A field of \( 2 \times 10^{3}\,\mathrm{A/m} \) produces a flux of \( 2.4 \times 10^{-5}\,\mathrm{Wb} \) in an iron bar having area of cross-section \( 0.2\,\mathrm{cm^2} \). The relative permeability of medium is:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → 478
The given field is the magnetic intensity \( H \). The flux gives the magnetic field \( B \) inside the bar, and the permeability is \( \mu = B/H \).
\( \displaystyle A = 0.2\,\mathrm{cm^2} = 0.2\times10^{-4}\,\mathrm{m^2} \)
\( \displaystyle B = \frac{\Phi}{A} = \frac{2.4\times10^{-5}}{0.2\times10^{-4}} = 1.2\,\mathrm{T} \)
\( \displaystyle \mu = \frac{B}{H} = \frac{1.2}{2\times10^{3}} = 6\times10^{-4}\,\mathrm{T\,m\,A^{-1}} \)
\( \displaystyle \mu_r = \frac{\mu}{\mu_0} = \frac{6\times10^{-4}}{4\pi\times10^{-7}} \approx 477.5 \approx 478 \)