An iron rod is placed parallel to magnetic field intensity 1000 A/m. The magnetic flux through the rod is 3 × 10⁻⁴ Wb and its cross-sectional area is 1.5 cm ². The magnetic permeability of rod in Wb/A-m is (a) 2 × 10⁻² (b) 2 × 10⁻³ (c) 2 × 10⁻⁴ (d) 1 × 10⁻²

Answer: (b). B = Φ/A = 3×10⁻⁴1.5×10⁻⁴ = 2 T μ = B/H = 2/1000 = 2×10⁻³ Wb/A-m So the answer is (b).

An iron rod is placed parallel to magnetic field intensity 1000 A/m… | MHT CET 2026 (15 Apr, Shift 2)

Q3MHT CET 2026 · 15 APR · SHIFT 2Easy

Magnetisation: B, H, M, χ & μ

An iron rod is placed parallel to magnetic field intensity 1000 A/m. The magnetic flux through the rod is $3 \times 10^{-4}$ Wb and its cross-sectional area is $1.5$ cm ${}^{2}$ . The magnetic permeability of rod in Wb/A-m is

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Answer: (b)

$B = \dfrac{\Phi}{A} = \dfrac{3\times10^{-4}}{1.5\times10^{-4}} = 2\text{ T}$
$\mu = \dfrac{B}{H} = \dfrac{2}{1000} = 2\times10^{-3}\text{ Wb/A-m}$
So the answer is (b).

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Magnetisation: B, H, M, χ & μ question from MHT CET 2026 (15 Apr, Shift 2)

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