The permeability of a metal is 0.1256 TmA⁻¹. Its relative permeability will be ((μ₀)/(4π) = 10⁻⁷ SI unit) ( π = 3.14) (a) 10⁵ (b) 3 × 10⁵ (c) 2 × 10⁶ (d) 10⁴

Answer: (a). μr = (μ)/(μ₀), with μ₀ = 4π×10⁻⁷ = 1.256×10⁻⁶. μr = 0.12561.256×10⁻⁶ = 10⁵ So the answer is (a) 10⁵.

The permeability of a metal is 0.1256 TmA⁻¹. Its relative… | MHT CET 2021 (22 Sep, Shift 1)

Q36MHT CET 2021 · 22 SEP · SHIFT 1Easy

Magnetisation: B, H, M, χ & μ

The permeability of a metal is $0.1256\text{ TmA}^{-1}$ . Its relative permeability will be $\left(\frac{\mu_0}{4\pi} = 10^{-7}\text{ SI unit}\right)$ ( $\pi = 3.14$ )

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Answer: (a)

$\mu_r = \dfrac{\mu}{\mu_0}$ , with $\mu_0 = 4\pi\times10^{-7} = 1.256\times10^{-6}$ .
$\mu_r = \dfrac{0.1256}{1.256\times10^{-6}} = 10^{5}$
So the answer is (a) $10^5$ .

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Magnetisation: B, H, M, χ & μ question from MHT CET 2021 (22 Sep, Shift 1)

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