Q. A radio wave of frequency 90 MHz enters a ferrite rod. If $\varepsilon_{r} = 10^{3}$ and $\mu_{r} = 10$, then the velocity and wavelength of the wave in ferrite are
A) $3 \times 10^{8}\ ms^{- 1};3.33 \times 10^{- 2}\ m$
B) $3 \times 10^{6}\ ms^{- 1};3.33 \times 10^{- 2}\ m$
C) $3 \times 10^{8}\ ms^{- 1};3.33 \times 10^{- 1}\ m$
D) $3 \times 10^{7}\ ms^{- 1};3.33 \times 10^{- 3}\ m$
Explanation
(b) $v_{ferrite} = \frac{c}{\sqrt{\mu_{r}\varepsilon_{r}}} = \frac{3 \times 10^{8}}{\sqrt{10 \times 10^{33}}} = 3 \times 10^{6}\ ms^{- 1}$
$\lambda_{ferrite} = \frac{v_{ferrite}}{v} = \frac{3 \times 10^{6}}{90 \times 10^{6}} = 3.33 \times 10^{- 2}\ m$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.