A field of $2 × 10^3 A/m$ produces a flux of $2.4 × 10^{-5} Wb$ in an iron bar having area of cross-section $0.2\, cm^2$. The relative permeability of medium is:

478

The correct answer is Option (1) → 478

Given:

Magnetic field strength $H = 2 \times 10^3 \, A/m$

Flux $ \Phi = 2.4 \times 10^{-5} \, Wb$

Area $A = 0.2 \, cm^2 = 0.2 \times 10^{-4} \, m^2 = 2 \times 10^{-5} \, m^2$

Flux density:

$B = \frac{\Phi}{A} = \frac{2.4 \times 10^{-5}}{2 \times 10^{-5}} = 1.2 \, T$

Relation: $B = \mu_0 \mu_r H$

$\mu_0 = 4\pi \times 10^{-7} \, H/m$

$\mu_r = \frac{B}{\mu_0 H}$

$\mu_r = \frac{1.2}{(4\pi \times 10^{-7})(2 \times 10^3)}$

$\mu_r = \frac{1.2}{8\pi \times 10^{-4}}$

$\mu_r \approx \frac{1.2}{0.002513} \approx 477.5$

Answer: Relative permeability $ \mu_r \approx 478$