A cylindrical magnetic rod has length $5 \mathrm{~cm}$ and diameter $1 \mathrm{~cm}$. It has uniform…

A cylindrical magnetic rod has length $5 \mathrm{~cm}$ and diameter $1 \mathrm{~cm}$. It has uniform magnetization $5 \cdot 3 \times 10^{3} \frac{\mathrm{A}}{\mathrm{m}^{3}}$. Its net magnetic dipole moment is nearly $\left(\pi=\frac{22}{7}\right)$
  1. $2 \cdot 5 \times 10^{-2} \frac{\mathrm{J}}{\mathrm{T}}$
  2. $0 \cdot 5 \times 10^{-2} \frac{\mathrm{J}}{\mathrm{T}}$
  3. $2 \times 10^{-2} \frac{\mathrm{J}}{\mathrm{T}}$
  4. $10^{-2} \frac{\mathrm{J}}{\mathrm{T}}$

Solution

$\begin{aligned} & \mathrm{M}=\mathrm{I} \times \mathrm{V} \text { where volume of cylinder } \mathrm{v}=\pi \mathrm{r}^2 \mathrm{l} \\ & \mathrm{M}=\mathrm{I} \pi r^2 \mathrm{l}=5.3 \times 10^3 \times \frac{2}{2} 7 \times\left(0.5 \times 10^{-2}\right)^2 \times\left(5 \times 10^{-2}\right) \\ & 2.08 \times 10^{-2} \mathrm{~J} / \mathrm{T}\end{aligned}$ /

Asked in: MHT CET 2020 (15 Oct Shift 2)

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