Question

In an electomegnetic wave, the amplitude of electric field is 10V//m. The frequency of wave is 5xx10^14Hz. The wave is propagating along Z-axis, find (i) the average energy density of electric field (ii) the average energy density of magnetic field (iii) the total average energy density of e.m. wave.

Answer 1

Given:

E₀ = 10 Vm^-1

v = 5 × 10^14 Hz

To find:

(i) the average energy density of electric field

(ii) the average energy density of magnetic field

(iii) the total average energy density of e.m. wave.

Solution:

(i) the average energy density of electric field

The average energy density of electric field is given by a formula,

$u_E=\dfrac{1}{2}\in _0 E_{rms}^2$

$u_E=\dfrac{1}{2}\in _0 \left ( \dfrac{E_0}{\sqrt 2} \right )^2$

$u_E=\dfrac{1}{4}\in _0 E_0^2$

$u_E=\dfrac{1}{4} \times (8.85 \times 10^{-12}) \times 10^2$

$\therefore u_E=2.21 \times 10^{-1}J/m^3$

(ii) the average energy density of magnetic field

$u_B=\dfrac{1}{2}\dfrac{B_{rms}^2}{\mu_0}$

$u_B=\dfrac{1}{2\mu_0}\left ( \dfrac{B_0}{\sqrt 2} \right )^2$

$u_B=\dfrac{1}{4}\left ( \dfrac{B_0^2}{\mu_0} \right )$

$u_B=\dfrac{1}{4} \dfrac{\left (\frac{E_0}{c} \right )^2}{\mu_0}$

$u_B=\dfrac{1}{4\mu_0}\dfrac{E_0^2}{c^2}$

$u_B=\dfrac{1}{4\mu_0}\dfrac{E_0^2}{\left ( \frac{1}{\mu_0\in _0} \right )}$

$u_B=\dfrac{1}{4\mu_0}\in _0E_0^2$

$\Rightarrow u_B = u_E$

$\therefore u_B=2.21 \times 10^{-1}J/m^3$

(iii) the total average energy density of e.m. wave.

$u=u_E+u_B$

$u=2.21 \times 10^{-10}+2.21 \times 10^{-10}$

$\therefore u = 4.42\times 10^{-10}J/m^3$

Answer 2

(i) The average energy density of electric field is 2.21 × 10⁻¹⁰ J/m³

(ii) The average energy density of magnetic field is 2.21 × 10⁻¹⁰ J/m³

(iii) The total average energy density of electromagnetic wave is 4.42 × 10⁻¹⁰ J/m³

Given:

E₀ = 10 Vm⁻¹

v = 5 × 10¹⁴ Hz

Explanation:

(i) The average energy density of electric field is given by the formula:

ue = 1/2 ε₀ (E₀/√2)²

ue = 1/4 ε₀ E₀²

On substituting the values, we get,

ue = 1/4 × 8.85 × 10⁻¹² × 10²

∴ ue = 2.21 × 10⁻¹⁰ J/m³

(ii) The average energy density of magnetic field is given by the formula:

um = ue

∴ um = 2.21 × 10⁻¹⁰ J/m³

(iii) Total average energy density of electromagnetic wave:

u = ue + um

On substituting the values, we get,

u = (2.21 × 10⁻¹⁰) + (2.21 × 10⁻¹⁰)

∴ u = 4.42 × 10⁻¹⁰ J/m³