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NEET-XII-Physics

37: Magnetic Properties of Matter

with Solutions - page 3

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  • #2-b
    If the magnetization I of the core is found to be 0.12 A m-1, find the susceptibility of the material of the rod.
    Ans : Magnetisation of the core, I = 0.12 A/m
    We know:
    `` I=\chi H``,
    where `` \chi `` is the susceptibility of the material of the rod.
    `` \therefore \chi =\frac{I}{H}=\frac{0.12}{1500}``
    `` =0.00008=8\times {10}^{-5}``
  • #2-c
    Is the material paramagnetic, diamagnetic or ferromagnetic?
    Ans : The material is paramagnetic.
    Page No 286:
  • Qstn #3
    The magnetic field inside a long solenoid of 50 turns cm-1 is increased from 2.5 × 10-3 T to 2.5 T when an iron core of cross-sectional area 4 cm2 is inserted into it. Find
    Ans : Given:
    Magnetic field strength without iron core, B1 = 2.5 × 10-3 T
    Magnetic field after introducing the iron core, B2 = 2.5 T
    Area of cross-section of the iron core, A = 4 × 10-4 m2
    Number of turns per unit length, n = 50 turns/cm = 5000 turns/m
  • #3-a
    the current in the solenoid (b) the magnetisation I of the core and (c) the pole strength developed in the core.
    Ans : Magnetic field produced by a solenoid `` \left(B\right)`` is given by,
    `` B={\,\mathrm{\,\mu \,}}_{0}ni``,
    where i = electric current in the solenoid
    2.5 × 10-3 = 4`` \,\mathrm{\,\pi \,}`` × 10-7 × 5000 × i
    `` \Rightarrow i=\frac{2.5\times {10}^{-3}}{4\,\mathrm{\,\pi \,}\times {10}^{-7}\times 5000}``
    `` =0.398\,\mathrm{\,A\,}=0.4\,\mathrm{\,A\,}``
    `` ``
    `` \left(b\right)\,\mathrm{\,Magneti\,}\text{sa}\,\mathrm{\,tion\,}\left(I\right)\,\mathrm{\,is\,}\,\mathrm{\,given\,}\,\mathrm{\,by\,}``
    `` I\mathit{=}\frac{B}{{\mu }_{\mathit{0}}}\mathit{-}H,``
    `` \text{w}\,\mathrm{\,here\,}B\,\mathrm{\,is\,}\,\mathrm{\,the\,}\,\mathrm{\,net\,}\,\mathrm{\,magnetic\,}\,\mathrm{\,field\,}\,\mathrm{\,after\,}\,\mathrm{\,introducing\,}\,\mathrm{\,the\,}\,\mathrm{\,core\,},\,\mathrm{\,i\,}.\,\mathrm{\,e\,}.B=2.5\,\mathrm{\,T\,}.``
    `` \,\mathrm{\,And\,}{\,\mathrm{\,\mu \,}}_{0}H\,\mathrm{\,will\,}\,\mathrm{\,be\,}\,\mathrm{\,the\,}\,\mathrm{\,magnetising\,}\,\mathrm{\,field\,},\,\mathrm{\,i\,}.\,\mathrm{\,e\,}.\,\mathrm{\,the\,}\,\mathrm{\,diffrence\,}\,\mathrm{\,between\,}\,\mathrm{\,the\,}\,\mathrm{\,two\,}\,\mathrm{\,magnetic\,}\,\mathrm{\,fields\,}\text{'}\,\mathrm{\,strengths\,}.``
    `` \Rightarrow I=\frac{2.5\times {10}^{-3}}{4\,\mathrm{\,\pi \,}\times {10}^{-7}}.\left({B}_{2}-{B}_{1}\right)``
    `` \Rightarrow I=\frac{2.5\left(1-{\displaystyle \frac{1}{1000}}\right)}{4\,\mathrm{\,\pi \,}\times {10}^{-7}}``
    `` \Rightarrow I\approx 2\times {10}^{6}\,\mathrm{\,A\,}/\,\mathrm{\,m\,}``
    `` \left(\,\mathrm{\,c\,}\right)\,\mathrm{\,Intensity\,}\,\mathrm{\,of\,}\,\mathrm{\,magneti\,}\text{s}\,\mathrm{\,ation\,}\left(I\right)\,\mathrm{\,is\,}\,\mathrm{\,given\,}\,\mathrm{\,by\,},``
    `` I\mathit{=}\frac{M}{V}``
    `` \mathit{\Rightarrow }I\mathit{=}\frac{m\times 2I}{A\times 2I}\mathit{=}\frac{m}{A}``
    `` \mathit{\Rightarrow }m\mathit{=}lA``
    `` \Rightarrow ``m= 2 × 106 × 4 × 10-4
    `` \Rightarrow ``m= 800 A-m
    Page No 286:
  • Qstn #4
    A bar magnet of length 1 cm and cross-sectional area 1.0 cm2 produces a magnetic field of 1.5 × 10-4 T at a point in end-on position at a distance 15 cm away from the centre. (a) Find the magnetic moment M of the magnet. (b) Find the magnetisation I of the magnet. (c) Find the magnetic field B at the centre of the magnet.
    digAnsr:   b
    Ans : Given:
    Distance of the observation point from the centre of the bar magnet, d = 15 cm = 0.15 m
    Length of the bar magnet, l = 1 cm = 0.01 m
    Area of cross-section of the bar magnet, A = 1.0 cm2 = 1 × 10-4 m2
    Magnetic field strength of the bar magnet, B = 1.5 × 10-4 T
    As the observation point lies at the end-on position, magnetic field `` \left(B\right)`` is given by,
    `` \stackrel{\to }{B}=\frac{{\,\mathrm{\,\mu \,}}_{0}}{4\,\mathrm{\,\pi \,}}\times \frac{2Md}{({d}^{2}-{l}^{2}{)}^{2}}``
    On substituting the respective values, we get:
    `` 1.5\times {10}^{-4}=\frac{{10}^{-7}\times 2\times M\times 0.15}{(0.0225-0.0001{)}^{2}}``
    `` \Rightarrow 1.5\times {10}^{-4}=\frac{3\times {10}^{-8}\times M}{5.01\times {10}^{-4}}``
    `` \Rightarrow M=\frac{1.5\times {10}^{-4}\times 5.01\times {10}^{-4}}{3\times {10}^{-8}}``
    `` =2.5\,\mathrm{\,A\,}``
    (b) Intensity of magnetisation (I) is given by,
    I = `` \frac{M}{V}``
    `` =\frac{2.5}{{10}^{-4}\times {10}^{-2}}``
    `` =2.5\times {10}^{6}\,\mathrm{\,A\,}/\,\mathrm{\,m\,}``
    (c)
    `` \,\mathrm{\,H\,}=\frac{\,\mathrm{\,M\,}}{4\,\mathrm{\,\pi \,}l{d}^{2}}``
    `` =\frac{2.5}{4\times 3.14\times 0.01\times (0.15{)}^{2}}``
    `` =\frac{2.5}{4\times 3.14\times 1\times {10}^{-2}\times 2.25\times {10}^{-2}}``
    Net H = HN + HS
    = 884.6 = 8.846 × 102
    = 314 T
    `` \stackrel{\to }{\,\mathrm{\,B\,}}`` = µ0 (H + 1)
    = π × 10-7 (2.5 × 106 + 2 × 884.6)
    = 3.14 T
    Page No 286:
  • Qstn #5
    The susceptibility of annealed iron at saturation is 5500. Find the permeability of annealed iron at saturation.
    Ans : Susceptibility of annealed iron, `` \chi `` = 5500
    The relation between permeability and susceptibility:
    Permeability, µ = µ0(1 + x)
    µ = 4`` \,\mathrm{\,\pi \,}`` × 10-7 (1 + 5500)
    `` \Rightarrow `` µ = 4 × 3.14 × 10-7 × 5501
    `` \Rightarrow ``µ = 69092.56 × 10-7
    `` \Rightarrow ``µ = 6.9 × 10-3
    Page No 286:
  • Qstn #6
    The magnetic field B and the magnetic intensity H in a material are found to be 1.6 T and 1000 A m-1, respectively. Calculate the relative permeability µr and the susceptibility χ of the material.
    Ans : Here,
    Magnetic field strength, B = 1.6 T
    Magnetising intensity in a material, H = 1000 A/m
    The relation between magnetic field and magnetising field is given by
    `` \mu \mathit{=}\frac{B}{H}``
    `` \Rightarrow \mu =\frac{1.6}{1000}``
    `` \Rightarrow \mu =1.6\times {10}^{-3}``
    `` \text{R}\,\mathrm{\,elative\,}\,\mathrm{\,perm\,}\text{ea}\,\mathrm{\,bility\,}\,\mathrm{\,is\,}\,\mathrm{\,defined\,}\,\mathrm{\,as\,}\,\mathrm{\,the\,}\,\mathrm{\,ratio\,}\,\mathrm{\,of\,}\,\mathrm{\,permeability\,}\,\mathrm{\,in\,}\text{a}\,\mathrm{\,medium\,}\,\mathrm{\,to\,}\,\mathrm{\,that\,}\,\mathrm{\,in\,}\,\mathrm{\,vacuum\,}.``
    `` \text{So,}{\mu }_{r}=\frac{\mu }{{\mu }_{0}}=\frac{1.6\times {10}^{-3}}{4\,\mathrm{\,\pi \,}\times {10}^{-7}}``
    `` \Rightarrow {\mu }_{r}=0.127\times {10}^{4}``
    `` \Rightarrow {\mu }_{r}=1.3\times {10}^{3}``
    `` \,\mathrm{\,Relative\,}\,\mathrm{\,permeability\,}\left({\mu }_{r}\right)\,\mathrm{\,is\,}\,\mathrm{\,given\,}\,\mathrm{\,by\,},``
    `` {\mu }_{r}=(1+\chi )``
    `` \Rightarrow \chi =1.3\times {10}^{3}-1``
    `` \Rightarrow \chi =1300-1``
    `` \Rightarrow \chi =1299=1.299\times {10}^{3}``
    `` \Rightarrow \chi \approx 1.3\times {10}^{3}``
    Page No 287:
  • Qstn #7
    The susceptibility of magnesium at 300 K is 1.2 × 10-5. At what temperature will the susceptibility increase to 1.8 × 10-5?
    Ans : Given,
    Susceptibility of magnesium at 300 K, `` {\chi }_{1}`` = 1.2 × 10-5
    Let T1 be the temperature at which susceptibility of magnesium is 1.2 × 10-5 and T2 be the temperature at which susceptibility of magnesium is 1.8 × 10-5.
    According to Curie's law,
    `` \chi =\frac{\,\mathrm{\,C\,}}{T},``
    `` ``
    where C is Curie's constant.
    `` \Rightarrow \frac{{\chi }_{1}}{{\chi }_{2}}=\frac{{T}_{2}}{{T}_{1}}``
    `` \Rightarrow \frac{1.2\times {10}^{-5}}{1.8\times {10}^{-5}}=\frac{{T}_{2}}{300}``
    `` \Rightarrow {T}_{2}=\frac{12}{18}\times 300``
    `` \Rightarrow {T}_{2}=\frac{2}{3}\times 300=200\,\mathrm{\,K\,}``
    Page No 287:
  • 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.
    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
  • #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:
  • Qstn #9
    The coercive force for a certain permanent magnet is 4.0 × 104 A m-1. This magnet is placed inside a long solenoid of 40 turns/cm and a current is passed in the solenoid to demagnetise it completely. Find the current.
    Ans : Given:
    Number of turns per unit length, n = 40 turns/cm = 4000 turns/m
    Magnetising field, H = 4 × 104 A/m
    Magnetic field inside a solenoid (B) is given by,
    B = µ0nI,
    where, n = number of turns per unit length.
    I = current through the solenoid.
    `` \therefore \frac{B}{{\mu }_{\mathit{0}}}\mathit{=}nI\mathit{=}H``
    `` \Rightarrow H\mathit{=}\frac{N}{l}I``
    `` \mathit{\Rightarrow }I\mathit{=}\frac{Hl}{N}\mathit{=}\frac{H}{n}``
    `` \Rightarrow I=\frac{4\times {10}^{4}}{4000}=10\,\mathrm{\,A\,}``
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