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Q.
A bar magnet has length $3 \,cm$, cross-sectional area $2 \, cm^{2}$ and magnetic moment $3 \, Am^2$. The intensity of magnetisation of bar magnet is
Length of the bar magnet $L =3 \,cm =3 \times 10^{-2} \,m$
Area of bar magnet $A =2\, cm ^{2}=2 \times 10^{-4}\, m ^{2}$
Thus volume of bar magnet $V = AL$
$\therefore V =3 \times 10^{-2} \times 3 \times 10^{-4}=6 \times 10^{-6} \,m ^{3}$
Magnetic moment $\mu=3 \,Am ^{2}$
Thus intensity of magnetisation $I=\frac{\mu}{V}$
$\therefore I =\frac{3}{6 \times 10^{-6}}=5 \times 10^{5} \,A / m$