Question

The electric field of a sinusoidal electromagnetic wave obeysthe equation

E = - (375 (V/m))sin ((5.97 * 10^15(rad/s)) t +(1.99 * 10^7(rad/m))x).

What is the amplitude of the electric field of this wave?

What is the amplitude of the magnetic field of this wave?

What is the frequency of the wave?

What is the wavelength of the wave?

What is the period of the wave?

What is the speed of the wave?

Answer 1

Answer with Explanation:

We are given that

`E=-(375V/m)sin(5.97* 10^(15)(rad/s)t+(1.99* 1067(rad/m)x)`

a.General equation of electric field wave

`E=E_0sin(\omega t+kx)`

Where
`E_0`=Amplitude of electric field wave

By comparing

`\omega=5.97* 10^(15)rad/s`

`k=1.99* 10^7rad/m`

a.Amplitude of electric field wave=
`E_0=375V/m`

b.Amplitude of magnetic field wave,
`B_0=(E_0)/(c)`

Where
`c=3* 10^8 m/s`

Amplitude of magnetic field wave=
`B_0=(375)/(3* 10^8)=125* 10^(-8) T`

c.Frequency of wave,
`f=(\omega)/(2\pi)=(5.97* 10^(15))/(2\pi)=0.95* 10^(15)Hz`

d.Wavelength,
`\lambda=(2\pi)/(k)`

`\lambda=(2\pi)/(1.99* 10^7)=3.16* 10^(-7) m`

e.Period of wave,
`T=(1)/(f)=(1)/(0.95* 10^(15))=1.05* 10^(-15) s`

f.Speed of wave,
`v=f\lambda=0.95* 10^(15)* 3.16* 10^(-7)=3.00* 10^8 m/s`