In short

Every electromagnetic-induction problem reduces to Faraday's law with a possible Lenz-sign check:

\varepsilon \;=\; -\frac{d\Phi_B}{dt}, \qquad \Phi_B \;=\; \int \vec{B}\cdot d\vec{A}.

For JEE Advanced:

Rotating-rod EMF. A rod of length L rotating about one end at angular velocity \omega in a uniform field B perpendicular to the plane of rotation sweeps out area at rate \tfrac{1}{2}L^2\omega — every second. So

\boxed{\;\varepsilon_{\text{rod}} \;=\; \tfrac{1}{2}B\,L^2\,\omega\;}

Same as integrating \int_0^L B\,v(r)\,dr with v(r) = r\omega.

Mutual inductance of coaxial solenoids. Wind a short inner coil (N_2 turns, area A_2) inside a long outer solenoid (n_1 turns/m carrying I_1). The outer solenoid makes B_1 = \mu_0 n_1 I_1 uniform inside; so

\boxed{\;M \;=\; \mu_0\,n_1\,N_2\,A_2\;}

Mutual inductance is always symmetric: M_{12} = M_{21}.

LC oscillations. Charge q(t) on a capacitor C in series with an inductor L (no resistance) obeys

L\,\frac{d^2 q}{dt^2} + \frac{q}{C} \;=\; 0 \qquad \Longrightarrow \qquad q(t) \;=\; Q_0\cos(\omega_0 t), \quad \omega_0 = \frac{1}{\sqrt{LC}}.

Natural frequency:

\boxed{\;f_0 \;=\; \frac{1}{2\pi\sqrt{LC}}\;}

The energy \tfrac{q^2}{2C} + \tfrac{1}{2}LI^2 is conserved and sloshes between capacitor (electric) and inductor (magnetic) twice per cycle — exactly like KE and PE in a mass-spring SHM, with the identification L \leftrightarrow m, 1/C \leftrightarrow k, I \leftrightarrow v, q \leftrightarrow x.

Coupled LC circuits. Two LC circuits linked by mutual inductance M split their shared energy into two normal modes at slightly different frequencies — a classical-mechanics two-body problem in disguise.

The JEE Advanced question on electromagnetic induction almost never asks for Faraday's law as a formula — it asks for Faraday's law applied to a situation where the geometry is just non-obvious enough that a student who only memorised \varepsilon = -d\Phi/dt gets stuck. A rod rotating about one end in a field. A loop that is partly inside and partly outside a region of magnetic field, moving at an angle to the boundary. A metal bar that slides down a pair of parallel rails at an angle to gravity, with a resistor across the ends. An LC circuit where the switch closes at t = 0 and the problem asks at what time the energy is shared 50–50 between capacitor and inductor. Each of these is the same physics as the introductory articles — but applied with the algebra executed cleanly.

This article is an advanced-problems article, in the same way that the elastic collisions article is a technique article in its going-deeper section. It assumes you have already read Faraday's law, Lenz's law, motional EMF, and self- and mutual-inductance. The goal here is not to re-state those laws; it is to work five canonical JEE Advanced setups rigorously, derive the LC-oscillation result cleanly from Kirchhoff, and show how the tools generalise to coupled circuits.

The test that this article has done its job: after reading it, you should be able to sit for a JEE Advanced paper, meet any induction problem in the book, and know within the first ten seconds which of the tools — motional EMF, flux-rate, mutual-inductance formula, LC analogy — is the right one to open the problem with.

Tool 1: Rotating-rod EMF

Consider a straight conducting rod of length L pivoted at one end, rotating about that end with angular velocity \omega in a plane perpendicular to a uniform magnetic field \vec{B}.

Rotating rod in a uniform magnetic fieldA horizontal conducting rod pivoted at one end rotates counterclockwise about the pivot in a plane perpendicular to a magnetic field B pointing out of the page (shown as dots). A small element dr of the rod at distance r from the pivot has velocity r omega; its contribution to the total motional EMF is B times r omega times dr. Integration from 0 to L gives one half B L squared omega. Labels identify the pivot, the rod, an arrow for omega, and a small element dr with its velocity arrow.B out of pagepivotωdrv = r ωrL
A rod of length $L$ rotating about a pivot at one end with angular velocity $\omega$ in a uniform field $\vec{B}$ perpendicular to the plane of motion. A small element at radius $r$ moves with velocity $v = r\omega$, so its motional-EMF contribution is $B\,r\omega\,dr$. Integrating from 0 to $L$ gives the total EMF $\tfrac{1}{2}BL^2\omega$ across the rod.

Different parts of the rod move at different speeds. The pivot is stationary, so the end at r = 0 has v = 0; the far end at r = L has v = L\omega. The naïve \varepsilon = B\ell v formula from motional EMF assumed a single velocity for the whole rod — but here v varies linearly with r. You cannot use one number for v.

The correct move is to integrate. Take a small element of the rod at radius r, of length dr. Its velocity is v(r) = r\omega. Its tiny EMF contribution is

d\varepsilon \;=\; B\,v(r)\,dr \;=\; B\,r\omega\,dr.

Why: the motional-EMF-per-unit-length, (\vec{v}\times\vec{B})\cdot d\vec{\ell}, for the small element is B v \,dr when \vec{v}, \vec{B}, and d\vec{\ell} are mutually perpendicular. Each element of the rod acts like a miniature sliding bar.

Integrate from the pivot (r=0) to the far end (r=L):

\varepsilon \;=\; \int_0^L B\,r\omega\,dr \;=\; B\omega \int_0^L r\,dr \;=\; B\omega \cdot \frac{L^2}{2}.
\boxed{\;\varepsilon \;=\; \tfrac{1}{2}\,B\,L^2\,\omega\;} \tag{1}

Why: \int_0^L r\,dr = L^2/2. The quadratic L^2 dependence is the telltale signature of the fact that the outer end moves faster than the inner end — doubling the rod length doubles the speed of the far end, so the EMF goes up four-fold.

Cross-check via flux. In time dt, the rod sweeps out an area. The far end traces an arc of length L\omega\,dt; the rod sweeps a triangle of area dA = \tfrac{1}{2} L \cdot L\omega\,dt = \tfrac{1}{2}L^2\omega\,dt. Multiply by B and divide by dt:

|\varepsilon| \;=\; \left|\frac{d\Phi}{dt}\right| \;=\; B\cdot \frac{dA}{dt} \;=\; B\cdot \tfrac{1}{2}L^2\omega,

which matches (1). Flux and Lorentz-force routes agree — as they must.

The rotating rod is the kernel of the Faraday disc (the first DC generator, built by Faraday in 1831: a copper disc rotating between the poles of a magnet, with brushes at the centre and the rim producing a DC EMF \tfrac{1}{2}BR^2\omega across them). A variant appears in JEE Advanced questions where a rod rotates on a pair of circular rails, closing a circuit — the EMF drives a current I = \tfrac{1}{2}BL^2\omega / R, which in turn produces a retarding torque on the rod. Mechanical power in equals electrical power out, exactly.

Tool 2: Mutual inductance of coaxial coils

Two coils linked by a magnetic flux are characterised by their mutual inductance M: if current I_1 flows in coil 1, the flux through coil 2 is \Phi_2 = M I_1. If I_1 changes, the induced EMF in coil 2 is \varepsilon_2 = -M\,dI_1/dt.

Computing M requires one flux integral — but you get to choose which coil acts as the source, since M is symmetric (M_{12} = M_{21}). Exploit the symmetry by letting the simpler geometry be the source.

The canonical JEE Advanced setup: a short coil inside a long solenoid.

Mutual inductance of a short coil inside a long solenoidA long outer solenoid is drawn as a series of vertical ellipses, with current I1 arrows indicating the winding direction and density n1 turns per metre. Inside the solenoid, at the centre, sits a short coil (drawn as a smaller rectangle) with N2 turns and cross-sectional area A2. The magnetic field B1 equals mu0 n1 I1 is uniform inside the outer solenoid and threads entirely through the inner coil.B₁ = μ₀ n₁ I₁inner coil (N₂ turns, area A₂)I₁ flows in outer solenoid (n₁ turns/m)
A short coil (red rectangle, $N_2$ turns, cross-sectional area $A_2$) placed inside a long solenoid (black ellipses, $n_1$ turns/m) carrying current $I_1$. The solenoid field $B_1 = \mu_0 n_1 I_1$ is uniform and axial over the whole inner volume; the flux through the inner coil is $N_2 B_1 A_2$, giving $M = \mu_0 n_1 N_2 A_2$.

Setup. An infinite solenoid ("long" means long compared to its radius) has n_1 turns per metre and carries current I_1. A short coil of N_2 turns and cross-sectional area A_2 sits inside it, coaxially, at the centre.

Route A: use the solenoid as the source.

From the magnetic field of common current configurations, the field inside a long solenoid is

B_1 \;=\; \mu_0 n_1 I_1,

uniform and axial. The flux through each turn of the inner coil is B_1 A_2 = \mu_0 n_1 I_1 A_2. Since there are N_2 turns, the total flux linkage is N_2 \mu_0 n_1 I_1 A_2. Using \Phi_2 = M I_1:

\boxed{\;M \;=\; \mu_0\,n_1\,N_2\,A_2.\;} \tag{2}

Why: this works because the solenoid's field is uniform, so computing the flux through the inner coil is trivially B_1 \times A_2. The inner coil's geometry — its radius, its number of turns — matters only through N_2 A_2.

Route B: use the inner coil as the source.

This route is much harder. The inner coil's field (outside it) is not uniform — it resembles a magnetic dipole at large distances. Computing the flux through the outer (infinite) solenoid from this dipole field requires integrating \vec{B}_{\text{inner}} over every turn of the outer solenoid. A painful calculation.

But the answer must be the same: the reciprocity theorem M_{12} = M_{21} (see the going-deeper section) guarantees it. So you can always pick the easier-source direction and quote M directly.

Takeaway. For JEE Advanced, when a mutual inductance is needed, (i) identify which coil has the simpler field inside the other, (ii) compute B from that source at the other coil's location, (iii) integrate only if B varies across the other coil's cross-section.

Tool 3: Inductance of coaxial cables

A coaxial cable — two long coaxial conductors, the inner of radius a and the outer of radius b, with current I flowing down the inner and back on the outer — carries a magnetic field only in the region a < r < b, between the conductors.

Inside this region, Ampère's law gives B(r) = \mu_0 I / (2\pi r) (field of an infinite wire). Outside (r > b) and inside the inner conductor (r < a), the field is zero (exterior: the net enclosed current is zero because the inner and outer carry opposite currents; interior of the inner conductor: by symmetry and Ampère's law applied to r < a).

Self-inductance per unit length.

Compute the flux through a rectangular "loop" of length \ell aligned with the cable axis, with one side on the inner conductor's surface and the other side on the outer:

\Phi \;=\; \int_a^b B(r)\,\ell\,dr \;=\; \int_a^b \frac{\mu_0 I}{2\pi r}\,\ell\,dr \;=\; \frac{\mu_0 I \ell}{2\pi}\ln\!\frac{b}{a}.

Why: the flux is \int \vec{B}\cdot d\vec{A}; the area element dA = \ell\,dr for a strip at radius r; the field B = \mu_0 I/(2\pi r) is constant along the \ell direction.

Self-inductance from \Phi = L I:

\boxed{\;L \;=\; \frac{\mu_0 \ell}{2\pi}\,\ln\!\frac{b}{a}.\;} \tag{3}

For a cable of \ell = 1 km = 10^3 m, b/a = 5 (typical TV coax ratio), this gives

L \;=\; \frac{4\pi \times 10^{-7} \times 10^3}{2\pi}\,\ln 5 \;=\; 2 \times 10^{-4}\,\ln 5 \;\approx\; 3.2 \times 10^{-4}\ \text{H} \;=\; 0.32\ \text{mH}.

Small, but not negligible at radio frequencies — which is why impedance-matching to coaxial cable requires knowing L and the capacitance per unit length together. JEE Advanced typically asks for the energy stored per unit length:

\frac{U}{\ell} \;=\; \frac{1}{2}\cdot\frac{L}{\ell}\cdot I^2 \;=\; \frac{\mu_0 I^2}{4\pi}\ln\!\frac{b}{a}.

The same result follows from integrating the magnetic-field energy density u = B^2/(2\mu_0) over the volume between the conductors — a useful consistency check that JEE Advanced often asks you to verify.

Tool 4: LC oscillations — derivation from Kirchhoff

Here is where advanced induction meets SHM. Wire an inductor L in series with a capacitor C, add a switch, and charge the capacitor to charge Q_0 before closing the switch. After closing, no battery, no resistance (for now). What does the current do?

LC oscillator circuitA simple LC loop circuit. On the left, a capacitor with charge Q_0 (plus on top, minus on bottom). Wire loops around to an inductor labelled L on the right. A switch between them is closed at t equals 0. Arrows indicate the instantaneous current I flowing around the loop, and another label shows the charge q(t) on the capacitor at an arbitrary instant.+q−qcapacitor CswitchLI = dq/dt
An ideal LC loop. At $t = 0$ the switch closes with the capacitor charged to $Q_0$. The capacitor discharges through the inductor; the current rises as charge flows, then falls as the capacitor recharges with the opposite polarity; then the cycle reverses. The charge and current oscillate sinusoidally at $\omega_0 = 1/\sqrt{LC}$.

Step 1. Apply Kirchhoff's voltage law around the loop. The sum of voltage drops is zero. For the capacitor, the voltage is V_C = q/C (where q is the instantaneous charge, with the sign chosen so that V_C drives current out of the positive plate). For the inductor, the voltage is V_L = L\,dI/dt (Faraday's law applied to the inductor: EMF opposes the changing current).

The current I equals the rate of charge flow, with a sign convention. If I is the current flowing out of the capacitor's positive plate, then I = -dq/dt (the charge on the positive plate decreases as current flows).

Why: sign conventions in oscillator circuits are treacherous. The physical statement is: as charge leaves the positive plate, it flows through the inductor; the inductor's self-EMF resists this flow. The minus sign in I = -dq/dt is necessary so that the differential equation below is \ddot q + \omega_0^2 q = 0, not \ddot q - \omega_0^2 q = 0.

Step 2. Around the loop (going from - plate of the capacitor, through the wire, through the inductor, back to the + plate):

\frac{q}{C} \;-\; L\,\frac{dI}{dt} \;=\; 0.

Substituting I = -dq/dt:

\frac{q}{C} \;+\; L\,\frac{d^2 q}{dt^2} \;=\; 0.

Rearrange:

\boxed{\;L\,\frac{d^2 q}{dt^2} \;+\; \frac{q}{C} \;=\; 0.\;} \tag{4}

Step 3. This is the equation of simple harmonic motion in disguise. Compare with the mass-spring equation m\ddot x + kx = 0:

LC circuit Mass-spring SHM
Inductance L Mass m
Inverse capacitance 1/C Spring constant k
Charge q Position x
Current I = \dot q Velocity v = \dot x
Magnetic energy \tfrac{1}{2}L I^2 Kinetic energy \tfrac{1}{2}m v^2
Electric energy \tfrac{1}{2}q^2/C Potential energy \tfrac{1}{2}k x^2

The correspondence is line-by-line. An inductor stores energy in its magnetic field exactly as a moving mass stores energy in its motion; a capacitor stores energy in its electric field exactly as a compressed spring stores energy in its deformation.

Step 4. Solve (4). It is the SHM equation with \omega_0^2 = 1/(LC):

q(t) \;=\; Q_0\cos(\omega_0 t + \phi), \qquad \omega_0 \;=\; \frac{1}{\sqrt{LC}}. \tag{5}

Why: any function whose second derivative is minus \omega_0^2 times itself is a linear combination of \cos(\omega_0 t) and \sin(\omega_0 t) — equivalently, \cos(\omega_0 t + \phi) for some amplitude and phase.

With initial conditions q(0) = Q_0 (fully charged) and I(0) = -\dot q(0) = 0 (no initial current), \phi = 0:

q(t) \;=\; Q_0\cos(\omega_0 t), \qquad I(t) \;=\; -\dot q(t) \;=\; Q_0\,\omega_0\,\sin(\omega_0 t).

The frequency:

\boxed{\;f_0 \;=\; \frac{\omega_0}{2\pi} \;=\; \frac{1}{2\pi\sqrt{LC}}.\;} \tag{6}

For L = 1 mH and C = 1\,\muF, f_0 = 5 kHz — a tuning frequency for AM-radio-era circuits. For L = 1\,\muH and C = 10 pF, f_0 = 50 MHz — FM-radio range. The LC-oscillator was the heart of every AM/FM radio receiver ever built, and its natural frequency f_0 is what you tuned your grandfather's 1970s transistor radio to when you turned the big dial.

Energy conservation in the LC loop

Total energy at any instant is

U(t) \;=\; \frac{q^2(t)}{2C} + \frac{1}{2}LI^2(t).

Substituting (5):

U \;=\; \frac{Q_0^2 \cos^2(\omega_0 t)}{2C} + \frac{1}{2}L\,Q_0^2\,\omega_0^2\,\sin^2(\omega_0 t) \;=\; \frac{Q_0^2}{2C}\bigl[\cos^2(\omega_0 t) + \sin^2(\omega_0 t)\bigr] \;=\; \frac{Q_0^2}{2C}.

Why: the key step is L\omega_0^2 = L/(LC) = 1/C. This makes the amplitudes of the two energy terms match, so their sum is (Q_0^2/2C)[\cos^2 + \sin^2] = Q_0^2/2C, a constant. Energy oscillates between U_C and U_L but their sum is conserved.

At t = 0: all energy in the capacitor, U_C = Q_0^2/(2C). At t = T/4 (a quarter period): all energy in the inductor, U_L = \tfrac{1}{2}L I_\text{max}^2 = Q_0^2/(2C) (with I_\text{max} = Q_0\omega_0 = Q_0/\sqrt{LC}). The energy trades back and forth twice per period — exactly like KE and PE in a pendulum.

Interactive: LC energy exchange

The figure below shows the electric energy U_C(t) (red) and magnetic energy U_L(t) (dark) over one period of an LC oscillation, with L = 1 mH, C = 1\ \muF, Q_0 = 10^{-6} C. Drag the red dot along the time axis to sample the energies at any moment; the readout shows the two values and their sum.

Interactive LC energy exchange A plot over two periods of an LC circuit. The red curve is the electric energy in the capacitor, U_C equal to Q_0 squared over 2C times cosine squared omega t. The dark curve is the magnetic energy in the inductor, U_L equal to Q_0 squared over 2C times sine squared omega t. The two oscillate at frequency 2 omega in antiphase; their sum is constant at Q_0 squared over 2C. A horizontal dashed line marks the conserved total energy. A draggable dot on the time axis lets the reader sample the energies at any time. time t (in units of period T) energy (J) 0 0.5T T 1.5T 2T 0 U₀ 2U₀ U_C + U_L = const U_C U_L drag along time axis
The electric energy $U_C$ (red) and magnetic energy $U_L$ (dark) over two periods of an LC oscillation. They are $90°$ out of phase and each reaches maximum twice per cycle. Their sum (the dashed horizontal line) is constant at $U_0 = Q_0^2/(2C) = 5\times 10^{-7}$ J, confirming energy conservation in the ideal LC loop.

Worked examples

Example 1: Rotating rod with a resistor

A rod of length L = 0.5 m rotates at \omega = 10 rad/s about one end in a horizontal plane, inside a vertical magnetic field B = 0.2 T. The pivot end and the far end are connected via sliding contacts to the two terminals of an external resistor R = 2\ \Omega (the rod itself has negligible resistance). Find (a) the EMF across the rod, (b) the current through the external resistor, (c) the power dissipated in the resistor, and (d) the torque required to keep the rod rotating at constant \omega.

Rotating rod connected through sliding contacts to an external resistorTop-down view. A rod of length 0.5 m rotates counterclockwise about a central pivot in a vertical magnetic field. Sliding contacts at the pivot and at the tip connect to a circuit with a 2 ohm resistor. The EMF, current direction, and retarding force on the rod are labelled.B outωRI = ε/Rretarding force on rod: F = BIL
Top-down view of the setup. The rod rotates counterclockwise in the plane of the page; the field $\vec{B}$ is out of the page. Sliding contacts at both ends feed the EMF through an external resistor $R$.

Step 1. Apply equation (1) directly:

\varepsilon \;=\; \tfrac{1}{2}BL^2\omega \;=\; \tfrac{1}{2} \times 0.2 \times (0.5)^2 \times 10 \;=\; 0.25\ \text{V}.

Why: the rotating-rod formula is exact; no approximation needed. Units: T × m² × rad/s = Wb/s = V.

Step 2. Current through the external resistor (the rod itself has no resistance, so the full EMF is across R):

I \;=\; \frac{\varepsilon}{R} \;=\; \frac{0.25}{2} \;=\; 0.125\ \text{A}.

Why: Ohm's law for a loop with one EMF source and one resistance. A real rod would have some small resistance r in series and the formula would become I = \varepsilon / (R + r), but the problem states the rod has negligible resistance.

Step 3. Electrical power dissipated in the resistor:

P \;=\; I^2 R \;=\; (0.125)^2 \times 2 \;=\; 0.0313\ \text{W} \;=\; 31.3\ \text{mW}.

Step 4. Torque required. The rod carries the current I in a magnetic field B; each element at radius r experiences a force dF = B I\,dr perpendicular to the rod and perpendicular to the rotation. The torque about the pivot is dr \cdot dF... wait: the lever arm is r, the force is B I\,dr, so the torque from a length element is dN = r \cdot B I\,dr. Integrate:

N \;=\; \int_0^L r\,B I\,dr \;=\; B I \frac{L^2}{2} \;=\; 0.2 \times 0.125 \times \frac{0.25}{2} \;=\; 3.13\ \text{mN·m}.

By Lenz's law, this torque opposes the rotation; the external agent applies an equal and opposite torque to keep \omega constant.

Why: the current in the rod, crossed with B, gives a force per unit length. Each element of the rod is at a different radius, so each contributes a different torque. Integrating gives the full retarding torque.

Step 5. Energy conservation check. Mechanical power supplied by the external agent:

P_{\text{mech}} \;=\; N \omega \;=\; 3.13 \times 10^{-3} \times 10 \;=\; 0.0313\ \text{W} \;=\; 31.3\ \text{mW}.

This exactly equals the electrical power dissipated — confirming energy is conserved. Nothing is lost, nothing is gained; the mechanical input is completely converted to heat in the resistor.

Result: \varepsilon = 0.25 V, I = 0.125 A, P = 31.3 mW, N = 3.13 mN·m. Mechanical power in = electrical power out, exactly.

What this shows: The rotating-rod-with-resistor is a miniature DC generator. Its behaviour — EMF quadratic in L and linear in \omega, torque quadratic in everything through B^2 L^4 \omega / R — is the prototype for every Faraday disc and every homopolar generator. The energy-balance check is the physicist's way of catching a sign error or a missing factor: if mechanical power in does not equal electrical power out, something is wrong.

Example 2: LC circuit — when is energy shared equally?

An LC circuit has L = 4 mH, C = 100 pF, and is initialised by charging the capacitor to Q_0 = 10\ \muC with the switch open. At t = 0 the switch closes. Find (a) the natural frequency f_0 and period T, (b) the first time at which U_C = U_L, (c) the current at that instant.

Sketch of U_C and U_L showing the equal-energy timeA plot of U_C equals Q_0 squared over 2C times cosine squared of omega t versus time. U_L equals Q_0 squared over 2C times sine squared of omega t, the complementary curve. The two curves intersect at omega t equals pi over 4, i.e. t equals T over 8. Both functions equal half the total energy at that intersection.tenergy0U₀/2U₀T/8T/4T/23T/4TU_C = U_L = U₀/2U_CU_L
Electric energy $U_C$ (red) and magnetic energy $U_L$ (dark) over one period. They cross at $t = T/8, 3T/8, 5T/8, 7T/8$ — the four instants per cycle where each energy is exactly half the total.

Step 1. Natural frequency from equation (6):

\omega_0 \;=\; \frac{1}{\sqrt{LC}} \;=\; \frac{1}{\sqrt{4 \times 10^{-3} \times 100 \times 10^{-12}}} \;=\; \frac{1}{\sqrt{4 \times 10^{-13}}} \;=\; \frac{1}{6.325 \times 10^{-7}} \;=\; 1.581 \times 10^{6}\ \text{rad/s}.
f_0 \;=\; \frac{\omega_0}{2\pi} \;=\; 252\ \text{kHz}, \qquad T \;=\; \frac{1}{f_0} \;=\; 3.97\ \mu\text{s}.

Why: \omega_0 = 1/\sqrt{LC} is the universal natural frequency of an ideal LC loop, independent of the initial charge amplitude. The period is about 4 microseconds — an AM-radio-type frequency.

Step 2. Equal-energy condition. U_C(t) = (Q_0^2/2C)\cos^2(\omega_0 t) and U_L(t) = (Q_0^2/2C)\sin^2(\omega_0 t). Setting them equal:

\cos^2(\omega_0 t) \;=\; \sin^2(\omega_0 t) \;\;\Longleftrightarrow\;\; \tan^2(\omega_0 t) \;=\; 1 \;\;\Longleftrightarrow\;\; \omega_0 t \;=\; \pi/4, \; 3\pi/4, \; \ldots

The first occurrence is at \omega_0 t = \pi/4, i.e. t = \pi/(4\omega_0) = T/8.

t_{\text{eq}} \;=\; \frac{T}{8} \;=\; \frac{3.97}{8}\ \mu\text{s} \;=\; 0.497\ \mu\text{s}.

Why: the equal-energy point is where the capacitor has discharged by half (in energy terms), which is when the charge has fallen to Q_0/\sqrt{2} — i.e., when \cos(\omega_0 t) = 1/\sqrt{2}, i.e., when \omega_0 t = \pi/4. This is one-eighth of a period because the energy frequency is twice the charge frequency: \cos^2(\omega_0 t) = (1 + \cos 2\omega_0 t)/2 oscillates at 2\omega_0.

Step 3. Current at t_{\text{eq}}. From the magnetic-energy condition U_L = U_0/2 = Q_0^2/(4C):

\tfrac{1}{2}L I^2 \;=\; \frac{Q_0^2}{4C} \;\;\Longrightarrow\;\; I \;=\; \frac{Q_0}{\sqrt{2LC}} \;=\; \frac{Q_0\,\omega_0}{\sqrt{2}}.

Numerically:

I \;=\; \frac{10 \times 10^{-6} \times 1.581 \times 10^{6}}{\sqrt{2}} \;=\; \frac{15.81}{\sqrt{2}} \;=\; 11.2\ \text{A}.

Why: at the equal-energy instant, the current is exactly 1/\sqrt{2} of its peak value I_{\max} = Q_0\omega_0 = Q_0/\sqrt{LC}. Here I_{\max} = 15.81 A, which is a very large peak current for a microcoulomb-scale charge — characteristic of high-Q tuned circuits.

Result: f_0 = 252 kHz, T = 3.97\ \mus. Equal-energy at t = T/8 = 0.497\ \mus. Current at that instant: I = 11.2 A.

What this shows: The LC oscillator's energy frequency is twice its charge frequency — a fact that pays off in many JEE Advanced problems. The peak current can be surprisingly large when the capacitance is small: small C means small \sqrt{LC}, and I_{\max} = Q_0/\sqrt{LC} is therefore large even for modest Q_0. This is why high-frequency LC tank circuits must use wire thick enough not to burn out — the conduction losses scale with I^2 R_{\text{wire}}.

Common confusions

If you came here to solve JEE Advanced induction problems, the five tools plus the worked examples are what you need. What follows is for readers who want the reciprocity theorem M_{12} = M_{21} derived from energy conservation, the two-mode structure of coupled LC circuits, and the treatment of an LCR loop's damped oscillation.

The reciprocity theorem: M_{12} = M_{21}

Why is mutual inductance symmetric? The result is not obvious — the flux through coil 2 due to coil 1 involves completely different geometry from the flux through coil 1 due to coil 2. Yet the two numbers are always equal. The proof uses energy conservation.

Start with both coils carrying zero current. Slowly ramp up I_1 to its final value while keeping I_2 = 0. Work done against the self-induction of coil 1: \tfrac{1}{2}L_1 I_1^2. Work against coil 2: zero (since I_2 = 0 throughout).

Now slowly ramp I_2 up from 0 to its final value, keeping I_1 fixed. Work against coil 2's self-induction: \tfrac{1}{2}L_2 I_2^2. But coil 1 feels a time-varying flux from coil 2 (via M_{12}): this induces an EMF -M_{12}\,dI_2/dt in coil 1, and to keep I_1 fixed the external source must do work M_{12} I_1 \,dI_2 per increment. Integrating gives extra work M_{12} I_1 I_2.

Total work: W_A = \tfrac{1}{2}L_1 I_1^2 + \tfrac{1}{2}L_2 I_2^2 + M_{12} I_1 I_2.

Now reverse the order: ramp I_2 first, then I_1. By identical reasoning, total work is W_B = \tfrac{1}{2}L_2 I_2^2 + \tfrac{1}{2}L_1 I_1^2 + M_{21} I_2 I_1.

Energy conservation requires W_A = W_B (the final state is the same, so the work to reach it must be the same regardless of path). Therefore M_{12} = M_{21}, written simply as M. The symmetry of mutual inductance is a deep consequence of energy conservation in linear magnetic systems.

Damped LC — the LCR loop

Add a resistor R to the LC loop. Kirchhoff becomes

L\,\ddot q + R\,\dot q + \frac{q}{C} \;=\; 0.

This is the damped-oscillator equation. Solutions depend on the damping ratio \zeta = R/(2\sqrt{L/C}):

  • \zeta < 1 (under-damped): q(t) = Q_0 e^{-Rt/(2L)}\cos(\omega' t + \phi), with \omega' = \sqrt{\omega_0^2 - (R/(2L))^2}. Oscillates with slowly decaying amplitude.
  • \zeta = 1 (critical): q(t) = (A + Bt) e^{-Rt/(2L)}. Returns to zero in minimum time without oscillating.
  • \zeta > 1 (over-damped): two different exponential decays, no oscillation.

The Q-factor (quality factor) Q = \omega_0 L/R = \sqrt{L/C}/R measures how many oscillations the circuit rings for: amplitude drops by e^{-\pi} (\approx 1/23) in Q/\pi cycles. A good AM-radio tuning coil has Q \sim 100; a superconducting RF cavity has Q \sim 10^9.

Coupled LC circuits and normal modes

Two LC circuits linked by a mutual inductance M form a coupled oscillator. Let (L_1, C_1, q_1) be one loop and (L_2, C_2, q_2) the other. Kirchhoff:

L_1 \ddot q_1 + M \ddot q_2 + \frac{q_1}{C_1} = 0, \qquad L_2 \ddot q_2 + M \ddot q_1 + \frac{q_2}{C_2} = 0.

Look for normal modes of the form q_j(t) = a_j \cos(\omega t). Substituting and writing as a matrix eigenvalue problem gives two mode frequencies \omega_\pm. For identical loops (L_1 = L_2 = L, C_1 = C_2 = C):

\omega_\pm^2 \;=\; \frac{1/C}{L \pm M}.
  • \omega_+ (symmetric mode): charges in the two loops swing in phase; effective inductance is L + M; frequency lower.
  • \omega_- (antisymmetric mode): charges swing 180° out of phase; effective inductance is L - M; frequency higher.

If you excite one loop and not the other, the energy will beat between the two loops at the difference frequency |\omega_+ - \omega_-|, exactly like two coupled pendulums. This is the physics of the electromagnetic-resonance transformer used in wireless charging (MIT's 2007 "WiTricity" demonstration sent 60 W across 2 m using two coupled LC loops at 10 MHz). Coupled LC circuits also underpin the bandpass-filter topology in every mobile phone's radio front end.

Rogowski's coil — a beautiful JEE-Advanced application

A toroidal coil of small cross-section wrapped tightly around a large straight wire measures the current in the wire by Faraday's law. The magnetic field of the straight wire at distance r is B = \mu_0 I/(2\pi r); the flux through the toroid's small cross-section is \Phi = \mu_0 I A_{\text{toroid}}/(2\pi r) (approximately, for small toroid radius); the induced EMF in the toroid's secondary winding measures dI/dt directly. Integrate this EMF in time (with an op-amp integrator) and you get I(t) — no direct electrical contact with the primary wire needed. Modern non-contact current probes (used at every BHEL substation and every HV test lab at CPRI Bengaluru) are Rogowski coils in essence.

Where this leads next