MPH-007 Solved Assignment 2024 | IGNOU

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MPH-007 Sample Solution

MPH-007 Solved Assignment 2024 | IGNOU MPH-007 Sample Solution

MPH-007 Solved Assignment

PART A

Question:-01

  1. a) Write Maxwell’s equations in differential and integral forms. How are these equations modified in vacuum (charge and current free region)? Explain how Maxwell modified Ampere’s law and proposed generalised Ampere’s law consistent with equation of continuity and obtain the expression for the generalised Ampere’s law in differential form.

Answer:

Maxwell’s Equations

Maxwell’s equations are a set of four fundamental equations that describe the behavior of electric and magnetic fields. They can be expressed in both differential and integral forms.

Differential Form:

  1. Gauss’s Law for Electricity:
    ∇ ⋅ E = ρ ϵ 0 ∇ ⋅ E = ρ ϵ 0 grad*E=(rho)/(epsilon_(0))\nabla \cdot \mathbf{E} = \frac{\rho}{\epsilon_0}∇⋅E=ρϵ0
    where E E E\mathbf{E}E is the electric field, ρ ρ rho\rhoρ is the charge density, and ϵ 0 ϵ 0 epsilon_(0)\epsilon_0ϵ0 is the permittivity of free space.
  2. Gauss’s Law for Magnetism:
    ∇ ⋅ B = 0 ∇ ⋅ B = 0 grad*B=0\nabla \cdot \mathbf{B} = 0∇⋅B=0
    where B B B\mathbf{B}B is the magnetic field.
  3. Faraday’s Law of Induction:
    ∇ × E = − ∂ B ∂ t ∇ × E = − ∂ B ∂ t grad xxE=-(delB)/(del t)\nabla \times \mathbf{E} = -\frac{\partial \mathbf{B}}{\partial t}∇×E=−∂B∂t
  4. Ampere’s Law (with Maxwell’s correction):
    ∇ × B = μ 0 J + μ 0 ϵ 0 ∂ E ∂ t ∇ × B = μ 0 J + μ 0 ϵ 0 ∂ E ∂ t grad xxB=mu_(0)J+mu_(0)epsilon_(0)(delE)/(del t)\nabla \times \mathbf{B} = \mu_0 \mathbf{J} + \mu_0 \epsilon_0 \frac{\partial \mathbf{E}}{\partial t}∇×B=μ0J+μ0ϵ0∂E∂t
    where J J J\mathbf{J}J is the current density, and μ 0 μ 0 mu_(0)\mu_0μ0 is the permeability of free space.

Integral Form:

  1. Gauss’s Law for Electricity:
    ∮ ∂ V E ⋅ d A = Q enc ϵ 0 ∮ ∂ V E ⋅ d A = Q enc ϵ 0 oint_(del V)E*dA=(Q_(“enc”))/(epsilon_(0))\oint_{\partial V} \mathbf{E} \cdot d\mathbf{A} = \frac{Q_{\text{enc}}}{\epsilon_0}∮∂VE⋅dA=Qencϵ0
    where Q enc Q enc Q_(“enc”)Q_{\text{enc}}Qenc is the total charge enclosed by the surface ∂ V ∂ V del V\partial V∂V.
  2. Gauss’s Law for Magnetism:
    ∮ ∂ V B ⋅ d A = 0 ∮ ∂ V B ⋅ d A = 0 oint_(del V)B*dA=0\oint_{\partial V} \mathbf{B} \cdot d\mathbf{A} = 0∮∂VB⋅dA=0
  3. Faraday’s Law of Induction:
    ∮ ∂ S E ⋅ d l = − d d t ∫ S B ⋅ d A ∮ ∂ S E ⋅ d l = − d d t ∫ S   B ⋅ d A oint_(del S)E*dl=-(d)/(dt)int _(S)B*dA\oint_{\partial S} \mathbf{E} \cdot d\mathbf{l} = -\frac{d}{dt} \int_S \mathbf{B} \cdot d\mathbf{A}∮∂SE⋅dl=−ddt∫SB⋅dA
  4. Ampere’s Law (with Maxwell’s correction):
    ∮ ∂ S B ⋅ d l = μ 0 ( I enc + ϵ 0 d d t ∫ S E ⋅ d A ) ∮ ∂ S B ⋅ d l = μ 0 I enc + ϵ 0 d d t ∫ S   E ⋅ d A oint_(del S)B*dl=mu_(0)(I_(“enc”)+epsilon_(0)(d)/(dt)int _(S)E*dA)\oint_{\partial S} \mathbf{B} \cdot d\mathbf{l} = \mu_0 \left( I_{\text{enc}} + \epsilon_0 \frac{d}{dt} \int_S \mathbf{E} \cdot d\mathbf{A} \right)∮∂SB⋅dl=μ0(Ienc+ϵ0ddt∫SE⋅dA)
    where I enc I enc I_(“enc”)I_{\text{enc}}Ienc is the total current passing through the surface S S SSS.

Modifications in Vacuum

In a vacuum, there are no free charges ( ρ = 0 ρ = 0 rho=0\rho = 0ρ=0) and no currents ( J = 0 J = 0 J=0\mathbf{J} = 0J=0). Thus, Maxwell’s equations in a vacuum are simplified as follows:
  1. Gauss’s Law for Electricity:
    ∇ ⋅ E = 0 ∇ ⋅ E = 0 grad*E=0\nabla \cdot \mathbf{E} = 0∇⋅E=0
  2. Gauss’s Law for Magnetism:
    ∇ ⋅ B = 0 ∇ ⋅ B = 0 grad*B=0\nabla \cdot \mathbf{B} = 0∇⋅B=0
  3. Faraday’s Law of Induction:
    ∇ × E = − ∂ B ∂ t ∇ × E = − ∂ B ∂ t grad xxE=-(delB)/(del t)\nabla \times \mathbf{E} = -\frac{\partial \mathbf{B}}{\partial t}∇×E=−∂B∂t
  4. Ampere’s Law (with Maxwell’s correction):
    ∇ × B = μ 0 ϵ 0 ∂ E ∂ t ∇ × B = μ 0 ϵ 0 ∂ E ∂ t grad xxB=mu_(0)epsilon_(0)(delE)/(del t)\nabla \times \mathbf{B} = \mu_0 \epsilon_0 \frac{\partial \mathbf{E}}{\partial t}∇×B=μ0ϵ0∂E∂t

Maxwell’s Modification of Ampere’s Law

Before Maxwell, Ampere’s law was written as:
∇ × B = μ 0 J . ∇ × B = μ 0 J . grad xxB=mu_(0)J.\nabla \times \mathbf{B} = \mu_0 \mathbf{J}.∇×B=μ0J.
However, this form is inconsistent with the continuity equation for charge, which states:
∇ ⋅ J + ∂ ρ ∂ t = 0. ∇ ⋅ J + ∂ ρ ∂ t = 0. grad*J+(del rho)/(del t)=0.\nabla \cdot \mathbf{J} + \frac{\partial \rho}{\partial t} = 0.∇⋅J+∂ρ∂t=0.
To see why, take the divergence of both sides of Ampere’s law:
∇ ⋅ ( ∇ × B ) = ∇ ⋅ ( μ 0 J ) . ∇ ⋅ ( ∇ × B ) = ∇ ⋅ ( μ 0 J ) . grad*(grad xxB)=grad*(mu_(0)J).\nabla \cdot (\nabla \times \mathbf{B}) = \nabla \cdot (\mu_0 \mathbf{J}).∇⋅(∇×B)=∇⋅(μ0J).
The left-hand side is always zero because the divergence of a curl is always zero:
∇ ⋅ ( ∇ × B ) = 0. ∇ ⋅ ( ∇ × B ) = 0. grad*(grad xxB)=0.\nabla \cdot (\nabla \times \mathbf{B}) = 0.∇⋅(∇×B)=0.
For the right-hand side, using the continuity equation:
∇ ⋅ J = − ∂ ρ ∂ t . ∇ ⋅ J = − ∂ ρ ∂ t . grad*J=-(del rho)/(del t).\nabla \cdot \mathbf{J} = -\frac{\partial \rho}{\partial t}.∇⋅J=−∂ρ∂t.
Therefore:
0 = μ 0 ( − ∂ ρ ∂ t ) . 0 = μ 0 − ∂ ρ ∂ t . 0=mu_(0)(-(del rho)/(del t)).0 = \mu_0 \left(-\frac{\partial \rho}{\partial t}\right).0=μ0(−∂ρ∂t).
This implies that charge conservation would require ∂ ρ ∂ t = 0 ∂ ρ ∂ t = 0 (del rho)/(del t)=0\frac{\partial \rho}{\partial t} = 0∂ρ∂t=0, which is not generally true in time-varying fields.

Maxwell’s Correction

Maxwell proposed adding an additional term to Ampere’s law to make it consistent with charge conservation. He added the term ϵ 0 ∂ E ∂ t ϵ 0 ∂ E ∂ t epsilon_(0)(delE)/(del t)\epsilon_0 \frac{\partial \mathbf{E}}{\partial t}ϵ0∂E∂t, which represents the displacement current density:
∇ × B = μ 0 J + μ 0 ϵ 0 ∂ E ∂ t . ∇ × B = μ 0 J + μ 0 ϵ 0 ∂ E ∂ t . grad xxB=mu_(0)J+mu_(0)epsilon_(0)(delE)/(del t).\nabla \times \mathbf{B} = \mu_0 \mathbf{J} + \mu_0 \epsilon_0 \frac{\partial \mathbf{E}}{\partial t}.∇×B=μ0J+μ0ϵ0∂E∂t.
To show this is consistent with the continuity equation, take the divergence of both sides of the corrected Ampere’s law:
∇ ⋅ ( ∇ × B ) = ∇ ⋅ ( μ 0 J + μ 0 ϵ 0 ∂ E ∂ t ) . ∇ ⋅ ( ∇ × B ) = ∇ ⋅ μ 0 J + μ 0 ϵ 0 ∂ E ∂ t . grad*(grad xxB)=grad*(mu_(0)J+mu_(0)epsilon_(0)(delE)/(del t)).\nabla \cdot (\nabla \times \mathbf{B}) = \nabla \cdot \left( \mu_0 \mathbf{J} + \mu_0 \epsilon_0 \frac{\partial \mathbf{E}}{\partial t} \right).∇⋅(∇×B)=∇⋅(μ0J+μ0ϵ0∂E∂t).
Again, the left-hand side is zero:
∇ ⋅ ( ∇ × B ) = 0. ∇ ⋅ ( ∇ × B ) = 0. grad*(grad xxB)=0.\nabla \cdot (\nabla \times \mathbf{B}) = 0.∇⋅(∇×B)=0.
For the right-hand side:
∇ ⋅ ( μ 0 J ) + ∇ ⋅ ( μ 0 ϵ 0 ∂ E ∂ t ) . ∇ ⋅ ( μ 0 J ) + ∇ ⋅ μ 0 ϵ 0 ∂ E ∂ t . grad*(mu_(0)J)+grad*(mu_(0)epsilon_(0)(delE)/(del t)).\nabla \cdot (\mu_0 \mathbf{J}) + \nabla \cdot \left( \mu_0 \epsilon_0 \frac{\partial \mathbf{E}}{\partial t} \right).∇⋅(μ0J)+∇⋅(μ0ϵ0∂E∂t).
Using the continuity equation:
∇ ⋅ J = − ∂ ρ ∂ t , ∇ ⋅ J = − ∂ ρ ∂ t , grad*J=-(del rho)/(del t),\nabla \cdot \mathbf{J} = -\frac{\partial \rho}{\partial t},∇⋅J=−∂ρ∂t,
we get:
0 = μ 0 ( − ∂ ρ ∂ t ) + μ 0 ϵ 0 ( ∂ ∂ t ( ∇ ⋅ E ) ) . 0 = μ 0 − ∂ ρ ∂ t + μ 0 ϵ 0 ∂ ∂ t ( ∇ ⋅ E ) . 0=mu_(0)(-(del rho)/(del t))+mu_(0)epsilon_(0)((del)/(del t)(grad*E)).0 = \mu_0 \left( -\frac{\partial \rho}{\partial t} \right) + \mu_0 \epsilon_0 \left( \frac{\partial}{\partial t} (\nabla \cdot \mathbf{E}) \right).0=μ0(−∂ρ∂t)+μ0ϵ0(∂∂t(∇⋅E)).
By Gauss’s law:
∇ ⋅ E = ρ ϵ 0 . ∇ ⋅ E = ρ ϵ 0 . grad*E=(rho)/(epsilon_(0)).\nabla \cdot \mathbf{E} = \frac{\rho}{\epsilon_0}.∇⋅E=ρϵ0.
Thus:
μ 0 ( − ∂ ρ ∂ t ) + μ 0 ϵ 0 ∂ ∂ t ( ρ ϵ 0 ) = 0 , μ 0 − ∂ ρ ∂ t + μ 0 ϵ 0 ∂ ∂ t ρ ϵ 0 = 0 , mu_(0)(-(del rho)/(del t))+mu_(0)epsilon_(0)(del)/(del t)((rho)/(epsilon_(0)))=0,\mu_0 \left( -\frac{\partial \rho}{\partial t} \right) + \mu_0 \epsilon_0 \frac{\partial}{\partial t} \left( \frac{\rho}{\epsilon_0} \right) = 0,μ0(−∂ρ∂t)+μ0ϵ0∂∂t(ρϵ0)=0,
μ 0 ( − ∂ ρ ∂ t ) + μ 0 ∂ ρ ∂ t = 0 , μ 0 − ∂ ρ ∂ t + μ 0 ∂ ρ ∂ t = 0 , mu_(0)(-(del rho)/(del t))+mu_(0)(del rho)/(del t)=0,\mu_0 \left( -\frac{\partial \rho}{\partial t} \right) + \mu_0 \frac{\partial \rho}{\partial t} = 0,μ0(−∂ρ∂t)+μ0∂ρ∂t=0,
0 = 0. 0 = 0. 0=0.0 = 0.0=0.
This consistency confirms that Maxwell’s addition is correct. Therefore, the generalized Ampere’s law in differential form is:
∇ × B = μ 0 J + μ 0 ϵ 0 ∂ E ∂ t . ∇ × B = μ 0 J + μ 0 ϵ 0 ∂ E ∂ t . grad xxB=mu_(0)J+mu_(0)epsilon_(0)(delE)/(del t).\nabla \times \mathbf{B} = \mu_0 \mathbf{J} + \mu_0 \epsilon_0 \frac{\partial \mathbf{E}}{\partial t}.∇×B=μ0J+μ0ϵ0∂E∂t.

Summary

Maxwell modified Ampere’s law to include the displacement current term μ 0 ϵ 0 ∂ E ∂ t μ 0 ϵ 0 ∂ E ∂ t mu_(0)epsilon_(0)(delE)/(del t)\mu_0 \epsilon_0 \frac{\partial \mathbf{E}}{\partial t}μ0ϵ0∂E∂t, ensuring consistency with the continuity equation. The complete Maxwell’s equations in differential form are:
  1. ∇ ⋅ E = ρ ϵ 0 ∇ ⋅ E = ρ ϵ 0 grad*E=(rho)/(epsilon_(0))\nabla \cdot \mathbf{E} = \frac{\rho}{\epsilon_0}∇⋅E=ρϵ0,
  2. ∇ ⋅ B = 0 ∇ ⋅ B = 0 grad*B=0\nabla \cdot \mathbf{B} = 0∇⋅B=0,
  3. ∇ × E = − ∂ B ∂ t ∇ × E = − ∂ B ∂ t grad xxE=-(delB)/(del t)\nabla \times \mathbf{E} = -\frac{\partial \mathbf{B}}{\partial t}∇×E=−∂B∂t,
  4. ∇ × B = μ 0 J + μ 0 ϵ 0 ∂ E ∂ t ∇ × B = μ 0 J + μ 0 ϵ 0 ∂ E ∂ t grad xxB=mu_(0)J+mu_(0)epsilon_(0)(delE)/(del t)\nabla \times \mathbf{B} = \mu_0 \mathbf{J} + \mu_0 \epsilon_0 \frac{\partial \mathbf{E}}{\partial t}∇×B=μ0J+μ0ϵ0∂E∂t.
In vacuum, where ρ = 0 ρ = 0 rho=0\rho = 0ρ=0 and J = 0 J = 0 J=0\mathbf{J} = 0J=0, these simplify to:
  1. ∇ ⋅ E = 0 ∇ ⋅ E = 0 grad*E=0\nabla \cdot \mathbf{E} = 0∇⋅E=0,
  2. ∇ ⋅ B = 0 ∇ ⋅ B = 0 grad*B=0\nabla \cdot \mathbf{B} = 0∇⋅B=0,
  3. ∇ × E = − ∂ B ∂ t ∇ × E = − ∂ B ∂ t grad xxE=-(delB)/(del t)\nabla \times \mathbf{E} = -\frac{\partial \mathbf{B}}{\partial t}∇×E=−∂B∂t,
  4. ∇ × B = μ 0 ϵ 0 ∂ E ∂ t ∇ × B = μ 0 ϵ 0 ∂ E ∂ t grad xxB=mu_(0)epsilon_(0)(delE)/(del t)\nabla \times \mathbf{B} = \mu_0 \epsilon_0 \frac{\partial \mathbf{E}}{\partial t}∇×B=μ0ϵ0∂E∂t.
b) Derive expressions for electric and magnetic fields in terms of scalar and vector potentials using the homogeneous Maxwell’s equations. Show that the potentials associated with a given magnetic field are not unique. Express the inhomogeneous Maxwell’s equations in terms of scalar and vector potentials.

Answer:

Expressing Electric and Magnetic Fields in Terms of Potentials

To derive the expressions for electric and magnetic fields in terms of scalar and vector potentials, we start with the homogeneous Maxwell’s equations:
  1. Faraday’s Law:
    ∇ × E = − ∂ B ∂ t ∇ × E = − ∂ B ∂ t grad xxE=-(delB)/(del t)\nabla \times \mathbf{E} = -\frac{\partial \mathbf{B}}{\partial t}∇×E=−∂B∂t
  2. Gauss’s Law for Magnetism:
    ∇ ⋅ B = 0 ∇ ⋅ B = 0 grad*B=0\nabla \cdot \mathbf{B} = 0∇⋅B=0

Magnetic Vector Potential A A A\mathbf{A}A

Gauss’s law for magnetism ( ∇ ⋅ B = 0 ∇ ⋅ B = 0 grad*B=0\nabla \cdot \mathbf{B} = 0∇⋅B=0) implies that the magnetic field B B B\mathbf{B}B can be written as the curl of a vector potential A A A\mathbf{A}A:
B = ∇ × A B = ∇ × A B=grad xxA\mathbf{B} = \nabla \times \mathbf{A}B=∇×A

Electric Scalar Potential Φ Φ Phi\PhiΦ

Substituting B = ∇ × A B = ∇ × A B=grad xxA\mathbf{B} = \nabla \times \mathbf{A}B=∇×A into Faraday’s law:
∇ × E = − ∂ ( ∇ × A ) ∂ t ∇ × E = − ∂ ( ∇ × A ) ∂ t grad xxE=-(del(grad xxA))/(del t)\nabla \times \mathbf{E} = -\frac{\partial (\nabla \times \mathbf{A})}{\partial t}∇×E=−∂(∇×A)∂t
Using the vector identity ∇ × ( ∇ × A ) = ∇ ( ∇ ⋅ A ) − ∇ 2 A ∇ × ( ∇ × A ) = ∇ ( ∇ ⋅ A ) − ∇ 2 A grad xx(grad xxA)=grad(grad*A)-grad^(2)A\nabla \times (\nabla \times \mathbf{A}) = \nabla (\nabla \cdot \mathbf{A}) – \nabla^2 \mathbf{A}∇×(∇×A)=∇(∇⋅A)−∇2A, we have:
∇ × E = − ∇ × ∂ A ∂ t ∇ × E = − ∇ × ∂ A ∂ t grad xxE=-grad xx(delA)/(del t)\nabla \times \mathbf{E} = -\nabla \times \frac{\partial \mathbf{A}}{\partial t}∇×E=−∇×∂A∂t
This suggests that E + ∂ A ∂ t E + ∂ A ∂ t E+(delA)/(del t)\mathbf{E} + \frac{\partial \mathbf{A}}{\partial t}E+∂A∂t is a gradient of some scalar function Φ Φ Phi\PhiΦ:
E = − ∇ Φ − ∂ A ∂ t E = − ∇ Φ − ∂ A ∂ t E=-grad Phi-(delA)/(del t)\mathbf{E} = -\nabla \Phi – \frac{\partial \mathbf{A}}{\partial t}E=−∇Φ−∂A∂t
Therefore, the electric and magnetic fields in terms of the scalar potential Φ Φ Phi\PhiΦ and the vector potential A A A\mathbf{A}A are:
E = − ∇ Φ − ∂ A ∂ t E = − ∇ Φ − ∂ A ∂ t E=-grad Phi-(delA)/(del t)\mathbf{E} = -\nabla \Phi – \frac{\partial \mathbf{A}}{\partial t}E=−∇Φ−∂A∂t
B = ∇ × A B = ∇ × A B=grad xxA\mathbf{B} = \nabla \times \mathbf{A}B=∇×A

Non-uniqueness of Potentials (Gauge Freedom)

The potentials Φ Φ Phi\PhiΦ and A A A\mathbf{A}A associated with a given electromagnetic field are not unique. This is because of gauge freedom: we can perform a gauge transformation without changing the physical electric and magnetic fields.
Consider the gauge transformation:
A ′ = A + ∇ Λ A ′ = A + ∇ Λ A^(‘)=A+grad Lambda\mathbf{A}’ = \mathbf{A} + \nabla \LambdaA′=A+∇Λ
Φ ′ = Φ − ∂ Λ ∂ t Φ ′ = Φ − ∂ Λ ∂ t Phi^(‘)=Phi-(del Lambda)/(del t)\Phi’ = \Phi – \frac{\partial \Lambda}{\partial t}Φ′=Φ−∂Λ∂t
where Λ Λ Lambda\LambdaΛ is an arbitrary scalar function. Under this transformation:
  1. The new magnetic field B ′ B ′ B^(‘)\mathbf{B}’B′ is:
    B ′ = ∇ × A ′ = ∇ × ( A + ∇ Λ ) = ∇ × A = B B ′ = ∇ × A ′ = ∇ × ( A + ∇ Λ ) = ∇ × A = B B^(‘)=grad xxA^(‘)=grad xx(A+grad Lambda)=grad xxA=B\mathbf{B}’ = \nabla \times \mathbf{A}’ = \nabla \times (\mathbf{A} + \nabla \Lambda) = \nabla \times \mathbf{A} = \mathbf{B}B′=∇×A′=∇×(A+∇Λ)=∇×A=B
  2. The new electric field E ′ E ′ E^(‘)\mathbf{E}’E′ is:
    E ′ = − ∇ Φ ′ − ∂ A ′ ∂ t E ′ = − ∇ Φ ′ − ∂ A ′ ∂ t E^(‘)=-gradPhi^(‘)-(delA^(‘))/(del t)\mathbf{E}’ = -\nabla \Phi’ – \frac{\partial \mathbf{A}’}{\partial t}E′=−∇Φ′−∂A′∂t
    Substituting the transformed potentials:
    E ′ = − ∇ ( Φ − ∂ Λ ∂ t ) − ∂ ∂ t ( A + ∇ Λ ) E ′ = − ∇ Φ − ∂ Λ ∂ t − ∂ ∂ t ( A + ∇ Λ ) E^(‘)=-grad(Phi-(del Lambda)/(del t))-(del)/(del t)(A+grad Lambda)\mathbf{E}’ = -\nabla \left( \Phi – \frac{\partial \Lambda}{\partial t} \right) – \frac{\partial}{\partial t} (\mathbf{A} + \nabla \Lambda)E′=−∇(Φ−∂Λ∂t)−∂∂t(A+∇Λ)
    E ′ = − ∇ Φ + ∇ ∂ Λ ∂ t − ∂ A ∂ t − ∂ ∇ Λ ∂ t E ′ = − ∇ Φ + ∇ ∂ Λ ∂ t − ∂ A ∂ t − ∂ ∇ Λ ∂ t E^(‘)=-grad Phi+grad(del Lambda)/(del t)-(delA)/(del t)-(del grad Lambda)/(del t)\mathbf{E}’ = -\nabla \Phi + \nabla \frac{\partial \Lambda}{\partial t} – \frac{\partial \mathbf{A}}{\partial t} – \frac{\partial \nabla \Lambda}{\partial t}E′=−∇Φ+∇∂Λ∂t−∂A∂t−∂∇Λ∂t
    Since the partial derivatives commute:
    E ′ = − ∇ Φ − ∂ A ∂ t = E E ′ = − ∇ Φ − ∂ A ∂ t = E E^(‘)=-grad Phi-(delA)/(del t)=E\mathbf{E}’ = -\nabla \Phi – \frac{\partial \mathbf{A}}{\partial t} = \mathbf{E}E′=−∇Φ−∂A∂t=E
Thus, the electric and magnetic fields remain unchanged under the gauge transformation, demonstrating the non-uniqueness of Φ Φ Phi\PhiΦ and A A A\mathbf{A}A.

Inhomogeneous Maxwell’s Equations in Terms of Potentials

Now, express the inhomogeneous Maxwell’s equations in terms of the potentials Φ Φ Phi\PhiΦ and A A A\mathbf{A}A:
  1. Gauss’s Law for Electricity:
    ∇ ⋅ E = ρ ϵ 0 ∇ ⋅ E = ρ ϵ 0 grad*E=(rho)/(epsilon_(0))\nabla \cdot \mathbf{E} = \frac{\rho}{\epsilon_0}∇⋅E=ρϵ0
  2. Ampere’s Law (with Maxwell’s correction):
    ∇ × B = μ 0 J + μ 0 ϵ 0 ∂ E ∂ t ∇ × B = μ 0 J + μ 0 ϵ 0 ∂ E ∂ t grad xxB=mu_(0)J+mu_(0)epsilon_(0)(delE)/(del t)\nabla \times \mathbf{B} = \mu_0 \mathbf{J} + \mu_0 \epsilon_0 \frac{\partial \mathbf{E}}{\partial t}∇×B=μ0J+μ0ϵ0∂E∂t
Substituting E = − ∇ Φ − ∂ A ∂ t E = − ∇ Φ − ∂ A ∂ t E=-grad Phi-(delA)/(del t)\mathbf{E} = -\nabla \Phi – \frac{\partial \mathbf{A}}{\partial t}E=−∇Φ−∂A∂t and B = ∇ × A B = ∇ × A B=grad xxA\mathbf{B} = \nabla \times \mathbf{A}B=∇×A into these equations:

Gauss’s Law for Electricity

∇ ⋅ E = ∇ ⋅ ( − ∇ Φ − ∂ A ∂ t ) = − ∇ 2 Φ − ∂ ∂ t ( ∇ ⋅ A ) ∇ ⋅ E = ∇ ⋅ − ∇ Φ − ∂ A ∂ t = − ∇ 2 Φ − ∂ ∂ t ( ∇ ⋅ A ) grad*E=grad*(-grad Phi-(delA)/(del t))=-grad^(2)Phi-(del)/(del t)(grad*A)\nabla \cdot \mathbf{E} = \nabla \cdot \left(-\nabla \Phi – \frac{\partial \mathbf{A}}{\partial t}\right) = -\nabla^2 \Phi – \frac{\partial}{\partial t} (\nabla \cdot \mathbf{A})∇⋅E=∇⋅(−∇Φ−∂A∂t)=−∇2Φ−∂∂t(∇⋅A)
Using the Coulomb gauge ( ∇ ⋅ A = 0 ∇ ⋅ A = 0 grad*A=0\nabla \cdot \mathbf{A} = 0∇⋅A=0):
− ∇ 2 Φ = ρ ϵ 0 − ∇ 2 Φ = ρ ϵ 0 -grad^(2)Phi=(rho)/(epsilon_(0))-\nabla^2 \Phi = \frac{\rho}{\epsilon_0}−∇2Φ=ρϵ0
So:
∇ 2 Φ = − ρ ϵ 0 ∇ 2 Φ = − ρ ϵ 0 grad^(2)Phi=-(rho)/(epsilon_(0))\nabla^2 \Phi = -\frac{\rho}{\epsilon_0}∇2Φ=−ρϵ0

Ampere’s Law (with Maxwell’s Correction)

∇ × B = ∇ × ( ∇ × A ) = μ 0 J + μ 0 ϵ 0 ∂ E ∂ t ∇ × B = ∇ × ( ∇ × A ) = μ 0 J + μ 0 ϵ 0 ∂ E ∂ t grad xxB=grad xx(grad xxA)=mu_(0)J+mu_(0)epsilon_(0)(delE)/(del t)\nabla \times \mathbf{B} = \nabla \times (\nabla \times \mathbf{A}) = \mu_0 \mathbf{J} + \mu_0 \epsilon_0 \frac{\partial \mathbf{E}}{\partial t}∇×B=∇×(∇×A)=μ0J+μ0ϵ0∂E∂t
Using the vector identity ∇ × ( ∇ × A ) = ∇ ( ∇ ⋅ A ) − ∇ 2 A ∇ × ( ∇ × A ) = ∇ ( ∇ ⋅ A ) − ∇ 2 A grad xx(grad xxA)=grad(grad*A)-grad^(2)A\nabla \times (\nabla \times \mathbf{A}) = \nabla (\nabla \cdot \mathbf{A}) – \nabla^2 \mathbf{A}∇×(∇×A)=∇(∇⋅A)−∇2A and the Coulomb gauge ( ∇ ⋅ A = 0 ∇ ⋅ A = 0 grad*A=0\nabla \cdot \mathbf{A} = 0∇⋅A=0):
∇ × ( ∇ × A ) = − ∇ 2 A ∇ × ( ∇ × A ) = − ∇ 2 A grad xx(grad xxA)=-grad^(2)A\nabla \times (\nabla \times \mathbf{A}) = -\nabla^2 \mathbf{A}∇×(∇×A)=−∇2A
Thus:
− ∇ 2 A = μ 0 J + μ 0 ϵ 0 ∂ ∂ t ( − ∇ Φ − ∂ A ∂ t ) − ∇ 2 A = μ 0 J + μ 0 ϵ 0 ∂ ∂ t − ∇ Φ − ∂ A ∂ t -grad^(2)A=mu_(0)J+mu_(0)epsilon_(0)(del)/(del t)(-grad Phi-(delA)/(del t))-\nabla^2 \mathbf{A} = \mu_0 \mathbf{J} + \mu_0 \epsilon_0 \frac{\partial}{\partial t} \left(-\nabla \Phi – \frac{\partial \mathbf{A}}{\partial t}\right)−∇2A=μ0J+μ0ϵ0∂∂t(−∇Φ−∂A∂t)
− ∇ 2 A = μ 0 J − μ 0 ϵ 0 ∇ ∂ Φ ∂ t − μ 0 ϵ 0 ∂ 2 A ∂ t 2 − ∇ 2 A = μ 0 J − μ 0 ϵ 0 ∇ ∂ Φ ∂ t − μ 0 ϵ 0 ∂ 2 A ∂ t 2 -grad^(2)A=mu_(0)J-mu_(0)epsilon_(0)grad(del Phi)/(del t)-mu_(0)epsilon_(0)(del^(2)A)/(delt^(2))-\nabla^2 \mathbf{A} = \mu_0 \mathbf{J} – \mu_0 \epsilon_0 \nabla \frac{\partial \Phi}{\partial t} – \mu_0 \epsilon_0 \frac{\partial^2 \mathbf{A}}{\partial t^2}−∇2A=μ0J−μ0ϵ0∇∂Φ∂t−μ0ϵ0∂2A∂t2
In the Coulomb gauge, ∇ ⋅ A = 0 ∇ ⋅ A = 0 grad*A=0\nabla \cdot \mathbf{A} = 0∇⋅A=0, simplifying the equations:
∇ 2 A − μ 0 ϵ 0 ∂ 2 A ∂ t 2 = − μ 0 J ∇ 2 A − μ 0 ϵ 0 ∂ 2 A ∂ t 2 = − μ 0 J grad^(2)A-mu_(0)epsilon_(0)(del^(2)A)/(delt^(2))=-mu_(0)J\nabla^2 \mathbf{A} – \mu_0 \epsilon_0 \frac{\partial^2 \mathbf{A}}{\partial t^2} = -\mu_0 \mathbf{J}∇2A−μ0ϵ0∂2A∂t2=−μ0J

Summary

The electric and magnetic fields can be expressed in terms of the scalar potential Φ Φ Phi\PhiΦ and the vector potential A A A\mathbf{A}A as:
E = − ∇ Φ − ∂ A ∂ t E = − ∇ Φ − ∂ A ∂ t E=-grad Phi-(delA)/(del t)\mathbf{E} = -\nabla \Phi – \frac{\partial \mathbf{A}}{\partial t}E=−∇Φ−∂A∂t
B = ∇ × A B = ∇ × A B=grad xxA\mathbf{B} = \nabla \times \mathbf{A}B=∇×A
The potentials Φ Φ Phi\PhiΦ and A A A\mathbf{A}A are not unique due to gauge freedom. The inhomogeneous Maxwell’s equations in terms of these potentials, using the Coulomb gauge, are:
  1. ∇ 2 Φ = − ρ ϵ 0 ∇ 2 Φ = − ρ ϵ 0 grad^(2)Phi=-(rho)/(epsilon_(0))\nabla^2 \Phi = -\frac{\rho}{\epsilon_0}∇2Φ=−ρϵ0
  2. ∇ 2 A − μ 0 ϵ 0 ∂ 2 A ∂ t 2 = − μ 0 J ∇ 2 A − μ 0 ϵ 0 ∂ 2 A ∂ t 2 = − μ 0 J grad^(2)A-mu_(0)epsilon_(0)(del^(2)A)/(delt^(2))=-mu_(0)J\nabla^2 \mathbf{A} – \mu_0 \epsilon_0 \frac{\partial^2 \mathbf{A}}{\partial t^2} = -\mu_0 \mathbf{J}∇2A−μ0ϵ0∂2A∂t2=−μ0J
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