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 figur

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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 of the larger coil is L. Neglecting edge effects and all the Ohmic resistances, the resonant frequency of the circuit is:

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Correct answer: C. 2 3\, LC

Let the larger coil be coil 1 and the smaller coil be coil 2. The self-inductance of the larger coil is given by: L_1 = _0 n_1^2 A_1 l_1 = _0 N^2 S d = L The self-inductance of the smaller coil is: L_2 = _0 n_2^2 A_2 l_2 = _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 n_1 n_2 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 Let i_1 be the current in the larger coil and i_2 be the current in the smaller coil. The voltage across the larger coil is: V = L_1 di_1 dt + M di_2 dt Since the smaller coil is short-circuited, the net voltage across it is zero: 0 = L_2 di_2 dt + M di_1 dt di_2 dt = - M L_2 di_1 dt Substituting this into the first equation, we get the effective inductance L_ eq : V = L_1 di_1 dt - M^2 L_2 di_1 dt = ( L_1 - M^2 L_2 ) di_1 dt L_ eq = L_1 - M^2 L_2 = L - (L/2)^2 L = L - L 4 = 3L 4 The resonant angular frequency of the circuit is: = 1 L_ eq C = 1 3L 4 C = 2 3LC Answer: 2 3\, LC

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