JEE PYQ: Electromagnetic Waves Question 10
Question 10 - 2021 (26 Feb 2021 Shift 1)
A radiation is emitted by 1000 W bulb and it generates an electric field and magnetic field at P$_1$ 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 $\epsilon_0 = 8.85 \times 10^{-12}$ C$^2$ N$^{-1}$ m$^{-2}$, $c = 3 \times 10^{8}$ ms$^{-1}$]
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Answer: 137
Solution
$I = \frac{1}{2} C \epsilon_0 E_0^2 = \frac{P}{4\pi r^2}$, solving gives $E_0 \approx 137 \times 10^{-1}$ V/m.
BETA
