΈπ». 2ππ π = βν0πΈππΈ
ππ‘ππ 2π = β
ππ
ππ‘
[we have to compute the integral over a closed surface, but the flux through the ends of the βcylinderβ (the plates) is zero ,
since π β₯ surface; the integral is negative because flux is in.]
This is just Poyntingβs theorem with π½ π = 0 : the rate at which
energy flows in = the rate at which the (electric in this case) field energy is increasing. Energy flows in from outside!
Q.5. EM Waves
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[Griffiths, Problem 9.9 ] Write down the (real) electric and magnetic fields for a monochromatic plane wave of amplitude πΈ0, frequency π, (and phase angle zero) that is: (a) travelling in the βπ₯-direction and polarized in the +π§-direction; (b) travelling in the direction from the origin to the point (1,1,1), with polarization parallel to the π₯π§-plane. In each case, sketch the wave, and give the explicit Cartesian
components of π and π .
Solution: (a) The frequency is specified so we should give π in
terms of π ; the wave direction is βπ₯ , so π = β π π π₯
Then π. π = β π π π₯ β π₯π₯ + π¦π¦ + π§π§ = β π π π₯
πΈ is in the given direction of polarization, π§
π΅ is in the direction π Γ π = βπ₯ Γ π§ = π¦
Q.5. Solution [continued]
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Thus the fields are
πΈ π₯, π‘ = πΈ0π§ cos π π π₯ + ππ‘
π΅ π₯, π‘ = πΈ0 π π¦ cos π π π₯ + ππ‘
(b) The unit vector from the origin to the
point 1,1,1 is π₯ + π¦ + π§ 3 , so
π = π π π₯ + π¦ + π§ 3 Polarization direction π is β₯ π₯π§-plane, so
must have the form πΌπ₯ + π½π§ ; π . π = 0, so must have πΌ = βπ½;
π is a unit vector, so πΌ = 1 2 and thus π = π₯ β π§ 2
This is the direction of πΈ ; the direction of π΅ is π Γ π :
π Γ π =1
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π₯ π¦ π§ 1 1 11 0 β1
=βπ₯ + 2π¦ β π§
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Q.5. Solution [continued]
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Thus the fields are
πΈ π₯, π¦, π§, π‘ = πΈ0 cos π π₯ + π¦+ π§
3πβ ππ‘
π₯ β π§
2
π΅ π₯, π‘ =πΈ0
πcos
π π₯ +π¦+ π§
3πβ ππ‘
βπ₯ + 2π¦ β π§
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This is what this wave would look like; sometimes this might be quite difficult to draw!
[Part (b) is far more intricate than any question you would get in a test or exam.]
Q.6. Energy in EM Waves
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The βsolar constantβ, the average intensity of radiation from the Sun at the top of the Earthβs atmosphere, is 1.34 kW/m2. If this radiation was a linearly polarized monochromatic (single frequency) plane wave (propagating in free space, of course),
find the amplitudes of πΈ and π΅ for this wave.
Solution: Intensity (in W m2 ) is the (time averaged) Poynting
vector: πΌ = π = 1
2 π ν0πΈ0
2 Inserting the values,
πΈ0 =2πΌ
πν0=
2 Γ 1.34 Γ 103
3.0 Γ 108 Γ 8.85 Γ 10β12= 1.0 Γ 103 V/m
πΈ0 = 1 kV/m ! The magnetic field is
π΅0 =πΈ0
π=
1.0Γ 103
3.0Γ 108= 3.3Γ 10β6 T = 3.3 ΞΌT