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Question: Answered & Verified by Expert
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
PhysicsMagnetic Properties of MatterMHT CETMHT CET 2021 (23 Sep Shift 2)
Options:
  • 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 $5 \times 10^5 \mathrm{~A} / \mathrm{m}$
Solution:
2257 Upvotes Verified Answer
The correct answer is: $5 \times 10^5 \mathrm{~A} / \mathrm{m}$
$\begin{aligned} & \mathrm{L}=3 \mathrm{~cm}=3 \times 10^{-2} \mathrm{~m}^2 \\ & \mathrm{~A}=2 \mathrm{~cm}^2=2 \times 10^{-4} \mathrm{~m}^2 \\ & \mathrm{M}=3 \mathrm{Am}^2 \\ & \mathrm{M}_{\mathrm{z}} \rightarrow \text { Intensity of magnetization } \\ & =\frac{\mathrm{M}}{\mathrm{L} \times \mathrm{A}}=\frac{3 \mathrm{Am}^2}{3 \times 2 \times 10^{-6} \mathrm{~m}^3}=\frac{1}{2} \times 10^6 \mathrm{~A} / \mathrm{m}=5 \times 10^5 \mathrm{~A} / \mathrm{m}\end{aligned}$

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