At room temperature, the relative permittivity of water is 80 and the speed of light in water is \( 2.25\times10^{8}\,\mathrm{m\,s^{-1}} \). What is the relative permeability of water?
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 0.022
In a material medium of permittivity \( \varepsilon \) and permeability \( \mu \), the speed of light is \( v=1/\sqrt{\mu\varepsilon} \). We write \( \varepsilon=\varepsilon_r\varepsilon_0 \) and \( \mu=\mu_r\mu_0 \), where \( \varepsilon_r \) is the relative permittivity and \( \mu_r \) the relative permeability. Using \( c=1/\sqrt{\mu_0\varepsilon_0} \):
\( \begin{aligned} v &= \frac{1}{\sqrt{\mu_r\mu_0\,\varepsilon_r\varepsilon_0}} \\ &= \frac{c}{\sqrt{\mu_r\varepsilon_r}} \end{aligned} \)
Solving for \( \mu_r \), with \( c=3\times10^{8}\,\mathrm{m\,s^{-1}} \):
\( \begin{aligned} \mu_r &= \frac{c^{2}}{v^{2}\,\varepsilon_r} \\ &= \frac{(3\times10^{8}\,\mathrm{m\,s^{-1}})^{2}}{(2.25\times10^{8}\,\mathrm{m\,s^{-1}})^{2}\times80} \\ &= \frac{1.78}{80} \\ &\approx 0.022 \end{aligned} \)
NCERT (2026-27):
Nature of Electromagnetic Waves – “… We have seen earlier that the total electric and magnetic fields inside a medium are described in terms of a permittivity ε and a magnetic permeability μ … These replace ε0 and μ0 in the description to electric and magnetic fields in Maxwell’s equations with the result that in a material medium of permittivity ε and magnetic permeability μ, the velocity of light becomes,
\( \displaystyle v=\frac{1}{\sqrt{\mu\varepsilon}} \)\( \qquad (8.11) \)
Thus, the velocity of light depends on electric and magnetic properties of the medium.”