- Home
- Standard 12
- Physics
A plane $EM$ wave travelling in vacuum along $z-$ direction is given by $\vec E = {E_0}\,\,\sin (kz - \omega t)\hat i$ and $\vec B = {B_0}\,\,\sin (kz - \omega t)\hat j$.
$(i)$ Evaluate $\int {\vec E.\overrightarrow {dl} } $ over the rectangular loop $1234$ shown in figure.
$(ii)$ Evaluate $\int {\vec B} .\overrightarrow {ds} $ over the surface bounded by loop $1234$.
$(iii)$ $\int {\vec E.\overrightarrow {dl} = - \frac{{d{\phi _E}}}{{dt}}} $ to prove $\frac{{{E_0}}}{{{B_0}}} = c$
$(iv)$ By using similar process and the equation $\int {\vec B} .\overrightarrow {dl} = {\mu _0}I + { \in _0}\frac{{d{\phi _E}}}{{dt}}$ , prove that $c = \frac{1}{{\sqrt {{\mu _0}{ \in _0}} }}$
Solution
Electromagnetic waves are propagating in $z$-direction. Let electric field vector $\overrightarrow{\mathrm{E}}$ and $\overrightarrow{\mathrm{B}}$ is in $x$-direction and magnetic field vector is in $y$-direction.
$\therefore \overrightarrow{\mathrm{E}}=\mathrm{E}_{0} \hat{i} \text { and } \overrightarrow{\mathrm{B}}=\mathrm{B}_{0} \hat{j}$
– As shown in figure line integration over square path 1234 ,
$\iint \overrightarrow{\mathrm{E}} \cdot \overrightarrow{d l}=\int_{1}^{2} \overrightarrow{\mathrm{E}} \cdot \overrightarrow{d l}+\int_{2}^{3} \overrightarrow{\mathrm{E}} \cdot \overrightarrow{d l}+\int_{3}^{4} \overrightarrow{\mathrm{E}} \cdot \overrightarrow{d l}+\int_{4}^{1} \overrightarrow{\mathrm{E}} \cdot \overrightarrow{d l}$
$=\int_{1}^{2} \mathrm{E} d l \cos 90^{\circ}+\int_{2}^{3} \mathrm{E} d l \cos 0^{\circ}+\int_{3}^{4} \mathrm{E} d l \cos 90^{\circ}+\int_{4}^{1} \mathrm{E} d l \cos 180^{\circ} $
$\left.\therefore \quad \int\right] \overrightarrow{\mathrm{E}} \cdot \overrightarrow{d l}=\mathrm{E}_{0} h$ $\left[\sin \left(k z_{2}-\omega t\right)-\sin \left(k z_{1}-\omega t\right)\right]\dots(1)$
$(ii)$ Let square 1234 is made up of very large no. of small strip $d s$, let area of one strip $=d s=h d z$
$\therefore \int \overrightarrow{\mathrm{B}} \cdot \overrightarrow{d s} =\int_{j} \mathrm{~B} d s \cos 0^{\circ}$
$=\int \mathrm{B} d s \quad\left[\because \cos 0^{\circ}=1\right]$
$=\int_{z_{1}}^{z_{2}} \mathrm{~B}_{0} \sin (k z-\omega t) h d z$
$[\because d s=h d z] =-\frac{\mathrm{B}_{0} h}{k}$
