The area of the hysteresis loop of a material | vedclass

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(A) The energy loss per unit volume per cycle is equal to the area of the hysteresis loop, which is $250 \, J/m^3$ (assuming the given value is per un

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$6 	imes 10^4 \, J$

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(A) The energy loss per unit volume per cycle is equal to the area of the hysteresis loop, which is $250 \, J/m^3$ (assuming the given value is per unit volume). Total energy loss $E = (	ext{Area}) 	imes (	ext{Volume}) 	imes (	ext{Frequency}) 	imes (	ext{Time})$. Volume $V = \frac{	ext{mass}}{	ext{density}} = \frac{10 \, kg}{7.5 \, g/cm^3} = \frac{10 	imes 10^3 \, g}{7.5 \, g/cm^3} = \frac{10^4}{7.5} \, cm^3 = \frac{10^4}{7.5} 	imes 10^{-6} \, m^3 = \frac{1}{750} \, m^3$. Frequency $n = 50 \, Hz$. Time $t = 1 \, hour = 3600 \, s$. $E = 250 	imes \frac{1}{750} 	imes 50 	imes 3600 = \frac{250}{750} 	imes 50 	imes 3600 = \frac{1}{3} 	imes 50 	imes 3600 = 50 	imes 1200 = 6 	imes 10^4 \, J$.

The area of the hysteresis loop of a material is equivalent to $250 \, J$. When $10 \, kg$ of the material is magnetized by an alternating field of $50 \, Hz$, what is the energy lost in one hour? (Given: density of the material is $7.5 \, g/cm^3$)

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