An electromagnetic wave of electric field `E=10 sin (omega t-Kx)N//C` is incident normal to the cross - sectional area of a cylinder of `10 cm^(2)` and having length 100 cm, lying along X - axis. Find (a) the energy density, (b) energy contained in the cylinder, (c ) the intensity of the wave, (d) momentum transferred to the cross - sectional area of the cylinder in 1 s, considering total absorption, (e ) radiation pressure. `[epsilon_(0)=8.854xx10^(12)C^(2)N^(-1)m^(-2), c=3xx10^(8)ms^(-1)]`

Explanation

For relative cylinder,
area `A=10cm^(2)=10xx10^(-4)m^(2)`
length l = 100 cm = 1 meter (unit length)
For the electromagnetic wave incident perpendicularly,
`E=10 sin(omega t-kx)N//C`
(a) Energy density `rho = epsilon_(0)E_(rms)^(2)`
`=epsilon_(0)(E_(0)^(2))/(2) " "(E_(0)=10 N//C)`
`therefore rho = (8.85xx10^(-12)xx100)/(2)=4.425xx10^(-10)J//m^(3)`
(b) Energy contained in the cylinder
U = energy density `xx` volume
`therefore U = 4.425xx10^(-10)xx10xx10^(-4)xx1`
`=4.425xx10^(-13)J`
(c ) Intensity `I = rho`
`therefore I=4.425xx10^(-10)xx3xx10^(8)`
`=1.3275xx10^(-1)Wm^(-2)`
(d) Momentum transferred to the cross - section area of the cylinder in 1 sec considering total absorption (means forces)
`P=(rho Al)/(c )=(4.425xx10^(-10)xx10^(-3)xx1)/(3xx10^(8))`
`=1.475xx10^(-21)N`
(e ) Radiation pressure `=("force")/("area")=(1.475xx10^(-21))/(10^(-3))`
`=1.475xx10^(-18)Nm^(-2)`.