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8. Electromagnetic waves — Physics STD 12 Science — Question

CBSE BoardEnglish MediumSTD 12 SciencePhysics8. Electromagnetic waves1 Mark
MCQ
The energy of electromagnetic wave in vacuum is given by the relation
  • A
    $\frac{{{E^2}}}{{2{\varepsilon _0}}} + \frac{{{B^2}}}{{2{\mu _0}}}$
  • B
    $\frac{1}{2}{\varepsilon _0}{E^2} + \frac{1}{2}{\mu _0}{B^2}$
  • C
    $\frac{{{E^2} + {B^2}}}{c}$
  • ✓
    $\frac{1}{2}{\varepsilon _0}{E^2} + \frac{{{B^2}}}{{2{\mu _0}}}$
✓

Answer

Correct option: D.
$\frac{1}{2}{\varepsilon _0}{E^2} + \frac{{{B^2}}}{{2{\mu _0}}}$
d
$\frac{1}{2}{\varepsilon _0}E_0^2$ is electric energy density. $\frac{{{B^2}}}{{2{\mu _0}}}$ is magnetic energy density. So, total energy $ = \frac{1}{2}{\varepsilon _0}E_0^2 + \frac{{B_0^2}}{{2{\mu _0}}}$

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