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Electromagnetic Induction · TOPIC LESSON

Electromagnetic Induction

Track changing flux, follow Lenz’s direction, and account for the energy in inductive circuits.

JEE Main · JEE Advanced · NEET UG75 min study guide9 indexed sections

BY THE END, YOU CAN

  • Calculate induced emf from field, area and orientation changes.
  • Explain motional emf, induced electric fields and mechanical energy conversion.
  • Derive self-inductance and magnetic stored energy under linear assumptions.
  • Solve RL growth/decay with correct continuity and final-state conditions.

BEFORE YOU START

  • Magnetic flux, Lorentz force and simple circuit equations.
  • Exponential functions, elementary derivatives and energy conservation.

Original educational lesson. Work through the examples before revealing the practice answers. Check the current official syllabus separately; historical question frequency does not establish eligibility or predict future questions.

CONNECT THE CONCEPT TO THE EXAM

Selected exam questions & worked solutions

These selected examples are separate from the original practice below. Each identifies its exam, year and wording. They do not represent every paper or every selected year.

No source-verified exam questions have been published for this lesson yet. The worked examples and practice below are original teaching material, not past-exam questions.

01 / UNDERSTAND & APPLY

1. A changing flux linkage creates an emf

Electromagnetic induction connects changing magnetic conditions to an electric driving effect around a circuit. For a loop with consistently chosen positive traversal and area normal, Faraday’s law gives ε = −dΦB/dt. For N turns linking the same flux, ε = −N dΦB/dt. Emf has units of volts and represents work per unit charge around the loop, not necessarily a potential difference between two uniquely defined points.

Magnetic flux is ΦB = ∫B·dA. It can change because the field changes, the loop area changes, its orientation changes, or the loop moves into a spatially different field. A large constant flux produces no induced emf in a fixed loop. Conversely, flux can be momentarily zero while changing rapidly and producing a substantial emf.

The minus sign is Lenz’s law: the induced current’s magnetic effect opposes the change in flux that drives it. It does not always oppose the original field. If outward flux is increasing, the induced field is inward; if outward flux is decreasing, the induced field is outward to resist that decrease. State the change first, then choose the current direction by the right-hand rule.

An induced emf can exist in an open loop, while a sustained conduction current requires a closed path. If resistance R dominates and self-inductance is negligible, I = ε/R. If appreciable inductance is present, changing current produces an additional emf and an RL equation is needed. Faraday’s law determines the driving effect, not automatically the final current.

ε = −d(NΦB)/dt

Flux linkage NΦB assumes identical linkage for every turn; otherwise sum the turn fluxes.

ΦB = BA cosθ

Uniform field through a plane area; θ is the angle to the area normal.

ORIGINAL WORKED EXAMPLE

A decreasing field

A 50-turn coil of area 0.020 m² is perpendicular to a uniform field that decreases from 0.30 T to 0.10 T in 0.10 s. Find average emf magnitude and induced-field direction.

  1. Flux-linkage change is N A ΔB = 50 × 0.020 × (−0.20) = −0.20 Wb-turn.
  2. Average emf magnitude is 0.20/0.10 = 2.0 V.
  3. The induced field points in the original field direction to oppose its decrease. The corresponding current sense depends on which side is viewed.

Answer 2.0 V; induced field reinforces the original direction.

02 / UNDERSTAND & APPLY

2. A moving conductor separates charge through magnetic force

Consider a rod of length l moving at speed v perpendicular to both its length and a uniform B. Charges in the rod experience qv × B and redistribute until an internal electric field balances that magnetic force in open-circuit equilibrium. The potential difference magnitude between ends is Blv. Its polarity follows the force on positive charge, not an arbitrary battery symbol.

If the rod slides on conducting rails to make a closed circuit, charges can circulate. For an ideal rail geometry with resistance R and negligible inductance, current magnitude is Blv/R. The current-carrying rod experiences a magnetic force opposing the motion. An external agent maintaining constant speed must supply mechanical power, which becomes electrical dissipation.

The same emf can be computed through the changing area of the loop: Φ = Blx, so |dΦ/dt| = Blv. This agreement links the force-on-carriers description to the flux rule. For a moving circuit, the general emf includes both the electric field and the magnetic force from local circuit motion; simply applying a stationary-loop electric-field argument can miss the motional term.

For a rod rotating about one end in uniform perpendicular B, speed varies with radius: v = ωr. Integrating dε = Bωr dr from 0 to l gives ε = Bωl²/2 in magnitude. Replacing every point’s speed by the tip speed would overestimate by a factor of two. End polarity follows the sign of B and rotation sense.

Full-size scientific diagram. Swipe or scroll horizontally; keyboard users can focus the diagram and use arrow keys.

Sliding rod: force, current and energy transferA rod moves right on rails in a field into the page. Current rises through the rod and circulates counterclockwise through a left-hand resistor. Magnetic force on the rod points left.RvIFmagB into page: ×Moving rod, length lCounterclockwise current; force opposes motion
Schematic. Length, velocity and field are mutually perpendicular. Constant-speed external power equals I²R when inductance is negligible.

ORIGINAL WORKED EXAMPLE

A sliding rod and energy balance

A 0.40 m rod moves at 5.0 m s⁻¹ across rails in a 0.50 T perpendicular field. Total circuit resistance is 2.0 Ω and inductance is negligible. Find emf, current, resisting force and required mechanical power.

  1. Emf is Blv = 0.50 × 0.40 × 5.0 = 1.0 V.
  2. Current is ε/R = 0.50 A. Magnetic force magnitude is BIl = 0.50 × 0.50 × 0.40 = 0.10 N opposite motion.
  3. Mechanical power needed at constant speed is Fv = 0.50 W.
  4. Electrical heating is I²R = 0.25 × 2.0 = 0.50 W, confirming energy balance.

Answer 1.0 V; 0.50 A; 0.10 N; 0.50 W.

03 / UNDERSTAND & APPLY

3. Induced electric fields are not electrostatic

For a stationary loop, Faraday’s law is ∮E·dl = −dΦB/dt. An induced electric field can circulate around a changing magnetic field even where no conducting wire is present. A wire merely provides mobile charges that reveal the effect. The field’s closed circulation means it cannot be represented globally by one electrostatic scalar potential alone.

If a uniform axial magnetic field occupies a circular region of radius R and changes in time, cylindrical symmetry makes the induced electric field azimuthal. For a circular path of radius r < R, E(2πr) = −πr² dB/dt with a signed tangential convention, so magnitude is (r/2)|dB/dt|. For r > R, only the field-filled area contributes flux, giving magnitude R²|dB/dt|/(2r).

The induced electric field is continuous at r = R in this ideal model, rising linearly inside and falling inversely outside. It can be nonzero outside the region where B itself is nonzero. This is a useful reminder that local field magnitude and enclosed flux change are different quantities.

In a circuit containing time-varying flux, voltage readings can depend on the paths of the measurement leads because the induced electric field has nonzero circulation. Elementary lumped circuit diagrams package this induction into explicit emf sources and inductors. Consistent modelling avoids treating all induced effects as ordinary electrostatic voltage drops.

ORIGINAL WORKED EXAMPLE

An induced field outside the changing-B region

A uniform field confined to radius 0.10 m changes at 4.0 T s⁻¹. Find induced electric-field magnitude at r = 0.20 m.

  1. The circular integration path encloses flux change rate πR² dB/dt, not πr² dB/dt.
  2. Use E = R²|dB/dt|/(2r) = (0.10)² × 4.0/(0.40).
  3. This gives 0.10 V m⁻¹, despite zero idealized magnetic field at the observation radius.

Answer 0.10 V m⁻¹.

04 / UNDERSTAND & APPLY

4. An inductor resists changes in current

A current creates magnetic flux through its own circuit. In a linear fixed-geometry system, flux linkage is proportional to current: NΦ = LI. The proportionality L is self-inductance, measured in henry. Changing current therefore induces emf εL = −L dI/dt. The opposition is to change in current, not necessarily to current itself.

For an ideal long solenoid of length l, area A and N turns in a linear medium of permeability μ, internal field is μNI/l. Flux linkage is N times BA, giving L = μN²A/l. Doubling turn count at fixed geometry quadruples inductance. This model neglects fringe fields, winding resistance and nonlinear core behaviour.

An ideal inductor with constant current has zero induced voltage, even if its magnetic field and stored energy are large. A rapidly changing small current can require a large voltage. With finite voltage, current through an ideal inductor cannot jump instantaneously because integrating vL = L dI/dt over an infinitesimal interval gives no finite current change.

At a switching instant, carry the pre-switch current into the new circuit. At long times in a DC circuit an ideal inductor behaves as a short, while its winding resistance remains a resistor in a real model. These are the reverse types of limiting behaviour from a capacitor, whose voltage is continuous and whose steady DC current vanishes.

NΦ = LI; εL = −L dI/dt

Linear inductance and fixed geometry. In passive circuit sign convention, the voltage drop is vL = L dI/dt.

Lsolenoid = μN²A/l

Long-solenoid approximation with linear permeability.

ORIGINAL WORKED EXAMPLE

Voltage during a current ramp

A 0.40 H ideal inductor carries current increasing uniformly from 1 A to 4 A in 0.020 s. Find induced emf magnitude.

  1. Current change rate is (4 − 1)/0.020 = 150 A s⁻¹.
  2. Induced emf magnitude is L|dI/dt| = 0.40 × 150 = 60 V.
  3. Its polarity opposes the increase; under the passive sign convention the applied inductor voltage drop is +60 V in the current direction.

Answer 60 V.

05 / UNDERSTAND & APPLY

5. The magnetic field stores the work of building current

To increase current against an inductor’s opposing emf, a source must supply power vI = LI dI/dt. Integrating from zero current gives stored energy U = LI²/2 for a linear inductor. This energy is recoverable in an ideal circuit; it is not automatically heat. Resistance in the winding or external circuit is what dissipates energy.

For a vacuum magnetic field, energy density is B²/(2μ₀). The long-solenoid energy can be found either from LI²/2 or by multiplying this density by the approximate field volume Al; the results agree. In a simple linear magnetic medium the corresponding expression uses its permeability under the appropriate linear assumptions.

Opening an inductive circuit can generate a large voltage as the inductor attempts to keep current continuous through whatever path remains, including parasitic capacitance or an arc. The ideal formula does not predict safe or infinite physical voltages without a complete circuit model. Practical circuits provide a controlled path for stored energy, such as a flyback diode in suitable DC coil circuits.

If an isolated ideal inductor begins with energy LI₀²/2 and discharges through a resistor with no other storage element, all that energy eventually becomes heat. The resistance changes the decay rate, not the total initial energy available. This is the magnetic counterpart of capacitor discharge.

ORIGINAL WORKED EXAMPLE

Stored energy and a changed current

A 2.0 H inductor carries 3.0 A. Find stored energy and the additional energy needed to raise current to 6.0 A.

  1. Initial energy is (1/2)(2.0)(3.0²) = 9.0 J.
  2. Final energy is (1/2)(2.0)(6.0²) = 36 J.
  3. Additional energy is 27 J, not 9 J, because stored energy scales as current squared.

Answer Initial 9.0 J; additional 27 J.

06 / UNDERSTAND & APPLY

6. Derive current growth and decay in an RL circuit

For a series resistor R and ideal inductor L connected to constant emf ε at t = 0 with zero initial current, the loop equation is ε = IR + L dI/dt. At the start, I = 0 and the entire source voltage appears across the inductor. As current grows, the resistor drop grows and the current’s rate of increase falls.

The solution is I(t) = (ε/R)[1 − exp(−tR/L)]. The time constant is τ = L/R, and the final current is ε/R. After one τ, current reaches 63.2% of its final value. The inductor voltage falls as ε exp(−t/τ), while resistor voltage rises toward ε. Unlike a capacitor, the inductor starts by opposing current establishment.

If the source is removed and the inductor discharges through a closed resistance R, current decays as I₀ exp(−t/τ). Its direction initially remains the same; the inductor’s terminal polarity reverses to drive the existing current. Energy decays as exp(−2t/τ), because it is proportional to I².

For a fixed linear one-inductor circuit with constant sources after switching, L > 0 and passive resistance Rseen > 0 seen by the inductor, use I(t) = I∞ + (I₀ − I∞)exp(−t/τ), with τ determined by the resistance seen by the inductor and L. Find I₀ from pre-switch continuity and I∞ from the new DC steady state. Do not use a source resistance from the old switch position to set the new decay timescale.

Full-size scientific diagram. Swipe or scroll horizontally; keyboard users can focus the diagram and use arrow keys.

The inductor needs a closed path during current decayA series resistor and inductor connect through a two-position switch. A connects the positive source terminal for growth. B removes the source but closes a resistor-inductor return path for decay. Current goes downward through the inductor and remains continuous at switching.ε+−A: current growthB: closed decay pathRLII is continuous; τ = L/R > 0
Ideal L>0 and passive total R>0. The switch shown in A never bridges the source to return; position B provides a closed decay loop.

ORIGINAL WORKED EXAMPLE

Current after one time constant

A 12 V source is connected to a 6 Ω resistor and 3 H inductor in series, initially with zero current. Find τ, final current and current at 0.50 s.

  1. Time constant is L/R = 3/6 = 0.50 s.
  2. Final current is ε/R = 2 A.
  3. At one time constant, I = 2(1 − e⁻¹) ≈ 1.264 A. Inductor voltage is 12/e ≈ 4.41 V.

Answer τ = 0.50 s; I∞ = 2 A; I(0.50 s) ≈ 1.26 A.

07 / UNDERSTAND & APPLY

7. Coupling circuits and converting mechanical motion

When changing current in one coil changes flux linkage through another, the coils have mutual inductance M. The magnitude of induced emf in coil 2 is |M| |dI₁/dt| for fixed linear geometry, with sign set by winding orientation and Lenz’s law. For passive coupled coils, |M| cannot exceed √(L₁L₂); a coupling coefficient k between zero and one quantifies the linked fraction in a common model.

Mutual inductance depends on geometry, turns and magnetic environment, not on the instantaneous current in a linear model. Moving coils or saturating a magnetic core can change that relation. A transformer uses mutual induction with alternating flux; its steady operation is analysed with AC relations in the alternating-current lesson.

For a coil of N turns and area A rotating at constant angular speed ω in uniform B, flux linkage can be written NBA cosωt after choosing time origin. Differentiating gives sinusoidal emf NBAω sinωt, with peak NBAω. The emf is greatest when flux passes through zero and zero when flux is at an extremum.

Eddy currents are circulating currents induced within bulk conductors by changing flux. They can dissipate energy and create magnetic braking. Laminating a core increases resistance to large circulating paths and reduces unwanted eddy-current loss. Induction heating instead uses such dissipation intentionally. In each case the energy comes from the source changing the field or producing motion, not from Lenz’s law creating energy.

08 / UNDERSTAND & APPLY

8. Original practice: change, direction and timescale

For direction questions, name the change in flux before naming the induced field. For transient questions, state which quantity must remain continuous.

ORIGINAL PRACTICE · 1

A stationary loop has constant magnetic flux of 0.50 Wb. What is induced emf from that flux?

Reveal answer & reasoning

Zero.

Faraday’s law depends on time derivative, not flux magnitude.

ORIGINAL PRACTICE · 2

Flux through each of 100 turns rises by 0.002 Wb in 0.050 s. Find average emf magnitude.

Reveal answer & reasoning

4 V.

NΔΦ/Δt = 100 × 0.002/0.050.

ORIGINAL PRACTICE · 3

Out-of-page flux through a loop is increasing. Viewed from the front, what induced current direction occurs?

Reveal answer & reasoning

Clockwise.

The induced field must point into the page to oppose the increase; clockwise current produces that direction.

ORIGINAL PRACTICE · 4

Out-of-page flux is decreasing instead. What current direction occurs?

Reveal answer & reasoning

Counterclockwise.

The induced field now points out of the page to oppose the decrease.

ORIGINAL PRACTICE · 5

A 0.50 m rod moves at 4 m s⁻¹ in a 0.20 T field, with its length, velocity and field mutually perpendicular. Find motional emf.

Reveal answer & reasoning

0.40 V.

Blv = 0.20 × 0.50 × 4.

ORIGINAL PRACTICE · 6

A rod of length l rotates about one end at angular speed ω in perpendicular B. Why is emf not Bωl²?

Reveal answer & reasoning

Speed varies along the rod.

Integrating Bωr dr gives Bωl²/2; using tip speed for the whole rod overcounts.

ORIGINAL PRACTICE · 7

A solenoid’s turn count doubles with fixed geometry and permeability. What happens to self-inductance?

Reveal answer & reasoning

It quadruples.

L ∝ N².

ORIGINAL PRACTICE · 8

An ideal inductor has constant 5 A current. What is its induced emf?

Reveal answer & reasoning

Zero.

dI/dt = 0 even though stored magnetic energy can be nonzero.

ORIGINAL PRACTICE · 9

A 0.20 H inductor carries 4 A. Find energy.

Reveal answer & reasoning

1.6 J.

U = LI²/2 = 0.20 × 16/2.

ORIGINAL PRACTICE · 10

A 2 H inductor discharges through 4 Ω. Find current half-life.

Reveal answer & reasoning

(0.5 ln2) s ≈ 0.347 s.

τ = L/R = 0.5 s and I/I₀ = e^(−t/τ).

ORIGINAL PRACTICE · 11

In an RL discharge, current becomes half its initial value. What fraction of stored energy remains?

Reveal answer & reasoning

One quarter.

U ∝ I².

ORIGINAL PRACTICE · 12

A rotating coil has peak flux linkage 0.10 Wb-turn and angular speed 200 rad s⁻¹. Find peak emf.

Reveal answer & reasoning

20 V.

Differentiating the sinusoidal linkage multiplies its amplitude by ω.

09 / UNDERSTAND & APPLY

9. A consistency test for every induction result

Check the changing quantity, the circuit’s continuity and the energy source. A stationary loop with constant flux cannot sustain an induction emf just because B is large. A moving rod in a closed resistive circuit must require mechanical input if it is delivering heat. An inductor’s current cannot change discontinuously under a finite voltage.

For signs, choose a surface normal and traversal direction once, or use Lenz’s law physically and report magnitude separately. For transients, test the initial slope, final current and L/R units. Finally compare energy lost from the magnetic field with energy delivered elsewhere. These checks turn several memorised formulas into one conservation-based account.

  • Induction responds to changing flux linkage, not merely magnetic field presence.
  • Lenz’s law opposes the change, which can mean reinforcing the original field.
  • Motional emf and induced electric circulation are complementary descriptions.
  • Self-inductance opposes current change, not current itself.
  • Inductor current is continuous when voltage is finite.
  • Stored magnetic energy is LI²/2 and must go somewhere during switching.

Full-size scientific diagram. Swipe or scroll horizontally; keyboard users can focus the diagram and use arrow keys.

Current changes continuously on the L/R timescaleSeparate normalized examples show current growth from zero to I∞ and decay from I0 to zero. At t=τ=L/R, growth reaches 0.632 and decay retains 0.368. Both curves start continuously from their stated initial currents.0123400.510.6320.368Growth: I/I∞ = 1 − e^(−t/τ)Decay: I/I₀ = e^(−t/τ)Normalized currentt/τ, with τ = L/R > 0
Quantitative curves for L>0, passive R>0 and constant post-switch sources. Growth and decay are separate circuits/initial conditions, not two simultaneous currents in one branch.

KEEP THE BIG IDEAS

Quick revision

  • A nonzero flux is not the same as a changing flux.
  • The induced effect resists the cause of the flux change in an energy-consistent direction.
  • An inductor stores magnetic energy and constrains current continuity.
  • RL time constants come from the post-switch circuit, not just the component labels.
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The teaching notes, worked explanations and original practice are written for this lesson. Separately labeled selected exam questions identify their exam, year, source and whether wording is paraphrased or an exact excerpt. Educational references do not certify current exam coverage.

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