JEE MAIN - Physics (2022 - 24th June Evening Shift - No. 16)
An electric bulb is rated as 200 W. What will be the peak magnetic field at 4 m distance produced by the radiations coming from this bulb? Consider this bulb as a point source with 3.5% efficiency.
1.19 $$\times$$ 10$$-$$8T
1.71 $$\times$$ 10$$-$$8T
0.84 $$\times$$ 10$$-$$8T
3.36 $$\times$$ 10$$-$$8T
Explanation
The total power (PT) of the light bulb is given as 200 W, but only 3.5% of this power is actually emitted as radiation, which we will call P.
So, Effective power output of the bulb
$$ \mathrm{P}=\frac{3 \cdot 5}{100} \times 200=7 \mathrm{~W} $$
$$ \begin{aligned} \text { Intensity } \mathrm{I} & =\frac{\mathrm{P}}{4 \pi r^2}=\frac{7}{4 \pi(4)^2} \mathrm{~W} / \mathrm{m}^2 \\\\ \text { Intensity } \mathrm{I} & =\text { Average energy density } \times c \\\\ & =\frac{1}{2} \in_0 \mathrm{E}_0^2 c \\\\ & =\frac{1}{2} \frac{\mathrm{B}_0^2}{\mu_0} c \end{aligned} $$
$$ \begin{aligned} \therefore \quad \mathrm{B}_0 & =\sqrt{\frac{2 \mu_0 \mathrm{I}}{c}} \\\\ & =\sqrt{\frac{2 \times 4 \pi \times 10^{-7} \times 7}{3 \times 10^8 \times 4 \pi \times 16}} \\\\ & =1.71 \times 10^{-8} \mathrm{~T} \end{aligned} $$
So, Effective power output of the bulb
$$ \mathrm{P}=\frac{3 \cdot 5}{100} \times 200=7 \mathrm{~W} $$
$$ \begin{aligned} \text { Intensity } \mathrm{I} & =\frac{\mathrm{P}}{4 \pi r^2}=\frac{7}{4 \pi(4)^2} \mathrm{~W} / \mathrm{m}^2 \\\\ \text { Intensity } \mathrm{I} & =\text { Average energy density } \times c \\\\ & =\frac{1}{2} \in_0 \mathrm{E}_0^2 c \\\\ & =\frac{1}{2} \frac{\mathrm{B}_0^2}{\mu_0} c \end{aligned} $$
$$ \begin{aligned} \therefore \quad \mathrm{B}_0 & =\sqrt{\frac{2 \mu_0 \mathrm{I}}{c}} \\\\ & =\sqrt{\frac{2 \times 4 \pi \times 10^{-7} \times 7}{3 \times 10^8 \times 4 \pi \times 16}} \\\\ & =1.71 \times 10^{-8} \mathrm{~T} \end{aligned} $$
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