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JEE Mains Physics · Electromagnetic Induction

Magnetic Flux, Faraday's Law and Lenz's Law

Flux is NBA cos θ with θ measured from the normal; the induced emf is the rate at which the flux changes, ε = −dΦ/dt, and the induced current always flows so as to oppose that change.

Why this matters

Thirty PYQs, eighteen of them multiple choice, and seven from 2026. Fifteen ask for an emf or a current from a flux that changes, through a field, an area or a flux formula in time; six ask for the charge, heat or power that follows; nine are about the direction of the induced current, or which changes induce an emf at all. Nearly all of them start by writing the flux down.

Concept 1 of 3: Induced emf from a changing magnetic flux

Flux counts the field lines through a loop: the field times the area, times the cosine of the angle between the field and the loop's NORMAL. The emf is how fast that number changes. So find what is changing, the field, the area or the angle, write the flux as a function of time, and differentiate. Only the field component along the normal counts.

Definition

  • Φ=NBAcos⁡θ\Phi = NBA\cos\theta, with θ\theta between B⃗\vec B and the normal. Plane perpendicular to B: θ=0\theta = 0, full flux. Plane parallel to B: zero flux.
  • ε=−dΦdt\varepsilon = -\dfrac{d\Phi}{dt}, and the current is I=ε/RI = \varepsilon/R. The minus sign is Lenz's law; for a size, take the magnitude.
  • Flux given as a polynomial in t: differentiate, then put in t. A constant term in Φ\Phi adds nothing.
  • Field changing, area fixed: ε=NAcos⁡θ dBdt\varepsilon = NA\cos\theta\,\dfrac{dB}{dt}. On a B–t graph, dB/dt is the slope of the segment. For B=B0sin⁡ωtB = B_0\sin\omega t, the largest emf is NAB0ωcos⁡θNAB_0\omega\cos\theta.
  • Field as a vector, loop in a coordinate plane: keep only the component along the loop's normal (a loop in the xy-plane sees only BzB_z).
  • Loop inside a long solenoid: B=μ0nIB = \mu_0 nI, and the area is the LOOP's area, not the solenoid's.
  • Area changing in a fixed field: a circle with dr/dtdr/dt gives ε=B⋅2πr drdt\varepsilon = B \cdot 2\pi r\,\dfrac{dr}{dt}. A circle reshaped into a square of the same perimeter loses area, so flux changes.
  • A finite change over a time: average emf =ΔΦ/Δt= \Delta\Phi/\Delta t.

Flux and Faraday's law

Φ=NBAcos⁡θε=−dΦdt\Phi = NBA\cos\theta \qquad \varepsilon = -\frac{d\Phi}{dt}

Worked example

A coil of 100 turns and area 30 cm230\ \text{cm}^{2} has its normal at 37∘37^{\circ} to a uniform field. The field rises steadily from 0.1 T0.1\ \text{T} to 0.6 T0.6\ \text{T} in 0.2 s0.2\ \text{s}. Find the induced emf. (Take cos⁡37∘=0.8\cos 37^{\circ} = 0.8.)
Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2026 · 6 Apr 2026 Shift 2 · Q14Moderate

Example 1 · Electromagnetic Induction · Magnetic Flux, Faraday's Law and Lenz's Law

A square loop of side 2 cm is placed in a time varying magnetic field with magnitude as B=0.4sin⁡(300t)B = 0.4\sin(300t) Tesla. The normal to the plane of loop makes an angle of 60∘60^{\circ} with the field. The maximum induced emf produced in the loop is ____\_\_\_\_ mV .

Angle measured from the plane

The cosine in NBA cos θ takes the angle between the field and the NORMAL. If a question gives the angle between the field and the plane of the loop, the cosine of that angle is the wrong factor: use its sine.

Using the solenoid's area for a loop inside it

A small loop inside a long solenoid links only the field over its own area. The flux is μ₀nI times the loop's area; the solenoid's larger cross-section does not enter.

Putting the time into the flux instead of its derivative

The emf at time t is dΦ/dt evaluated at t, not Φ(t) divided by t. Differentiate first, then substitute.

Concept 2 of 3: Charge, heat and power from an induced current

The charge that flows depends only on how much the flux changed, never on how fast: a quick change gives a big current for a short time, a slow one a small current for a long time, and the product is the same. Heat and power do depend on the rate, through ε²/R. When the field reverses, the flux goes from +NBA to −NBA, a change of twice NBA.

Definition

  • Charge: Q=N ΔΦRQ = \dfrac{N\,\Delta\Phi}{R} when Φ\Phi is the flux through one turn. Coil pulled out of the field: ΔΦ=BA\Delta\Phi = BA. Field reversed or coil flipped through 180°: ΔΦ=2BA\Delta\Phi = 2BA.
  • Average emf over a finite change: εˉ=N ΔΦ/Δt\bar\varepsilon = N\,\Delta\Phi/\Delta t.
  • Heat: H=∫ε2R dtH = \displaystyle\int \frac{\varepsilon^{2}}{R}\,dt; for a constant emf, ε2t/R\varepsilon^{2}t/R.
  • A sinusoidal emf of amplitude ε0\varepsilon_0: average power ε022R\dfrac{\varepsilon_0^{2}}{2R}, and energy per period is that power times 2π/ω2\pi/\omega.
  • Scaling a short-circuited coil: ε∝NA\varepsilon \propto NA; its wire's resistance ∝\propto (wire length)/(wire cross-section) ∝NA/rw2\propto N\sqrt{A}/r_w^{2}. So P=ε2/R∝NA3/2rw2P = \varepsilon^{2}/R \propto NA^{3/2}r_w^{2}.

Induced charge and power

Q=N ΔΦRP=ε2RQ = \frac{N\,\Delta\Phi}{R} \qquad P = \frac{\varepsilon^{2}}{R}

Worked example

A coil of 50 turns and area 40 cm240\ \text{cm}^{2} lies with its plane perpendicular to a field of 0.8 T0.8\ \text{T}. The total resistance of its circuit is 5 Ω5\ \Omega. Find the charge that flows when (a) the coil is pulled right out of the field, (b) the coil is instead turned through 180∘180^{\circ}.
Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2023 · 13 April 2023 · Q115Moderate

Example 2 · Electromagnetic Induction · Magnetic Flux, Faraday's Law and Lenz's Law

An insulated copper wire of 100 turns is wrapped around a wooden cylindrical core of the cross-sectional area 24 cm224{\text{ }cm}^{2}. The two ends of the wire are connected to a resistor. The total resistance in the circuit is 12Ω12\Omega. If an externally applied uniform magnetic field in the core along its axis changes from 1.5 T1.5\text{ }T in one direction to 1.5 T1.5\text{ }T in the opposite direction, the charge flowing through a point in the circuit during the change of magnetic field will be___ mCmC.

Reversing the field is not 'no change'

When a field of size B turns to the opposite direction, the flux goes from +NBA to −NBA. The change is 2NBA, so the charge is twice that of simply removing the field.

Charge does not depend on the time taken

Q = NΔΦ/R has no time in it. A time given in such a question is there only for the average emf or current, not for the charge.

Power of a sinusoidal emf uses half the square of the peak

The average of sin² over a cycle is one half, so the mean power is ε₀²/2R. Using ε₀²/R doubles the answer.

Concept 3 of 3: Lenz's law and the direction of the induced current

The induced current makes a field of its own, and that field always fights the CHANGE in flux, not the flux itself. If the flux through a loop is growing, the loop's own field points against it; if it is shrinking, the loop's field points with it, trying to keep it up. The same opposition shows up as a force: a magnet pushed at a loop is repelled, a magnet falling through a metal tube is slowed.

Definition

  • Four steps: (1) which way does the external flux through the loop point? (2) is it growing or shrinking? (3) the induced field points against a growth and along a shrinkage; (4) curl the fingers of the right hand around that field to get the current.
  • A field out of the page reverses every direction in the table below.
  • An emf needs a CHANGE of flux: a change of B, of area, of angle (rotation) or a reversal of B. Moving a coil through a uniform field, at any speed, changes nothing.
  • In a solid conductor the induced currents are eddy currents. They drag on the motion that causes them: a magnet falling in a long copper tube reaches a steady speed, and a swinging metal plate between magnet poles stops quickly.
  • Coaxial coils: field of an anticlockwise current points towards the viewer. Moving a coil closer raises its field at a neighbour; moving it away lowers it.
SituationWhat the flux doesInduced current or effect
Field into the page, increasingFlux into the page growsAnticlockwise, so its own field points out of the page
Field into the page, decreasingFlux into the page fallsClockwise, so its own field points into the page
North pole pushed towards a loopFlux from the magnet growsNear face of the loop becomes a north pole; magnet repelled
North pole pulled away from a loopFlux from the magnet fallsNear face becomes a south pole; magnet attracted back
Bar magnet passing right through a loopRises as it enters, falls as it leavesTwo emf pulses of opposite sign, with a gap while it is inside
Coil moved through a uniform fieldUnchangedNo emf and no current
Coil rotated in a uniform fieldChanges with the angleAlternating emf
Field reversed in directionChanges by twice BAEmf while it reverses
Magnet dropped down a long copper tubeChanges in every ring of the tubeEddy currents brake it; it falls at a nearly constant speed
A non-magnetic bar of the same size falls freely and arrives first.
The induced current opposes the change in flux, never the flux itself.
Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2026 · 22 Jan 2026 Shift 1 · Q20Moderate

Example 3 · Electromagnetic Induction · Magnetic Flux, Faraday's Law and Lenz's Law

Three identical coils C1,C2C_{1},C_{2} and C3C_{3} are closely placed such that they share a common axis. C2C_{2} is exactly midway. C1C_{1} carries current II in anti-clockwise direction while C3C_{3} carries current II in clockwise direction. An induced current flows through C2C_{2} will be in clockwise direction when

Opposing the flux instead of its change

A growing flux into the page gives an induced field out of the page, but a shrinking flux into the page gives an induced field INTO the page. The current opposes the change, so it can point the same way as the external field.

Motion alone does not induce an emf

A coil translated through a uniform, steady field, even with changing speed, keeps the same flux. Only a change of field, area, angle or direction induces an emf.

Eddy currents need a conductor

An insulator carries no eddy currents, so it feels no magnetic braking. The drag on a falling magnet comes from currents in the metal tube around it.

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (2)

Reference tables (1)

Lenz's law and the direction of the induced current9 rows
SituationWhat the flux doesInduced current or effect
Field into the page, increasingFlux into the page growsAnticlockwise, so its own field points out of the page
Field into the page, decreasingFlux into the page fallsClockwise, so its own field points into the page
North pole pushed towards a loopFlux from the magnet growsNear face of the loop becomes a north pole; magnet repelled
North pole pulled away from a loopFlux from the magnet fallsNear face becomes a south pole; magnet attracted back
Bar magnet passing right through a loopRises as it enters, falls as it leavesTwo emf pulses of opposite sign, with a gap while it is inside
Coil moved through a uniform fieldUnchangedNo emf and no current
Coil rotated in a uniform fieldChanges with the angleAlternating emf
Field reversed in directionChanges by twice BAEmf while it reverses
Magnet dropped down a long copper tubeChanges in every ring of the tubeEddy currents brake it; it falls at a nearly constant speed
A non-magnetic bar of the same size falls freely and arrives first.
The induced current opposes the change in flux, never the flux itself.

Watch out for (9)

Test yourself on Electromagnetic Induction

20 past JEE Mains questions from this chapter, timed at 48 minutes and marked the way the exam marks it. You see your score and every answer the moment you finish. Free to start.