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authorprashantsinalkar2017-10-10 12:27:19 +0530
committerprashantsinalkar2017-10-10 12:27:19 +0530
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+//Example18.6//
+
+a=8.9*10^4;//(amperes/m)(webers/m^2) // Area
+mprintf("a= %e (amperes.webers)/m^3",a)
+//one ampere weber is equal to 1joule. The area is then a volume density of energy ,or
+e=8.9*10^4//J/m^3 //energy loss
+b= 10^-3; //As 1Kilogram = 10^3 gram
+e1=e*b
+mprintf("\ne1 = %i kJ/m^3 (per cycle) (As 1Kilogram = 10^3 garm)",e1)