Maxwell's fourth equation is the differential form of Ampere's circuital law.
$\overrightarrow{\nabla} \times \overrightarrow{H} = \overrightarrow{J} + \frac{\partial{\overrightarrow{D}}}{\partial{t}}$
Derivation:
According to Ampere's circuital law
$\oint_{l} \overrightarrow{B}. \overrightarrow{dl}=\mu_{0} i \qquad(1)$
According to Stroke's theorem-
$\oint_{l} \overrightarrow{B}.\overrightarrow{dl}=\oint_{S} (\overrightarrow{\nabla} \times \overrightarrow{B}). \overrightarrow{dS} \qquad(2)$
From equation$(1)$ and equation$(2)$
$\oint_{S} (\overrightarrow{\nabla} \times \overrightarrow{B}). \overrightarrow{dS} = \mu_{0} i \qquad(3)$
where $i=\oint_{S} \overrightarrow{J}. \overrightarrow{dS} \qquad(4)$
So from equation$(3)$ and equation$(4)$
$\oint_{S} (\overrightarrow{\nabla} \times \overrightarrow{B}). \overrightarrow{dS} = \mu_{0} \oint_{S} \overrightarrow{J}. \overrightarrow{dS}$
$\oint_{S} (\overrightarrow{\nabla} \times \overrightarrow{B}). \overrightarrow{dS} - \mu_{0} \oint_{S} \overrightarrow{J}. \overrightarrow{dS}=0$
$\oint_{S} [(\overrightarrow{\nabla} \times \overrightarrow{B}) - \mu_{0} \overrightarrow{J}]. \overrightarrow{dS}=0$
$(\overrightarrow{\nabla} \times \overrightarrow{B}) - \mu_{0} \overrightarrow{J}=0$
$\overrightarrow{\nabla} \times \overrightarrow{B} = \mu_{0} \overrightarrow{J}$
We know that $\overrightarrow{B}=\mu_{0} \overrightarrow{H}$. So the above equation can be again written as-
$\overrightarrow{\nabla} \times \overrightarrow{H} = \overrightarrow{J}$
Modified Maxwell's fourth equation:
The modified Maxwell's fourth equation is the differential form of the modified Ampere's circuital law.
We know the modified Ampere's circuital law-
$\oint_{l} \overrightarrow{B}. \overrightarrow{dl}=\mu_{0} i + i_{d}$
Where $i_{d}$ - Displacement current
So the derivation of the modified Maxwell equation is similar to the above derivation. Therefore the modified Maxwell's fourth equation can be written as-
$\overrightarrow{\nabla} \times \overrightarrow{H} = \overrightarrow{J} + \overrightarrow{J_{d}} \qquad(1)$
Where $\overrightarrow{J_{d}}$ - Displacement current density
And the value of $\overrightarrow{J_{d}}$ is
$\overrightarrow{J_{d}}=\epsilon_{0} \frac{\partial{\overrightarrow{E}}}{\partial{t}}$
$\overrightarrow{J_{d}}=\frac{\partial{\overrightarrow{D}}}{\partial{t}} \qquad(\because \overrightarrow{D}=\epsilon_{0} \overrightarrow{E})$
Now substitute the value of $\overrightarrow{J_{d}}$ in equation $(1)$, then
$\overrightarrow{\nabla} \times \overrightarrow{H} = \overrightarrow{J} + \frac{\partial{\overrightarrow{D}}}{\partial{t}}$
This is modified by Maxwell's fourth equation.
Angle of Acceptance → If incident angle of light on the core for which the incident angle on the core-cladding interface equals the critical angle then incident angle of light on the core is called the "Angle of Acceptance. Transmission of light when the incident angle is equal to the acceptance angle If the incident angle is greater than the acceptance angle i.e. $\theta_{i}>\theta_{0}$ then the angle of incidence on the core-cladding interface will be less than the critical angle due to which part of the incident light is transmitted into cladding as shown in the figure below Transmission of light when the incident angle is greater than the acceptance angle If the incident angle is less than the acceptance angle i.e. $\theta_{i}<\theta_{0}$ then the angle of incidence on the core-cladding interface will be greater than the critical angle for which total internal reflection takes place inside the core. As shown in the figure below Transmission of lig
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