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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);
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