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8.Electromagnetic waves
hard
A radiation is emitted by $1000\, W$ bulb and it generates an electric field and magnetic field at $P$, placed at a distance of $2\, m$. The efficiency of the bulb is $1.25 \%$. The value of peak electric field at $P$ is $x \times 10^{-1} \,V / m$. Value of $x$ is. (Rounded-off to the nearest integer)
[Take $\varepsilon_{0}=8.85 \times 10^{-12} C ^{2} N ^{-1} m ^{-2}, c =3 \times 10^{8}$ $ms ^{-1}$ ]
A
$137$
B
$149$
C
$164$
D
$121$
(JEE MAIN-2021)
Solution
$I _{ avg }=\frac{1}{2} \varepsilon_{0} E _{0}^{2} C$
$\frac{1.25}{100} \times \frac{1000}{4 \pi(2)^{2}}=\frac{1}{2} \times 8.85 \times 10^{-12} \times 3\times 10^{8} \times E _{0}^{2}$
$E _{0}^{2}=187.4$
$\therefore E _{0}=13.689 V / m$
$=136.89 \times 10^{-1} V / m$
$\therefore x =136.89$
Rounding off to nearest integer $x =137$
Standard 12
Physics
