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+
+//To Find the Magnetic Field on Axis of Solenoid
+
+//Example 36.1
+
+clear;
+
+clc;
+
+i=10;//Current carried by Solenoid in Amperes
+
+r=1*10^-2;//Radius of Solenoid in metres
+
+A=%pi*r^2;//Area of Cross Section of Solenoid in metre^2
+
+u=i*A;//Dipole Moment of each turn
+
+l=10*10^-2;//Length of Solenoid in metres
+
+R=10*10^-2;//Distance of point P from the centre of solenoid
+
+n=200;//Number of turns in Solenoid
+
+d=l/n;//Seperation between two consecutive turns
+
+m=u/d;//Pole Strength for each Current Loop
+
+k=1*10^-7;//Constant (u0/(4*pi))
+
+Rn=R-(l/2);//Distance of point P from North Pole
+
+Bn=k*m/Rn^2;//Magnetic Field at P due to North Pole
+
+Rs=R+(l/2);//Distance of point P from South Pole
+
+Bs=k*m/(Rs)^2;//Magnetic Field at P due to South Pole
+
+B=Bn-Bs;//Resultant Magnetic Field at point P
+
+printf("Magnetic field at a point on the axis of Solenoid at a distance of 10cm from centre = %.1f*10^-4 T away from the solenoid",B*10^4);