In a plane electromagnetic wave, the electric field oscillates sinusoidally at a frequency of `2.0xx10^10` Hz and amplitude 48 V. `m^(-1)`.
Show that the average energy density of the electric field equals the average energy density of the magnetic field. `[c=3xx10^8 m.s^(-1)]`

In a plane electromagnetic wave, the electric field oscillates sinusoidally at a frequency of 2.0xx10^10 Hz and amplitude 48 V. m^(-1) . What is the wavelength of the wave?

In a plane electromagnetic wave, the electric field oscillates sinusoidally at a frequency of 2.0xx10^10 Hz and amplitude 48 V. m^(-1) . What is the amplitude of the oscillating magnetic field?

In a plane electromagnetic wave, the electric field oscillates sinusoidally at a frequency of 2.0xx10^10Hz and amplitude 48V/m. What is the amplitude of the oscillating magnetic field?

In a plane electromagnetic wave, the electric field sinusoidally oscillates at a frequency 2 xx 10^10 Hz and amplitude 45 V . m^(-1) . (i) What is the wavelength of the wave? (ii) What is the amplitude of the oscillating magnetic field? (iii) Find out the relation between average energy density of the electric field vecE(u_E) and the average energy density of the magnetic field vecB(u_B)

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

Show that , the energy density in an electromagnetic field, u= in_0 E_0^2=1/(mu_0)B_0^2

If in air medium the amplitude of the electric field is 5 xx 10^-4v//m , then what will be the amplitude of the magnetic field?

In an electromagnetic wave, the electric field is E=5 sin (10^12t-x/(3xx10^3))V.m^(-1) . Calculate the intensity of that wave. Given mu_0=4pixx10^(-7) H. m^(-1)

The equation of the electric field associated with an electromagnetic wave is E=0.001 sin (10^14t-1/3xx10^(-8) x)V.m^(-1) . What is the (i) frequency, (ii) wavelength of this wave? Establish the equation of the associated magnetic field.