An iron rod is placed parallel to magnetic field of intensity 2000 A/m. The magnetic flux through the rod is 6 × 10⁻⁴ Wb and its cross-sectional area is 3 cm². The magnetic permeability of the rod in Wb/A-m is (a) 10⁻¹ (b) 10⁻² (c) 10⁻⁴ (d) 10⁻³

Answer: (d). B = Φ/A = 6×10⁻⁴3×10⁻⁴ = 2 T μ = B/H = 2/2000 = 10⁻³ Wb/A-m So the answer is (d) 10⁻³.

An iron rod is placed parallel to magnetic field of intensity 2000… | MHT CET 2020 (16 Oct, Shift 1)

Q40MHT CET 2020 · 16 OCT · SHIFT 1Easy

Magnetisation: B, H, M, χ & μ

An iron rod is placed parallel to magnetic field of intensity $2000\text{ A/m}$ . The magnetic flux through the rod is $6 \times 10^{-4}\text{ Wb}$ and its cross-sectional area is $3\text{ cm}^2$ . The magnetic permeability of the rod in $\frac{\text{Wb}}{\text{A-m}}$ is

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Answer: (d)

$B = \dfrac{\Phi}{A} = \dfrac{6\times10^{-4}}{3\times10^{-4}} = 2\text{ T}$
$\mu = \dfrac{B}{H} = \dfrac{2}{2000} = 10^{-3}\text{ Wb/A-m}$
So the answer is (d) $10^{-3}$ .

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Magnetisation: B, H, M, χ & μ question from MHT CET 2020 (16 Oct, Shift 1)

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