Consider a circuit consisting of a capacitor of capacitance C and a coil with N turns per unit length, cross sectional area S and length d , where d^2 S . There is another coil of length d/2 , cross sectional area S/2 and 2N turns per unit length completely inside the larger coil, as shown in the figure. The ends of this smaller coil are connected with each other by an insulated conducting wire. The self-inductance o

Step-by-step solution

Let the larger coil be coil 1 and the smaller coil be coil 2. The self-inductance of the larger coil is given by: L₁ = ₀ n₁^2 A₁ l₁ = ₀ N^2 S d = L The self-inductance of the smaller coil is: L₂ = ₀ n₂^2 A₂ l₂ = ₀ (2N)^2 ( S 2 ) ( d 2 ) = ₀ (4N^2) ( Sd 4 ) = ₀ N^2 S d = L The mutual inductance between the two coils is: M = ₀ n₁ n₂ A_ common l_ common Since the smaller coil is completely inside the larger coil, the common area is S/2 and the common length is d/2 . M = ₀ (N) (2N) ( S 2 ) ( d 2 ) = 1 2 ₀ N^2 S d = L 2