Physics Electromagnetic Waves

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.

$l=\frac{1}{2}c{\epsilon }_0{E}_0^2\Rightarrow \frac{8}{4\pi {\left(10\right)}^2}*\frac{1}{2}=\frac{1}{4}*c*\frac{1}{c^2{\mu }_0}*E_0^2\Rightarrow E_0=\frac{2}{10}\sqrt[]{\frac{{\mu }_{0}c}{\pi }}\Rightarrow x=2$

The EM waves originate from an accelerating charge. The charge moving with uniform velocity produces steady state magnetic field.

According to modified Ampere's law

$?B.dI={\mu }_0\left(I_C+I_D\right)$

For Loop  $L_1{I}_C\ne 0$ and $I_D=0$

For Loop  $L_2{I}_C=0$ and $I_D\ne 0$

Due to  $KCL{I}_C=I_D$

$SpeedoflightC=\frac{w}{k}=\frac{1.5*10^{11}}{0.5*10^3}=3*10^{8}m/s$

So, E₀ = B₀ C

= 2 * 10^-8 * 3 * 10⁸

= 6v/m

Direction will be along z – axis

${\lambda }_1=6MHz=6*10^{6}Hz$

${\lambda }_1=\frac{C}{{\upsilon }_1}=\frac{3*10^8}{6*10^6}=50m$

${\lambda }_2=\frac{C}{{\upsilon }_2}=10*10^6=\frac{3*10^8}{10*10^6}=30m$

Wavelength bandwidth

$=\left|{\lambda }_1-{\lambda }_2\right|=20m$

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

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

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

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

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

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

Work done is equal to change in K.E.

$So{W}_1+W_2=\frac{1}{2}M{\left(0.8\sqrt[]{gh}\right)}^2-0$ W₁ ^-> work done by mg

$mgh+W_2=\frac{1}{2}m*0.64gh$  W₂ -> work done by air friction

$W_2=0.32mgh-mgh=-0.68mgh$              

W₂ = -0.68 mgh