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In a plane electromagnetic wave, the ele...

In a plane electromagnetic wave, the electric field oscillates sinusoidally at a frequency of `2.0 xx 10^(10) Hz` and amplitude `48 V m^(–1)`.
(a) What is the wavelength of the wave?
(b) What is the amplitude of the oscillating magnetic field?
(c) Show that the average energy density of the E field equals the average energy density of the B field. `[c = 3 xx 10^(8) m s^(-1).]`

Text Solution

Verified by Experts

(a) `lamda =(c//v) =1.5 xx10^(-2) m`
(b) `B_(0)=(E_(0)//c)=1.6 xx10^(-7) T`
(c ) Energy density in E field: `u_(E) =(1//2)epsi_(0) E^(2)`
Using `E= cB, and c=(1)/(sqrt(mu_(0) epsi_(0)), u_(E)=u_(B)`
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Knowledge Check

  • In a plane electromagnetic wave, the electric field (E) having amplitude of 48 . V m^(-1) oscillates at a frequency of 2.0 xx 10^10 Hz. The amplitude of the oscillating magnetic field (B) is

    A
    `3.2xx10^(-8)`T
    B
    `3xx10^7` T
    C
    `16xx10^(-7)` T
    D
    `1.6xx10^(-7)` T
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