NEET Physics Electromagnetic Waves Notes

Electromagnetic Waves

Maxwell modified the Ampere’s circuital law \(\oint \overrightarrow{\mathrm{B}} \cdot \mathrm{d} \vec{\ell}=\mu_0 \mathrm{I}\) to explain the continuity of current in a circuit containing a capacitor.

⇒ \(\phi=\mathrm{EA}=\frac{\sigma}{\mathrm{s}_0} \mathrm{~A}=\frac{\mathrm{Q}}{\mathrm{A} \varepsilon_0} \mathrm{~A}\)

⇒ \(\frac{\mathrm{d} \phi}{\mathrm{dt}}=\frac{1}{\varepsilon_0} \frac{\mathrm{dQ}}{\mathrm{dt}}\)

⇒ \(\mathrm{I}_{\mathrm{d}}=\frac{\mathrm{dQ}}{\mathrm{dt}}=\varepsilon_0 \frac{\mathrm{dQ}}{\mathrm{dt}}\)

 

This was the missing term suggested by Maxwell called displacement current Id.

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Displacement current is due to the rate of change of electric field between the plates of the capacitor

Conduction current and the displacement in a circuit may not be continuous but their sum is always continuous.

A modified form of Ampere’s circuital law / Ampere–Maxwell law is,

⇒ \(\oint \overrightarrow{\mathrm{B}} \cdot \mathrm{d} \vec{\ell}-\mu_0 \mathrm{I}+\varepsilon \mu_0 \frac{\mathrm{d} \phi_1}{\mathrm{dt}}\)

 

Maxwell’s Equations

 

\(1. \oint \vec{E} \cdot d \vec{A}=\frac{Q}{e_0} (Gauss’ law of electricity)\)

 

\(2. \oint \overrightarrow{\mathrm{B}} \cdot \mathrm{d} \overrightarrow{\mathrm{A}}=0 (Gauss’ law for magnetism)\)

 

\(3. \oint \overrightarrow{\mathrm{E}} \cdot \mathrm{d} \vec{\ell}=-\frac{\mathrm{d} \phi_{\mathrm{B}}}{\mathrm{dt}} (Faraday’s law)\)

 

\(4. \oint \overrightarrow{\mathrm{B}} \cdot \mathrm{d} \vec{\ell}-\mu_0 \mathrm{i}_{\mathrm{c}}+\mu_0 \varepsilon_0 \frac{\mathrm{d} \phi}{\mathrm{dt}} (Ampere – Maxwell law)\)

 

Representation of An Electromagnetic Wave:

\(\begin{aligned}
& E_{\mathrm{z}}=\mathrm{E}_0 \sin (\mathrm{kz}-\omega \mathrm{t}) \\
& \mathrm{B}_{\mathrm{y}}=\mathrm{B}_0 \sin (\mathrm{kz}-\omega \mathrm{t})
\end{aligned}\)

The direction of the EM wave is given by \(\overrightarrow{\mathrm{E}} \times \overrightarrow{\mathrm{B}}\)

\(\begin{aligned}
\frac{\mathrm{E}_0}{\mathrm{~B}_0} & =\mathrm{c} \\
\mathrm{k} & =\frac{2 \pi}{\lambda}, \omega=2 \pi \mathrm{f} \\
\frac{\omega}{\mathrm{k}} & =\frac{2 \pi \mathrm{f}}{2 \pi} \lambda=\mathrm{c} \\
\mathrm{c} & =\frac{1}{\sqrt{\mu_0 \varepsilon_0}}
\end{aligned}\)

The velocity of light in a medium is given by,

\(\begin{aligned}
& \mathrm{v}=\frac{1}{\sqrt{\mu \varepsilon}}=\frac{1}{\sqrt{\mu_0 \mu_{\mathrm{r}} \varepsilon_0 \varepsilon_{\mathrm{r}}}} \\
& \mathrm{v}=\frac{1}{\sqrt{\mu_0 \varepsilon_0}} \times \frac{1}{\sqrt{\mu_{\mathrm{r}} \varepsilon_{\mathrm{t}}}} \\
& \mathrm{v}=\frac{\mathrm{c}}{\sqrt{\mu_z \varepsilon_{\mathrm{r}}}}
\end{aligned}\)

The rate of energy transported per unit area by EM wave is given by,

Pointing vector, \(\overrightarrow{\mathrm{S}}=\frac{\mathrm{EB}}{\mu_0}=\frac{\mathrm{E}^2}{\mu_0 \mathrm{c}}\)

Energy density of the electric field is given by,

\(u_E=\frac{1}{2} \varepsilon_0 E^2\)

Energy density of the magnetic field is given by,

\(\mathrm{u}_{\mathrm{B}}=\frac{\mathrm{B}^2}{2 \mu_0}\)

• Average electric energy density = average magnetic energy density.
• The intensity of the EM wave is given by,

\(\mathrm{I}=\mathrm{U}_{\mathrm{m}} \mathrm{c}=\frac{\mathrm{B}_0^2}{2 \mu_0} \mathrm{c}=\frac{1}{2} \varepsilon_0 \mathrm{E}_0^2 \mathrm{c}\)

• The electric vector is responsible for all optical effects. This vector is also known as light vector.
• The pressure exerted by an EM wave is given by,

\(\mathrm{P}=\frac{\mathrm{I}}{\mathrm{c}}\)

• The momentum carried by an EM wave is given by,

\(\mathrm{P}=\frac{\mathrm{U}}{\mathrm{c}}\)