The constant in the Biot-Savart law is \(\mu_0/4\pi\). The constant in Coulomb’s law is \(1/4\pi\varepsilon_0\). The speed of light in vacuum is \(c=3\times10^{8}\ \mathrm{m\,s^{-1}}\). The ratio \(\dfrac{\mu_0/4\pi}{1/4\pi\varepsilon_0}\) is
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \(1.1\times10^{-17}\ \mathrm{s^{2}\,m^{-2}}\)
Written out, the ratio of the two constants is the product \( \varepsilon_0\mu_0 \). This product is linked to the speed of light by \( \varepsilon_0\mu_0=1/c^{2} \).
\( \begin{aligned} \frac{\mu_0/4\pi}{1/4\pi\varepsilon_0} &= \frac{\mu_0}{4\pi}\times4\pi\varepsilon_0 \\ &= \varepsilon_0\mu_0 \\ &= \frac{1}{c^{2}} \\ &= \frac{1}{(3\times10^{8})^{2}} \\ &= 1.1\times10^{-17}\ \mathrm{s^{2}\,m^{-2}} \end{aligned} \)
NCERT (2026-27):
Magnetic Field Due to a Current Element, Biot-Savart Law – “There is an interesting relation between ε0, the permittivity of free space; μ0, the permeability of free space; and c, the speed of light in vacuum:”
\( \varepsilon_0\mu_0=(4\pi\varepsilon_0)\frac{\mu_0}{4\pi}=\frac{1}{9\times10^{9}}(10^{-7})=\frac{1}{(3\times10^{8})^{2}}=\frac{1}{c^{2}} \)