Semester II, 2017-18
Department of Physics, IIT Kanpur
PHY103A: Lecture # 20
(Text Book: Intro to Electrodynamics by Griffiths, 3rd Ed.)
Anand Kumar Jha
16-Feb-2018
Notes
β’ Mid-sem: Feb 22, Thursday, 8-10 am
β’ Course coverage up to class on Monday (Feb 12)
β’ Q1: 10; Q2=10; Midsem: 80; Endsem: 120
β’ Regraded Quiz # 1 will be returned after midsem
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Summary of Lecture # 19:
β’ Magnetic Vector Potential
βππππππ = ππ0ππ
One can always redefine the vector potential such that its divergence is zero ππ β ππ = 0
ππ β ππ = 0
β ππ(π«π«) = ππ0
4ππ οΏ½
ππ(π«π«π«) r πππππ«
β’ Magnetostatic Boundary Conditions (Consequences of the fundamental laws):
π΅π΅ο
above= π΅π΅ο
below π΅π΅ο·
above β π΅π΅ο·
below = ππ0πΎπΎ ππ = ππ Γ ππ
β
β
π΄π΄ο
above= π΄π΄ο
below π΄π΄ο·
above = π΄π΄ο·
below
ππ π«π« = ππ0πΌπΌ 4ππ
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ππ οΏ½ πππ₯π₯β² + ππ0πΌπΌ 4ππ
1
ππ2 οΏ½ ππβ² cosπΌπΌ πππ₯π₯β²
β’ Multipole Expansion of vector potential
Prob. 5.23 (Griffiths, 3rd Ed. ): What is the current density ππ that would produce the magnetic potential ππ = πππποΏ½
Magnetic Vector Potential:
ππ = πππποΏ½
ππ = ππ Γ ππ
ππ = 1
ππ0 ππ Γ ππ
= 1 π π
ππ
πππ π π π ππ π³π³οΏ½ = ππ π π π³π³οΏ½
= ππ ππ0π π 2 πποΏ½
= 1
ππ0 β ππ
πππ π ππ
π π πποΏ½
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Prob. 5.22 (Griffiths, 3rd Ed. ): Find the magnetic vector potential due to a finite wire and using that find the magnetic field.
Magnetic Vector Potential:
ππ = ππ0 4ππ οΏ½
πΌπΌ r
π§π§2
βπ§π§1
ππππ π³π³οΏ½ = ππ0πΌπΌ 4ππ οΏ½
1 π π 2 + ππ2
π§π§2
βπ§π§1
ππππ π³π³οΏ½
= ππ0πΌπΌ
4ππ ln ππ2 + ππ22 + π π 2
βππ1 + ππ12 + π π 2 π³π³οΏ½
ππ = ππ Γ ππ = βπππ΄π΄π§π§
πππ π πποΏ½
ππ(π«π«) = ππ0 4ππ οΏ½
ππ(π«π«π«) r πππππ«
Magnetostatics in matter (magnetic field in matter)
What is Polarization? - dipole moment per unit volume
β’ The dipole moment is caused either by stretch of an atom/molecule or by rotation of polar molecules
What is Magnetization? - magnetic dipole moment per unit volume
β’ The magnetic dipole moment is caused by electric charges in motion:
(i) electrons orbiting around nuclei & (ii) electrons spinning about their own axes.
β’ Polarization in the direction parallel to the applied electric field
β’ In some material, magnetization is in the direction parallel to ππ (Paramagnets).
β’ In some other material, magnetization is opposite to ππ (Diamagnets).
β’ In other, there can be magnetization even in the absence of ππ (Ferromagnets).
ο If a ferromagnetic material is exposed to strong-enough magnetic field, the magnetization in the material does not return to zero after the field is
removed and this way the material becomes a permanent magnet.
ο Magnetization in Ferromagnetic material is much higher.
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The Field of a Magnetized Object:
= ππ0
4ππ οΏ½ ππ π«π«β² Γ ππβ² 1
r ππππ«
vol [ Using ππβ² 1r = rΜ r2 ]
[ Using ππ Γ ππππ = ππ ππ Γ ππ β ππ Γ (ππππ) ]
= ππ0 4ππ οΏ½
1
r [ππβ²Γ ππ π«π«β² ]πππβ² vol
= ππ0 4ππ οΏ½
1
r [ππβ²Γ ππ π«π«β² ]πππβ² β ππ0
4ππ οΏ½ ππβ² Γ ππ π«π«β²
r ππππ«
vol vol
= ππ0 4ππ οΏ½
ππ π«π«β² Γ rΜ r2 ππππ« vol
ππdip π«π« = ππ0 4ππ
π¦π¦ Γ π«π«οΏ½
ππ2
[ Griffiths Prob. 1.60 (b) + ππ0
4ππ οΏ½ 1
r [ππ π«π«β² Γ πππππ«]
π π π π π π π π
Ampereβs law in magnetized material:
ππππ π«π«β² = ππβ² Γ ππ π«π«β² ππππ π«π«β² = ππ π«π«β² Γ π§π§οΏ½
Volume current Surface current Total volume current is
ππ = πππ π + ππππ
What happens to the Ampereβs law?
ππ Γ ππ = ππ0ππ = ππ0(πππ π + ππππ) = ππ0(πππ π + ππ Γ ππ)
ππ Γ ππ
ππ0 β ππ = πππ π
ππ Γ ππ = πππ π ππ β‘ ππ
ππ0 β ππ
οΏ½ ππ β πππ₯π₯ = πΌπΌπ π enc Or,
Define:
So, Ampereβs law in magnetized material (differential form) Ampereβs law in magnetized material (integral form) And,
ππ β‘ ππ0(ππ + ππ) Which means
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Magnetic Susceptibility and Permeability In paramagnetic and diamagnetic materials, the magnetization is proportional to the applied field
ππ = ππππ ππ
β’ ππππ is called the magnetic susceptibility
β’ ππππ is a dimensionless quantity
β’ ππππ is positive for paramagnetic and negative for diamagnets
β’ Typical values of ππππ are around 10β5 The magnetic field thus becomes
All the best for midsem
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