Two coil A `&` B having 15 turns and 20 turns respectively are placed close to each other. When 4A current is passing through the coil A then flux linked with coil A is `1.6xx10^(-4)` weber per turn. If 5A current is passed through coil B, then calculate the flux linked per turn of coil B.

Explanation

Given : No. of turns In A coil `N_(1)=15`
No. of turns on B coil `N_(2)=20`
Current flowing through `A=A.I_(1)`
Current flowing through `B=5A.I_(2)`
Flux linked with coil B `=1.6xx10^(-4)`weber per turn `=phi_(2)`
To find :
Flux linked with coil A `=phi_(1)=?`
Flux linked with `A,phi_(1)=muI_(2)`
Flux linked with `B,phi_(2)=muI_(1)`

Soution :
`(phi_(2))/(phi_(1))=(I_(1))/(I_(2))rArr(20xx1.6xx10^(-4))/(15xxphi)=(4)/(5)`
`15xx4xxphi=20xx5xx1.6xx10^(-4)`
`phi_(1)=(20xx5xxcancel(1.6)^(0.4)xx10^(-4))/(cancel(15)_(3)xxcancel(4))`
`=(8xx10^(-4))/(3)`
`phi=2.66xx10^(-4)Wb.`