The magnetic potential due to a magnetic dipole at a point on its axis situated at a distance of 20 cm from its center is 1.5 × 10⁻⁵ T m. The magnetic moment of the dipole is _________ A m². (Given: (μ₀)/(4π) = 10⁻⁷ TmA⁻¹) Numerical answer type. Work it out, then open the answer.

Answer: 6. Magnetic potential on the axis: V = (μ₀)/(4π)M/r². M = (V r²)/(μ₀/4π) = 1.5 × 10⁻⁵ × (0.2)²10⁻⁷ = 6 So the magnetic moment is 6 A m².

The magnetic potential due to a magnetic dipole at a point on its… | JEE Main 2024 (29 Jan, Shift 1)

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Field of a Magnet & Magnet in a Uniform Field

The magnetic potential due to a magnetic dipole at a point on its axis situated at a distance of $20\text{ cm}$ from its center is $1.5 \times 10^{-5}\text{ T m}$ . The magnetic moment of the dipole is _________ $\text{A m}^2$ . (Given: $\frac{\mu_0}{4\pi} = 10^{-7}\text{ TmA}^{-1}$ )

Numerical answer type. Work it out, then open the answer.

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Answer: 6

Magnetic potential on the axis: $V = \dfrac{\mu_0}{4\pi}\dfrac{M}{r^2}$ .
$M = \dfrac{V r^2}{\mu_0/4\pi} = \dfrac{1.5 \times 10^{-5} \times (0.2)^2}{10^{-7}} = 6$
So the magnetic moment is $6\ \text{A m}^2$ .

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