MHT CET · Physics · Magnetic Properties of Matter
A bar magnet has length 4 cm , cross-sectional area \(2 \mathrm{~cm}^2\) and magnetic moment \(6 \mathrm{Am}^2\). The intensity of magnetisation of bar magnet is
- A \(9 \times 10^5 \mathrm{~A} / \mathrm{m}\)
- B \(7.5 \times 10^5 \mathrm{~A} / \mathrm{m}\)
- C \(4.5 \times 10^5 \mathrm{~A} / \mathrm{m}\)
- D \(3.0 \times 10^5 \mathrm{~A} / \mathrm{m}\)
Answer & Solution
Correct Answer
(B) \(7.5 \times 10^5 \mathrm{~A} / \mathrm{m}\)
Step-by-step Solution
Detailed explanation
Intensity of magnetization
\(\begin{aligned}
& =\frac{M_{\text {net }}}{\text { Volume }} \\
& =\frac{M_{\text {net }}}{\text { length } \times \text { area of cross-section }} \\
& =\frac{6}{\left(4 \times 10^{-2}\right) \times\left(2 \times 10^{-4}\right)}=7.5 \times 10^5 \mathrm{~A} / \mathrm{m}
\end{aligned}\)
\(\begin{aligned}
& =\frac{M_{\text {net }}}{\text { Volume }} \\
& =\frac{M_{\text {net }}}{\text { length } \times \text { area of cross-section }} \\
& =\frac{6}{\left(4 \times 10^{-2}\right) \times\left(2 \times 10^{-4}\right)}=7.5 \times 10^5 \mathrm{~A} / \mathrm{m}
\end{aligned}\)
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