A cylindrical magnetic rod has length 5 cm and diameter 1 cm. It has uniform magnetization 5.3 × 10³ A/m. Its net magnetic dipole moment is nearly (a) 1 × 10⁻² J/T (b) 0.5 × 10⁻² J/T (c) 2.5 × 10⁻² J/T (d) 2 × 10⁻² J/T

Answer: (d). Dipole moment m = M × V, with V = π r² l. V = π (0.5 × 10⁻²)² × 5 × 10⁻² ≈ 3.93 × 10⁻⁶ m³ m = 5.3 × 10³ × 3.93 × 10⁻⁶ ≈ 2.08 × 10⁻² J/T So the answer is (d) 2 × 10⁻² J/T.

A cylindrical magnetic rod has length 5 cm and diameter 1 cm. It has… | MHT CET 2023 (11 May, Shift 1)

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Other: Induction & Displacement Current

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

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Answer: (d)

Dipole moment $m = M \times V$ , with $V = \pi r^2 l$ .
$V = \pi (0.5 \times 10^{-2})^2 \times 5 \times 10^{-2} \approx 3.93 \times 10^{-6}\text{ m}^3$
$m = 5.3 \times 10^{3} \times 3.93 \times 10^{-6} \approx 2.08 \times 10^{-2}$ J/T
So the answer is (d) $2 \times 10^{-2}$ J/T.

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