Electromagnetic Waves

JEE Main level, single correct, medium

113 of 165

Question 113 of 165

The energy associated with electric field is (UE)\left({U}_{E}\right) and with magnetic field is (UB)\left({U}_{B}\right) for an electromagnetic wave in free space. Then:

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Answer: (B) UE=UB{U}_{E}={U}_{B}

The average energy density of electric field is given by
uE=12ϵE02{u}_{E}=\dfrac{1}{2}\epsilon {{E}_{0}}^{2}
Here, E0{E}_{0} is the maximum value of electric field.
The average energy density of magnetic field is given by
uB=B022μ{u}_{B}=\dfrac{{{B}_{0}}^{2}}{2\mu }
Here, B0{B}_{0} is the maximum value of magnetic field.
The speed of light is given by c2=1μϵ{c}^{2}=\dfrac{1}{\mu \epsilon }
Also, E0=cB0{E}_{0}=c{B}_{0}
So, uE=12c2μ(cB0)2=B022μ=uB{u}_{E}=\dfrac{1}{2{c}^{2}\mu }{\left(c{B}_{0}\right)}^{2}=\dfrac{{{B}_{0}}^{2}}{2\mu }={u}_{B}
As the energy density of electric field and magnetic field are equal, the energy associated for equal volume will also be equal.

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