The efficiency of a bulb of power 60 W is 16 | vedclass

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(D) Given: Power of the bulb $P = 60 \ W$,Efficiency $\eta = 16 \% = 0.16$,Distance $r = 2 \ m$.First,calculate the intensity of the electromagnetic r

Vedclass · Accepted Answer

$12$

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(D) Given: Power of the bulb $P = 60 \ W$,Efficiency $\eta = 16 \% = 0.16$,Distance $r = 2 \ m$.First,calculate the intensity of the electromagnetic radiation emitted by the bulb at distance $r$:The power radiated as electromagnetic waves is $P_{rad} = \eta 	imes P = 0.16 	imes 60 = 9.6 \ W$.The intensity $I$ at distance $r$ is given by $I = \frac{P_{rad}}{4 \pi r^2} = \frac{9.6}{4 \pi (2)^2} = \frac{9.6}{16 \pi} = \frac{0.6}{\pi} \ W/m^2$.The intensity of an electromagnetic wave is also related to the peak electric field $E_0$ by $I = \frac{1}{2} c \epsilon_0 E_0^2$.Equating the two expressions for intensity: $\frac{0.6}{\pi} = \frac{1}{2} c \epsilon_0 E_0^2$.We know that $c = \frac{1}{\sqrt{\mu_0 \epsilon_0}}$ and $\frac{1}{4 \pi \epsilon_0} = 9 	imes 10^9$,so $\epsilon_0 = \frac{1}{36 \pi 	imes 10^9}$.Using $c = 3 	imes 10^8 \ m/s$,we have $E_0^2 = \frac{2 	imes I}{c \epsilon_0} = \frac{2 	imes 0.6 / \pi}{3 	imes 10^8 	imes (1 / (36 \pi 	imes 10^9))}$.$E_0^2 = \frac{1.2}{\pi} 	imes \frac{36 \pi 	imes 10^9}{3 	imes 10^8} = 1.2 	imes 12 	imes 10 = 144$.Therefore,$E_0 = \sqrt{144} = 12 \ Vm^{-1}$.

The efficiency of a bulb of power $60 \ W$ is $16 \%$. The peak value of the electric field produced by the electromagnetic radiation from the bulb at a distance of $2 \ m$ from the bulb is $\left(\frac{1}{4 \pi \epsilon_0}=9 \times 10^9 \ Nm^2 C^{-2}\right)$ (in $Vm^{-1}$)

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