Q 12-05-063JEE MainJEE Main 2020 (5 Sep, Shift 2)Medium
An iron rod of volume $10^{-3}\ \text{m}^3$ and relative permeability $1000$ is placed as core in a solenoid with $10$ turns cm$^{-1}$. If a current of $0.5$ A is passed through the solenoid, then the magnetic moment of the rod will be:
Answer: (B) $5\times10^2\ \text{A m}^2$
$n = 10\ \text{cm}^{-1} = 1000\ \text{m}^{-1}$, so $H = nI = 500$ A/m.
Magnetisation $M = \chi H = (\mu_r - 1)H = 999\times500 \approx 5\times10^5$ A/m.
Magnetic moment $= M\times V = 5\times10^5\times10^{-3} = 5\times10^2\ \text{A m}^2$.
Solution by Sreeraj P, M.Sc Physics