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JEE Mains · Physics · STD 12 - Electromagnetic waves
  • JEE Mains 2022
  • Hard
  • MCQ

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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Answer & Solution

Correct Answer

(B) \(1.71\)

Step-by-step Solution

Detailed explanation

\(\frac{\eta P }{4 \pi r ^{2}}=\frac{ cB _{0}^{2}}{2 \mu_{0}}\)

\(B _{0}=\sqrt{\frac{\mu_{0}}{4 \pi} \frac{\eta P }{ c } \frac{1}{ r }}\)

\(\Rightarrow B _{0}=\frac{1}{4} \sqrt{\frac{10^{-7} \times 4 \times 3.5}{3 \times 10^{8}}}=1.71 \times 10^{-8} \,T\)

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