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JEE Mains · Physics · STD 12 - Magnetism and matter
  • JEE Mains 2024
  • Hard
  • Numerical

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

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Answer & Solution

Correct Answer

(A) \(6\)

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Detailed explanation

\( \mathrm{V}=\frac{\mu_0}{4 \pi} \frac{\mathrm{M}}{\mathrm{r}^2} \)

\( \Rightarrow 1.5 \times 10^{-5}=10^{-7} \times \frac{\mathrm{M}}{\left(20 \times 10^{-2}\right)^2} \)

\( \Rightarrow \mathrm{M}=\frac{1.5 \times 10^{-5} \times 20 \times 20 \times 10^{-4}}{10^{-7}} \)

\( \mathrm{M}=1.5 \times 4=6\)

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