A magnetising field of 5000 A/m produces a magnetic flux of 4 × 10⁻⁵ Wb in an iron rod of cross- sectional area 0.4 cm ². The permeability of the rod (in Wb/Am) is (a) 4 × 10⁻⁶ (b) 1 × 10⁻³ (c) 2 × 10⁻⁴ (d) 3 × 10⁻⁵

Answer: (c). B = Φ/A and μ = B/H. B = 4×10⁻⁵0.4×10⁻⁴ = 1 T μ = 1/5000 = 2×10⁻⁴ Wb/Am So the answer is (c).

A magnetising field of 5000 A/m produces a magnetic flux of 4 × 10⁻⁵… | MHT CET 2026 (15 May, Shift 1)

Q1MHT CET 2026 · 15 MAY · SHIFT 1Easy

Magnetisation: B, H, M, χ & μ

A magnetising field of $5000$ A/m produces a magnetic flux of $4 \times 10^{-5}$ Wb in an iron rod of cross- sectional area $0.4$ cm ${}^{2}$ . The permeability of the rod (in Wb/Am) is

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Answer: (c)

$B = \Phi/A$ and $\mu = B/H$ .
$B = \dfrac{4\times10^{-5}}{0.4\times10^{-4}} = 1\text{ T}$
$\mu = \dfrac{1}{5000} = 2\times10^{-4}\text{ Wb/Am}$
So the answer is (c).

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