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 c m 2 . The magnetic permeability of the rod in W b A - 1 m - 1 is

μ = ϕ A × 1 H = 6 × 10 − 4 3 × 10 − 4 × 1 2 × 10 3 = 10 − 3 Wb A - 1 m - 1

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Physics - Magnetism Question with Solution | TestHub

PhysicsMagnetismMagnetic properties of matterEasy2 minPYQ_2020

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An iron rod is placed parallel to magnetic field of intensity $2000 A m$ . The magnetic flux through the rod is $6 \times 10^{- 4} Wb$ and its cross-sectional area is $3 c m^{2}$ . The magnetic permeability of the rod in $W b A^{- 1} m^{- 1}$ is

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Solution:

$μ = \frac{ϕ}{A} \times \frac{1}{H} = \frac{6 \times 10^{- 4}}{3 \times 10^{- 4}} \times \frac{1}{2 \times 10^{3}} = 10^{- 3} Wb A^{- 1} m^{- 1}$

Stream:NTA_ABHYAS_NEETSubject:PhysicsTopic:MagnetismSubtopic:Magnetic properties of matter

⏱ 2mℹ️ Source: PYQ_2020

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