Question

In an electromagnetic wave, the amplitude of electric field is $10{\text{Vm}}^{-1}$. The frequency of the wave is $5\times {10}^{14}\text{Hz}$ and the wave is propagating along $z–$axis, then total average energy density of E.M. wave is

• A. $4.42\times {10}^{-10}{\text{Jm}}^{-3}$
• B. $2.21\times {10}^{-10}{\text{Jm}}^{-3}$
• C. $3.31\times {10}^{-10}{\text{Jm}}^{-3}$
• D. $1.11\times {10}^{-10}{\text{Jm}}^{-3}$

Answer 1

The correct option is A $4.42\times {10}^{-10}{\text{Jm}}^{-3}$

Given,
$E_0=10{\text{Vm}}^{-1};f=5\times {10}^{14}\text{Hz}$

$\because C=\frac{E_0}{B_0}$
$\Rightarrow B_0=\frac{E_0}{C}=\frac{10}{3\times {10}^8}=0.33\text{nT}$

The average energy density of the electric field is given by,

$U_E=\frac{1}{4}{\epsilon }_0{E}^2=\frac{1}{4}\times 8.85\times {10}^{-12}\times {10}^2$
$=2.2125\times {10}^{-10}{\text{Jm}}^{-3}$

The average energy density of the electric field is given by,

$U_B=U_E=2.2125\times {10}^{-10}{\text{Jm}}^{-3}$

Therefore, the average energy density of the E.M. wave is given by,

$U=U_E+U_B=4.425\times {10}^{-10}{\text{Jm}}^{-3}$

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, $\left(B\right)$ is the correct answer.