Maxwell fourth equation

The Maxwell fourth equation is called "the modified Ampere's circuital law".

Statement:
It states that the line integral of the magnetic field H around any closed part or circuit is equal to the current enclosed by the path.

Differential form (without modification):
That is,
$$\int H.dL = I$$
Let the current be distributed through the current with current density J, then:
$$I = \int J. ds$$
This implies that:
$$\int H.dL = \int J.ds$$ .........(9)
Applying Stokes theorem to the LHS of equ(9) to change line integral to surface integral we have:
$$\int_{s} (\nabla X H).ds = \int_{s} J.ds$$
Since, two surface integrals are equal only if their integrands are equal.
Thus,
$$\nabla X H =J$$ .........(10)
Equ(10) is the differential form of Maxwell fourth equation (without modification)
Take divergence of equ(10)
$$\nabla.(\nabla X H) = \nabla.J$$
Since, the divergence of the curl of a vector is zero.  Therefore,
$$\nabla.(\nabla X H) = 0$$
It means that
$$\nabla.J=0$$
This is the equation of continuity for steady current not for time varying field. The equation for time varying field is given as:
$$\nabla.J = - \frac{d\rho}{dt}$$
This is the reason why Maxwell modify his Ampere's circuital law.


Modification of Ampere's circuital law:
Differential form (with modification):
The Ampere's circuital law was modified by the given concept of displacement current D and also the concept of displacement current density $$J_d$$ for time varying fields.
Equ(10) for time varying field should be rewritten as:
$$\nabla X H = J + J_d$$ ......(11)
By taking divergence of equ(11)
$$\nabla.(\nabla X H) = \nabla J + \nabla J_d$$
As divergence of the curl of a vector is always zero, therefore:
$$\nabla.(\nabla X H) = 0$$
This means that:
$$\nabla.(J + J_d) = 0$$
or
$$\nabla J = - \nabla J_d$$ ..........(α)
But from equation of continuity for time varying fields
$$\nabla.J = \frac{d\rho}{dt}$$ ..........(β)
By comparing equ(α) and equ(β)
$$\nabla.J_d = \frac{d}{dt}(\nabla.D)$$ .....(12)

Because from Maxwell first equation

$$\nabla.D = \rho$$

Since, the divergence of two vectors are equal only if the vectors are equal.
Then:
$$J_d = \frac{dD}{dt}$$
Substitute the above equation into equ(11)
$$\nabla X H = J + \frac{dD}{dt}$$ .....(13)
Here,
$$\frac{dD}{dt}=J_d$$
J = conduction current
D = displacement current.
Equ(13) is the differential form of the Maxwell fourth equation.

Integral form:
Taking surface integral of both sides of equ(13), we have:
$$\int (\nabla X H).ds = \int (J + \frac{dD}{dt}).ds$$
Applying Stokes law to the LHS
$$\int (\nabla X H).ds = \int_{l} H.dL$$
Comparing the above two equation, we get:
$$\int H.dL = \int (J + \frac{dD}{dt}).ds$$ ......(14)
Equ(14) is the integral form of Maxwell fourth equation.