A bar magnet has length 4 text{ cm},cross-sec | vedclass

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(B) The intensity of magnetisation $(I)$ is defined as the magnetic moment per unit volume of the material.Formula: $I = \frac{M}{V} = \frac{M}{A 	im

Vedclass · Accepted Answer

$7.5 	imes 10^5 	ext{ A/m}$

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(B) The intensity of magnetisation $(I)$ is defined as the magnetic moment per unit volume of the material.Formula: $I = \frac{M}{V} = \frac{M}{A 	imes L}$Given:Magnetic moment $(M)$ = $6 	ext{ Am}^2$Length $(L)$ = $4 	ext{ cm} = 4 	imes 10^{-2} 	ext{ m}$Area $(A)$ = $2 	ext{ cm}^2 = 2 	imes 10^{-4} 	ext{ m}^2$Calculation:Volume $(V)$ = $A 	imes L = (2 	imes 10^{-4} 	ext{ m}^2) 	imes (4 	imes 10^{-2} 	ext{ m}) = 8 	imes 10^{-6} 	ext{ m}^3$$I = \frac{6 	ext{ Am}^2}{8 	imes 10^{-6} 	ext{ m}^3} = 0.75 	imes 10^6 	ext{ A/m} = 7.5 	imes 10^5 	ext{ A/m}$

$A$ bar magnet has length $4 \text{ cm}$,cross-sectional area $2 \text{ cm}^2$ and magnetic moment $6 \text{ Am}^2$. The intensity of magnetisation of the bar magnet is

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