Total energy density of electromagnetic waves in vacuum is given by the relation

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$\frac{1}{2} \cdot \frac{E^{2}}{ε_{0}} + \frac{B^{2}}{2 μ_{0}}$

$\frac{E^{2} + B^{2}}{c}$

$\frac{1}{2} ε_{0} E^{2} + \frac{1}{2} μ_{0} B^{2}$

$\frac{1}{2} ε_{0} E^{2} + \frac{B^{2}}{2 μ_{0}}$

answer is D.

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Detailed Solution

The energy in EM waves is divided equally between the electric and magnetic fields.

The total energy per unit volume, $u = u_{e} + u_{m}$

$= \frac{1}{2} ε_{0} E^{2} + \frac{1}{2} \frac{B^{2}}{μ_{0}}$

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