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MHT CET 2026 (13 Apr, Shift 2) · Magnetisation: B, H, M, χ & μ

Magnetisation: B, H, M, χ & μ question from MHT CET 2026 (13 Apr, Shift 2)

Q12MHT CET 2026 · 13 APR · SHIFT 2Difficult

Magnetisation: B, H, M, χ & μ

A rod with a circular cross-section area 22 cm2{}^{2} and length 4040 cm is wound uniformly with 400400 turns of an insulated wire. If a current 0.40.4 A flows in the wire winding, the total magnetic flux produced inside the winding is 4π×10−64\pi \times 10^{-6} Wb. The relative permeability of the rod is

(Given permeability of vacuum μ0=4π×10−7\mu_0 = 4\pi \times 10^{-7} N m A−2{}^{-2})

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Answer: (c)

  1. B=ΦA=4π×10−62×10−4=2π×10−2 TB = \dfrac{\Phi}{A} = \dfrac{4\pi \times 10^{-6}}{2 \times 10^{-4}} = 2\pi \times 10^{-2}\ \text{T}
  2. n=4000.4=1000 m−1n = \dfrac{400}{0.4} = 1000\ \text{m}^{-1}, so μ0nI=4π×10−7×1000×0.4=1.6π×10−4 T\mu_0 nI = 4\pi \times 10^{-7} \times 1000 \times 0.4 = 1.6\pi \times 10^{-4}\ \text{T}
  3. μr=Bμ0nI=2π×10−21.6π×10−4=125\mu_r = \dfrac{B}{\mu_0 nI} = \dfrac{2\pi \times 10^{-2}}{1.6\pi \times 10^{-4}} = 125
  4. So the answer is (c) 125.
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