MHT CET · Physics · Magnetic Properties of Matter
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
- A \(7 \cdot 2 \times 10^{9} \mathrm{Am}^{-1}\)
- B \(7 \cdot 2 \times 10^{5} \mathrm{Am}^{-1}\)
- C \(7 \cdot 2 \times 10^{12} \mathrm{Am}^{-1}\)
- D \(7 \cdot 2 \times 10^{3} \mathrm{Am}^{-1}\)
Answer & Solution
Correct Answer
(B) \(7 \cdot 2 \times 10^{5} \mathrm{Am}^{-1}\)
Step-by-step Solution
Detailed explanation
(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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