The intensity of magnetization of a bar magne | vedclass

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(A) The intensity of magnetization $I$ is defined as the magnetic moment $M$ per unit volume $V$,i.e.,$I = M/V$.Therefore,the magnetic moment is given

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$0.6$

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(A) The intensity of magnetization $I$ is defined as the magnetic moment $M$ per unit volume $V$,i.e.,$I = M/V$.Therefore,the magnetic moment is given by $M = I 	imes V$.The volume $V$ of the bar magnet is the product of its magnetic length $l$ and its cross-sectional area $A$.Given:$I = 5.0 	imes 10^{4} 	ext{ A m}^{-1}$$l = 12 	ext{ cm} = 0.12 	ext{ m}$$A = 1 	ext{ cm}^{2} = 1 	imes 10^{-4} 	ext{ m}^{2}$Volume $V = l 	imes A = 0.12 	imes 10^{-4} 	ext{ m}^{3} = 1.2 	imes 10^{-5} 	ext{ m}^{3}$.Now,calculating the magnetic moment $M$:$M = (5.0 	imes 10^{4} 	ext{ A m}^{-1}) 	imes (1.2 	imes 10^{-5} 	ext{ m}^{3})$$M = 6.0 	imes 10^{-1} 	ext{ A m}^{2} = 0.6 	ext{ A m}^{2}$.

The intensity of magnetization of a bar magnet is $5.0 \times 10^{4} \text{ A m}^{-1}$. The magnetic length and the area of cross-section of the magnet are $12 \text{ cm}$ and $1 \text{ cm}^{2}$ respectively. The magnitude of the magnetic moment of this bar magnet is (in $SI$ unit):

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