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JEE Mains · Physics · STD 12 - Electromagnetic waves
  • JEE Mains 2014
  • Hard
  • MCQ

A lamp emits monochromatic green light uniformly in all directions. The lamp is \(3\%\) efficient in converting electrical power to electromagnetic waves and consumes \(100\,W\) of power . The amplitude of the electric field associated with the electromagnetic radiation at a distance of \(5\,m\) from the lamp will be nearly.......\(V/m\)

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Answer & Solution

Correct Answer

(B) \(2.68\)

Step-by-step Solution

Detailed explanation

Wavelength of monochromatic green light \(=5.5 \times 10^{-5} \mathrm{cm}\)

Intensity  \(I\,=\,\frac{\text { Power }}{\text { Area }}\)

\( = \frac{{100 \times (3/100)}}{{4\pi {{(5)}^2}}}\)

\(=\frac{3}{100 \pi} \mathrm{Wm}^{-2}\)

Now, half of this intensity ( \(\mathrm{I}\) ) belongs to electric field and half of that to magnetic field, therefore,

\(\frac{1}{2}=\frac{1}{4} \varepsilon_{0} E_{0}^{2} C\)

or \(\mathrm{E}_{0}=\sqrt{\frac{2 \mathrm{I}}{\varepsilon_{0} \mathrm{C}}}\)

\( = \sqrt {\frac{{2 \times \left( {\frac{3}{{100}}\pi } \right)}}{{\left( {\frac{1}{{4\pi  \times 9 \times {{10}^9}}}} \right) \times \left( {3 \times {{10}^8}} \right)}}} \)

\(=\sqrt{\frac{6}{25} \times 30}=\sqrt{7.2}\)

\(\therefore \mathrm{E}_{0}=2.68 \mathrm{V} / \mathrm{m}\)

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