Electromagnetic Waves

I=½ε? E? ²c ⇒ E? =√ (2I/ε? c)
Erms = E? /√2 = √ (I/ε? c) = √ (315/π / (8.86e-12 * 3e8) = 194

x = 1/√ (µ? ε? ) = speed ⇒ [x] = [L¹T? ¹]
y = E/B = speed ⇒ [y] = [L¹T? ¹]
z = l/ (RC) = l/τ ⇒ [z] = [L¹T? ¹]
So, x, y, z all have the same dimensions.

Energies of given Radiation can have
The following relation
Eγ-Rays > EX-Rays > Emicrowave > EAM Radiowaves
∴ λγ-Rays < X-Rays < microwave < AM Radiowaves
According to tres.
(a) Microwave → 10? ³ m
(b) Gamma Rays → 10? ¹? m (ii)
(c) AM Radio wve → 100 m (i)
(d) X-Rays → 10? ¹? m

$\mathrm{m}=\mathrm{L},\mathrm{b}=\mathrm{R},\mathrm{k}=\frac{1}{\mathrm{c}}$ (speed of light in vacuum)

$\mathrm{}\frac{{\mathrm{E}}_0}{{\mathrm{B}}_0}=\mathrm{C}$

So, ${\mathrm{E}}_0={\mathrm{B}}_0\mathrm{C}=3*{10}^{-8}*3*{10}^8=9\mathrm{N}/\mathrm{C}$

f = 3GHz = 3 * 10⁹ Hz

${\lambda }_{vacuum}=\frac{v}{f}=\frac{3*10^{8}m/s}{3*10^9/s}=10^{-1}m=0.1m$

${\lambda }_{vacuum}=\frac{{\lambda }_{vacuum}}{{\mu }_{medium}}=\frac{0.1}{{\mu }_{medium}}$

${\mu }_{medium}=\sqrt[]{{\mu }_r{\in }_r}=\sqrt[]{1*2.25}=1.5$

$\therefore {\mu }_{medium}=\frac{0.1m}{1.5}=0.0667m=6.67cm=667*10^{-2}cm$

(A) Source of microwave frequency   => Magnetron

(B) Source of infrared frequency                =>Vibration of atoms and molecules

(C) Source of Gamma rays                           => Radioactive decay of nucleus

(D) Source of x-rays                                      => inner shell electrons.

$l=\frac{1}{2}C{\epsilon }_0{E}_0^2=\frac{1}{2}C{\epsilon }_0{\left(C{B}_0\right)}^2=\frac{1}{2}{C}^3{\epsilon }_0{B}_0^2$

$B_0=\sqrt[]{\frac{2l}{{\epsilon }_0{C}^3}}$

$=\sqrt[]{\frac{2*0.092}{8.85*10^{-12}*27*10^{24}}}$

$=2.27*10^{-8}T$

As we know that direction of propagation $\left(\widehat{p}\right)$ $\stackrel{}{}$ of wave is given as

$\widehat{p}=\widehat{E}*\widehat{B}=\widehat{i}*\widehat{k}=\left(-\widehat{j}\right)$

Force on q₁ :

$F_{q_1}=q\left(V*B\right)=4\pi \left(0.5c\widehat{i}\right)*\frac{B_0}{\sqrt[]{2}}\left(cos\left(kz-\omega t\right)\widehat{i}+cos\left(kz-wt\right)\widehat{j}\right)$

$=4\pi \left(0.5c\right)\left(\frac{B_0}{\sqrt[]{2}}\right)cos\left(k\frac{\pi }{k}\right)\widehat{k}$      

For on q₂:

= $2\pi \left(0.5c\right)\left(\frac{B_0}{\sqrt[]{2}}\right)cos\left(k\frac{3\pi }{k}\right)\widehat{k}$

So ratio will be 2 : 1

$V=\frac{coefficientoftime}{coefficientofx}=\frac{\omega }{k}$

$\Rightarrow V=\frac{10*10^{10}}{500}=2*10^8=\frac{2*c}{3}$