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A compass needle made of pure iron (with density 8000 kg//m^2) has a length 5 cm, width 1.0 mm and thickness 0.50 mm. The magnitude of magnetic dipole moment of an iron is mu_"Fe"=2 xx 10^(-23) J/T. If magnetisation of needle is equivalent to the alignment of 10% of the atoms in the needle, what is the magnitude of the neddle's magnetic dipole momentmu ? (mass of iron per mole =0.05 kg//"mole", N_A=6xx10^23) |
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Answer» `2.4xx10^(-3)` J/T |
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