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MHT CET 2026 (14 May, Shift 2) · Magnetisation: B, H, M, χ & μ

Magnetisation: B, H, M, χ & μ question from MHT CET 2026 (14 May, Shift 2)

Q8MHT CET 2026 · 14 MAY · SHIFT 2Medium

Magnetisation: B, H, M, χ & μ

A paramagnetic substance in the form of a cube with sides 11 cm has a magnetic dipole moment 20×10−620 \times 10^{-6} J/T when a magnetic intensity 60×10360 \times 10^{3} A/m is applied. Its magnetic susceptibility is

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Answer: (a)

  1. Magnetisation M=mV=20×10−6(10−2)3=20 A m−1M = \dfrac{m}{V} = \dfrac{20 \times 10^{-6}}{(10^{-2})^3} = 20\ \text{A m}^{-1}
  2. Susceptibility χ=MH=2060×103=3.3×10−4\chi = \dfrac{M}{H} = \dfrac{20}{60 \times 10^{3}} = 3.3 \times 10^{-4}
  3. So the answer is (a) 3.3×10−43.3 \times 10^{-4}.
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