Saved Bookmarks
| 1. |
A magnetising field of 1600 Am^(-1) produces a magnetic flux of2.4 xx 10^(-5) weber in a bar of iron of cross section 0.2 cm^(2) . Calculate permeability and susceptibility of the bar . |
|
Answer» Solution :MAGNETIC induction `, B = (phi)/(A) = (2.4 xx 10^(-5))/(0.2 xx 10^(-4))` ` =1.2 Wb//m^(2)` i) Permeability , `mu=(B)/(H) = (1.2)/(1600) = 7.5 xx 10^(-4) TA^(-1)m` II)As `mu=mu_0(1+ chi_m) ` then Susceptibility , ` chi_m =(mu)/(mu_0)-1 =(7.5 xx 10^(-4))/( 4 pi xx 10^(-7)) -1` `=596.1 ` |
|