Question

An electromagnetic wave of electric field E = 10 sin (ωt — kx)N/C is incident normal to the cross-sectional area of a cylinder of l0 cm² 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.[Take : ε₀=8.854*10⁻¹² C² N⁻¹ m⁻², c=3*10⁸ ms⁻¹] [Ans: (a) 4.427*10⁻¹⁰ J m⁻³, (b) 4.427*10⁻¹³ J, (c) 1.3278*10⁻¹ Wm⁻² (d) 1.475*10⁻²¹ N (e) 1.475*10⁻¹⁸ N m⁻².]

Answer 1

Answer:

(a) $Energy \ density = 4.425 \times 10^{-10} \ Jm^{-3}$

(b) $Energy = 4.425 \times 10^{-13}$

(c) $Intensity = 0.13275 \ \frac{W}{m^{2}}$

(d) $Momentum = 4.425 \times 10^{-10} \ N-s$

(e) $Radiation \ pressure = 4.425 \times 10^{-10} pa$

Explanation:

As per the question,

Given electric field E = 10 sin (ωt — kx)N/C

$E_{0} = 10 \frac{V}{m}$

cross sectional area = 10 cm-square  

Length = 100 cm

Therefore,

Volume = area × length

= 10 × 100 = 1000 cm-cube

∴ Volume = $10^{-3} \ m^{3}$

(a) Formula used for electric density of an electromagnetic wave :

$Energy \ density= \frac{1}{2} \epsilon_{0}E_{0}^{2}$

where $\epsilon_{0} = 8.85 \times 10^{-12}$

on putting the values we get,

Energy density $= 4.425 \times 10^{-10} \ Jm^{-3}$

(b) Energy contained in the cylinder is given by:

U = energy density × volume of the cylinder

Put the value of volume and energy density in this formula we grt,

$U = 4.425 \times 10^{-10} \ Jm^{-3} \times 10^{-3} \ m^{3}$

$U = 4.425 \times 10^{-13}$

(c) The intensity of the wave is given by :

I =energy density × speed of wave

where speed of wave $= 3 \times 10^{8} \frac{m}{s}$

∴ $I = 4.425 \times 10^{-10} \ Jm^{-3} \times 3 \times 10^{8} \frac{m}{s}$

$I = 0.13275 \ \frac{W}{m^{2}}$

(d) Momentum transferred to the cross—sectional area of the cylinder in 1 s, considering total absorption is given by :

Momentum = energy density × cross section

$Momentum = 4.425 \times 10^{-10} \ Jm^{-3} \times 10 \times 10^{-4}$

$Momentum = 4.425 \times 10^{-10} \ N-s$

(e) Radiation pressure is given by:

$Radiation \ pressure = \frac{Intensity}{Speed of the wave}$

$Radiation \ pressure = \frac{0.13275 \ \frac{W}{m^{2}}}{3 \times 10^{8} \frac{m}{s}}$

$Radiation \ pressure = 4.425 \times 10^{-10} pa$