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An LC circuit whose coil holds a shorted inner coil of half the area and half the length: self and mutual inductance, flux exclusion and the resonant frequency
JEE Advanced 2026
Paper 1 · Question 2
Physics
Self and mutual inductance of coaxial solenoids - a shorted coil lowers the effective inductance and raises the LC resonance
Moderate
A glowing LC bench: a long solenoid with a shorted inner coil of half its area and half its length. Close the switch and the inner coil's current pushes the magnetic field out of its core - a quarter of the volume - so the effective inductance falls to 3L/4 and the resonance rises to 2/sqrt(3LC), watched on an oscilloscope and a frequency sweep, with every option audited.
Open the interactive simulationSub-topics covered
- Self-inductance of a long solenoid, L = mu0 n^2 A l
- Mutual inductance of coaxial solenoids, M = mu0 n1 n2 (common area)(common length)
- A resistance-free shorted coil keeps its flux linkage fixed: L2 dI2/dt + M dI1/dt = 0
- Effective inductance L_eff = L1 - M^2/L2 = L1 (1 - k^2)
- LC resonance, omega = 1/sqrt(L_eff C)
What it teaches
- The inner coil has 2N turns per length over d/2 - the same N d turns - on area S/2, so L2 = mu0 (2N)^2 (S/2)(d/2) = L
- The outer field mu0 N I1 threads the inner coil's turns over its area: M = mu0 N (2N)(S/2)(d/2) = L/2
- With no resistance the shorted coil's EMF is zero, so I2 = -(M/L2) I1 = -I1/2, and the field inside it vanishes
- The field is expelled from (S/2)(d/2) = a quarter of the volume, so L_eff = L - M^2/L2 = 3L/4
- omega = 1/sqrt(L_eff C) = 2/sqrt(3LC); for coaxial solenoids L_eff = L(1 - a l) whatever the inner turns
Simulation ID
- ADV-2026-P1-PHY-Q02 · revision 1
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