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NEET PHYSICSEasy

An iron rod of susceptibility 599599 is subjected to a magnetizing field of 1200 A m−11200 \text{ A m}^{-1}. The permeability of the material of the rod is: (μ0=4π×10−7 T m A−1)(\mu_0 = 4\pi \times 10^{-7} \text{ T m A}^{-1})

A

8.0×10−5 T m A−18.0 \times 10^{-5} \text{ T m A}^{-1}

B

2.4π×10−5 T m A−12.4\pi \times 10^{-5} \text{ T m A}^{-1}

C

2.4π×10−7 T m A−12.4\pi \times 10^{-7} \text{ T m A}^{-1}

D

2.4π×10−4 T m A−12.4\pi \times 10^{-4} \text{ T m A}^{-1}

Step-by-Step Solution

  1. Concept: The magnetic permeability (μ\mu) of a material is related to its magnetic susceptibility (χ\chi) and the permeability of free space (μ0\mu_0).
  2. Formula: The relative permeability is given by μr=1+χ\mu_r = 1 + \chi. The absolute permeability is given by μ=μ0μr=μ0(1+χ)\mu = \mu_0 \mu_r = \mu_0 (1 + \chi).
  3. Given Data:
  • Susceptibility, χ=599\chi = 599.
  • Permeability of free space, μ0=4π×10−7 T m A−1\mu_0 = 4\pi \times 10^{-7} \text{ T m A}^{-1}.
  • Magnetizing field H=1200 A m−1H = 1200 \text{ A m}^{-1} (This data is extra and not required to find permeability).
  1. Calculation: μr=1+599=600\mu_r = 1 + 599 = 600 μ=4π×10−7×600\mu = 4\pi \times 10^{-7} \times 600 μ=2400π×10−7\mu = 2400\pi \times 10^{-7} μ=2.4π×10−4 T m A−1\mu = 2.4\pi \times 10^{-4} \text{ T m A}^{-1}
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