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JEE Main 201411 AprilOnline
Exam
JEE Main
Year
2014
Date
11 April
Mode
Online
Subject
Physics
Chapter
Electromagnetic Waves

JEE Main 2014 · 11 April · Online · Physics · Electromagnetic Waves

An electromagnetic wave of frequency 1 × 10^ 14 hertz is...

An electromagnetic wave of frequency 1×10141 \times 10^{14} hertz is propagating along z\mathrm{z}-axis. The amplitude of electric field is 4 V/m4 \mathrm{~V} / \mathrm{m}. If ε0=8.8×10−12C2/N−\varepsilon_0=8.8 \times 10^{-12} \mathrm{C}^2 / \mathrm{N}- m2\mathrm{m}^2, then average energy density of electric field will be:

Options

  1. A35.2×10−10 J/m335.2 \times 10^{-10} \mathrm{~J} / \mathrm{m}^3
  2. B35.2×10−11 J/m335.2 \times 10^{-11} \mathrm{~J} / \mathrm{m}^3
  3. C35.2×10−12 J/m335.2 \times 10^{-12} \mathrm{~J} / \mathrm{m}^3
  4. D35.2×10−13 J/m335.2 \times 10^{-13} \mathrm{~J} / \mathrm{m}^3

Correct answer

C. 35.2×10−12 J/m335.2 \times 10^{-12} \mathrm{~J} / \mathrm{m}^3

Step-by-step solution

Given: Amplitude of electric field, E_0=4 v\mathrm{v}/ m\mathrm{m}Absolute permitivity, \varepsilon_0=8.8 ×\times10^{-12} \mathrm{c}^2 / \mathrm{N}-\mathrm{m}^2 Average energy density uE=u_E= ? Applying formula, Average energy density uE=14ε0E2u_E=\frac{1}{4} \varepsilon_0 E^2⇒uE=14×8.8×10−12×(4)2=35.2×10−12 J/m3\begin{aligned}&\Rightarrow \quad u_E=\frac{1}{4} \times8.8 \times10^{-12} \times(4)^2 \\ &=35.2 \times10^{-12} \mathrm{~J}/ \mathrm{m}^3 \end{aligned}$
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