A bar magnet has length 3 cm, cross-sectional area $2\ \text{cm}^2$ and magnetic moment $3\ \text{A m}^2$. The intensity of magnetisation of the bar magnet is
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An alternative equivalent definition for the intensity of magnetization is pole strength per unit cross-sectional area ($I = \frac{m_{\text{pole}}}{A}$). Both approaches yield identical results, but always make sure to convert metric prefixes like $\text{cm}^2$ carefully: remember that $1\ \text{cm}^2 = 10^{-4}\ \text{m}^2$, not $10^{-2}\ \text{m}^2$.
Step 1: Define intensity of magnetisation. The intensity of magnetisation $I$ is the magnetic moment packed into each unit of volume: $$I = \frac{M}{V}.$$ Step 2: Express the volume of the bar. For a uniform bar magnet, $$V = A \times L,$$ where $A$ is the cross-sectional area and $L$ is the length. Step 3: Convert the data to SI units. $M = 3\ \text{A m}^2$, $L = 3\ \text{cm} = 3\times 10^{-2}\ \text{m}$, and $A = 2\ \text{cm}^2 = 2\times 10^{-4}\ \text{m}^2.$ Step 4: Compute the volume. $$V = (2\times 10^{-4})(3\times 10^{-2}) = 6\times 10^{-6}\ \text{m}^3.$$ Step 5: Divide moment by volume. $$I = \frac{3}{6\times 10^{-6}}.$$ Step 6: Simplify the result. $$I = 0.5\times 10^{6} = 5\times 10^{5}\ \text{A m}^{-1}.$$ \[ \boxed{I = 5\times 10^{5}\ \text{A m}^{-1}} \]