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QJEEM22Phc57d: JEE Main 2022 Physics - Class 12 JEEM22Phc57d Solution: An electric bulb is rated as 200 W. What will be the peak…

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.

Multiple Choice Options

  • A 1.19 ×\times 10 −- 8T
  • B 1.71 ×\times 10 −- 8T
  • C 0.84 ×\times 10 −- 8T
  • D 3.36 ×\times 10 −- 8T

Step-by-Step Verified Solution

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

P=3⋅5100×200=7 W\mathrm{P}=\frac{3 \cdot 5}{100} \times 200=7 \mathrm{~W}

 Intensity Iamp;=P4πr2=74π(4)2 W/m2 Intensity Iamp;= Average energy density ×camp;=12∈0E02camp;=12B02μ0c\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}

∴B0amp;=2μ0Icamp;=2×4π×10−7×73×108×4π×16amp;=1.71×10−8 T\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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