Electromagnetic Induction | ISC Class 12 Physics Notes

This note covers electromagnetic induction, magnetic flux, Faraday’s laws, Lenz’s law, motional electromotive force, mechanical and electrical power, eddy currents, self-induction, mutual induction, transformers, energy losses and transmission of electrical power.

What observations establish electromagnetic induction?

Electromagnetic induction is the production of an electromotive force by a changing magnetic flux. An electromotive force, abbreviated emf, represents energy supplied per unit charge. When the conducting circuit is closed, the induced emf can drive an electric current.

A coil consists of turns of conducting wire insulated from one another. A galvanometer is an instrument that detects small electric currents. Connecting a coil to a galvanometer allows changes in current direction to be observed as opposite pointer deflections.

What happens when a magnet moves near a coil?

Pushing the north pole of a bar magnet towards the coil produces a deflection. Holding the magnet stationary produces no deflection. Withdrawing it produces a deflection in the opposite direction. Moving it faster produces a larger deflection.

Using the south pole reverses the deflections for corresponding movements. Moving the coil towards or away from a fixed magnet produces the same effects. These observations connect the induced current with relative motion in this arrangement.

A bar magnet has its north pole facing a coil labelled C₁. An arrow shows the magnet moving towards the coil. The coil is connected to a galvanometer labelled G; magnetic field lines pass through it.

See Fig. 6.1 in your NCERT textbook

Is mechanical motion essential?

Place two stationary coils near each other. Connect one to a galvanometer and the other to a battery through a key, which is a switch. Closing the key gives a momentary deflection; keeping it closed gives no continuing deflection. Opening it gives an opposite momentary deflection.

The battery coil’s current changes during switching, changing the magnetic field through the neighbouring coil. Inserting an iron rod along the coils’ common axis increases the deflection dramatically. Relative motion is not an absolute requirement: changing magnetic flux is the common feature.

Two separated coils, C₁ and C₂, share a dotted horizontal axis. C₁ connects to galvanometer G. C₂ connects to a battery and tapping key K. The circuits are separate.

See Fig. 6.3 in your NCERT textbook

What is magnetic flux and how is it measured?

Magnetic flux, denoted by Φ, measures the magnetic field passing through a specified surface. For a plane surface in a uniform magnetic field, let B be magnetic flux density, A the area and θ the angle between the field and the perpendicular to the surface.

Φ = BA cos θ

The area vector has magnitude A and points normally to the surface, meaning perpendicular to it. The angle θ is measured from this normal, not from the plane itself. Flux is a scalar, although its sign depends on the chosen normal.

The SI unit of magnetic flux is weber, symbol Wb. 1 Wb = 1 T m², where T denotes tesla, the unit of magnetic flux density, and m denotes metre. Thus, a uniform perpendicular field of one tesla through one square metre gives one weber.

How does orientation affect flux?

For a curved surface or a non-uniform field, divide the surface into small area elements. Calculate the field’s normal contribution through each element and add these contributions. The simple expression BA cos θ applies directly when the field and surface orientation are uniform.

Parallel magnetic field lines pass through a tilted plane. The field is labelled B and the area vector is labelled A. The marked angle θ lies between the field direction and the area vector.

See Fig. 6.4 in your NCERT textbook

For a closely wound coil with N turns experiencing the same flux per turn, flux linkage is NΦ. N is a dimensionless count. Distinguish flux through one turn from total flux linkage before applying an induction formula.

Which units and dimensions are used?

In dimensional expressions below, M, L, T and A denote the base dimensions mass, length, time and electric current respectively. They are dimensional labels, not the inductance symbols introduced later.

How do Faraday’s laws determine induced emf and current?

Faraday’s first law identifies the condition for induction: an emf is induced when the magnetic flux linked with a circuit changes. Faraday’s second law relates the magnitude of this emf to the rate of change of flux linkage.

Let ε denote induced emf and t denote time in seconds, symbol s. The notation dΦ/dt means instantaneous rate of change of flux. For N turns with equal flux per turn, ε = −N dΦ/dt. The negative sign expresses Lenz’s law.

For a finite interval Δt, where Δ means final value minus initial value, the magnitude of the average emf is |ε| = N|ΔΦ|/Δt. Vertical bars indicate magnitude. A steady rate of flux change makes this average equal to the instantaneous magnitude throughout the interval.

The SI unit of emf is volt. 1 V = 1 Wb/s. For a closed circuit of resistance R, measured in ohms, symbol Ω, the induced current magnitude is I = |ε|/R, when resistance determines the current and effects of the circuit’s own changing magnetic flux can be neglected.

The SI unit of resistance is ohm. 1 Ω = 1 V/A. Resistance has dimensions [M L² T⁻³ A⁻²]; time has dimensions [T]. An open circuit can have induced emf across its ends without a circulating conduction current.

Which changes can produce induction?

From the flux expression, changing B, A or θ can change flux. Examples are changing a nearby coil’s current, changing a loop’s shape or rotating the loop in a field. A large flux that remains constant gives no induced emf.

Worked example 1. A square loop of side 10 cm and resistance 0.5 Ω stands vertically in the east-west plane. A uniform field of 0.10 T points north-east and falls steadily to zero in 0.70 s. Find the emf and current magnitudes.

Formula: Φ = BA cos θ; |ε| = |ΔΦ|/Δt; I = |ε|/R. The angle between the field and the loop normal is 45°.

Substitute: A = (0.10)² = 0.010 m²; initial Φ = 0.10 × 0.010/√2 Wb; final Φ = 0.

Answer: |ε| = (0.001/√2)/0.70 = 1.0 mV = 0.001 V, and I = 0.001/0.5 = 2 mA = 0.002 A, rounded as shown. Here mV means millivolt and mA means milliampere, each one thousandth of its base unit.

Worked example 2. A circular coil has radius 10 cm, 500 turns and resistance 2 Ω. Its plane is initially perpendicular to Earth’s horizontal magnetic field, 3.0 × 10⁻⁵ T. It rotates through 180° about its vertical diameter in 0.25 s. Estimate the average emf and current magnitudes.

Formula: A = πr²; |ε| = N|ΔΦ|/Δt; I = |ε|/R. Here r is the coil radius, measured in metres, and π is the ratio of a circle’s circumference to its diameter.

Substitute: A = π × 10⁻² m²; initial Φ = 3π × 10⁻⁷ Wb; final Φ = −3π × 10⁻⁷ Wb.

Answer: |ε| = 500 × 6π × 10⁻⁷/0.25 = 3.8 × 10⁻³ V = 0.0038 V; I = 1.9 × 10⁻³ A = 0.0019 A. These are estimated average values; instantaneous values depend on the rotational speed at that instant.

How does Lenz’s law express conservation of energy?

Definition: Lenz’s law states that the polarity of induced emf is such that it tends to produce a current which opposes the change in magnetic flux that produced it.

Polarity identifies which end or terminal is at higher potential. The law opposes the change in flux, not necessarily the existing field. If external flux increases, the induced field opposes that increase. If external flux decreases, the induced field supports the original flux direction.

How is the current direction found?

1. Identify the direction of the external magnetic field through the loop.
2. Decide whether the flux in that direction increases or decreases.
3. Choose an induced magnetic field that opposes this particular change.
4. Use the right-hand rule: curl the fingers in the current direction so that the thumb points along the field through the loop.

When a north pole approaches a closed coil, the near face becomes a north pole and repels it. Viewed from the magnet’s side, the induced current is anticlockwise. When that north pole recedes, the near face becomes south and the current is clockwise.

Where does the electrical energy come from?

In either motion, the magnetic interaction opposes the magnet’s movement. An external agent must do work to keep moving it. In a resistive loop, the induced current converts this supplied mechanical energy into heat through Joule heating, heating caused by current through resistance.

If the induced current instead helped an approaching magnet move faster, both its kinetic energy and the electrical heating could increase without an energy supply. This would violate conservation of energy, the principle that energy cannot be created or destroyed.

Note: Specify the observer’s side when stating clockwise or anticlockwise current. The same circulation appears reversed when viewed from the opposite side of the loop.

How is motional emf produced in a moving conductor?

Motional emf is induced by a conductor’s motion through a magnetic field. Consider a straight rod of length l moving at speed v on conducting rails. Let a uniform, constant magnetic field B be perpendicular to the plane of the rails.

The rod is perpendicular to its velocity, and both are perpendicular to the magnetic field. Length l is measured in metres and has dimension [L]; speed v is measured in metres per second, m/s, and has dimensions [L T⁻¹]. Assume smooth electrical contact and no frictional energy loss.

Derivation: Emf across a sliding rod

1. Let x be the distance from the fixed end of the rectangular circuit to the rod, measured in metres. Its enclosed area is lx.
2. The flux through the circuit is Φ = Blx. For a fixed rod length and constant magnetic field, only x changes with time.
3. Faraday’s law gives ε = −dΦ/dt = −Bl dx/dt. Here dx/dt is the signed rate at which the circuit length changes.
4. If the rod moves towards the fixed end at speed v, dx/dt = −v. Substitution gives ε = Blv for the chosen orientation.

|ε| = Blv is the motional emf magnitude under these perpendicular conditions.

The vertical rod PQ joins two horizontal rails. The fixed side SR is on the left, with SP and RQ completing the rectangle. Crosses mark the magnetic field into the page. The velocity arrow points left; current arrows run clockwise around the shrinking circuit.

See Fig. 6.10 in your NCERT textbook

How does charge separation explain the effect?

The magnetic force on mobile charges pushes them along the rod, leaving opposite charges at its ends. This separation creates a potential difference. A closed external path permits current; without such a path, the ends can still acquire different potentials.

The flux method also shows why motion alone is insufficient. A rigid loop translating wholly within a uniform, steady field at fixed orientation has constant linked flux and no net induced emf around the complete loop. Entering or leaving a bounded field region changes the area exposed to it.

Worked example 3. A rectangular loop of sides 8 cm and 2 cm has a small cut. It moves out of a uniform 0.3 T field perpendicular to its plane at 1 cm/s, normal to its longer side. Find the emf across the cut during exit.

Formula: |ε| = Blv. Use the side perpendicular to the motion as l.

Substitute: l = 0.08 m and v = 0.01 m/s.

Answer: |ε| = 0.3 × 0.08 × 0.01 = 2.4 × 10⁻⁴ V = 0.00024 V. The cut prevents a circulating current but does not prevent an induced emf.

How are mechanical power and electrical heating related?

In the sliding-rod arrangement, let the total circuit resistance be R and let the rod move steadily at speed v. The current magnitude is I = Blv/R. Assume that the magnetic field, rod length and resistance remain constant and neglect friction and self-inductive transients.

The rod experiences a magnetic force opposing its motion. Its magnitude is F = BIl, where F denotes force. The SI unit of force is newton, N, with dimensions [M L T⁻²]. This unit symbol N is distinct from N used as a turn count.

Derivation: Power supplied to the rod

1. To maintain constant speed, the external pulling force balances the opposing magnetic force, so its magnitude is BIl.
2. The mechanical power, denoted by P, is the work supplied per unit time: P = Fv = BIlv.
3. Since Blv is the emf magnitude, this becomes P = |ε|I.
4. Substituting I = Blv/R gives P = (Blv)²/R. Substituting |ε| = IR gives the equivalent heating rate I²R.

P = (Blv)²/R = I²R

The SI unit of power is watt, symbol W. 1 W = 1 J/s, where J denotes joule, the unit of work or energy. Power has dimensions [M L² T⁻³]; energy has dimensions [M L² T⁻²].

The equality between mechanical power input and electrical heating makes energy conservation explicit. At constant field, length and resistance, increasing speed increases emf and current; the heating rate depends on the square of speed. The external agent supplies the energy dissipated in the circuit.

What are eddy currents and why are laminations used?

Eddy currents are circulating induced currents within the bulk of a conductor when the magnetic flux through it changes. A solid metal body provides closed conducting paths even though it has not been wound into a wire coil.

In a transformer’s iron core, the alternating magnetic flux induces such currents. Alternating means varying periodically with reversal of direction or polarity. The core has electrical resistance, so eddy currents produce heating and consume part of the input energy.

How does a laminated core reduce heating?

A laminated core is built from thin sheets electrically insulated from each other. The insulation interrupts large current paths across the core and increases resistance to those paths. This reduces eddy currents and the associated energy loss.

The core still provides a path for magnetic flux linking the windings. Lamination therefore reduces unwanted electrical circulation within the magnetic material while retaining the core’s magnetic function. It reduces eddy-current heating rather than eliminating every transformer loss.

Which loss is being reduced?

Distinguish heating in the solid core from heating in the winding wire. Lamination addresses eddy-current loss in the core. Using thicker winding wire addresses resistive loss in high-current windings. These remedies act on different conducting paths and should not be interchanged.

What is self-induction and what determines self-inductance?

Self-induction occurs when a changing current in a coil changes its own linked flux and induces emf in that same coil. For a fixed geometry and a medium in which flux linkage is proportional to current, NΦ = LI.

Here L is self-inductance, the ratio of the coil’s flux linkage to its current. It depends on the coil’s geometry and the magnetic properties of the medium. Inductance is a scalar; its dimensions are [M L² T⁻² A⁻²].

At constant L, Faraday’s law gives ε = −L dI/dt. The induced emf opposes either an increase or a decrease in current and is called back emf. Inductance plays a role analogous to inertia because it resists changes in current.

The SI unit of self-inductance is henry, symbol H. 1 H = 1 V s/A. A coil has self-inductance one henry if a current change of one ampere per second induces an emf of one volt in it.

Derivation: Self-inductance of a long solenoid

A solenoid is a long coil of closely wound turns. Let its length be l, cross-sectional area A and total turn count N. Define n = N/l as turns per unit length, measured in m⁻¹. Let μ₀ denote permeability of free space, measured in H/m.

1. For a long air-cored solenoid, neglecting edge effects, the magnetic field inside is B = μ₀nI. Treat this field as uniform over the cross-section.
2. The flux through each turn is Φ = BA = μ₀nIA.
3. The total turn count is N = nl, so the flux linkage is NΦ = μ₀n²AlI.
4. Divide this expression by I and use L = NΦ/I to obtain the self-inductance.

L = μ₀n²Al = μ₀N²A/l

For a core of relative permeability μᵣ, the expression becomes L = μᵣμ₀n²Al. Relative permeability is the dimensionless ratio of the medium’s permeability to μ₀. Permeability has dimensions [M L T⁻² A⁻²]. The long-solenoid approximation neglects the weaker, non-uniform field near its ends.

Worked example 4. Current in a circuit falls from 5.0 A to zero in 0.1 s, inducing an average emf magnitude of 200 V. Estimate its self-inductance.

Formula: |ε| = L|ΔI|/Δt; L = |ε|Δt/|ΔI|.

Substitute: L = 200 × 0.1/5.0.

Answer: L = 4 H. The 200 V induced emf acts to oppose the fall in current, tending to maintain its original direction.

How does mutual induction link two separate circuits?

Mutual induction occurs when a changing current in one coil changes the magnetic flux linked with another coil and induces emf in it. Direct electrical contact between the coils is unnecessary. Their magnetic fields provide the coupling.

Let I₁ be current in coil 1, N₂ the turn count of coil 2 and Φ₂ the flux per turn of coil 2 due to I₁. For fixed geometry and proportional flux response, N₂Φ₂ = MI₁, where M is their mutual inductance.

The SI unit of mutual inductance is henry. Its dimensions are [M L² T⁻² A⁻²]. One henry gives an induced emf magnitude of one volt in the second coil when the first coil’s current changes at one ampere per second.

For constant M, ε₂ = −M dI₁/dt, where ε₂ denotes the emf in coil 2. Mutual inductance depends on separation, relative orientation, geometry and the intervening medium. Interchanging the roles of the two coils gives the same mutual inductance.

How is mutual inductance found for coaxial solenoids?

Coaxial means sharing a common axis. Consider two long, air-cored solenoids of the same length l, with turn counts N₁ and N₂. Let A be the cross-sectional area of the inner solenoid. Neglect edge effects and the field outside the inner solenoid.

A current I₁ in the inner solenoid gives B = μ₀N₁I₁/l. The flux through each turn of the outer solenoid is BA, using the inner area where the field is effectively confined. Thus N₂Φ₂ = μ₀N₁N₂AI₁/l.

M = μ₀N₁N₂A/l

Equivalently, M = μ₀n₁N₂A, where n₁ = N₁/l is the inner solenoid’s turns per unit length. A uniform medium of relative permeability μᵣ introduces a factor μᵣ. The long-solenoid assumptions justify treating the field as uniform and neglecting leakage in this calculation.

Worked example 5. A long air-cored solenoid has 15 turns per cm. A single small loop of area 2.0 cm² lies inside, with its plane normal to the axis. The solenoid current rises steadily from 2.0 A to 4.0 A in 0.1 s. Find the induced emf. Use μ₀ = 4π × 10⁻⁷ H/m.

Formula: B = μ₀nI; |ε| = Aμ₀n|ΔI|/Δt.

Substitute: n = 1500 m⁻¹, A = 2.0 × 10⁻⁴ m² and |ΔI|/Δt = 20 A/s.

Answer: |ε| = 2.0 × 10⁻⁴ × 4π × 10⁻⁷ × 1500 × 20 = 7.54 × 10⁻⁶ V = 0.00000754 V, approximately.

Worked example 6. Two adjacent coils have mutual inductance 1.5 H. Current in one changes from zero to 20 A in 0.5 s. Find the change of flux linkage with the other and the average induced emf magnitude.

Formula: Δ(N₂Φ₂) = MΔI₁; |ε₂| = M|ΔI₁|/Δt.

Substitute: Δ(N₂Φ₂) = 1.5 × 20; |ε₂| = 1.5 × 20/0.5.

Answer: The flux linkage increases by 30 Wb and the average induced emf magnitude is 60 V. Flux linkage includes the turns already, so no additional turn factor is inserted.

How does a transformer change alternating voltage?

A transformer changes an alternating voltage through mutual induction. It has two insulated windings on a soft-iron core. Often the primary winding is the input coil and the secondary winding is the output coil. In the arrangement considered here, the primary receives the applied alternating voltage and the secondary supplies the transformed voltage.

An alternating primary current produces alternating core flux. This changing flux links the secondary and induces an emf there. It also induces a back emf in the primary. A steady direct current, meaning current constant in direction and magnitude, cannot maintain this continuously changing flux.

Both drawings show a rectangular soft-iron core labelled with primary and secondary windings. In arrangement (a), the coils are wound one over the other. In arrangement (b), the windings occupy separate limbs of the core.

See Fig. 7.16 in your NCERT textbook

What assumptions define the ideal calculation?

Assume negligible winding resistance, no energy losses and complete magnetic coupling: the same flux links every turn of both windings. Let Nₚ and Nₛ be their turn counts. Subscripts p and s identify primary and secondary respectively.

Let Vₚ and Vₛ be corresponding voltage magnitudes, and Iₚ and Iₛ corresponding current magnitudes, using the same amplitude convention on both sides. For power calculations below, use a resistive load and root mean square values, the effective alternating values giving the same heating as direct current.

Because each turn experiences the same rate of flux change, Vₛ/Vₚ = Nₛ/Nₚ. Under the lossless power assumption, VₚIₚ = VₛIₛ, giving Iₛ/Iₚ = Nₚ/Nₛ.

A transformer increases voltage at the expense of current, or increases current at the expense of voltage. It does not increase the available power. The simple terminal-voltage relation is a good approximation when winding voltage drops and flux leakage are small.

Worked example 7. An ideal transformer has 100 primary turns and 200 secondary turns. Its input is 220 V at 10 A. For a resistive load, find the output voltage and current using effective alternating values.

Formula: Vₛ = (Nₛ/Nₚ)Vₚ; Iₛ = (Nₚ/Nₛ)Iₚ.

Substitute: Vₛ = (200/100) × 220; Iₛ = (100/200) × 10.

Answer: Vₛ = 440 V and Iₛ = 5.0 A. This is a step-up transformer. Input and output powers are equal under the ideal assumption.

Why do transformers lose energy and help power transmission?

Efficiency, denoted by η, is the ratio of useful output power to input power. It is dimensionless. Writing Pᵢₙ for input power and Pₒᵤₜ for output power, η = Pₒᵤₜ/Pᵢₙ. Multiply this ratio by 100 to express efficiency as a percentage.

An ideal lossless transformer has 100% efficiency. Some energy is always lost in a real transformer, although a well-designed transformer may have an efficiency of more than 95%. The ideal model remains a useful approximation when these losses are small.

Which losses can be reduced?

Hysteresis is the lag of a material’s magnetic response behind changes in the applied magnetising field. Repeated reversal therefore wastes energy in the core. Flux leakage concerns incomplete magnetic linking; winding resistance concerns electrical heating. They are distinct limitations of an actual transformer.

Why transmit electricity at high voltage?

Transmission wires have resistance and lose power as I²R. For a fixed transmitted power and the same power factor, raising the voltage reduces current and hence this heating. Power factor is the ratio of real average power to the product of effective voltage and current.

A step-up transformer raises the generator’s voltage for long-distance transmission. Near consumers, step-down transformers reduce voltage at substations and distribution points. This combination limits transmission losses while providing a lower voltage for use.

Glossary

• Electromagnetic induction — Production of an electromotive force when the magnetic flux linked with a circuit changes.
• Magnetic flux — Scalar measure of magnetic field through a surface, including the effect of surface orientation.
• Area vector — Vector perpendicular to a surface whose magnitude equals the area of that surface.
• Flux linkage — Sum of the magnetic fluxes linked with all turns of a coil.
• Induced emf — Electromotive force produced through changing magnetic flux, capable of driving current in a closed circuit.
• Lenz’s law — Rule giving induced polarity so that the resulting current opposes the flux change producing it.
• Motional emf — Electromotive force generated across a conductor through its motion in a magnetic field.
• Eddy currents — Circulating currents induced within the bulk of conducting material by changing linked magnetic flux.
• Self-inductance — Ratio of a coil’s own flux linkage to its current under proportional flux response.
• Mutual inductance — Flux linkage of one coil per unit current in the other under fixed coupling conditions.
• Back emf — Self-induced electromotive force that opposes an increase or decrease of current in a circuit.
• Transformer — Device using mutual induction between windings to increase or decrease an alternating voltage.
• Laminated core — Magnetic core made of insulated thin sheets that reduce circulating eddy currents and heating.
• Transformer efficiency — Ratio of useful output power to input power, often expressed as a percentage.

Common errors and misconceptions

• Misconception: A very strong, steady field must induce current in a stationary loop. Correct: Induction requires changing flux linkage; field strength alone is insufficient.
• Misconception: The angle in BA cos θ is measured from the loop’s plane. Correct: It is measured between the field and the area vector perpendicular to that plane.
• Misconception: Lenz’s law requires the induced field to oppose the original field in every case. Correct: It opposes the change; a decreasing original flux is supported.
• Misconception: No current means no induced emf. Correct: An open circuit can have emf across its ends while preventing a circulating conduction current.
• Misconception: Flux linkage must be multiplied by the turn count again in Faraday’s law. Correct: Flux linkage already includes every turn; multiply by N only when starting with flux per turn.
• Misconception: A self-induced emf opposes the existing current even while it decreases. Correct: It opposes the decrease and therefore tends to sustain that current.
• Misconception: A step-up transformer increases both voltage and available power. Correct: Its ideal output current falls as voltage rises, with output power equal to input power.
• Misconception: Lamination removes every transformer loss. Correct: It reduces eddy-current loss; winding resistance, hysteresis and flux leakage require their own measures.

Exam-style questions with model answers

Q1. State Faraday’s law for a closely wound coil of N turns, each linked by magnetic flux Φ, and explain its negative sign. [2 marks]

1. The induced emf is ε = −N dΦ/dt, where ε is emf and dΦ/dt is the instantaneous rate of change of flux through one turn.
2. The negative sign expresses Lenz’s law: induced polarity tends to drive current opposing the change in flux that caused it.

Q2. An external agent moves a north pole towards a closed resistive conducting coil. State the near-face polarity and current direction as seen from the magnet, and explain the energy transfer. [3 marks]

1. The near face becomes a north pole. This produces repulsion that opposes the approach and the associated increase in magnetic flux through the coil.
2. Viewed from the approaching magnet, the induced current is anticlockwise. This circulation gives the required north polarity of the facing coil surface.
3. An external agent must do work against the repulsion to maintain the motion. In a resistive coil, the induced current dissipates that supplied energy as heat.

Q3. A circular coil of radius 10 cm has 500 turns and resistance 2 Ω. Its plane is initially perpendicular to a uniform horizontal field of 3.0 × 10⁻⁵ T. It turns through 180° about its vertical diameter in 0.25 s. Neglect its self-induced emf. Find the estimated average emf and current magnitudes, and distinguish these from instantaneous values. [5 marks]

1. The radius is 0.10 m, so the area of each turn is A = πr² = π × 10⁻² m². Initially, the area normal is parallel to the field.
2. The initial flux per turn is Φ = BA = 3π × 10⁻⁷ Wb. After the half-turn, the normal reverses and the final flux is −3π × 10⁻⁷ Wb.
3. The magnitude of the flux change per turn is therefore 6π × 10⁻⁷ Wb. The average emf magnitude is N|ΔΦ|/Δt = 3.8 × 10⁻³ V, approximately.
4. Using the resistance, the estimated average current magnitude is I = |ε|/R = 3.8 × 10⁻³/2 = 1.9 × 10⁻³ A.
5. These values describe the interval as a whole. Instantaneous emf and current depend on the rate of flux change, including the rotational speed, at the particular instant.

Q4. Derive the emf magnitude across a straight rod of length l sliding at speed v on conducting rails in a uniform, steady field B. The rod, velocity and field are mutually perpendicular. Let x be the distance to the fixed end of the rectangular circuit. [4 marks]

1. The rectangular circuit has enclosed area lx, because its width is the rod length l and its changing length is x.
2. The field is perpendicular to the circuit, so its magnetic flux is Φ = Blx. Both B and l remain constant.
3. Faraday’s law gives ε = −dΦ/dt = −Bl dx/dt. If the rod moves towards the fixed end, dx/dt = −v.
4. Substitution gives emf magnitude |ε| = Blv. The sign for any chosen circuit orientation follows from Lenz’s law and opposes the flux change.

Q5. A circuit’s current falls from 5.0 A to zero in 0.1 s and produces an average induced emf magnitude of 200 V. Estimate its self-inductance and explain the action of the induced emf. [3 marks]

1. The magnitude of the average current-change rate is |ΔI|/Δt = 5.0/0.1 = 50 A/s. The current decreases, although this rate is written as a magnitude.
2. Using |ε| = L|ΔI|/Δt, the self-inductance is L = 200/50 = 4 H, where H denotes henry.
3. The induced emf opposes the reduction of current. It therefore tends to maintain current in the original direction while that current is falling.

Q6. Two fixed adjacent coils have constant mutual inductance 1.5 H. Current in the first changes from zero to 20 A in 0.5 s. Find the change of flux linkage with the second and its average induced emf magnitude. [3 marks]

1. Flux linkage with the second coil is proportional to the first coil’s current. Its change is MΔI = 1.5 × 20 = 30 Wb.
2. The average induced emf magnitude is the flux-linkage change divided by the interval: |ε| = 30/0.5 = 60 V.
3. The linkage already includes all turns of the second coil, so no extra turn count is needed. The induced polarity opposes the change that produces it.

Q7. An ideal transformer has 100 primary turns and 200 secondary turns. Its effective input voltage and current are 220 V and 10 A with a resistive secondary load. Explain its principle, calculate output voltage and current, and state its power relation. [5 marks]

1. The transformer works by mutual induction. Alternating current in the primary creates changing core flux linking the secondary, inducing an alternating emf there without electrical contact between windings.
2. For the ideal case, the same flux links every turn of both windings and losses are neglected. The voltage ratio therefore equals the turns ratio.
3. The secondary voltage is Vₛ = (Nₛ/Nₚ)Vₚ = (200/100) × 220 = 440 V. More secondary turns make this a step-up transformer.
4. The ideal current ratio is inverse to the turns ratio. Hence Iₛ = (Nₚ/Nₛ)Iₚ = (100/200) × 10 = 5.0 A.
5. Input power equals output power: VₚIₚ = VₛIₛ. The increased voltage is accompanied by reduced current, so the ideal transformer does not create additional electrical power.

Q8. Explain four distinct limitations of a real transformer and give one measure to reduce each. [4 marks]

1. Flux leakage means some primary flux does not link the secondary. Winding the two coils one over the other improves magnetic coupling and reduces leakage.
2. Winding resistance produces electrical heating. Thick wire reduces this loss in the high-current, low-voltage winding.
3. Alternating core flux induces eddy currents that heat the iron. A laminated core reduces these circulating currents and their heating effect.
4. Repeated magnetisation reversal causes hysteresis loss in the core. A magnetic material with low hysteresis loss reduces the energy dissipated during these reversals.

Key takeaways

• Induction depends on changing magnetic flux linkage, which can result from changing field strength, loop area or orientation.
• For a uniform field through a plane surface, magnetic flux is BA cos θ, with θ measured from the surface normal.
• Faraday’s law determines the emf magnitude; Lenz’s law determines the polarity that opposes the change producing it.
• A rod moving under mutually perpendicular field, length and velocity conditions develops a motional emf of magnitude Blv.
• The mechanical power supplied to a steadily moving rod equals resistive heating under the stated loss-free mechanical assumptions.
• Self-induction acts within the same coil; mutual induction transfers the effect of changing current to another magnetically linked coil.
• Ideal transformers conserve power while changing voltage and current in inverse ratios determined by their winding turn counts.
• Laminated cores, suitable magnetic materials, thick winding wires and improved magnetic coupling reduce different transformer limitations.

Test yourself

Does a stationary loop in a very strong but constant magnetic field necessarily carry an induced current?

No. With fixed area and orientation, its linked flux remains constant, so this arrangement produces no induced emf.

A plane loop is parallel to a uniform magnetic field. What is its magnetic flux?

Its area normal is perpendicular to the field, so θ = 90° and the magnetic flux is zero.

What happens in a neighbouring stationary coil when a battery coil’s current settles to a steady value?

The changing flux ceases. The momentary induced emf and current disappear once the linked magnetic flux becomes constant.

Why can an open moving conductor have emf without a circulating conduction current?

Charges can separate towards its ends and establish a potential difference even though there is no complete conducting path.

What does the back emf do when current through a coil decreases?

It opposes the decrease and tends to sustain the current in its original direction, consistently with Lenz’s law.

For two long coaxial solenoids of unequal radii, which cross-sectional area enters the mutual-inductance expression?

The inner solenoid’s area enters, under the approximation that its magnetic field is confined within that area.

Why is a steady direct current unsuitable for continuous transformer action?

After switching transients, a steady current produces steady flux, which cannot sustain an induced secondary emf.

Why does raising transmission voltage reduce heating for fixed power and power factor?

It lowers the current, and resistive heating in the transmission wires depends on the square of that current.