The Total solution for NCERT class 6-12
Suppose that the electric fieldamplitude of an electromagnetic wave is┬аE0┬а= 120 N/C and thatits frequency is┬а╬╜┬а= 50.0 MHz.(a) Determine,┬аB0,┬а╧Й,┬аk,┬аand ╬╗. (b) Find expressions for┬аE┬аand┬аB.
Electric field amplitude,┬аE0┬а= 120 N/C
Frequencyof source,┬а╬╜┬а= 50.0 MHz = 50 ├Ч106┬аHz
Speed oflight,┬аc┬а= 3 ├Ч 108┬аm/s
Angular frequency of source isgiven as:
╧Й┬а= 2╧А╬╜
= 2╧А ├Ч 50├Ч 106
= 3.14 ├Ч108┬аrad/s
Propagationconstant is given as:
Wavelength of wave is given as:
(b)┬аSupposethe wave is propagating in the positive┬аx┬аdirection.Then, the electric field vector will be in the positive┬аy┬аdirection and the magnetic fieldvector will be in the positive┬аz┬аdirection.This is because all three vectors are mutually perpendicular.
Equation of electric field vectoris given as:
And, magnetic field vector isgiven as: