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(C) The energy dissipated per unit volume per cycle is equal to the area of the hysteresis loop. Total energy dissipated per second is given by the fo

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$1.19 	imes 10^{-3} \, 	ext{cal}$

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(C) The energy dissipated per unit volume per cycle is equal to the area of the hysteresis loop. Total energy dissipated per second is given by the formula: $E = V 	imes A 	imes n 	imes t$. Given: Volume $V = 10^{-3} \, m^3$ Area of hysteresis loop $A = 0.1 \, J/m^3$ Frequency $n = 50 \, Hz$ Time $t = 1 \, 	ext{sec}$ Substituting the values: $E = 10^{-3} 	imes 0.1 	imes 50 	imes 1 = 5 	imes 10^{-3} \, J$. To convert Joules to calories, we use the conversion factor $1 \, 	ext{cal} \approx 4.2 \, J$, so $1 \, J \approx 0.2388 \, 	ext{cal}$. $E = 5 	imes 10^{-3} 	imes 0.2388 \approx 1.194 	imes 10^{-3} \, 	ext{cal}$. Thus, the energy dissipated is $1.19 	imes 10^{-3} \, 	ext{cal}$.

$A$ ferromagnetic material of volume $10^{-3} \, m^3$ is subjected to a magnetic field with a frequency of $50 \, Hz$. The area of the hysteresis loop is $0.1 \, M.K.S. \, \text{units}$. Calculate the energy dissipated in $1 \, \text{sec}$.

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