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Derivation of Maxwell’s Equation

1.    Let us consider a surface S bounded by a volume V in a dielectric medium. In a dielectric medium total charge consists of free charge plus polarization charge.

Let ρ and ρp be the charge densities of free charge and polarization charge at a point in small volume element dV. Then Gauss law can be expressed as
    ∫s E. dS = 1/ε0 ∫(ρ + ρp) dV. But ρp = - div P
 . :    ∫s ε0 E.dS = ∫v ρ dV - ∫v div P dV
But from Gauss divergence theorem
    ∫s ε0 E.dS = ∫v div (ε0 E) dv

Therefore Eq. (i) gives
    = ∫v div (ε0 E)dV = ∫v ρdV - ∫v div P dV

i.e.,    ∫v div (ε0 E + P)dv = ∫v ρ dV

i.e.,     ∫v div D dV = ∫v ρ dV                            (since D = ε0 E + P)

i.e.,      ∫v (div D – ρ) dV = 0

As volume in arbitrary, we must have
    div D – ρ = 0 i.e., div D = ρ                                … (1)

2.    We know that isolated magnetic poles do not exist. This fact is also expressed by saying that magnetic lines of force are, in general, closed curves. Because of this the magnetic flux leaving and entering any closed surface is always same. In other words, the total magnetic induction flux over any closed surface is always zero, i.e,
s = B. dS = 0
Using Gauss’s theorem we change surface integral into a volume integral.
    . :        ∫v (div B)dv = 0
Since the above statement is true for any arbitrary volume, V, it is only possible if the integrand itself is zero, i.e.,
            div B = 0
or            ∇ . B = 0

3.    By Faraday’s law of electromagnetic induction, the induced emf in a closed loop is
e = - ∂Φ/∂t                                … (1)

By definition, the magnetic flux Φ over any arbitrary surface S is,
            Φ = ∫s B. dS

Therefore, e = - ∂/∂t [∫s B. ds] = - ∫s ∂B/∂t. dS                                ... (2)

(. : surface S is fixed and hence only B is a function of time).
Again e.m.f., by definition, is the work done in carrying a unit charge round a closed loop. If C is the loop constituting the surface S, then
            e = ∫c E.dl                                … (3)
where E is the electric field intensity at a point where elementary loop dl is located.

Comparing Equations (2) and (3), we get
        ∫ E.dl = - ∫s ∂B/∂t . dS                                … (4)
From Stoke’s theorem, ∫E. dl ∫s curl E.dS
Eq. (4) becomes,     ∫s curl E. dS = - ∫s ∂B/∂t . dS

Since surface is arbitrary, curl E = - ∂B/∂t
Or        ∇ x E = - ∂B/∂t                                    … (5)

4.    According to this law, the work done in carrying a unit magnetic pole once around a closed arbitrary path linked with the current is expressed by
         Φ H dl = I = ∫ J.ds                                … (1)

However, the use of Stoke’s theorem gives
        ∫s (∇ x H).dS = ∫ J.ds
or            ∇ x H = J                                … (2)

This equation is incomplete and accounts for steady currents only. But for varying electric field, the current density should be modified. The difficulty with above equation is that if we take divergence of above equation, then,
        ∇. (∇ x H) = ∇.J = 0                                … (3)
However, this fact is disallowed by Continuity Equation according to which
        ∇. J = - ∂ρ/∂t
On this basis Maxwell added another current density J to J to make Eq. (2) as
        ∇ x H = J + J’                                    … (4)
So that continuity equation too is satisfied. Now divergence of Eq. (4)  gives
        ∇. (∇ x H) = ∇. J + ∇. J’ = 0
Or        ∇.J’ = ∇.J = ∂ρ/∂t                                … (5)

We know that    ∇.D  = ρ

Eq. (5) becomes,     ∇.J = ∂ (∇.D)/∂t
Or        ∇.J’ = ∇. ∂D/∂t
Or        J’ = ∂D/∂t                                    … (6)

In this way we observe that the additional current density J’ is due to time variations of electric displacement D. It is termed as Displacement Current density. According to Maxwell, it is just as effective as J in producing magnetic field.
Thus Eq. (4) is written as
            ∇ x H = J + ∂D/∂t                            … (7)

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