MHT CET · Physics · Magnetic Properties of Matter
A bar magnet has length \(3 \mathrm{~cm}\), cross-sectional area \(2 \mathrm{~cm}^2\) and magnetic moment \(3 \mathrm{Am}^2\). The intensity of magnetization of bar magnet is
- A \(2 \times 10^5 \mathrm{~A} / \mathrm{m}\)
- B \(3 \times 10^5 \mathrm{~A} / \mathrm{m}\)
- C \(4 \times 10^5 \mathrm{~A} / \mathrm{m}\)
- D \(5 \times 10^5 \mathrm{~A} / \mathrm{m}\)
Answer & Solution
Correct Answer
(D) \(5 \times 10^5 \mathrm{~A} / \mathrm{m}\)
Step-by-step Solution
Detailed explanation
\(\mathrm{L}=3 \mathrm{~cm}=3 \times 10^{-2} \mathrm{~m}^2 \)
\( \mathrm{~A}=2 \mathrm{~cm}^2=2 \times 10^{-4} \mathrm{~m}^2 \)
\( \mathrm{M}=3 \mathrm{Am}^2 \)
\( \mathrm{M}_{\mathrm{z}} \rightarrow \text { Intensity of magnetization } \)
\( =\frac{\mathrm{M}}{\mathrm{L} \times \mathrm{A}}=\frac{3 \mathrm{Am}^2}{3 \times 2 \times 10^{-6} \mathrm{~m}^3}=\frac{1}{2} \times 10^6\)\(\mathrm{~A} / \mathrm{m}=5 \times 10^5 \mathrm{~A} / \mathrm{m}\)
\( \mathrm{~A}=2 \mathrm{~cm}^2=2 \times 10^{-4} \mathrm{~m}^2 \)
\( \mathrm{M}=3 \mathrm{Am}^2 \)
\( \mathrm{M}_{\mathrm{z}} \rightarrow \text { Intensity of magnetization } \)
\( =\frac{\mathrm{M}}{\mathrm{L} \times \mathrm{A}}=\frac{3 \mathrm{Am}^2}{3 \times 2 \times 10^{-6} \mathrm{~m}^3}=\frac{1}{2} \times 10^6\)\(\mathrm{~A} / \mathrm{m}=5 \times 10^5 \mathrm{~A} / \mathrm{m}\)
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