An iron rod is placed parallel to magnetic field of intensity H. The magnetic flux through the rod is Φ and its cross-sectional area is A. The magnetic permeability ( μ) of the rod is (a) Φ AH (b) (AH)/(Φ) (c) (Φ)/(AH) (d) A/Φ H

Answer: (c). B = Φ/A and B = μ H. μ = B/H = (Φ)/(AH) So the answer is (c).

An iron rod is placed parallel to magnetic field of intensity H. The… | MHT CET 2026 (13 May, Shift 1)

Q2MHT CET 2026 · 13 MAY · SHIFT 1Easy

Magnetisation: B, H, M, χ & μ

An iron rod is placed parallel to magnetic field of intensity H. The magnetic flux through the rod is $\Phi$ and its cross-sectional area is A. The magnetic permeability ( $\mu$ ) of the rod is

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Answer: (c)

$B = \Phi/A$ and $B = \mu H$ .
$\mu = \dfrac{B}{H} = \dfrac{\Phi}{AH}$
So the answer is (c).

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Magnetisation: B, H, M, χ & μ question from MHT CET 2026 (13 May, Shift 1)

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