The ratio of the electric field energy density to the magnetic field energy density of an EM wave is (Symbols have their usual meaning)
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 1:1
The energy densities of the electric and magnetic fields are \( \frac{1}{2}\varepsilon_0E^{2} \) and \( B^{2}/2\mu_0 \). In an electromagnetic wave the amplitudes are linked by E = cB, with \( c = 1/\sqrt{\mu_0\varepsilon_0} \).
\( \displaystyle \frac{u_E}{u_B} = \frac{\frac{1}{2}\varepsilon_0 E^{2}}{B^{2}/2\mu_0} = \mu_0\varepsilon_0\,\frac{E^{2}}{B^{2}} \)
\( \displaystyle \frac{u_E}{u_B} = \mu_0\varepsilon_0\,c^{2} = \mu_0\varepsilon_0 \times \frac{1}{\mu_0\varepsilon_0} = 1 \)
So the energy of the wave is shared equally between its electric and magnetic fields: \( u_E : u_B = 1 : 1 \).
Note: This question is from NCERT (2026-27). NTA sets numerical questions from NCERT too, so do not ignore NCERT while practising numericals.
NCERT (2026-27):
Exercises – …
