Flux-Exclusion Resonance Lab — a shorted coil inside an LC circuit — JEE Advanced 2026 Paper 1 Q.2

QMission brief — the question, verbatim

JEE (Advanced) 2026, Paper 1, Physics, Section 1 — four options, only one correct: +3 if only the correct option is chosen, 0 if unanswered, −1 in all other cases.

Q.2 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 d² ≫ 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:

(A)4/√(15 LC)
(B)6/√(5 LC)
(C)2/√(3 LC)
(D)√(2/(3 LC))

L₂ · M

L_eff

ω√(LC)

The mystery. A dead-short coil sits quietly inside the big coil — it is not even connected to the circuit. Can it change the resonant frequency? The lab switches it in, watches what its current does to the magnetic field, and times the oscillation.

01Interactive experiment

A flux-exclusion resonance bench: a glass long solenoid wound in copper, the silver inner coil with its shorting switch, the capacitor and the AC source, a phosphor oscilloscope and a frequency sweep. Press START EXPERIMENT: the circuit → the coil's field → the inner coil → mutual inductance → close the short → the field leaves the core → L_eff → oscillation → resonance sweep → audit.

Ready

01 Circuit02 Coil field03 Inner coil 04 Mutual M05 Short it06 L_eff 07 Oscillate08 Resonance09 Audit10 Final answer

Set your values or use the question defaults.

Answer——

Flux-exclusion resonance bench

step 1 of 8

A capacitor, an AC source and a long coil with a second coil hidden inside it. The question's values are loaded — press START EXPERIMENT.

✓ QUESTION VALUES LOADED

Customize experiment — optional

Instrument panel

I₁ outer coil

—

I₂ inner coil

—

B outside core

—

B inside core

—

L_eff / L

—

ω√(LC)

—

energy E · B

—

progress

0 %

option

—

B is shown as a fraction of μ₀NI₁; currents as a fraction of the peak I₁. Time is in units of √(LC) and frequency in units of 1/√(LC), so every number here is a pure number.

Nothing needs to be chosen. Press START EXPERIMENT.

Charge on the capacitor

against t/√(LC) — inner coil shorted (green) and open (grey)

Resonance sweep

response against drive frequency ω√(LC), with the four options marked

Result audit — the four options

(A)4/√(15 LC)

not yet examined

UNEVALUATED

(B)6/√(5 LC)

not yet examined

UNEVALUATED

(C)2/√(3 LC)

not yet examined

UNEVALUATED

(D)√(2/(3 LC))

not yet examined

UNEVALUATED

🧲Solenoid Sage

🔗Mutual Master

🛡️Flux Guardian

⚡Inductance Hunter

🪤Trap Spotter

🏆Physics Master

Reasoning log — input → action → change → observation → conclusion

Nothing investigated yet.

Evidence table (0)

02Coupled-coil explorer

Change the inner coil — its area, its length, its turns — and short it or open it. Watch where the field is allowed to live. The effective inductance is the fraction of the volume the field still fills: L_eff = L(1 − a·ℓ). The turns do not matter.

Area fraction a = S₂/S: 0.50

Length fraction ℓ = d₂/d: 0.50

Turns ratio m = n₂/N: 2

03Detailed solution

The complete worked solution — the inductances, the shorted coil, the effective inductance and the frequency.

Understand the concept

Inductance of a long solenoid

Inside a long solenoid B = μ₀nI, uniform. The flux linkage is (n l)(B A), so L = μ₀n²A l.

L = μ₀ N² S d

Mutual inductance

The outer coil's field μ₀NI₁ threads each of the inner coil's 2N·(d/2) turns over the inner area S/2.

M = μ₀ N (2N)(S/2)(d/2) = L/2

A shorted coil guards its flux

With no resistance its total EMF must be zero, so its flux linkage cannot change: a current appears that cancels the field inside it.

L₂ dI₂/dt + M dI₁/dt = 0

Less field, less inductance

The field is pushed out of a quarter of the volume, so the energy stored for a given current — and L — drops by a quarter.

L_eff = L − M²/L₂ = 3L/4

Step 1 — Self-inductances

L₁ = μ₀N²S d = L · L₂ = μ₀(2N)²(S/2)(d/2) = μ₀N²S d = L

The inner coil has twice the turns per length on half the length — the same N d turns — on half the area.
Step 2 — Mutual inductance

Flux linkage of the inner coil from I₁: (2N · d/2) turns × μ₀NI₁ × S/2 = (L/2)·I₁.

M = μ₀ · N · 2N · (S/2) · (d/2) = ½ μ₀N²S d = L/2
Step 3 — The shorted coil

Zero resistance means zero EMF round the inner coil, so its flux linkage stays at zero: its current makes the field inside it vanish.

L₂ dI₂/dt + M dI₁/dt = 0 ⇒ I₂ = −(M/L₂) I₁ = −I₁/2
Step 4 — Effective inductance

V = L₁ dI₁/dt + M dI₂/dt = (L₁ − M²/L₂) dI₁/dt ⇒ L_eff = L − (L/2)²/L = 3L/4

Energy check: for the same I₁ the field fills 1 − (S/2)(d/2)/(S d) = 3/4 of the volume, so ½L_effI₁² = ¾ · ½LI₁².

Step 5 — Resonant frequency

The short lowers the inductance, so the resonance moves up by 2/√3.

ω = 1/√(L_effC) = 1/√(3LC/4) = 2/√(3LC) → option —

Option	ω√(LC)	L_eff it implies	What went wrong
(A) 4/√(15 LC)	1.033	15L/16	removes only 1/16 of the volume — the area and length fractions squared
(B) 6/√(5 LC)	2.683	5L/36	no flux argument gives it
(C) 2/√(3 LC)	1.155	3L/4	correct
(D) √(2/(3 LC))	0.816	3L/2 = L + M	takes the induced current as aiding the flux (Lenz's law reversed)

Traps: the inner coil is not in the circuit, yet it matters; its extra turns do not matter (L_eff = L(1 − a·ℓ) for any turns); "resonant frequency" here is the angular frequency ω.

Key takeaways

Long solenoid: L = μ₀n²A l; for two coaxial coils M = μ₀ n₁ n₂ (common area)(common length).
A closed, resistance-free loop keeps its flux linkage fixed: L₂İ₂ + Mİ₁ = 0.
A shorted coil coupled to a circuit lowers its inductance: L_eff = L₁ − M²/L₂ = L₁(1 − k²).
For coaxial solenoids k² is the fraction of the outer coil's volume the inner coil fills — here 1/4 — whatever its turns.
Less inductance, higher resonance: ω = 1/√(L_effC).

Independent verification

Symbolic. sympy builds L₁, L₂ and M from μ₀n²Al and μ₀n₁n₂Al, imposes zero EMF on the shorted coil and gets L_eff = 3L/4, ω = 2/√(3LC); only option C equals it.
Circuit integration. The two coupled coil equations integrated with scipy (no L_eff used): the measured period gives ω√(LC) = 1.154701, I₂ = −I₁/2 throughout, energy conserved.
Field energy. Flux conservation fixes I₂; Ampère's law gives B = 0 inside the inner coil and μ₀NI₁ elsewhere; the energy integral gives L_eff = 3L/4.
Controls. Open inner coil: ω√(LC) = 1. Other turns, areas and lengths: always 1/√(1 − a·ℓ).
Solution website. MathonGo's published solution (their Q18 = Physics Q.2) gives (C) by the same steps.
This page. The bench computes L₂, M and L_eff from the geometry, integrates the circuit, and works the field energy; the answer is the option that matches.
Answer key. Checked against the official key printed with the paper: Q.2 → C.

04How to use this simulation

For a student opening it for the first time. The whole investigation runs by itself after one press.

STEP 1

Read the mission brief

An LC circuit whose coil hides a second, shorted coil. What is the resonant frequency?

STEP 2

Check the values

The question's coil is loaded. You do not need to change anything. The answer is shown above the experiment.

STEP 3

Optional: customize

Change the inner coil's turns, area or length, or leave it open. The lab says honestly when you have left the question. ↺ RESTORE QUESTION CONDITIONS puts everything back.

STEP 4

Start

Press ▶ START EXPERIMENT under the experiment screen. Nothing else to click (about a minute; SLOW MOTION or 2× changes the pace).

STEP 5

Watch the field

Blue particles are the magnetic field along the axis. When the switch closes, the inner coil's current (orange) drives them out of its core.

STEP 6

Read the scope

The oscilloscope compares the charge on C with the coil open (grey) and shorted (green): the shorted circuit rings faster. The sweep finds the resonance.

STEP 7

Explore

In the Coupled-coil explorer change the inner coil and watch L_eff. Then read the Detailed solution.

STEP 8

Experiment again

Press ▶ REPLAY EXPERIMENT, or change a value and press ⚙ RUN WITH CUSTOM VALUES. CLASSROOM MODE enlarges everything for a projector.

Classification

Exam	JEE Advanced 2026, Paper 1, Physics, Section 1, Q.2
Chapter	Electromagnetic Induction
Topic	Self and mutual inductance of coaxial solenoids - a shorted coil lowers the effective inductance and raises the LC resonance
Difficulty	Moderate
Verification	script-verified (automated pipeline, no human review)
Answer	— run the experiment

Flux-Exclusion Resonance Lab — what does a shorted coil do to an LC circuit?

QMission brief — the question, verbatim

01Interactive experiment

Flux-exclusion resonance bench

Customize experiment — optional

Instrument panel

Charge on the capacitor

Resonance sweep

Result audit — the four options

02Coupled-coil explorer

03Detailed solution

The solution is still locked

Understand the concept

Inductance of a long solenoid

Mutual inductance

A shorted coil guards its flux

Less field, less inductance

Step 1 — Self-inductances

Step 2 — Mutual inductance

Step 3 — The shorted coil

Step 4 — Effective inductance

Step 5 — Resonant frequency

Key takeaways

Independent verification

04How to use this simulation

Read the mission brief

Check the values

Optional: customize

Start

Watch the field

Read the scope

Explore

Experiment again

Classification