Electromagnetic Waves · Question 2

A cube of unit volume contains 35 107 photons of frequency 1015 Hz. If the energy of all the photons is viewed as the average energy being contained in the electromagnetic waves within the same volume, then the amplitude of the magnetic field is 10-9 T. Taking permeability of free space 0=4 10-7 Tm/A, Planck's constant h=6 10-34 Js and = 227, the value of is

Answer

21 to 25

Solution

Average energy density of an EM wave: u= B022 0.

u=Nh =35 107 6 10-34 1015=2.1 10-10 J m-3

B02=2 0u=2 4 227 10-7 2.1 10-10=52.8 10-17=5.28 10-16

B0=2.30 10-8 T=23 10-9 T, so 23

Answer: 23 (official range 21 to 25)