Q 12-05-032JEE MainJEE Main 2024 (29 Jan, Shift 1)Easy
The magnetic potential due to a magnetic dipole at a point on its axis situated at a distance of $20\ \text{cm}$ from its centre is $1.5\times10^{-5}\ \text{T m}$. The magnetic moment of the dipole is ______ $\text{A m}^2$. (Given: $\dfrac{\mu_0}{4\pi} = 10^{-7}\ \text{T m A}^{-1}$)
Numerical value type. Enter your answer.
Answer: 6
On the axis of a short magnetic dipole:
$$V = \frac{\mu_0}{4\pi}\frac{M}{r^2} \;\Rightarrow\; M = \frac{1.5\times10^{-5}\times(0.2)^2}{10^{-7}} = 6\ \text{A m}^2$$
Solution by Sreeraj P, M.Sc Physics