Question

Poynting vectors $\overrightarrow{S}$ is defined as $\overrightarrow{S}=\frac{1}{{\mu }_0}\overrightarrow{E}\times \overrightarrow{B}$. The average value of $\overrightarrow{S}$ over a single period T is given by

• A. $\frac{E_0^2}{2c{\mu }_0}$
• B. $\frac{E_0^2}{c{\mu }_0}$
• C. $\frac{2E_0^2}{2c{\mu }_0}$
• D. $\frac{2E_0^2}{2c{\mu }_0^2}$

Answer 1

The correct option is A $\frac{E_0^2}{2c{\mu }_0}$

Poynting vector is given by, $\overrightarrow{S}=\frac{\overrightarrow{E}\times \overrightarrow{B}}{{\mu }_0}$

Average value of magnitude of radiant flux density over one cycle is

$S_{av}=\frac{1}{{\mu }_0}\mid {\overrightarrow{E}}_0\times {\overrightarrow{B}}_0|(\frac{1}{T}){\int }_0^{T}co{s}^2(kx-\omega t)dt$

$[\because E=E_{0}cos(kx-\omega t),B=B_{0}cos(kx-\omega t\left)\right]$

$=\frac{E_0{B}_0}{{\mu }_0}(\frac{1}{T})(\frac{T}{2})$ $[\because {\int }_0^{T}co{s}^2(kx-\omega t)dt=\frac{T}{2}]$

$=\frac{E_0{B}_0}{2{\mu }_0}$

$\therefore S_{av}=\frac{E_0\left(E_0/c\right)}{2{\mu }_0}=(\frac{1}{2c{\mu }_0})E_0^2$ $(\because B_0=\frac{E_0}{c})$

So, option $\left(A\right)$ is correct