JEE Advanced 2026 — Paper 1 — PHY-I solution

JEE Advanced2026Paper 1PHY-I

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
$\frac{4}{15 L C}$
B
$\frac{6}{5 L C}$
C
$\frac{2}{3 L C}$
D
$\frac{2}{3 L C}$

Solution

Inductances of the two coils. For the large coil, $L = μ_{0} N^{2} S d$ . For the small coil with double turn density, half the cross-section, and half the length, $L_{small} = μ_{0} (2 N)^{2} (S /2) (d /2) = μ_{0} N^{2} S d = L$ .

Mutual inductance. Flux from the outer coil through the inner coil gives $M = μ_{0} N (2 N) (S /2) (d /2) = \frac{μ _{0} N ^{2} S d}{2} = \frac{L}{2}$ .

Shorted inner coil constraint. Since the inner coil is shorted, its net flux is zero, which links the two currents by $L_{small} i^{'} + M i = 0$ , giving $i^{'} = - i /2$ in magnitude (induced current opposes change).

Effective inductance of the outer coil. The net EMF in the outer coil is $E = - L \frac{d i}{d t} + M \frac{d i ^{'}}{d t} = - L \frac{d i}{d t} - \frac{L}{2} \cdot \frac{1}{2} \frac{d i}{d t} = - \frac{3 L}{4} \frac{d i}{d t}$ .

Hence $L_{eq} = \frac{3 L}{4}$ .

Resonant frequency. $ω = \frac{1}{L _{eq} C} = \frac{1}{( 3 L /4 ) C} = \frac{2}{3 L C}$ . Option (C).

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