Verified Solution • 2025–26 Session moderate

QJEEM23Phaf7a: JEE Main 2023 Physics - Class 12 JEEM23Phaf7a Solution: The energy density associated with electric field and magnetic…

The energy density associated with electric field E⃗\vec{E} and magnetic field B⃗\vec{B} of an electromagnetic wave in free space is given by (ϵ0−\left(\epsilon_{0}-\right. permittivity of free space, μ0−\mu_{0}- permeability of free space)

Multiple Choice Options

  • A UE=ϵ0E22,UB=B22μ0U_{E}=\frac{\epsilon_{0} E^{2}}{2}, U_{B}=\frac{B^{2}}{2 \mu_{0}}
  • B UE=E22ϵ0,UB=μ0B22U_{E}=\frac{E^{2}}{2 \epsilon_{0}}, U_{B}=\frac{\mu_{0} B^{2}}{2}
  • C UE=ϵ0E22,UB=μ0B22U_{E}=\frac{\epsilon_{0} E^{2}}{2}, U_{B}=\frac{\mu_{0} B^{2}}{2}
  • D UE=E22ϵ0,UB=B22μ0U_{E}=\frac{E^{2}}{2 \epsilon_{0}}, U_{B}=\frac{B^{2}}{2 \mu_{0}}

Step-by-Step Verified Solution

The energy density associated with the electric field E⃗\vec{E} and magnetic field B⃗\vec{B} of an electromagnetic wave in free space is given by:

For the electric field: UE=12ϵ0E2U_{E} = \frac{1}{2} \epsilon_{0} E^2

For the magnetic field: UB=12B2μ0U_{B} = \frac{1}{2} \frac{B^2}{\mu_{0}}

These expressions match Option A:

UE=ϵ0E22,UB=B22μ0U_{E} = \frac{\epsilon{0} E^{2}}{2}, U_{B} = \frac{B^{2}}{2 \mu{0}}

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