Electromagnetic Induction ISC Class 12: Notes, Derivations and Numericals

What Electromagnetic Induction is about

Weightage: Electromagnetic Induction is part of the unit Magnetic Effects of Current and Magnetism; Electromagnetic Induction and Alternating Currents, which carries 16 of the 70 marks in the ISC Physics Paper 1 (Theory), shared with 3 other chapters. Questions on it can appear in any section of the paper.Source: CISCE ISC Physics syllabus, examination year 2027, distribution of marks

Key concepts

Faraday's laws of electromagnetic induction

Magnetic flux through a coil of N turns: flux linkage $N\varphi =NBA\cos \theta$. First law: whenever the magnetic flux linked with a circuit changes, an emf is induced in it; it lasts only while the flux is changing. Second law: the induced emf equals the rate of change of flux linkage: $\epsilon =-N\frac{d\varphi }{dt}$; for a steady change, $\epsilon =-N\frac{\Delta \varphi }{\Delta t}$. Flux can change by changing B, the area A, or the angle θ (rotating the coil). If the circuit has resistance R, the induced current is $I=\epsilon /R$ and the total charge that flows is $q=\frac{N\Delta \varphi }{R}$, independent of how fast the change happens.

Worked example

A coil of 200 turns and area $0.01m^2$ is perpendicular to a field that falls uniformly from 0.5 T to 0.1 T in 0.2 s. The coil's resistance is 10 Ω. Find the induced emf, the current and the charge that flows.

Try it first, then show the solutionSolution

1. $\Delta \varphi =\Delta B\times A=0.4\times 0.01=4\times 10^{-3}Wb$
2. $\epsilon =N\frac{\Delta \varphi }{\Delta t}=\frac{200\times 4\times 10^{-3}}{0.2}=4V$; $I=\frac{4}{10}=0.4A$
3. $q=I\Delta t=0.4\times 0.2=0.08C$ (or $\frac{N\Delta \varphi }{R}=\frac{0.8}{10}$)

Answer: 4 V; 0.4 A; 0.08 C

Lenz's law and conservation of energy

The direction of the induced current is such that it opposes the change in flux that produces it. If a magnet's N pole approaches a coil, the induced current makes the near face of the coil an N pole, repelling the magnet; as the magnet recedes, the near face becomes S, attracting it. Either way, you must do work against the opposing force, and that work appears as electrical energy (then heat). If the induced current helped the change instead, the magnet would accelerate by itself, creating energy from nothing; so Lenz's law follows from conservation of energy. For a moving conductor, the direction is also given by Fleming's right-hand rule (thumb motion, first finger field, middle finger induced current).

Worked example

A bar magnet is dropped, N pole first, through a horizontal copper ring. (a) Seen from above, what is the direction of the induced current as the magnet approaches the ring? (b) Is the magnet's acceleration equal to, more than or less than g?

Try it first, then show the solutionSolution

1. (a) Downward flux through the ring increases. The induced current must create an upward field inside the ring (making the top face N), which means anticlockwise as seen from above.
2. (b) The ring opposes the motion both while the magnet approaches (repulsion) and while it leaves (attraction), so the magnet's acceleration is less than g.

Answer: (a) Anticlockwise seen from above (b) less than g

Motional emf and power

A rod of length l slides with speed v on rails across a uniform field B perpendicular to the plane. In time dt the area of the circuit changes by $lvdt$, so $\frac{d\varphi }{dt}=Blv$ and $\epsilon =Blv$. (Equivalently, each free electron feels $evB$ along the rod; the work per unit charge in moving along l is $vBl$.) With total resistance R: $I=\frac{Blv}{R}$; the rod feels a retarding force $F=BIl=\frac{B^2{l}^{2}v}{R}$ (Lenz). To keep v constant an external force must supply power $P=Fv=\frac{(Blv{)}^2}{R}$, which equals the electrical power $I^{2}R$ dissipated as heat: mechanical energy becomes electrical energy.

Worked example

A 0.5 m rod moves at $4m{s}^{-1}$ on rails perpendicular to a 0.2 T field; the circuit resistance is 2 Ω. Find the emf, current, force needed and power.

Try it first, then show the solutionSolution

1. $\epsilon =Blv=0.2\times 0.5\times 4=0.4V$; $I=0.4/2=0.2A$
2. $F=BIl=0.2\times 0.2\times 0.5=0.02N$
3. $P=Fv=0.02\times 4=0.08W$; check: $\frac{(Blv{)}^2}{R}=\frac{0.16}{2}=0.08W$

Answer: 0.4 V; 0.2 A; 0.02 N; 0.08 W

Eddy currents (qualitative)

When the flux through a bulk piece of metal changes, induced currents circulate inside it in closed loops: eddy currents. By Lenz's law they oppose the motion or change that causes them, and they heat the metal. Uses: electromagnetic (eddy-current) brakes in trains, damping of galvanometer coils (wound on a metal frame) so the pointer settles quickly, induction furnaces and induction cooktops, metal detectors. Disadvantages: heating and energy loss in transformer cores and motor armatures. Reduction: make cores of thin laminated sheets insulated from each other (lamination cuts the paths available to eddy currents), or use high-resistivity alloys.

Worked example

A copper plate swinging between the poles of a strong magnet stops quickly, but a similar plate with many slots cut in it swings for much longer. Explain.

Try it first, then show the solutionSolution

1. As the plate moves through the field, the flux through parts of it changes, inducing eddy currents that, by Lenz's law, oppose the motion and convert its kinetic energy to heat.
2. Slots break the large closed paths, so the eddy currents are much smaller, and the damping is weak.

Answer: Eddy currents brake the solid plate; slots (like laminations) cut the eddy-current paths, so damping is reduced.

Self-induction and the inductance of a solenoid

The flux linked with a coil due to its own current is proportional to that current: $N\varphi =LI$, where L is the coefficient of self-inductance. A changing current induces a 'back emf' $\epsilon =-L\frac{dI}{dt}$ that opposes the change (Lenz). So $L=\frac{\epsilon }{dI/dt}$; unit henry: $1H=1Vs{A}^{-1}$ (= 1 Wb A⁻¹). A coil has an inductance of 1 H if a current changing at 1 A s⁻¹ induces 1 V in it. Long solenoid (N turns, length l, area A): $B={\mu }_0\frac{N}{l}I$, flux linkage $NBA=\frac{{\mu }_0{N}^{2}A}{l}I$, so $L=\frac{{\mu }_0{N}^{2}A}{l}={\mu }_0{n}^{2}Al$. L depends only on geometry and the core (an iron core multiplies it by ${\mu }_r$). The spark when an inductive circuit is switched off is caused by the large back emf from the rapid fall of current.

Worked example

A solenoid of 500 turns is 25 cm long with cross-section $4\times 10^{-4}{m}^2$. Find its self-inductance and the emf induced when its current falls from 2 A to zero in 0.01 s.

Try it first, then show the solutionSolution

1. $L=\frac{{\mu }_0{N}^{2}A}{l}=\frac{4\pi \times 10^{-7}\times 500^2\times 4\times 10^{-4}}{0.25}=5.03\times 10^{-4}H$
2. $\epsilon =L\frac{\Delta I}{\Delta t}=5.03\times 10^{-4}\times \frac{2}{0.01}=0.10V$

Answer: $L\approx 0.50mH$; emf about 0.10 V

Mutual induction and coaxial solenoids

When the current $I_1$ in one coil changes, the flux it links with a second coil changes and an emf is induced in the second coil. Flux linkage $N_2{\varphi }_2=M{I}_1$ and ${\epsilon }_2=-M\frac{d{I}_1}{dt}$; M is the coefficient of mutual inductance (henry), the same whichever coil is the primary. For two coaxial solenoids (outer: $N_1$ turns over length l; inner: $N_2$ turns, area A), the field of the outer, $B_1={\mu }_0\frac{N_1}{l}{I}_1$, passes through the inner, so $N_2{B}_{1}A=\frac{{\mu }_0{N}_1{N}_{2}A}{l}{I}_1$, giving $M=\frac{{\mu }_0{N}_1{N}_{2}A}{l}$. M increases with the number of turns, the common area, close coupling and an iron core.

Worked example

A secondary of 200 turns is wound tightly over the middle of a 0.5 m long primary solenoid of 1000 turns and cross-section $10^{-3}{m}^2$. Find M and the emf in the secondary when the primary current changes at $50A{s}^{-1}$.

Try it first, then show the solutionSolution

1. $M=\frac{{\mu }_0{N}_1{N}_{2}A}{l}=\frac{4\pi \times 10^{-7}\times 1000\times 200\times 10^{-3}}{0.5}=5.03\times 10^{-4}H$
2. ${\epsilon }_2=M\frac{d{I}_1}{dt}=5.03\times 10^{-4}\times 50=0.025V$

Answer: $M\approx 0.50mH$; 25 mV

The transformer and power transmission

Principle: mutual induction. Two coils, primary ($N_p$) and secondary ($N_s$), are wound on a laminated soft-iron core. An alternating voltage on the primary sets up an alternating flux in the core that links the secondary. Each turn has the same emf, so for an ideal transformer $\frac{V_s}{V_p}=\frac{N_s}{N_p}$; with no losses $V_p{I}_p=V_s{I}_s$, so $\frac{I_s}{I_p}=\frac{N_p}{N_s}$. Step-up: $N_s>N_p$ (voltage up, current down). Efficiency $\eta =\frac{outputpower}{inputpower}$. Losses and remedies: copper ($I^{2}R$) loss: thick copper wire; eddy-current loss: laminated core; hysteresis loss: soft iron (narrow loop); flux leakage: windings on top of each other or on a shell-type core. Transmission: power P sent at voltage V needs current $I=P/V$; line loss $I^{2}R=\frac{P^{2}R}{V^2}$, so stepping the voltage up greatly reduces loss; it is stepped down near consumers.

Worked example

100 kW is sent through lines of total resistance 5 Ω. Find the power lost if it is sent at (a) 1000 V and (b) 10 000 V.

Try it first, then show the solutionSolution

1. (a) $I=\frac{100000}{1000}=100A$; loss $=I^{2}R=100^2\times 5=50kW$
2. (b) $I=10A$; loss $=10^2\times 5=500W$

Answer: (a) 50 kW (half the power!) (b) 500 W: ten times the voltage cuts the loss a hundredfold

Physics in real life: Electromagnetic Induction

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Topics in this chapter

• Self-Inductance and Mutual Inductance
• Motional Emf
• The Transformer and Power Transmission

Electromagnetic Induction formula sheet

Magnetic flux and linkage

$\varphi =BA\cos \theta ,fluxlinkage=N\varphi$

θ between B and the normal.

Faraday-Lenz law

$\epsilon =-N\frac{d\varphi }{dt}$

Minus sign: Lenz's law (opposition).

Induced current and charge

$I=\frac{\epsilon }{R},q=\frac{N\Delta \varphi }{R}$

Charge does not depend on the time taken.

Motional emf

$\epsilon =Blv$

B, l and v mutually perpendicular.

Force and power for a moving rod

$F=\frac{B^2{l}^{2}v}{R},P=\frac{(Blv{)}^2}{R}$

Mechanical power in = electrical power out.

Self-inductance

$N\varphi =LI,\epsilon =-L\frac{dI}{dt}$

Unit henry = V s A⁻¹.

Inductance of a long solenoid

$L=\frac{{\mu }_0{N}^{2}A}{l}={\mu }_0{n}^{2}Al$

With a core: multiply by ${\mu }_r$.

Mutual inductance

$N_2{\varphi }_2=M{I}_1,{\epsilon }_2=-M\frac{d{I}_1}{dt}$

Same M both ways.

Coaxial solenoids

$M=\frac{{\mu }_0{N}_1{N}_{2}A}{l}$

A = area of the inner coil; l = length of the outer.

Ideal transformer

$\frac{V_s}{V_p}=\frac{N_s}{N_p}=\frac{I_p}{I_s}$

Efficiency $\eta =\frac{V_s{I}_s}{V_p{I}_p}$.

Transmission loss

$P_{loss}=I^{2}R=\frac{P^{2}R}{V^2}$

Loss ∝ 1/V² for the same power delivered.

Electromagnetic Induction: diagrams to practise

Practise each one on paper: draw it from the checklist, then tick off every label.

How the board asks Electromagnetic Induction

The question patterns to expect, and the method that works for each.

Tricks to remember Electromagnetic Induction

Short memory aids. Say each one aloud once, then use it on a question the same day.

• Lenz is lazy

  The induced current always resists change: 'Lenz hates change'. Approach: repel. Recede: attract.

  Use it when: Direction of induced current and coil polarity.

• BLV like 'believe'

  Motional emf = B l v: 'believe' (B-l-v).

  Use it when: Recalling ε = Blv quickly.

• Right hand generates, left hand motors

  Fleming's right-hand rule for induced current (generator); left-hand rule for force on a current (motor).

  Use it when: Not confusing the two rules, a slip examiners report.

• L goes as N squared

  Self-inductance ∝ N² (each turn links the flux of all turns); mutual ∝ N₁N₂.

  Use it when: How-will-L-change questions.

• Transformer: voltage follows turns, current goes against

  $V\propto N$; $I\propto 1/N$; power stays (nearly) the same.

  Use it when: Turns-ratio numericals.

• Four losses, four cures

  'Copper, Eddy, Hysteresis, Leakage' pairs with 'thick wire, laminate, soft iron, wind close'. Say each loss with its cure in the same breath.

  Use it when: Listing transformer losses with remedies.

Electromagnetic Induction: board-level solved questions

Ordered from board level to HOTS. Cover the solution, write your own, then compare step by step.

Case-study practice (Section D style)

Read the situation, then answer every part. In the board paper this is a 5-mark question split into 1 + 2 + 1 + 1 marks.

Where the marks are

How a typical answer is split into step marks. Show every step: a right final answer alone rarely earns full marks.

Where students lose marks in Electromagnetic Induction

The slips examiners see most often, and what to write instead.

• ✗Mistake: Naming 'self-inductance' or 'induction' instead of self-induction

  CISCE examiners' comments, 2024: asked to name the phenomenon when a solenoid's falling current induces an emf, candidates wrote 'induction', 'self-inductance' or 'mutual induction'.

  Wrong working

  The phenomenon is self-inductance.

  Right working

  Self-induction (the phenomenon); self-inductance L is the coefficient.

  ✓Do this: Phenomenon names end in '-tion'; the coefficient names end in '-tance'.

  Usually costs: 1 mark.

• ✗Mistake: Ignoring milliseconds or using the wrong formula for L

  CISCE examiners' comments, 2024 and 2025: in self-inductance numericals candidates ignored the $10^{-3}$ in milliseconds, used the wrong formula, or gave the wrong unit.

  Wrong working

  $L=\frac{500}{5/20}=2000H$ (20 ms taken as 20 s).

  Right working

  $L=\frac{\epsilon \Delta t}{\Delta I}=\frac{500\times 0.020}{5}=2H$.

  ✓Do this: Convert every time to seconds before substituting; write the unit H.

  Usually costs: 1-2 marks.

• ✗Mistake: Not finding the velocity from a position-time equation

  CISCE examiners' comments, 2025: when the rod's position was given as a function of time, candidates did not differentiate to find v, did not know the motional emf formula, and confused Fleming's left- and right-hand rules.

  Wrong working

  Takes x itself as the speed in ε = Blv.

  Right working

  $v=dx/dt$ first, then $\epsilon =Blv$; direction by Lenz's law or Fleming's right-hand rule.

  ✓Do this: Any time a position function appears, differentiate it.

  Usually costs: 1-2 marks.

• ✗Mistake: Forgetting the number of turns in Faraday's law

  A frequent slip in coil numericals: the flux change is computed for one turn and N is left out.

  Wrong working

  $\epsilon =\frac{0.4\times 0.01}{0.2}=0.02V$ for a 200-turn coil.

  Right working

  $\epsilon =N\frac{\Delta \varphi }{\Delta t}=200\times 0.02=4V$.

  ✓Do this: Write $\epsilon =N\Delta \varphi /\Delta t$ every time.

  Usually costs: 1 mark.

• ✗Mistake: Saying a transformer works on d.c.

  Candidates sometimes state that a transformer steps up any voltage; a steady d.c. gives no changing flux, so no secondary emf.

  Wrong working

  A transformer converts 12 V d.c. to 230 V d.c.

  Right working

  A transformer needs a.c. (changing current) in the primary; with steady d.c. the secondary emf is zero.

  ✓Do this: Link the principle (mutual induction) to the need for changing flux.

  Usually costs: 1 mark.

Practice questions on Electromagnetic Induction

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

1. Q1 · 1 mark · MCQBoard level

   The emf induced in a coil depends on:
   (a) the flux linked with it
   (b) the rate of change of flux linked with it
   (c) the resistance of the coil
   (d) the material of the core only

   Hint

   Faraday's second law.

   Answer

   (b)

2. Q2 · 1 mark · MCQBoard level

   The core of a transformer is laminated to reduce:
   (a) copper loss
   (b) eddy-current loss
   (c) hysteresis loss
   (d) flux leakage

   Hint

   Insulated thin sheets cut current paths.

   Answer

   (b) eddy-current loss

3. Q3 · 1 mark · MCQBoard level

   In a step-up transformer:
   (a) $N_s<N_p$ and $I_s>I_p$
   (b) $N_s>N_p$ and $I_s<I_p$
   (c) $N_s>N_p$ and $I_s>I_p$
   (d) $N_s=N_p$

   Hint

   Voltage follows turns; current goes the other way.

   Answer

   (b)

4. Q4 · 1 mark · Very short answerBoard level

   Name the law that gives the direction of the induced current.

   Hint

   It opposes the change.

   Answer

   Lenz's law.

5. Q5 · 2 marks · Short answerBoard level

   A 200-turn coil of area $0.01m^2$ is perpendicular to a field that falls from 0.5 T to 0.1 T in 0.2 s. Find the induced emf and the current if the coil's resistance is 10 Ω.

   Hint

   $\epsilon =N\Delta \varphi /\Delta t$.

   Answer

   ε = 4 V; I = 0.4 A

6. Q6 · 2 marks · Short answerHarder board level

   In the previous question, find the charge that flows. Would it change if the field fell in 0.02 s instead? Explain.

   Hint

   $q=N\Delta \varphi /R$.

   Answer

   $q=\frac{200\times 4\times 10^{-3}}{10}=0.08C$. No: q depends only on the total flux change and R; a faster change gives a larger current for a shorter time.

7. Q7 · 2 marks · Short answerHarder board level

   A 200-turn secondary is wound over the middle of a 0.5 m, 1000-turn solenoid of cross-section $10^{-3}{m}^2$. Find the mutual inductance and the emf in the secondary when the primary current changes at $50A{s}^{-1}$.

   Hint

   $M={\mu }_0{N}_1{N}_{2}A/l$.

   Answer

   $M=5.03\times 10^{-4}H$; ε = 25 mV

8. Q8 · 2 marks · Short answerHOTS / competency

   A long solenoid has self-inductance L. Find its new self-inductance if (a) the number of turns is doubled over the same length, (b) the length is doubled with the same number of turns (same area).

   Hint

   $L={\mu }_0{N}^{2}A/l$.

   Answer

   (a) 4L (b) L/2

9. Q9 · 3 marks · Short answerHarder board level

   A 0.5 m rod moves at $4m{s}^{-1}$ on rails perpendicular to a 0.2 T field; total resistance 2 Ω. Find the emf, the current, the force needed to keep it moving steadily and the power.

   Hint

   ε = Blv, F = BIl.

   Answer

   0.4 V; 0.2 A; 0.02 N; 0.08 W

10. Q10 · 3 marks · Short answerHarder board level

    100 kW is to be delivered through a line of resistance 5 Ω. Compare the line losses at 1 kV and 10 kV and explain why transformers are essential for power distribution.

    Hint

    $P_{loss}=(P/V{)}^{2}R$.

    Answer

    1 kV: I = 100 A, loss 50 kW. 10 kV: I = 10 A, loss 500 W. Transformers step voltage up for transmission (low current, low loss) and down for safe use; only a.c. can be transformed this way.

11. Q11 · 3 marks · Short answerHarder board level

    Derive an expression for the mutual inductance of two long coaxial solenoids, and state two factors on which it depends.

    Hint

    Start from the field of the outer solenoid.

    Answer

    $B_1={\mu }_0{N}_1{I}_1/l$; flux linkage with inner coil $=N_2{B}_{1}A=\frac{{\mu }_0{N}_1{N}_{2}A}{l}{I}_1=M{I}_1$, so $M=\frac{{\mu }_0{N}_1{N}_{2}A}{l}$. Depends on the numbers of turns, the common area, the length, and the core material (×μr).

12. Q12 · 5 marks · Long answerHarder board level

    (a) State Faraday's laws of electromagnetic induction. [2]
    (b) Explain how Lenz's law follows from conservation of energy, using a magnet moved towards a coil. [1]
    (c) A square loop of side 10 cm and resistance 0.5 Ω lies perpendicular to a field B = 0.02t tesla. Find the induced emf and current. [2]

    Hint

    dB/dt is constant.

    Answer

    (a) A changing flux linked with a circuit induces an emf, which lasts while the flux changes; ε = −N dφ/dt. (b) The coil's induced current makes the near face N, repelling the approaching magnet; work done against this repulsion becomes electrical energy, so energy is conserved. (c) ε = A dB/dt = 0.01 × 0.02 = $2\times 10^{-4}V$; I = $4\times 10^{-4}A$.

13. Q13 · 2 marks · Short answerHarder board level

    An ideal transformer steps 220 V down to 11 V; its secondary has 100 turns and supplies 4 A. Find the number of primary turns and the primary current.

    Hint

    $V\propto N$, $I\propto 1/N$.

    Answer

    $N_p=2000$; $I_p=4\times \frac{11}{220}=0.2A$

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1. 1.State Faraday's second law.Show answer

   The induced emf equals the rate of change of magnetic flux linkage: $\epsilon =-Nd\varphi /dt$.

2. 2.State Lenz's law.Show answer

   The induced current flows in a direction that opposes the change in flux producing it.

3. 3.Expression for motional emf?Show answer

   $\epsilon =Blv$.

4. 4.Define self-inductance.Show answer

   The flux linkage per unit current in a coil ($L=N\varphi /I$), or the emf induced per unit rate of change of its own current.

5. 5.Self-inductance of a long solenoid?Show answer

   $L=\frac{{\mu }_0{N}^{2}A}{l}$.

6. 6.Mutual inductance of coaxial solenoids?Show answer

   $M=\frac{{\mu }_0{N}_1{N}_{2}A}{l}$.

7. 7.What are eddy currents? Give one use.Show answer

   Induced circulating currents in bulk metal in a changing flux; used in induction furnaces, eddy-current brakes and damping.

8. 8.Why is power transmitted at high voltage?Show answer

   For the same power the current is smaller, so the $I^{2}R$ loss in the lines is much smaller.

Quick revision checklist

The night before the exam: can you do each of these without looking?

• ☐Can you state Faraday's two laws and Lenz's law?
• ☐Can you find the direction of induced current when a magnet approaches or leaves a coil?
• ☐Can you derive ε = Blv and show mechanical power = electrical power?
• ☐Can you differentiate a position function to get v?
• ☐Can you explain eddy currents, give two uses and say how they are reduced?
• ☐Can you derive $L={\mu }_0{N}^{2}A/l$ and $M={\mu }_0{N}_1{N}_{2}A/l$?
• ☐Can you do transformer turns, current and efficiency numericals?
• ☐Can you list four transformer losses with remedies?

Electromagnetic Induction: frequently asked questions