Q133 / 150 · MHT CET 2023 (11 May Shift 1) · Physics

A cylindrical magnetic rod has length $5 \mathrm{~cm}$ and diameter $1 \mathrm{~cm}$. It has uniform magnetization $5.3 \times 10^3 \mathrm{~A} / \mathrm{m}^3$. Its net magnetic dipole moment is nearly

  • A
    $1 \times 10^{-2} \mathrm{~J} / \mathrm{T}$
  • B
    $0.5 \times 10^{-2} \mathrm{~J} / \mathrm{T}$
  • C
    $2.5 \times 10^{-2} \mathrm{~J} / \mathrm{T}$
  • Correct answer
    D
    $2.08 \times 10^{-2} \mathrm{~J} / \mathrm{T}$

Solution

$\begin{aligned} & \text { Magnetisation, } M=\frac{\mathrm{m}_{\text {net }}}{\mathrm{V}} \\ & \therefore \quad \mathrm{m}_{\text {net }}=\mathrm{M} \times \mathrm{V}=\mathrm{M} \times\left(\pi \mathrm{r}^2 l\right)=\mathrm{M} \times \pi \times \frac{\mathrm{d}^2}{4} \times l \\ & =5.3 \times 10^3 \times 3.142 \times\left(\frac{1 \times 10^{-2}}{2}\right)^2 \times 5 \times 10^{-2} \\ & =2.08 \times 10^{-2} \mathrm{~J} / \mathrm{T} \\ & \end{aligned}$

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

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