GATE problems in EMT

pb. An Electro Magnetic Wave Propagates through a loss less insulator with a velocity 1.5 X 10^{10} Cm/sec . Calculate the electric magnetic properties of the insulator if its intrinsic impedance is 90\pi Ohms.

Ans:- Given loss less insulator  \Rightarrow \alpha =0

v_{p} =1.5 X 10^{10} Cm/Sec

v_{p} = \frac{1}{\sqrt{\mu _{o}\epsilon _{o}\mu _{r}\epsilon _{r}}}

1.5 X 10^{10}= \frac{1}{\sqrt{\mu _{o}\epsilon _{o}\mu _{r}\epsilon _{r}}}

1.5 X 10^{10}= \frac{3 X 10^{10}}{\sqrt{\mu _{r}\epsilon _{r}}}

{\sqrt{\mu _{r}\epsilon _{r}}} = 2 ------------EQNI

from \eta =\sqrt{\frac{\mu _{o}\mu _{r }}{\epsilon _{o}\epsilon _{r }}}

90\pi = \sqrt{\frac{\mu _{o}\mu _{r }}{\epsilon _{o}\epsilon _{r }}}

90\pi = \sqrt{\frac{\mu _{r }}{\epsilon _{r }}} . 120 \pi

\sqrt{\frac{\mu _{r }}{\epsilon _{r }}} =\frac{3}{4}--------EQNII

from Equations I and II \mu _{r} = \sqrt{\frac{\mu _{r }}{\epsilon _{r }}} \sqrt{\mu _{r }\epsilon _{r}}

\mu _{r} = 2 X \frac{3}{4}

\mu _{r} = 1.5

\sqrt{\mu _{r }\epsilon _{r }} = 2

\sqrt{1.5 \epsilon _{r }} = 2

\epsilon _{r} = 2.66

1 Star2 Stars3 Stars4 Stars5 Stars (1 votes, average: 5.00 out of 5)


Author: Lakshmi Prasanna

Completed M.Tech in Digital Electronics and Communication Systems and currently working as a faculty.

Leave a Reply

This site uses Akismet to reduce spam. Learn how your comment data is processed.