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+// Estimating the hot spot temperature
+clc;
+disp('Example 4.7, Page No. = 4.14')
+// Given Data
+l = 1;// Length of mean turn in meter
+Sf = 0.56;// Space Factor
+p = 120;// Total loss in the coil in Watt
+pi = 8;// Thermal resistivity in ohm*meter
+A = 100*50;// Area of cross-section in mm square
+t = 50*10^(-3);// Thickness of coil in meter
+// Calculation of the temperature of the hot spot
+pe = pi*(1-Sf^(1/2));// Effective thermal resistivity in ohm*meter
+V = A*l*10^(-6);// Volume of coil(in meter cube)
+q = p/V;// Heat dissipated in Watt per meter cube
+T0 =q*pe*t*t/8;// Assuming equal inward and outward heat flows
+disp(T0,'Temperature of the hot spot (degree celsius)=');
+//in book answers is 15 degree celsius. The answers vary due to round off error
+