The solenoids `S_(1)andS_(2)` are wound on an iron-core of relative permeability 900. The area of their cross-section and their length are the same and are 4 cm 2 and 0.04 m respectively. If the number of turns in `S_(1)` is 200 and that in `S_(2)` is 800, calculate the mutual inductance between the coils. The current in solenoid 1 is increased form 2 A to 8A 0.04 second. Calculate the induced emf in solenoid 2.

Explanation

Given : Relative permeability of an iron are `=mu_(r)=900` ltbrtgt No. of turns in solenoid `S_(1)N_(1)=200`
No. of turns in solenoid `S_(2)N_(2)=800`
Area of cross section `A_(2)=4cm^(2)=(4xx10^(-2))m^(2)`
Length of `S_(1)l_(1)=0.04m`
Current in solenoid `I_(c)=I_(2)-I_(1)=(8-2)=6A`
Time taken `t=0.04s`
To find :
Induced emf in soleniod `S_(2)=?`
`=-M.(dI)/(dt)`
Solution :
Mutusl inductance `M=mu_(0)mu_(r)(N_(1))/(l).N_(2)A_(2)`
`M=(4xx3.4xx10^(-7)xx900xx200xx800xx4xxcancel(10^(-2)))/(cancel(0.04))`
M = 1.81 H
Induced emf `e=-1.81xx(6)/(0.04)`
e = 271.5 V.