Question

The electric field intensity of a plane electromagnetic wave travelling in a medium with $ϵ={ϵ}_0,\mu ={\mu }_{0}and\sigma =1.5\times {10}^{-5}S/m$ is given by $\overrightarrow{E}=8e^{az}\sin ({10}^6\pi t-\beta z){\hat{a}}_y,V/m$ How much faster does the wave propagates if $\sigma =0$?

• A. 1.25% faster in dielectric than in free space
• B. 3% faster in dielectric than in free space
• C. 3% faster in free space than in dielectric
• D. 2.5% faster in free space than in dielectric

Answer 1

The correct option is C 3% faster in free space than in dielectric

$\omega ={10}^6\pi r/s$
$\mu ={\mu }_0$
$\in ={\in }_o$
$\sigma =1.5\times {10}^{-5}S/m$
$\text{Phase shift constant}.,\beta =\omega \sqrt{\frac{{\mu }_o{\in }_o}{2}}(\sqrt{1+\frac{{\sigma }^2}{{\omega }^2{\in }_0^2}}+1)$
$\beta ={10}^6\pi \sqrt{\frac{{\mu }_o{\in }_o}{2}[\sqrt{1+{\left(\frac{1.5\times {10}^{-5}}{{10}^6\pi \times 8.854\times {10}^{-12}}\right)}^2}+1]}$
$=\frac{{10}^6\pi \times \frac{1}{\sqrt{2}}}{3\times {10}^8}\sqrt{[\sqrt{1+{\left(\frac{1.5\times {10}^{-5}}{{10}^6\times 8.854\times {10}^{-12}}\right)}^2}+1]}$
$\beta =0.0108rad/m$
Velocity of wave in lossy dielectric is given by
$v_d=\frac{\omega }{{\beta }_d}=\frac{{10}^6\pi }{0.0108}=2.91\times {10}^{8}m/sec$
If $\sigma =0$ , then the lossy dielectric becomes free space.
Velocity of wave in free space is given by
$v_o=\frac{1}{\sqrt{{\mu }_0{\in }_0}}=3\times {10}^{8}m/sec$
Percentage change in velocity in free space is
$\frac{v_o-v_d}{v_o}\times 100$
$=\frac{(3-2.91)\times {10}^4}{3\times {10}^8}=0.03\times 100$
= 3%
Therefore the wave travels 3% faster in free space than in lossy dielectric (dielectric)