The magnetic permeability of a ferromagnetic material is \(2.512\times10^{-3}\ \mathrm{T\,m\,A^{-1}}\). The magnetic susceptibility of the material is (take \(\pi=3.14\))
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → 1999
The relative permeability is the ratio \( \mu_r=\mu/\mu_0 \). The susceptibility is related to it by \( \mu_r=1+\chi \).
Permeability, \( \mu=2.512\times10^{-3}\ \mathrm{T\,m\,A^{-1}} \)
Permeability of free space, \( \mu_0=4\pi\times10^{-7}\ \mathrm{T\,m\,A^{-1}} \)
\( \begin{aligned} \mu_r &= \frac{\mu}{\mu_0} \\ &= \frac{2.512\times10^{-3}}{4\times3.14\times10^{-7}} \\ &= 2000 \end{aligned} \)
\( \begin{aligned} \chi &= \mu_r-1 \\ &= 2000-1 \\ &= 1999 \end{aligned} \)
NCERT (2026-27):
Magnetisation and Magnetic Intensity – “… The magnetic permeability of the substance is μ and it has the same dimensions and units as μ0; μ = μ0μr = μ0 (1+χ).”