1.

Question 8.11: Suppose that the electric field part of an electromagnetic wave in vacuum is E = {(3.1 N/C) cos [(1.8 rad/m) y + (5.4 × 10 6 rad/s)t]} . (a) What is the direction of propagation? (b) What is the wavelength λ? (c) What is the frequency ν? (d) What is the amplitude of the magnetic field part of the wave? (e) Write an expression for the magnetic field part of the wave.Class 12 - Physics - Electromagnetic Waves Electromagnetic Waves Page-287

Answer»

(a) \bf{E=[(3.1N/C)]cos[(1.8rad/m)y+(5.4\times10^8rad/s)t]\hat{i}}
As it is clear from the above electric FIELD vector that electric field is in the negative x direction. Therefore, the direction of propagation of the vector will be in negative y-direction i.e. –j.

(b) ) As we know that the general equation for electric field vector in the POSITIVE x-direction is: E=E_0cos(kx-\omega t)\hat{j}
here, K is propagation constant.
so, propagation constant , k = 1.8 = 2π/wavelength
wavelength = 2π/1.8 = 3.49 m

(c) angular velocity, \omega=5.4 × 10^8 rad/s
so, angular frequency = \omega/2π
=5.4× 10^8/2π = 86 MHZ

(d) AMPLITUDE of magnetic field part ,\frac{E_0}{B_0}=c
so, B_0=\frac{E_0}{c}=\frac{3.1}{3\times10^8}\\B_0=1.03\times10^{-8}T

(e) expression for magnetic field part of wave
\vec{B_z}=B_0cos(ky+\omega t)\hat{k}\\=1.03\times10^{-8}cos(1.8y+5.4\times10^8t)\hat{k}


Discussion

No Comment Found

Related InterviewSolutions