Saved Bookmarks
| 1. |
A solenoid a core of a material with relative permeability 400. The windings of the solenoid are insulated from the core and carry a current of 2A. If the number of turns is 1000 per metre, calculate (a) H, (b) B and (c) the magnetising current I_m. |
|
Answer» Solution : a. The field H is dependent of the material of the core, and is `H=nI=1000xx2.0=2xx10^(3)A//m` b. The magnetic field B is given by, `B=mu,mu_0H=400xx4pixx10^(-7)(N//A^2)xx2xx10^3(A//m)=1.0T` c. MAGNETISATION is given by, `M=((B-mu_0H))/(mu_0)=(mu_rmu_0H-mu_0H)/(mu_0)=(mu_r-1)=399xxH` d. The magnetising current `I_M` is the ADDITIONAL current that needs to be passed through the windings of the solenoid in the ABSENCE of the core which would give a B value as in the presence of the core . Thus, `B=mu_(r)n_(0)(I+I_(M))`. USING I = 2A , b = 1T , we get `I_M=794A`. |
|