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JEE Main 2026 (4 Apr, Shift 2) · Magnetisation: B, H, M, χ & μ

Magnetisation: B, H, M, χ & μ question from JEE Main 2026 (4 Apr, Shift 2)

Q7JEE Main 2026 · 4 APR · SHIFT 2Easy

Magnetisation: B, H, M, χ & μ

A solenoid has a core made of material with relative permeability 400. The magnetic field produced in the interior of solenoid is 1.0 T. The magnetic intensity in SI units is α×105\alpha \times 10^{5}. The value of α\alpha is ______.

(Free space permeability μ0=4π×10−7\mu_0 = 4\pi \times 10^{-7} SI units.)

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Answer: (b)

  1. Inside the core B=μ0μrHB = \mu_0 \mu_r H, so H=Bμ0μrH = \dfrac{B}{\mu_0 \mu_r}.
  2. H=1.04π×10−7×400=1041.6π=116π×105 A m−1H = \dfrac{1.0}{4\pi \times 10^{-7} \times 400} = \dfrac{10^{4}}{1.6\pi} = \dfrac{1}{16\pi} \times 10^{5}\ \text{A m}^{-1}
  3. So the answer is (b) α=116π\alpha = \dfrac{1}{16\pi}.
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