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એક વિકિરણ માટે વિદ્યુતક્ષેત્ર નીચે મુજબ આપવામાં આવે છે.
$\vec E = 2{E_0}\,\hat i\,\cos\, kz\,\cos\, \omega t$
તો તેના માટે ચુંબકીયક્ષેત્ર $\vec B$ કેટલું હશે?
$\frac{{2{E_0}}}{c}\hat j\,\sin\, kz\,\cos\, \omega t$
$-\frac{{2{E_0}}}{c}\hat j\,\sin\, kz\,\sin\, \omega t$
$\frac{{2{E_0}}}{c}\hat j\,\sin\, kz\,\sin\, \omega t$
$\frac{{2{E_0}}}{c}\hat j\,\cos\, kz\,\cos\, \omega t$
Solution
Given, Electric field component of monochromatic radiation,
$(\overrightarrow{\mathrm{E}})=2 \mathrm{E}_{0} \hat{\mathrm{i}} \cos \mathrm{kz} \cos \omega \mathrm{t}$
We know that, $\frac{d E}{d z}=-\frac{d B}{d t}$
$\frac{\mathrm{dE}}{\mathrm{dz}}=-2 \mathrm{E}_{0} \mathrm{k} \sin \mathrm{kz} \cos \omega \mathrm{t}=-\frac{\mathrm{dB}}{\mathrm{dt}}$
$\mathrm{dB}=+2 \mathrm{E}_{0} \mathrm{k} \,\sin \mathrm{kz} \cos\, \omega \mathrm{td}\, \mathrm{t}$ …. $(i)$
Integrating $eq^n$. $(i)$, we have
$B=+2 E_{0} k \sin k z \int \cos \omega t d t$
Magnetic field is given by,
$=+2 \mathrm{E}_{0} \frac{\mathrm{k}}{\omega} \sin \mathrm{kz} \sin \,\omega \mathrm{t}$
We also know that,
$\frac{E_{0}}{B_{0}}=\frac{\omega}{k}=c$
Magnetic field vector,
$\overrightarrow{\mathrm{B}}=\frac{2 \mathrm{E}_{0}}{\mathrm{c}} \hat{\mathrm{j}} \sin \mathrm{kz} \sin \omega \mathrm{t}$
