JEE Advanced · 2026 · 17 May 2026 · Paper 1 · Alternating Current

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

2 N

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:

(A) $\frac{4}{15 LC}$
(B) $\frac{6}{5 LC}$
(C) $\frac{2}{3 LC}$ ✓ Correct
(D) $\frac{2}{3 LC}$

Answer: (C)

Solution

The larger coil has self-inductance

L = μ_{0} SN^{2} d

. The smaller coil has self-inductance

L^{'} = μ_{0} \frac{S}{2} (2 N)^{2} \frac{d}{2} = L

.

The mutual inductance is

M = \frac{μ _{0} SN ^{2} d}{2} = \frac{L}{2}

.

When current changes in the larger coil, it induces emf in the smaller coil, which in turn affects the larger coil. The effective inductance becomes

L_{eq} = \frac{3 L}{4}

.

The resonant frequency is

ω = \frac{1}{L _{eq} C} = \frac{1}{\frac{3 L}{4} \cdot C} = \frac{2}{3 LC}

JEE Advanced · 2026 · 17 May 2026 · Paper 1 · Single correct option

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JEE Advanced · 2026 · 17 May 2026 · Paper 1 · Alternating Current

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