- Home
- Standard 12
- Physics
8.Electromagnetic waves
medium
A point source of electromagnetic radiation has an average power output of $800\,W$ . The maximum value of electric field at a distance $3.5\,m$ from the source will be.....$V/m$
A
$56.7$
B
$62.6$
C
$39.3$
D
$47.5$
Solution
Intensity of electromagnetic wave given is by
${\mathrm{I}=\frac{\mathrm{P}_{\mathrm{av}}}{4 \pi \mathrm{r}^{2}}=\frac{\mathrm{E}_{0}^{2}}{2 \mu_{0} \mathrm{c}}} $
${\mathrm{E}=\sqrt{\frac{\mu_{0} \mathrm{cP}_{\mathrm{av}}}{2 \pi \mathrm{r}^{2}}}} $
${=\sqrt{\frac{\left(4 \pi \times 10^{-7}\right) \times\left(3 \times 10^{8}\right) \times 800}{2 \pi \times(3.5)^{2}}}=62.6 \mathrm{\,V} / \mathrm{m}}$
Standard 12
Physics
