Question

Find the average energy density(in SI unit) of The electric field of a plane electromagnetic wave varies with time of amplitude $2 \, V m^{- 1}$ propagating along $z$ -axis.

• A. $13.29 \, \times 10^{- 12}$
• B. $8.86\times 10^{- 12}$
• C. $17.72\times 10^{- 12}$
• D. $4.43\times 10^{- 12}$

Answer 1

Answer: (B)

Amplitude of electric field and magnetic field are related by the relation
$\frac{E_{0}}{B_{0}}=c$
Average energy density of the magnetic field is
$u_{B}=\frac{1}{4}\frac{B_{0}^{2}}{\left(\mu \right)_{0}}=\frac{1}{4}\frac{E_{0}^{2}}{\left(\mu \right)_{0} c^{2}} \, \, \, \left(\because \, B_{0} = \frac{E_{0}}{c}\right)$
$=\frac{1}{4}\left(\epsilon \right)_{0}E_{0}^{2} \, \left(\because \, c = \frac{1}{\sqrt{\left(\mu \right)_{0} \left(\epsilon \right)_{0}}}\right)$
$=\frac{1}{4}\times 8.854\times \left(10\right)^{- 12}\times \left(2\right)^{2}=8.854\times \left(10\right)^{- 12} \, J \, m^{- 3}$
$\approx8.86\times 10^{- 12} \, J \, m^{- 3}$