Question

Suppose that the electric field of an electromagnetic wave in vacuum is E={(3.0 N //C) cos [1.8 rad//m) y +(5.4 xx 10^6 rad // s)t]}hat i (a) What is the direction of propagation of wave? (b) What is the wavelength lambda ? (c) What is the frequency f ? (d) What is the amplitude of the magnetic field of the wave (e) write an expression for the magnetic field of the wave.

Answer 1

Given :

E={(3.0 N /C) cos [1.8 rad//m) y +(5.4 x 10⁶ rad /s)t]}

To find  :

a) The direction of propagation of wave

b) wavelength

c) frequency

d) amplitude of the magnetic field of wave

e) expression for the magnetic field of wave

Solution :

• a) From the given equation we can say that the where is travelling in negative direction

• b) By comparing the given equation with  E=E₀cos(ky+ωt

     k=1.8rad/s

     ω=5.4×108rad/s

     E₀=3.1N/C

     λ=(2π/k)

        =2×3.14/1.8   =3.5m

   The wavelength of the wave is 3.5m

• c) v=(ω/2π)    

             =5.4×10⁸/2×3.14    =86MHz

          The frequency of the wave is  86MHz

• d)  B₀=E₀/C

                 =3.1/3×10⁸    =10.3nT

          The amplitude of the magnetic field of the wave is 10.3nT

• e) B=B₀cos(ky+ωt)k

           =(10.3nT)cos[(1.8rad/m)y+(5.4×10⁸rad/s)t]k

The expression for the magnetic field is (10.3nT)cos[(1.8rad/m)y+(5.4×10⁸rad/s)t]k