The energy dissipated per unit volume per cyc | vedclass

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(B) $1$. Calculate the volume of the iron sample: $V = \frac{	ext{mass}}{	ext{density}} = \frac{10 \, kg}{7500 \, kg \, m^{-3}} = \frac{1}{750} \, m

Vedclass · Accepted Answer

$48000$

Vedclass · Answer

(B) $1$. Calculate the volume of the iron sample: $V = \frac{	ext{mass}}{	ext{density}} = \frac{10 \, kg}{7500 \, kg \, m^{-3}} = \frac{1}{750} \, m^3$.$2$. Energy dissipated per cycle: $E_{cycle} = (	ext{Energy per unit volume per cycle}) 	imes V = 200 \, J \, m^{-3} \, cycle^{-1} 	imes \frac{1}{750} \, m^3 = \frac{200}{750} \, J \, cycle^{-1} = \frac{4}{15} \, J \, cycle^{-1}$.$3$. Energy dissipated per second (Power loss): $P = E_{cycle} 	imes 	ext{frequency} = \frac{4}{15} \, J \, cycle^{-1} 	imes 50 \, cycle \, s^{-1} = \frac{200}{15} \, J \, s^{-1} = \frac{40}{3} \, J \, s^{-1}$.$4$. Energy dissipated per hour $(3600 \, s)$: $E_{hour} = P 	imes 3600 \, s = \frac{40}{3} 	imes 3600 \, J = 40 	imes 1200 \, J = 48000 \, J$.

The energy dissipated per unit volume per cycle in the hysteresis of an iron sample of mass $10 \, kg$ is $200 \, J \, m^{-3} \, cycle^{-1}$. The density of iron is $7500 \, kg \, m^{-3}$. The loss of energy per hour at $50 \, cycle \, s^{-1}$ is: (in $J$)

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