Derive an expression for the total magnetic energy stored in two coils with inductances `L_(1)` and `L_(2)` and mutual inductance `M`, when the currents in the coils are `I_(1)` and `I_(2)`, respectively.

Explanation

When the currents are increaseing in the circuit, we have emfs
`E_(1) = - L_(1) (dI_(1))/(dt) +- M (dI_(2))/(dt)`,`E_(2) = - L_(2) (dI_(2))/(dt) +- M(dI_(1))/(dt)`
where `+-` sign appears conistently in both equations and depends on the geometry of the coils and the sense of current. Work done in pushing charges `dq_(1)` and `dq_(2)` through each circuit. respectively, is
`dW = - E_(1) dq - E_(2) dq_(2)`
`= L_(1) (dI_(1))/(dt) dq_(1) bar+ M (dI_(2))/(dt)dq_(1) + L_(2) (dI_(2))/(dt) dq_(2) bar+ M (dI_(1))/(dt) dq_(2)`
`I_(1) = (dq_(1))/(dt),dq_(1) = I_(1)dt, I_(2) = (dq_(2))/(dt),dq_(2) = I_(2) dt`
`dW = L_(1)I_(1)dI_(1) + L_(2)I_(2)dI_(2) bar+(MI_(1) dI_(2) + MI_(2) dI_(1))`
`= L_(1)I_(1)dI_(1) + L_(2)I_(2)dI_(2) bar+Md(I_(1)I_(2))`
On integating the above expression from `0` to final current, we have
`U = int dW = L_(1) int_(0)^(I_(1)) I_(1) dI_(1) + L_(2) int_(0)^(I_(2)) I_(2) dI_(2) bar+ M int_(0)^((I_(1)I_(2))) d(I_(1)I_(2))`
`= (1)/(2) L_(1)L_(1)^(2) + (1)/(2) L_(2)I_(2)^(2) bar+MI_(1)I_(2)`