NEET-XII-Physics
37: Magnetic Properties of Matter
- #4A 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: bAns : 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
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