8.Electromagnetic waves
hard

किसी एकवर्णीय विकिरण के वैद्युत क्षेत्र घटक को निम्न प्रकार से व्यक्त किया जा सकता है

$\overrightarrow{ E }=2 E _{0} \;\hat{i} \;\cos\; k z \;\cos \omega t$

उसके चुम्बकीय क्षेत्र $\overrightarrow{ B }$ का मान होगा

A

$\frac{{2{E_0}}}{c}\hat j\,\sin\, kz\,\cos\, \omega t$

B

$-\frac{{2{E_0}}}{c}\hat j\,\sin\, kz\,\sin\, \omega t$

C

$\frac{{2{E_0}}}{c}\hat j\,\sin\, kz\,\sin\, \omega t$

D

$\frac{{2{E_0}}}{c}\hat j\,\cos\, kz\,\cos\, \omega t$

(JEE MAIN-2017)

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,

$\overrightarrow{\mathrm{B}}=\frac{2 \mathrm{E}_{0}}{\mathrm{c}} \hat{\mathrm{j}} \sin \mathrm{kz} \sin \omega \mathrm{t}$

Standard 12
Physics

Similar Questions

Start a Free Trial Now

Confusing about what to choose? Our team will schedule a demo shortly.