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STD 12 - 8. Electromagnetic waves — NEET STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceNEETSTD 12 - 8. Electromagnetic waves4 Marks
MCQ
For the plane electromagnetic wave given by $\mathrm{E}=\mathrm{E}_0 \sin (\omega \mathrm{t}-\mathrm{kx})$ and $\mathrm{B}=\mathrm{B}_0 \sin (\omega \mathrm{t}-\mathrm{kx})$, the ratio of average electric energy density to average magnetic energy density is
  • ✓
    $1$
  • B
    $\frac{1}{2}$
  • C
    $2$
  • D
    $4$
✓

Answer

Correct option: A.
$1$
a
$\frac{\text { Electric energy density }}{\text { Magnetic energy density }}=\frac{\frac{1}{2} \in_0 \mathrm{E}_{\mathrm{rms}}^2}{\left(\frac{\mathrm{B}_{\mathrm{rms}}^2}{2 \mu_0}\right)}$

$=\left(\frac{\mathrm{E}_{\mathrm{rms}}}{\mathrm{B}_{\mathrm{rms}}}\right)^2 \cdot \mu_0 \in_0 \quad\left[\mathrm{C}=\frac{1}{\mu_0 \in_0}\right]$

$=\frac{\mathrm{C}^2}{\mathrm{C}^2}=1$

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