Q120 / 150 · MHT CET 2021 (23 Sep Shift 2) · Physics

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
    $4 \times 10^5 \mathrm{~A} / \mathrm{m}$
  • Correct answer
    D
    $5 \times 10^5 \mathrm{~A} / \mathrm{m}$

Solution

$\begin{aligned} & \mathrm{L}=3 \mathrm{~cm}=3 \times 10^{-2} \mathrm{~m}^2 \\ & \mathrm{~A}=2 \mathrm{~cm}^2=2 \times 10^{-4} \mathrm{~m}^2 \\ & \mathrm{M}=3 \mathrm{Am}^2 \\ & \mathrm{M}_{\mathrm{z}} \rightarrow \text { Intensity of magnetization } \\ & =\frac{\mathrm{M}}{\mathrm{L} \times \mathrm{A}}=\frac{3 \mathrm{Am}^2}{3 \times 2 \times 10^{-6} \mathrm{~m}^3}=\frac{1}{2} \times 10^6 \mathrm{~A} / \mathrm{m}=5 \times 10^5 \mathrm{~A} / \mathrm{m}\end{aligned}$

Hence, the correct answer is option: 'D'.

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