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JEE Advanced 2026 · Paper 1, Physics question 2

Consider a circuit consisting of a capacitor of capacitance CC and a coil with NN turns per unit length, cross sectional area SS and length dd, where d2≫Sd^2\gg S. There is another coil of length d/2d/2, cross sectional area S/2S/2 and 2N2N 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 LL. Neglecting edge effects and all the Ohmic resistances, the resonant frequency of the circuit is:

A long coil of length d connected in a loop with a capacitor C and an AC source. Inside it, drawn dashed, sits a shorter coil of length d/2 whose two ends are joined by a wire outside the large coil (shorted).
Choose one option
Answer and worked solution

Official answer: C

Inductances. L=μ0N2SdL=\mu_0N^2Sd. For the inner coil:

  • self-inductance L2=μ0(2N)2S2⋅d2=μ0N2Sd=LL_2=\mu_0(2N)^2\frac S2\cdot\frac d2=\mu_0N^2Sd=L;
  • mutual inductance M=μ0N(2N)S2⋅d2=L2M=\mu_0N(2N)\frac S2\cdot\frac d2=\frac L2 (the outer coil's field fills the inner coil).

Shorted inner coil. With no resistance, its emf must vanish: L2i˙2+Mi˙1=0L_2\dot i_2+M\dot i_1=0, so i2=−ML2i1i_2=-\frac{M}{L_2}i_1.

Effective inductance of the outer coil:
Leff=L−M2L2=L−L4=3L4.L_{\text{eff}}=L-\frac{M^2}{L_2}=L-\frac L4=\frac{3L}4.

Resonance.
ω0=1LeffC=23LC.\omega_0=\frac1{\sqrt{L_{\text{eff}}C}}=\frac2{\sqrt{3LC}}.

Topics:Mutual inductanceSelf inductanceLC oscillations
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