A straight solenoid has 50 turns per cm in primary and total 200 turns in the secondary. The area of cross section of the solenoids is `4cm^2`. Calculate the mutual inductance. Primary is tightly kept in side the secondary.

Explanation

The magnetic field at any point inside the straight solenoid of primary with `n_(1)` turns per unit length carrying a current `i_(1)` is given by the relation,
`B=mu_(0)n_(1)i_(1)`
The magnetic flux through the secondary of `N_(2)` turns each of area S is given as
`N_(2)phi_(2)=N_(2)(BS)=mu_(0)n_(1)N_(2)i_(1)S`
`therefore" "M=(N_(2)phi_(2))/(i_(1))=mu_(0)n_(1)N_(2)S`
Substituting the values,
`M=(4pixx10^(-7))((50)/(10^(-2)))(200)(4xx10^(-4))`
`=5xx10^(-4)H`