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A domain in a ferromagnetic substance is in the form of a cube of side $1 \mu \mathrm{m}$. If it
contains $8 \times 10^{10}$ atoms and each atomic dipole has a dipole moment of $9 \times 10^{-24}$
$\mathrm{A} / \mathrm{m}^{2}$, then the magnetisation of the domain is
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contains $8 \times 10^{10}$ atoms and each atomic dipole has a dipole moment of $9 \times 10^{-24}$
$\mathrm{A} / \mathrm{m}^{2}$, then the magnetisation of the domain is
Solution:
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Verified Answer
The correct answer is:
$7 \cdot 2 \times 10^{5} \mathrm{Am}^{-1}$
(A)
Volume of cube $=\ell^{3}=\left(10^{-6}\right)^{3}=10^{-18} \mathrm{~m}^{3}$
Total magnetic moment $\mathrm{M}=8 \times 10^{10} \times 9 \times 10^{-24}$
$=72 \times 10^{-14} \mathrm{~A}-\mathrm{m}^{2}$
Magnitization $=\frac{\mathrm{M}}{\mathrm{V}}=\frac{72 \times 10^{-14}}{10^{-18}}=7.2 \times 10^{5} \mathrm{~A} / \mathrm{m}$
Volume of cube $=\ell^{3}=\left(10^{-6}\right)^{3}=10^{-18} \mathrm{~m}^{3}$
Total magnetic moment $\mathrm{M}=8 \times 10^{10} \times 9 \times 10^{-24}$
$=72 \times 10^{-14} \mathrm{~A}-\mathrm{m}^{2}$
Magnitization $=\frac{\mathrm{M}}{\mathrm{V}}=\frac{72 \times 10^{-14}}{10^{-18}}=7.2 \times 10^{5} \mathrm{~A} / \mathrm{m}$
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