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JEE Main 20146 April
Exam
JEE Main
Year
2014
Date
6 April
Subject
Physics
Chapter
Electromagnetic Waves

JEE Main 2014 · 6 April · Physics · Electromagnetic Waves

During the propagation of electromagnetic wave in a particular...

During the propagation of electromagnetic wave in a particular medium :-

Options

  1. AElectric energy density is double of the magnetic energy density
  2. BElectric energy density is half of the magnetic energy density
  3. CElectric energy density is equal to the magnetic energy density
  4. DBoth electric and magnetic energy densities are zero

Correct answer

C. Electric energy density is equal to the magnetic energy density

Step-by-step solution

(UV)elec=12∈0E2{\left(\dfrac{U}{V}\right)}_{elec} = \dfrac{1}{2}{\in}_{0}{E}^{2} ....(i)

(UV)mag=12B2/μ0{\left(\dfrac{U}{V}\right)}_{mag} = \dfrac{1}{2}{B}^{2} / {\mu}_{0} ....(ii)

B=EC\text{B} = \dfrac{\text{E}}{\text{C}} (for EMW) ....(iii)

and C=1μ0ε0C = \dfrac{1}{\sqrt{{\mu}_{0} {ε}_{0}}} .....(iv)

∴ devide eqn (i) by (ii)

(U/V)elec(U/V)mag=12ε0E212B2/μ0\dfrac{{( U / V )}_{elec}}{{( U / V )}_{mag}} = \dfrac{\dfrac{1}{2}{ε}_{0}{E}^{2}}{\dfrac{1}{2}{B}^{2} / {\mu}_{0}}

=(EB)2μ0ε0= {( \dfrac{E}{B} )}^{2} {\mu}_{0} {ε}_{0}

by using eq (iii) & (iv)

=C2C2=1= \dfrac{{C}^{2}}{{C}^{2}} = 1

⇒(U/V)elec.=(UV)mag⇒ {( U / V )}_{elec .} = {( \dfrac{U}{V} )}_{mag}

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