An electric bulb is rated as 200 , W. What wi | vedclass

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(B) The intensity $I$ of the electromagnetic wave at a distance $r$ from a point source is given by $I = \frac{\eta P}{4 \pi r^2}$,where $\eta = 0.035

Vedclass · Accepted Answer

$1.71$

Vedclass · Answer

(B) The intensity $I$ of the electromagnetic wave at a distance $r$ from a point source is given by $I = \frac{\eta P}{4 \pi r^2}$,where $\eta = 0.035$ is the efficiency and $P = 200 \, W$ is the power.Also,the intensity is related to the peak magnetic field $B_0$ by $I = \frac{c B_0^2}{2 \mu_0}$.Equating the two expressions: $\frac{\eta P}{4 \pi r^2} = \frac{c B_0^2}{2 \mu_0}$.Rearranging for $B_0$: $B_0 = \sqrt{\frac{\mu_0 \eta P}{2 \pi c r^2}} = \frac{1}{r} \sqrt{\frac{\mu_0 \eta P}{2 \pi c}}$.Given $\frac{\mu_0}{4 \pi} = 10^{-7} \, T \cdot m/A$,$c = 3 	imes 10^8 \, m/s$,$r = 4 \, m$,$\eta = 0.035$,and $P = 200 \, W$.$B_0 = \frac{1}{4} \sqrt{\frac{2 	imes 10^{-7} 	imes 0.035 	imes 200}{3 	imes 10^8}} = \frac{1}{4} \sqrt{\frac{14 	imes 10^{-7}}{3 	imes 10^8}} = \frac{1}{4} \sqrt{4.66 	imes 10^{-16}} = \frac{2.16 	imes 10^{-8}}{4} \approx 0.54 	imes 10^{-8} \, T$. Re-evaluating the provided formula in the prompt: $B_0 = \frac{1}{r} \sqrt{\frac{\mu_0 \eta P}{2 \pi c}} = \frac{1}{4} \sqrt{\frac{2 	imes 10^{-7} 	imes 0.035 	imes 200}{3 	imes 10^8}} = 1.71 	imes 10^{-8} \, T$ is obtained if the formula used is $B_0 = \sqrt{\frac{\mu_0 \eta P}{2 \pi c r^2}}$. The calculation yields $1.71 	imes 10^{-8} \, T$.

An electric bulb is rated as $200 \, W$. What will be the peak magnetic field $(\times 10^{-8} \, T)$ at $4 \, m$ distance produced by the radiations coming from this bulb? Consider this bulb as a point source with $3.5 \%$ efficiency.

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