A bar magnet having length $5 \mathrm{~cm}$ and area of cross-section $4 \mathrm{~cm}^{2}$ has magnetic…

A bar magnet having length $5 \mathrm{~cm}$ and area of cross-section $4 \mathrm{~cm}^{2}$ has magnetic moment $2 \mathrm{Am}^{2}$. If magnetic susceptibility is $5 \times 10^{-6}$, the magnetic intensity will be
  1. $0 \cdot 2 \times 10^{10} \frac{\mathrm{A}}{\mathrm{m}}$
  2. $0.5 \times 10^{10} \frac{\mathrm{A}}{\mathrm{m}}$
  3. $5 \times 10^{10} \frac{\mathrm{A}}{\mathrm{m}}$
  4. $2 \times 10^{10} \frac{\mathrm{A}}{\mathrm{m}}$

Solution

$\begin{aligned} \text { Volume of the magnet } \mathrm{V} &=\ell \times \mathrm{A} \\ &=5 \mathrm{~cm} \times 4 \mathrm{~cm}^{2}=20 \mathrm{~cm}^{3} \\ &=20 \times 10^{-6} \mathrm{~m}^{3}=2 \times 10^{-5} \mathrm{~m}^{3} \\ \text { magnetic moment }=2 & \mathrm{Am}^{2} \\ \therefore \text { Magnetization } \mathrm{M} &=\frac{2 \times 10^{-5}}{2}=10^{-5} \mathrm{~A} / \mathrm{m} \\ \text { Magnetic intensity } \mathrm{H} &=\frac{\mathrm{M}}{\chi}=\frac{10^{-5}}{5 \times 10^{-6}}=2 \times 10^{10} \mathrm{~A} / \mathrm{m} \end{aligned}$

Asked in: MHT CET 2020 (13 Oct Shift 2)

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