The area of hysteresis loop of a material is equivalent to 250 J / m 3. When 10 kg material is magnestized by an alternating field of 50 Hz, then energy lost in one hour will be β joule. Find (β/1000). (Density of material is 7.5 g / cm 3 )

Question

The area of hysteresis loop of a material is equivalent to 250 J / m 3. When 10 kg material is magnestized by an alternating field of 50 Hz, then energy lost in one hour will be β joule. Find (β/1000). (Density of material is 7.5 g / cm 3 ) (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

Energy lost in 1 hour = Area × Volume × Time × Frequency =(250 × 10 × 3600 × 50/7500) =6 × 104 J

Q. The area of hysteresis loop of a material is equivalent to $250 J / m^{3}$ . When $10 k g$ material is magnestized by an alternating field of $50 Hz$ , then energy lost in one hour will be $β$ joule. Find $\frac{β}{1000}$ . (Density of material is $7.5 g / c m^{3}$ )

Energy lost in $1$ hour
$=$ Area $\times$ Volume $\times$ Time $\times$ Frequency
$= \frac{250 \times 10 \times 3600 \times 50}{7500}$
$= 6 \times 1 0^{4} J$

Q. The area of hysteresis loop of a material is equivalent to 250J/m3. When 10kg material is magnestized by an alternating field of 50Hz, then energy lost in one hour will be β joule. Find 1000β​. (Density of material is 7.5g/cm3 )

Energy lost in 1 hour = Area × Volume × Time × Frequency =7500250×10×3600×50​ =6×104J

Q. The area of hysteresis loop of a material is equivalent to $250 J / m^{3}$ . When $10 k g$ material is magnestized by an alternating field of $50 Hz$ , then energy lost in one hour will be $β$ joule. Find $\frac{β}{1000}$ . (Density of material is $7.5 g / c m^{3}$ )

Energy lost in $1$ hour
$=$ Area $\times$ Volume $\times$ Time $\times$ Frequency
$= \frac{250 \times 10 \times 3600 \times 50}{7500}$
$= 6 \times 1 0^{4} J$