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

JEE Main · 2019 · 09 Jan 2019 · Shift 2 · EM Waves

The energy associated with electric field is

(U_{E})

and with magnetic field is

(U_{B})

for an electromagnetic wave in free space. Then:

(A) $U_{E} = \frac{U _{B}}{2}$
(B) $U_{E} > U_{B}$
(C) $U_{E} < U_{B}$
(D) $U_{E} = U_{B}$ ✓ Correct

Answer: (D)

Solution

In an electromagnetic wave travelling through vacuum the two fields are tied together by

B = \frac{E}{c}

. The electric energy density is

u_{E} = \frac{1}{2} ε_{0} E^{2}

and the magnetic one is

u_{B} = \frac{B ^{2}}{2 μ _{0}} = \frac{E ^{2}}{2 μ _{0} c ^{2}}

. Replacing

\frac{1}{c ^{2}}

μ_{0} ε_{0}

turns

u_{B}

into

\frac{1}{2} ε_{0} E^{2}

, exactly

u_{E}

. So the energy is shared equally,

U_{E} = U_{B}

JEE Main · 2019 · 09 Jan 2019 · Shift 2 · Single correct option

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