The energy density associated with electric field \overline{ {\E}} and magnetic field \overline{ {\B}} of an electromagnetic wave in free space is given by (\epsilon_0-. permittivity of free space, \mu_0 - permeability of free space) [6-Apr-2023 shift 2]

Correct Answer is : UE=∈0E22,UB=B22µ0

{\U}_{ {\E}}=\frac{\in_0 {\E}^2}{2}, {\U}_{ {\B}}=\frac{ {\B}^2}{2 \mu_0}

{\U}_{ {\E}}=\frac{ {\E}^2}{2 E_0}, {\U}_{ {\B}}=\frac{\mu_0 {\B}^2}{2}

{\U}_{ {\E}}=\frac{ {\E}^2}{2 E_0}, {\U}_{ {\B}}=\frac{ {\B}^2}{2 \mu_0}

{\U}_{ {\E}}=\frac{\in_0 {\E}^2}{2}, {\U}_{ {\B}}=\frac{\mu_0 {\B}^2}{2}

JEE Mains 6-Apr-2023 Shift 2 Solved Paper

Section: Physics

Question : 39 of 90

Marks: +1, -0

The energy density associated with electric field E and magnetic field B of an electromagnetic wave in free space is given by (ε0−. permittivity of free space, µ0 - permeability of free space)

[6-Apr-2023 shift 2]

UE=‌
∈0E2
2
,UB=‌
B2
2µ0
UE=‌
E2
2E0
,UB=‌
µ0B2
2
UE=‌
E2
2E0
,UB=‌
B2
2µ0
UE=‌
∈0E2
2
,UB=‌
µ0B2
2

Validate

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