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NEET-XII-Physics

37: Magnetic Properties of Matter

with Solutions - page 3
Qstn# iv-8 Prvs-QstnNext-Qstn
  • #8
    Assume that each iron atom has a permanent magnetic moment equal to 2 Bohr magnetons (1 Bohr magneton equals 9.27 × 10-24 A m2). The density of atoms in iron is 8.52 × 1028 atoms m-3. (a) Find the maximum magnetisation I in a long cylinder of iron (b) Find the maximum magnetic field B on the axis inside the cylinder.
    Ans : Given:
    No of atoms per unit volume, f = 8.52 × 1028 atoms/m3
    Magnetisation per atom, M = 2 × 9.27 × 10-24 A-m2 (a) Intensity of magnetisation, I = `` \frac{M}{V}``
    `` \Rightarrow ``I = 2 × 9.27 × 10-24 × 8.52 × 1028
    `` \Rightarrow ``I = 1.58 × 106 A/m. (b) For maximum magnetisation ,the magnetising field will be equal to the intensity of magnetisation.
    So, I = H
    Magnetic field (B) will be,
    B = 4`` \,\mathrm{\,\pi \,}`` × 10-7 × 1.58 × 106
    `` \Rightarrow ``B ≈ 19.8 × 10-1 = 2.0 T.
    Page No 287:
  • #8-a
    Find the maximum magnetisation I in a long cylinder of iron
    Ans : Intensity of magnetisation, I = `` \frac{M}{V}``
    `` \Rightarrow ``I = 2 × 9.27 × 10-24 × 8.52 × 1028
    `` \Rightarrow ``I = 1.58 × 106 A/m.
  • #8-b
    Find the maximum magnetic field B on the axis inside the cylinder.
    Ans : For maximum magnetisation ,the magnetising field will be equal to the intensity of magnetisation.
    So, I = H
    Magnetic field (B) will be,
    B = 4`` \,\mathrm{\,\pi \,}`` × 10-7 × 1.58 × 106
    `` \Rightarrow ``B ≈ 19.8 × 10-1 = 2.0 T.
    Page No 287: