Meet Your Tutor
Electromagnetic Induction becomes predictable when every problem begins with one question: what is changing the magnetic flux? I will help you identify that change, apply Faraday’s and Lenz’s laws with the correct sign, and check the direction and units before accepting a numerical answer.
Here is the single sentence that unlocks the whole of Electromagnetic Induction: the circuit is lazy, and it hates change. Not magnetic fields — a coil is perfectly happy sitting inside the strongest field you can build. What it objects to is the field changing. Every EMF, every induced current, every mysterious minus sign in this chapter is the system complaining about a change and trying to undo it. Hold that thought and the chapter stops being a list of formulas and becomes one idea told eight different ways.
These Electromagnetic Induction Class 12 Physics notes cover the complete official 2026-27 scope — magnetic flux, Faraday’s laws, induced EMF and current, Lenz’s law, and self and mutual induction — with solved examples for every idea, a set of Electromagnetic Induction numericals with solutions, a formula list for quick revision, and a practice worksheet with answers you can hide and reveal. Everything is built around one diagnostic you will use on every single problem: which knob is turning?
Two things worth knowing before you start. First, this chapter sits in Unit IV — Electromagnetic Induction and Alternating Currents. The CBSE marks table groups Unit III and Unit IV together at 17 marks, so there is no separate official figure for Chapter 6 alone — be suspicious of anyone who quotes you one. Second, this chapter leans hard on the previous two. If the phrase “magnetic field into the page” still makes you pause, spend twenty minutes on Moving Charges and Magnetism first, and keep Magnetism and Matter nearby for how materials respond. If even circuits feel shaky, the Electricity chapter from Class 10 Science is the gentler place to warm up.
The through-line for the whole chapter
Flux by itself does nothing. Rate of change of flux does everything. Whenever you are stuck, stop calculating and ask two questions: what is changing, and how fast. That is genuinely the entire chapter.
What You’ll Learn
What You’ll Learn
- Magnetic Flux — The Idea Everything Rests On
- Faraday’s Laws of Electromagnetic Induction
- Lenz’s Law and the Direction of Induced Current
- Motional EMF — Induced EMF in a Moving Conductor
- Self-Induction and Self-Inductance
- Mutual Induction and Mutual Inductance
- Energy Stored in an Inductor
- Eddy Currents — A Scope Note and the Applications
- Electromagnetic Induction Class 12 Physics Numericals With Solutions
- Formula List and Quick Revision
Your Game Plan
Six sittings, roughly forty minutes each. Do them in this order — the chapter is genuinely sequential, and skipping ahead to inductance before flux is solid is the most common way students lose marks here.
- Sitting 1 — own the flux formula. Learn Φ = BA cosθ and the B–A–θ knob-box below until you can name the knob in any problem within five seconds.
- Sitting 2 — Faraday, then Lenz. Get the magnitude first (Faraday), then the direction (Lenz). Never both at once. Practise the 4-step direction routine on ten sketches.
- Sitting 3 — motional EMF. Derive ε = Blv two ways: from Faraday’s law and from the force on the free charges. Both derivations are examinable.
- Sitting 4 — self and mutual inductance. These are the two named items in the syllabus text, so they are near-certain question material. Learn the definitions as ratios, not as spells.
- Sitting 5 — energy. Half-L-I-squared, the energy density of the field, and the reason the minus sign exists at all.
- Sitting 6 — the worksheet at the bottom. Answers hidden. Attempt first, reveal second. Anything you get wrong, go back to the relevant section rather than re-reading the answer.
Study Notes
Magnetic Flux — The Idea Everything Rests On
The B–A–θ knob-box: every induction problem in the chapter is one of these three knobs being turned. Ask “which knob?” first, every time.
Magnetic flux is a counting idea. Imagine the magnetic field as a bundle of field lines threading through a loop of wire. Magnetic flux is how many of those lines get through. A big loop held square-on to a strong field catches a lot; the same loop turned edge-on catches nothing, even though the field has not changed at all.
Definition — magnetic flux
For a flat loop of area A in a uniform field B:
Φ = B A cosθ = B · A
Here θ is the angle between B and the normal (the perpendicular) to the loop’s surface — not the angle with the loop’s plane. SI unit: the weber (Wb), where 1 Wb = 1 T m². Flux is a scalar, and it can be positive, negative or zero depending on which way you choose to call “normal”.
Φ = B A cosθ = B · A
Here θ is the angle between B and the normal (the perpendicular) to the loop’s surface — not the angle with the loop’s plane. SI unit: the weber (Wb), where 1 Wb = 1 T m². Flux is a scalar, and it can be positive, negative or zero depending on which way you choose to call “normal”.
For a coil of N turns wound so that every turn encloses the same flux, what matters electrically is the flux linkage NΦ, measured in weber-turns. This is why coils have many turns: you multiply the effect without needing a bigger magnet.
The memory device — B–A–θ: Bend it, Area it, Angle it
Look at Φ = BA cosθ. There are exactly three symbols on the right, so there are exactly three knobs you can turn to change the flux:
B — Bend it: change the field strength (move a magnet, change a current).
A — Area it: change how much loop area sits in the field (slide a rod, stretch the loop, pull the coil out).
θ — Angle it: tilt or rotate the loop.
No fourth option exists. So on every problem in this chapter your first move is to ask: which knob is turning? That single question tells you which formula you need.
B — Bend it: change the field strength (move a magnet, change a current).
A — Area it: change how much loop area sits in the field (slide a rod, stretch the loop, pull the coil out).
θ — Angle it: tilt or rotate the loop.
No fourth option exists. So on every problem in this chapter your first move is to ask: which knob is turning? That single question tells you which formula you need.
The angle trap
Problems love to say “the field makes an angle of 30° with the plane of the coil.” That is not θ. The normal is perpendicular to the plane, so θ = 90° − 30° = 60°, and cosθ = 0.5, not 0.866. Read the word “plane” versus “normal” every single time. Half the flux marks lost in board exams are lost right here.
Example 1 — flux through a loop held square-on
A rectangular loop measuring 12 cm × 8 cm lies in a uniform field of 0.25 T with its plane perpendicular to the field. Find the flux.
Which knob? None — nothing is changing, this is just a flux calculation.
Plane perpendicular to B means the normal is parallel to B, so θ = 0° and cosθ = 1.
A = 0.12 × 0.08 = 9.6 × 10−3 m²
Φ = 0.25 × 9.6 × 10−3 × 1 = 2.4 × 10−3 Wb = 2.4 mWb
Which knob? None — nothing is changing, this is just a flux calculation.
Plane perpendicular to B means the normal is parallel to B, so θ = 0° and cosθ = 1.
A = 0.12 × 0.08 = 9.6 × 10−3 m²
Φ = 0.25 × 9.6 × 10−3 × 1 = 2.4 × 10−3 Wb = 2.4 mWb
Example 2 — the same loop, tilted
The loop of Example 1 is now tilted so that its normal makes 60° with the field. Find the new flux, and comment.
Φ = 2.4 × 10−3 × cos 60° = 2.4 × 10−3 × 0.5 = 1.2 × 10−3 Wb = 1.2 mWb
Comment: the field did not change and the loop did not change — only the angle knob turned, and the flux halved. If that tilt happened over time, an EMF would appear. Because it happened once and stopped, no EMF survives. Change, then stop, then nothing.
Φ = 2.4 × 10−3 × cos 60° = 2.4 × 10−3 × 0.5 = 1.2 × 10−3 Wb = 1.2 mWb
Comment: the field did not change and the loop did not change — only the angle knob turned, and the flux halved. If that tilt happened over time, an EMF would appear. Because it happened once and stopped, no EMF survives. Change, then stop, then nothing.
Example 3 — flux linkage of a multi-turn coil
A coil of 200 turns, each of area 5 cm², sits in a field of 0.4 T with the normal at 30° to the field. Find the flux per turn and the total flux linkage.
A = 5 cm² = 5 × 10−4 m² (remember: 1 cm² = 10−4 m²)
cos 30° = 0.8660
Φ per turn = 0.4 × 5 × 10−4 × 0.8660 = 1.732 × 10−4 Wb
NΦ = 200 × 1.732 × 10−4 = 3.464 × 10−2 Wb-turns
Why it works: each turn is its own little loop catching the same field lines, so the coil as a whole links 200 times as much. Nothing about the field changed — you just gave it 200 chances to be counted.
A = 5 cm² = 5 × 10−4 m² (remember: 1 cm² = 10−4 m²)
cos 30° = 0.8660
Φ per turn = 0.4 × 5 × 10−4 × 0.8660 = 1.732 × 10−4 Wb
NΦ = 200 × 1.732 × 10−4 = 3.464 × 10−2 Wb-turns
Why it works: each turn is its own little loop catching the same field lines, so the coil as a whole links 200 times as much. Nothing about the field changed — you just gave it 200 chances to be counted.
Unit conversions that cost marks
1 cm² = 10−4 m², not 10−2. 1 mm² = 10−6 m². 1 mWb = 10−3 Wb. 1 gauss = 10−4 T. Convert before you substitute, never after, and write the unit next to every number as you go.
Faraday’s Laws of Electromagnetic Induction
A schematic of Faraday’s original coil-and-magnet experiment. Note what the galvanometer is really measuring: not the flux, but the rate at which the flux is changing.
Faraday spent a decade chasing one question: can a magnet make electricity? His answer, arrived at in 1831, is the most quietly consequential result in this book — every power station on Earth runs on it. And the answer came with a condition attached: the magnet has to be moving.
Faraday’s first law
Whenever the magnetic flux linked with a closed circuit changes, an EMF is induced in the circuit. The induced EMF lasts only as long as the change lasts.
Faraday’s second law
The magnitude of the induced EMF equals the rate of change of magnetic flux linkage:
ε = − N dΦ/dt
For an average value over a finite interval, εav = − N ΔΦ/Δt. The minus sign is Lenz’s law, folded into the equation.
ε = − N dΦ/dt
For an average value over a finite interval, εav = − N ΔΦ/Δt. The minus sign is Lenz’s law, folded into the equation.
Read the formula out loud as a sentence and it stops being intimidating: the EMF is how fast the flux linkage is changing. Big flux, slowly changing, gives a small EMF. Tiny flux, violently changing, gives a huge EMF. This is why a lazily waved magnet does almost nothing and a snapped-open inductive circuit can produce a visible spark.
The minus sign is a price tag
Students memorise the minus sign and forget what it is for. It is energy conservation, written in shorthand. If the induced effect helped the change instead of opposing it, the smallest nudge would grow without limit and you would have electrical energy for free. Nature charges for it: you pay mechanical work at the input, and you collect electrical energy at the output. The minus sign is the receipt. Say that in an exam and you have the conceptual mark.
Example 4 — changing field through a fixed coil
A 50-turn coil of area 0.02 m² lies with its plane perpendicular to a magnetic field which increases steadily from 0.1 T to 0.5 T in 0.2 s. Find the average induced EMF.
Which knob? B — bend it. Area and angle are fixed, so pull them out of the derivative.
ε = NA ΔB/Δt
ΔB/Δt = (0.5 − 0.1)/0.2 = 2.0 T s−1
ε = 50 × 0.02 × 2.0 = 2.0 V
Which knob? B — bend it. Area and angle are fixed, so pull them out of the derivative.
ε = NA ΔB/Δt
ΔB/Δt = (0.5 − 0.1)/0.2 = 2.0 T s−1
ε = 50 × 0.02 × 2.0 = 2.0 V
Example 5 — current and charge in the same coil
The coil of Example 4 has a total circuit resistance of 4 Ω. Find the induced current and the total charge that flows during the 0.2 s.
I = ε/R = 2.0/4 = 0.5 A
q = I Δt = 0.5 × 0.2 = 0.1 C
The elegant route: combine the two steps algebraically and the time cancels:
q = ∫I dt = ∫(N/R)(dΦ/dt) dt = N ΔΦ/R = 50 × 0.02 × 0.4 / 4 = 0.4/4 = 0.1 C ✓
Why this matters: the charge depends only on the total flux change and the resistance — not on how fast it happened. Move the magnet fast or slow: you get a brief big current or a long small one, but the same charge. This is exactly how a ballistic galvanometer measures flux, and it is a favourite one-mark trap. Contrast it with steady-current work in Current Electricity, where charge and time are inseparable.
I = ε/R = 2.0/4 = 0.5 A
q = I Δt = 0.5 × 0.2 = 0.1 C
The elegant route: combine the two steps algebraically and the time cancels:
q = ∫I dt = ∫(N/R)(dΦ/dt) dt = N ΔΦ/R = 50 × 0.02 × 0.4 / 4 = 0.4/4 = 0.1 C ✓
Why this matters: the charge depends only on the total flux change and the resistance — not on how fast it happened. Move the magnet fast or slow: you get a brief big current or a long small one, but the same charge. This is exactly how a ballistic galvanometer measures flux, and it is a favourite one-mark trap. Contrast it with steady-current work in Current Electricity, where charge and time are inseparable.
Example 6 — flux given as a function of time
The flux through a single-turn loop varies as Φ = (3t² + 2t + 5) × 10−3 Wb, with t in seconds. Find the induced EMF at t = 2 s, and the current if the loop resistance is 7 Ω.
ε = − dΦ/dt = −(6t + 2) × 10−3 V
At t = 2 s: ε = −(12 + 2) × 10−3 = −14 mV, so the magnitude is 14 mV.
I = 14 × 10−3 / 7 = 2.0 × 10−3 A = 2 mA
Notice the constant 5 mWb did nothing. It differentiated away. The circuit genuinely does not care about the flux it already has — only about the change. That is the laziness principle in algebra.
ε = − dΦ/dt = −(6t + 2) × 10−3 V
At t = 2 s: ε = −(12 + 2) × 10−3 = −14 mV, so the magnitude is 14 mV.
I = 14 × 10−3 / 7 = 2.0 × 10−3 A = 2 mA
Notice the constant 5 mWb did nothing. It differentiated away. The circuit genuinely does not care about the flux it already has — only about the change. That is the laziness principle in algebra.
Example 7 — rotating the coil through a quarter turn
A 100-turn coil of area 0.05 m² sits in a uniform 0.2 T field with its normal along the field. It is rotated through 90° in 0.1 s. Find the average EMF.
Which knob? θ — angle it.
Initial linkage = NBA cos 0° = 100 × 0.2 × 0.05 = 1.0 Wb-turns
Final linkage = NBA cos 90° = 0
εav = Δ(NΦ)/Δt = 1.0/0.1 = 10 V
Which knob? θ — angle it.
Initial linkage = NBA cos 0° = 100 × 0.2 × 0.05 = 1.0 Wb-turns
Final linkage = NBA cos 90° = 0
εav = Δ(NΦ)/Δt = 1.0/0.1 = 10 V
Where the AC generator went
NCERT prints the AC generator at the end of this chapter, but the official CBSE 2026-27 syllabus lists AC generator under Chapter 7, Alternating Current, alongside the transformer. So it is examinable — just from the next chapter. Learn the continuously rotating coil, ε = NBAω sin ωt, when you get there. Here, stick to finite rotations and average EMFs like Example 7.
Lenz’s Law and the Direction of Induced Current
Run these four steps in order and you will never guess a current direction again. Step 2 is the one students skip — and it is the only step that carries the minus sign.
Faraday tells you how big. Lenz tells you which way. Keep the two jobs separate and this stops being the hardest part of the chapter.
Lenz’s law
The induced EMF drives a current in whichever direction opposes the change in flux that produced it. It is the physical content of the minus sign in Faraday’s law, and it is a direct consequence of the conservation of energy.
The memory device — OPPOSE the CHANGE, not the FIELD
This is the one line to carry into the exam hall. The induced current does not always oppose the applied field. It opposes whatever the flux is doing.
• Flux growing? The induced field points against the applied field, trying to hold it down.
• Flux shrinking? The induced field points along the applied field, trying to prop it up.
Half of all direction errors come from students who learned “oppose the field” instead of “oppose the change”. The field is not the enemy. The change is.
• Flux growing? The induced field points against the applied field, trying to hold it down.
• Flux shrinking? The induced field points along the applied field, trying to prop it up.
Half of all direction errors come from students who learned “oppose the field” instead of “oppose the change”. The field is not the enemy. The change is.
Now the routine. Four steps, in this order, every time — the flowchart above is your reference:
- Find the direction of B through the loop as it stands. Into the page, out of the page, left, right. Sketch it.
- Decide whether Φ is growing or shrinking. This is the step students skip, and it is the only step that carries the minus sign. Look at which knob is turning and which way.
- Work out which way the induced B must point inside the loop to fight that change — against the applied field if Φ is growing, along it if Φ is shrinking.
- Curl your right hand. Point your right thumb along the induced field inside the loop; your curled fingers give the sense of the induced current. Read it off as clockwise or anticlockwise as seen by you, and say which side you are viewing from.
The mechanical shortcut
For anything moving, Lenz’s law always shows up as a brake. Approaching magnet? The coil repels it. Receding magnet? The coil attracts it. Loop being pulled out of a field? The force resists the pull. If your answer ever comes out as a helpful push, you have made a sign error — a helpful push would be a perpetual-motion machine. Use this as a free sanity check on every direction answer you write.
Example 8 — bar magnet approaching a coil (reasoning)
The north pole of a bar magnet is pushed towards the near face of a circular coil. Which way does the induced current flow as seen from the magnet’s side, and what force does the magnet feel?
Step 1: the north pole sends field lines out of itself, so through the coil the field points away from the magnet.
Step 2: as the magnet gets closer, that field gets stronger — Φ is growing.
Step 3: the induced field inside the coil must therefore point back towards the magnet, to hold the growing flux down. That makes the near face of the coil behave as a north pole.
Step 4: right thumb pointing back towards the magnet (i.e. towards the viewer) ⇒ fingers curl anticlockwise as seen from the magnet’s side.
Force: north facing north, so the coil repels the magnet. You must push against that repulsion, and the work you do is exactly the electrical energy that appears in the coil. Sanity check passed: it is a brake, not a push.
Step 1: the north pole sends field lines out of itself, so through the coil the field points away from the magnet.
Step 2: as the magnet gets closer, that field gets stronger — Φ is growing.
Step 3: the induced field inside the coil must therefore point back towards the magnet, to hold the growing flux down. That makes the near face of the coil behave as a north pole.
Step 4: right thumb pointing back towards the magnet (i.e. towards the viewer) ⇒ fingers curl anticlockwise as seen from the magnet’s side.
Force: north facing north, so the coil repels the magnet. You must push against that repulsion, and the work you do is exactly the electrical energy that appears in the coil. Sanity check passed: it is a brake, not a push.
Example 9 — numbers on a shrinking field
A square loop of side 20 cm and resistance 0.1 Ω lies flat on the page in a field directed into the page. The field falls steadily from 0.6 T to 0.2 T in 0.5 s. Find the induced current and its direction.
A = (0.20)² = 0.04 m²
|ΔB/Δt| = (0.6 − 0.2)/0.5 = 0.8 T s−1
|ε| = A |ΔB/Δt| = 0.04 × 0.8 = 0.032 V = 32 mV
I = 0.032/0.1 = 0.32 A
Direction, by the routine: B is into the page (step 1); Φ into the page is shrinking (step 2); so the induced field inside the loop must also point into the page, to prop the flux up (step 3); right thumb into the page ⇒ fingers curl clockwise as you look at the page (step 4).
Note how “oppose the field” would have given you anticlockwise — the wrong answer. Oppose the change.
A = (0.20)² = 0.04 m²
|ΔB/Δt| = (0.6 − 0.2)/0.5 = 0.8 T s−1
|ε| = A |ΔB/Δt| = 0.04 × 0.8 = 0.032 V = 32 mV
I = 0.032/0.1 = 0.32 A
Direction, by the routine: B is into the page (step 1); Φ into the page is shrinking (step 2); so the induced field inside the loop must also point into the page, to prop the flux up (step 3); right thumb into the page ⇒ fingers curl clockwise as you look at the page (step 4).
Note how “oppose the field” would have given you anticlockwise — the wrong answer. Oppose the change.
Three Lenz mistakes that keep reappearing
1. Answering “clockwise” without saying which side you are viewing from. Clockwise from the front is anticlockwise from behind; an unqualified answer can be marked wrong.
2. Curling the right hand around the applied field instead of the induced field. Step 3 must be finished before step 4 begins.
3. Forgetting that a loop moving through a uniform field, entirely inside it, has constant flux and therefore no induced current. Motion alone is not enough — the flux has to change.
2. Curling the right hand around the applied field instead of the induced field. Step 3 must be finished before step 4 begins.
3. Forgetting that a loop moving through a uniform field, entirely inside it, has constant flux and therefore no induced current. Motion alone is not enough — the flux has to change.
Motional EMF — Induced EMF in a Moving Conductor
This is the “A — Area it” knob in its purest form. Picture a straight conducting rod of length l sliding on two frictionless rails, the whole circuit sitting in a uniform field B perpendicular to the plane of the circuit. As the rod slides with speed v, the enclosed area grows, so the flux grows, so an EMF appears.
Motional EMF
ε = B l v (for B, l and v mutually perpendicular)
More generally ε = Blv sin φ, where φ is the angle between the velocity and the field. If the rod moves along the field, no area is swept across field lines and the EMF is zero.
More generally ε = Blv sin φ, where φ is the angle between the velocity and the field. If the rod moves along the field, no area is swept across field lines and the EMF is zero.
Derivation 1 — straight from Faraday. Let x be the distance of the rod from the closed end. Enclosed area = lx, so Φ = Blx. Then
ε = − dΦ/dt = −Bl dx/dt = −Blv, whose magnitude is Blv.
Derivation 2 — from the force on the free charges. Every free electron in the rod is being carried sideways at speed v, so it feels a magnetic force qvB along the rod. Charge piles up at the ends until the electrostatic field it creates balances that push: qE = qvB, so E = vB, and the potential difference across the rod is El = Blv. Both derivations are examinable, and the second one is where this chapter shakes hands with the magnetic force work in Moving Charges and Magnetism.
EMF is not the same thing as potential difference
The rod is the seat of the EMF — it is doing the work of separating charge, exactly as a cell does. The terminal potential difference across the rest of the circuit is smaller than Blv whenever the rod itself has resistance, by the usual V = ε − Ir. That distinction is worth revising from Current Electricity before you attempt the numericals.
Example 10 — rod on rails — the full energy audit
A rod of length 0.5 m slides at 4 m s−1 on rails in a perpendicular field of 0.3 T. The circuit resistance is 1.5 Ω and friction is negligible. Find the EMF, the current, the force needed to keep the rod moving steadily, and check the power balance.
ε = Blv = 0.3 × 0.5 × 4 = 0.6 V
I = ε/R = 0.6/1.5 = 0.4 A
Retarding force on the rod = BI l = 0.3 × 0.4 × 0.5 = 0.06 N, so the applied force needed for constant speed is 0.06 N (Lenz: it is a brake).
Mechanical power in = Fv = 0.06 × 4 = 0.24 W
Electrical power out = εI = 0.6 × 0.4 = 0.24 W
Heat dissipated = I²R = (0.4)² × 1.5 = 0.24 W ✓
This is the whole chapter in one example. Every joule of electrical energy was paid for by mechanical work against the induced brake. That balance is the minus sign.
ε = Blv = 0.3 × 0.5 × 4 = 0.6 V
I = ε/R = 0.6/1.5 = 0.4 A
Retarding force on the rod = BI l = 0.3 × 0.4 × 0.5 = 0.06 N, so the applied force needed for constant speed is 0.06 N (Lenz: it is a brake).
Mechanical power in = Fv = 0.06 × 4 = 0.24 W
Electrical power out = εI = 0.6 × 0.4 = 0.24 W
Heat dissipated = I²R = (0.4)² × 1.5 = 0.24 W ✓
This is the whole chapter in one example. Every joule of electrical energy was paid for by mechanical work against the induced brake. That balance is the minus sign.
Example 11 — the aeroplane wingtip EMF
An aeroplane with a wingspan of 25 m flies horizontally at 200 m s−1 where the vertical component of the Earth’s magnetic field is 5 × 10−5 T. Find the EMF induced between the wingtips.
Only the field component perpendicular to both the wing and the velocity contributes — for horizontal flight and a horizontal wing, that is the vertical component.
ε = BV l v = 5 × 10−5 × 25 × 200 = 0.25 V
Why no current flows: there is no closed circuit through the sky. You get a steady potential difference of a quarter of a volt across the wings and nothing else — harmless, and a nice reminder that EMF can exist without current.
Only the field component perpendicular to both the wing and the velocity contributes — for horizontal flight and a horizontal wing, that is the vertical component.
ε = BV l v = 5 × 10−5 × 25 × 200 = 0.25 V
Why no current flows: there is no closed circuit through the sky. You get a steady potential difference of a quarter of a volt across the wings and nothing else — harmless, and a nice reminder that EMF can exist without current.
Example 12 — rod rotating about one end
A rod of length 1.0 m rotates in a horizontal plane about one end at 20 rad s−1, in a field of 0.5 T perpendicular to the plane of rotation. Find the EMF between the ends.
Each element at distance r moves at v = ωr, so it contributes dε = Bωr dr. Integrating from 0 to L:
ε = ½ B ω L²
ε = ½ × 0.5 × 20 × (1.0)² = 5.0 V
Sanity check: the tip moves at ωL = 20 m s−1 and the pivot at zero, so the average speed is 10 m s−1; Bl × average v = 0.5 × 1.0 × 10 = 5.0 V. The factor of ½ is just that average. Same answer, and now the formula is not something you have to trust blindly. Equivalent route: the rod sweeps area at the rate ½ωL² per second, and ε = B × (rate of area swept).
Each element at distance r moves at v = ωr, so it contributes dε = Bωr dr. Integrating from 0 to L:
ε = ½ B ω L²
ε = ½ × 0.5 × 20 × (1.0)² = 5.0 V
Sanity check: the tip moves at ωL = 20 m s−1 and the pivot at zero, so the average speed is 10 m s−1; Bl × average v = 0.5 × 1.0 × 10 = 5.0 V. The factor of ½ is just that average. Same answer, and now the formula is not something you have to trust blindly. Equivalent route: the rod sweeps area at the rate ½ωL² per second, and ε = B × (rate of area swept).
The “perpendicular” you must actually check
Blv needs B, the rod and v to be mutually perpendicular. If the rod slides along the field lines, the swept area lies in the field and the flux never changes — the EMF is zero no matter how fast it moves. Before you use Blv, resolve the field and keep only the component that is perpendicular to the plane in which the rod is sweeping area.
Self-Induction and Self-Inductance
Here the coil starts objecting to itself. A coil carrying a current makes its own magnetic field, and therefore links its own flux. Change the current and you change that flux, so the coil induces an EMF in itself — one that fights the change you are trying to make. This is self-induction, and the laziness metaphor has never been more literal: the coil resists being switched on and resists being switched off.
Self-inductance, defined as a ratio
For a coil with no magnetic material nearby, the flux linkage is proportional to the current:
NΦ = L i, so L = NΦ/i
and by Faraday’s law the self-induced (back) EMF is
ε = − L di/dt
SI unit: the henry (H). 1 H = 1 Wb A−1 = 1 V s A−1. L is a purely geometric-and-material property: it depends on the number of turns, the size and shape, and the core — never on the current.
NΦ = L i, so L = NΦ/i
and by Faraday’s law the self-induced (back) EMF is
ε = − L di/dt
SI unit: the henry (H). 1 H = 1 Wb A−1 = 1 V s A−1. L is a purely geometric-and-material property: it depends on the number of turns, the size and shape, and the core — never on the current.
Two sentences worth memorising as definitions. A coil has a self-inductance of one henry if a current of one ampere through it links a flux of one weber-turn. Or equivalently: if a current changing at one ampere per second induces a back EMF of one volt. Examiners accept either; the second is easier to say under pressure.
Self-inductance of a long solenoid — the standard derivation
Inside a long solenoid of N turns, length l and cross-section A, the field is B = μ0ni with n = N/l.
Flux per turn = BA = μ0(N/l)iA
Flux linkage = N × that = μ0N²iA/l
Therefore L = μ0N²A/l = μ0n²A l. With a core of relative permeability μr, replace μ0 by μ0μr — which is precisely why iron-cored coils are so much more inductive, as Magnetism and Matter explains.
Flux per turn = BA = μ0(N/l)iA
Flux linkage = N × that = μ0N²iA/l
Therefore L = μ0N²A/l = μ0n²A l. With a core of relative permeability μr, replace μ0 by μ0μr — which is precisely why iron-cored coils are so much more inductive, as Magnetism and Matter explains.
Look at the N². Doubling the turns doubles the field and doubles the number of turns linking it, so the inductance quadruples. That squared dependence is a favourite short-answer question.
Example 13 — back EMF from a changing current
The current in a 0.4 H coil rises steadily from 2 A to 6 A in 0.05 s. Find the magnitude of the self-induced EMF.
di/dt = (6 − 2)/0.05 = 80 A s−1
|ε| = L di/dt = 0.4 × 80 = 32 V
Direction: the current is growing, so the back EMF opposes the growth — it acts against the driving source. This is why a large inductor takes time to reach full current, and why breaking such a circuit can produce a spark far above the supply voltage: di/dt becomes enormous when the switch opens.
di/dt = (6 − 2)/0.05 = 80 A s−1
|ε| = L di/dt = 0.4 × 80 = 32 V
Direction: the current is growing, so the back EMF opposes the growth — it acts against the driving source. This is why a large inductor takes time to reach full current, and why breaking such a circuit can produce a spark far above the supply voltage: di/dt becomes enormous when the switch opens.
Example 14 — self-inductance of a solenoid from its dimensions
A solenoid has 500 turns wound over a length of 0.25 m, with a cross-sectional area of 4 cm² and an air core. Find its self-inductance.
A = 4 cm² = 4 × 10−4 m²
L = μ0N²A/l = (4π × 10−7) × (500)² × 4 × 10−4 / 0.25
Numerator: 1.2566 × 10−6 × 2.5 × 105 = 0.31416; × 4 × 10−4 = 1.2566 × 10−4
L = 1.2566 × 10−4 / 0.25 = 5.03 × 10−4 H = 0.503 mH
Reality check: a fraction of a millihenry is exactly right for a small air-cored coil. Iron-cored coils reach henries; if your air-core answer comes out in henries, you have almost certainly forgotten the cm² conversion.
A = 4 cm² = 4 × 10−4 m²
L = μ0N²A/l = (4π × 10−7) × (500)² × 4 × 10−4 / 0.25
Numerator: 1.2566 × 10−6 × 2.5 × 105 = 0.31416; × 4 × 10−4 = 1.2566 × 10−4
L = 1.2566 × 10−4 / 0.25 = 5.03 × 10−4 H = 0.503 mH
Reality check: a fraction of a millihenry is exactly right for a small air-cored coil. Iron-cored coils reach henries; if your air-core answer comes out in henries, you have almost certainly forgotten the cm² conversion.
Example 15 — reading L backwards from the flux linkage
A 2 H coil carries a steady current of 3 A. Find its flux linkage. Then state the EMF induced in it.
NΦ = Li = 2 × 3 = 6 Wb-turns
The current is steady, so di/dt = 0 and the induced EMF is zero.
The point: a big flux linkage sitting there does nothing at all. An inductor carrying a perfectly steady current behaves as a plain piece of wire. Laziness again — no change, no complaint.
NΦ = Li = 2 × 3 = 6 Wb-turns
The current is steady, so di/dt = 0 and the induced EMF is zero.
The point: a big flux linkage sitting there does nothing at all. An inductor carrying a perfectly steady current behaves as a plain piece of wire. Laziness again — no change, no complaint.
L does not depend on the current
A very common error is to write “L = NΦ/i, so doubling the current halves L.” No: doubling the current doubles Φ too, and the ratio is unchanged. Self-inductance is fixed by geometry and the core material. The same logic applies to capacitance in Electric Charges and Fields-style problems — a defining ratio is a constant, not a variable.
Mutual Induction and Mutual Inductance
Now put two coils near each other. Change the current in the first and you change the flux it sends through the second, so an EMF appears in the second coil even though nothing there moved and nothing there is connected. This is mutual induction, and it is the whole working principle of the transformer, the induction hob and every wireless charger you have ever used.
Mutual inductance, defined as a ratio
N2Φ2 = M i1, so M = N2Φ2/i1
and the EMF induced in coil 2 by a changing current in coil 1 is
ε2 = − M di1/dt
Unit: the henry, again. And crucially, M12 = M21 = M — the reciprocity theorem. The pair of coils has one mutual inductance, no matter which one you drive.
and the EMF induced in coil 2 by a changing current in coil 1 is
ε2 = − M di1/dt
Unit: the henry, again. And crucially, M12 = M21 = M — the reciprocity theorem. The pair of coils has one mutual inductance, no matter which one you drive.
That reciprocity result is not obvious and it is worth stating clearly in an answer: whichever coil you push the current through, the same M appears. It follows from the geometry of the coupling, and it is a frequent one-mark question.
Two coaxial solenoids — the standard result
For a short solenoid of N2 turns wound over a long solenoid of N1 turns, length l and common cross-section A:
M = μ0N1N2A/l
Coupling is also written M = k√(L1L2), where k is the coefficient of coupling, 0 ≤ k ≤ 1. Perfect coupling (k = 1) means every field line from one coil threads the other — the target a transformer core is built to hit.
M = μ0N1N2A/l
Coupling is also written M = k√(L1L2), where k is the coefficient of coupling, 0 ≤ k ≤ 1. Perfect coupling (k = 1) means every field line from one coil threads the other — the target a transformer core is built to hit.
Example 16 — EMF in the secondary coil
Two coils have a mutual inductance of 50 mH. The current in the first coil is switched from 0 to 10 A in 0.1 s. Find the average EMF induced in the second coil.
M = 50 mH = 0.05 H
di1/dt = 10/0.1 = 100 A s−1
|ε2| = M × 100 = 0.05 × 100 = 5.0 V
Note that the second coil need not be connected to anything for this EMF to exist. Complete its circuit and a current flows; leave it open and you simply have 5 V across its ends.
M = 50 mH = 0.05 H
di1/dt = 10/0.1 = 100 A s−1
|ε2| = M × 100 = 0.05 × 100 = 5.0 V
Note that the second coil need not be connected to anything for this EMF to exist. Complete its circuit and a current flows; leave it open and you simply have 5 V across its ends.
Example 17 — mutual inductance of two coaxial solenoids
A solenoid of 300 turns is wound over a length of 0.30 m; a second coil of 400 turns is wound tightly over it. The common cross-sectional area is 5 cm². Find M, assuming ideal coupling and an air core.
A = 5 × 10−4 m²
M = μ0N1N2A/l = (4π × 10−7)(300)(400)(5 × 10−4)/0.30
1.2566 × 10−6 × 1.2 × 105 = 0.15080; × 5 × 10−4 = 7.540 × 10−5
M = 7.540 × 10−5 / 0.30 = 2.51 × 10−4 H = 0.251 mH
A = 5 × 10−4 m²
M = μ0N1N2A/l = (4π × 10−7)(300)(400)(5 × 10−4)/0.30
1.2566 × 10−6 × 1.2 × 105 = 0.15080; × 5 × 10−4 = 7.540 × 10−5
M = 7.540 × 10−5 / 0.30 = 2.51 × 10−4 H = 0.251 mH
Example 18 — finding the coefficient of coupling
Two coils of self-inductance 8 mH and 18 mH have a mutual inductance of 9.6 mH. Find the coefficient of coupling, and the largest M these two coils could possibly have.
√(L1L2) = √(8 × 18) = √144 = 12 mH
k = M/√(L1L2) = 9.6/12 = 0.80
Maximum possible M is at k = 1, i.e. 12 mH.
Interpretation: 80% of the flux from one coil is being captured by the other. The missing 20% is leakage flux — field lines that escape into the air instead of threading the second coil. Wrapping both coils on a closed iron core is how engineers push k towards 1.
√(L1L2) = √(8 × 18) = √144 = 12 mH
k = M/√(L1L2) = 9.6/12 = 0.80
Maximum possible M is at k = 1, i.e. 12 mH.
Interpretation: 80% of the flux from one coil is being captured by the other. The missing 20% is leakage flux — field lines that escape into the air instead of threading the second coil. Wrapping both coils on a closed iron core is how engineers push k towards 1.
| Point of comparison | Self-induction | Mutual induction |
|---|---|---|
| How many coils | One coil, acting on itself | Two coils, one acting on the other |
| What causes the EMF | A change in the coil’s own current | A change in the current in the other coil |
| Defining ratio | L = NΦ/i | M = N2Φ2/i1 |
| Induced EMF | ε = −L di/dt | ε2 = −M di1/dt |
| Standard solenoid result | L = μ0N²A/l | M = μ0N1N2A/l |
| SI unit | henry (H) | henry (H) |
| Depends on | Geometry, number of turns, core material | Geometry of both coils, their separation and relative orientation, core material |
| Where you meet it | Chokes, inductors, the spark on a switch | Transformers, induction cooktops, wireless charging |
| Called | Back EMF | Mutually induced EMF |
The orientation mark nobody claims
M depends on relative orientation. Turn the second coil so its plane contains the axis of the first and the linked flux drops to zero — M becomes zero and the coils stop talking to each other entirely. If a question asks how to reduce coupling without moving the coils apart, “rotate one of them through 90°” is the answer being looked for.
Energy Stored in an Inductor
If the coil fights you every time you try to raise its current, then raising the current must cost work. Where does that work go? Into the magnetic field. An inductor is an energy store, exactly as a capacitor is — it just stores in a magnetic field instead of an electric one.
Energy stored in an inductor
Against a back EMF of L di/dt, the source delivers power P = εi = Li di/dt. Integrating from zero current to a final current I:
U = ∫0I Li di = ½ L I² (joules)
And the energy per unit volume of the field itself is
uB = B²/(2μ0) (J m−3)
U = ∫0I Li di = ½ L I² (joules)
And the energy per unit volume of the field itself is
uB = B²/(2μ0) (J m−3)
The parallel with electrostatics is worth writing down side by side, because examiners like the comparison and because it makes both easier to remember. A capacitor stores ½CV² with an energy density of ½ε0E²; an inductor stores ½LI² with an energy density of B²/2μ0. If the electrostatic half of that pairing is hazy, Electric Charges and Fields is where it is set up.
Example 19 — straightforward energy store
Find the energy stored in a 0.5 H inductor carrying 4 A.
U = ½LI² = ½ × 0.5 × (4)² = ½ × 0.5 × 16 = 4.0 J
Halve the current and the stored energy falls to a quarter — the square matters. This is also why switching a high-current inductor off is dangerous: those 4 joules have to go somewhere in a very short time.
U = ½LI² = ½ × 0.5 × (4)² = ½ × 0.5 × 16 = 4.0 J
Halve the current and the stored energy falls to a quarter — the square matters. This is also why switching a high-current inductor off is dangerous: those 4 joules have to go somewhere in a very short time.
Example 20 — energy density of a magnetic field
Find the energy density in a region where the magnetic field is 0.02 T (μ0 = 4π × 10−7 T m A−1).
uB = B²/(2μ0) = (0.02)² / (2 × 1.2566 × 10−6)
= 4.0 × 10−4 / 2.5133 × 10−6 = 159 J m−3
Roughly 159 joules in every cubic metre. Magnetic fields are a fairly thin energy store, which is why practical inductors need either many turns, an iron core, or both.
uB = B²/(2μ0) = (0.02)² / (2 × 1.2566 × 10−6)
= 4.0 × 10−4 / 2.5133 × 10−6 = 159 J m−3
Roughly 159 joules in every cubic metre. Magnetic fields are a fairly thin energy store, which is why practical inductors need either many turns, an iron core, or both.
Example 21 — building the current up — a power audit
The current in a 20 mH inductor is ramped linearly from 0 to 5 A in 2.0 ms. Find the back EMF during the ramp, the final stored energy, and the average power delivered to the field.
L = 0.020 H, di/dt = 5/(2.0 × 10−3) = 2500 A s−1
|ε| = L di/dt = 0.020 × 2500 = 50 V (constant, because the ramp is linear)
U = ½LI² = ½ × 0.020 × 25 = 0.25 J
Average power = U/Δt = 0.25/(2.0 × 10−3) = 125 W
Cross-check: instantaneous power is εi = 50i, which rises linearly from 0 to 250 W. Its average is 50 × 2.5 = 125 W. ✓ Note the 50 V back EMF from a supply that may only be a few volts — inductors generate voltages of their own, and 2 milliseconds is a very fast change.
L = 0.020 H, di/dt = 5/(2.0 × 10−3) = 2500 A s−1
|ε| = L di/dt = 0.020 × 2500 = 50 V (constant, because the ramp is linear)
U = ½LI² = ½ × 0.020 × 25 = 0.25 J
Average power = U/Δt = 0.25/(2.0 × 10−3) = 125 W
Cross-check: instantaneous power is εi = 50i, which rises linearly from 0 to 250 W. Its average is 50 × 2.5 = 125 W. ✓ Note the 50 V back EMF from a supply that may only be a few volts — inductors generate voltages of their own, and 2 milliseconds is a very fast change.
Why the minus sign has to be there — say this in the exam
Suppose Faraday’s law had a plus sign. Then an induced current would reinforce the flux change that created it, which would induce a bigger current, which would reinforce it further. Energy would appear out of nothing and grow without bound. The minus sign is the only thing standing between this chapter and a perpetual-motion machine. That is why it is not a convention you may drop — it is the statement that you pay mechanical work to get electrical energy.
Eddy Currents — A Scope Note and the Applications
Scope note — read this first
Eddy currents are not named in the official CBSE Class 12 Physics syllabus text for the 2026-27 session. Chapter 6 is listed as “Electromagnetic induction; Faraday’s laws, induced EMF and current; Lenz’s Law, Self and mutual induction.” The topic is kept here because it is genuinely useful for understanding Lenz’s law and because it explains several applications you will be asked about elsewhere — but do not spend heavy revision time on it at the expense of the named topics, and confirm with your teacher whether your school is examining it this session. Syllabus documents can be revised; your teacher has the final word.
An eddy current is Lenz’s law happening in a solid block instead of a wire. Take a sheet of metal and change the flux through it. There is no neat single loop for the induced current to follow, so it swirls in closed circulating whirlpools inside the bulk of the metal — hence the name, after eddies in water. And by Lenz’s law those swirls oppose the change, which means they always act as a brake and always dissipate energy as heat.
Two consequences, and they pull in opposite directions. Sometimes the heat and the braking are the whole point; sometimes they are a nuisance to be designed out.
- Useful — electromagnetic braking. Metro trains and heavy vehicles drop a strong electromagnet near a spinning metal disc or rail. Eddy currents in the metal oppose the motion, and the train slows with no physical contact and no brake pads to wear out. Note that the braking force fades as the speed falls, which is why these systems are paired with friction brakes for the final stop.
- Useful — induction heating and induction cooktops. A rapidly changing field drives eddy currents in the base of a steel pan; the pan’s own resistance turns them into heat. The hob surface itself stays comparatively cool, which is also why a copper or glass vessel will not work on one.
- Useful — metal detectors and coin validators. A changing field induces eddy currents in nearby metal; those currents produce their own field, which the detector picks up. Different metals respond differently, which is how a vending machine tells a real coin from a slug.
- Useful — damping in moving-coil instruments. Galvanometer coils are wound on light metal frames so that eddy currents kill the needle’s oscillations and it settles quickly on a reading instead of swinging about.
- Unwanted — heating in transformer and motor cores. Here eddy currents are pure waste. The fix is lamination: build the core from thin sheets separated by insulating varnish, so the circulating paths are chopped into short high-resistance loops. Less current circulates, and far less energy is lost as heat.
Example 22 — why lamination works (reasoning, worth 2–3 marks)
Question: Explain why a transformer core is laminated rather than made from one solid block of iron.
Answer to write: A changing flux through a solid core induces eddy currents that circulate in wide loops through the bulk of the metal. Those wide loops enclose a large area, so a large EMF is induced round them, and they run through a large cross-section, so they meet very little resistance — the result is a large current and a large I²R heat loss.
Laminating the core into thin insulated sheets, with the plane of each sheet parallel to the flux, does two things at once. It reduces the area of each available loop, so the induced EMF round it is smaller; and it forces the current through a much narrower cross-section, so the resistance of each path is much larger. Both effects cut the eddy current, so the wasteful heating drops sharply while the useful mutual induction between the windings is left untouched.
Marking note: the marks are for reduced loop area and increased resistance of the path. Saying only “to reduce eddy currents” restates the question and earns little.
Answer to write: A changing flux through a solid core induces eddy currents that circulate in wide loops through the bulk of the metal. Those wide loops enclose a large area, so a large EMF is induced round them, and they run through a large cross-section, so they meet very little resistance — the result is a large current and a large I²R heat loss.
Laminating the core into thin insulated sheets, with the plane of each sheet parallel to the flux, does two things at once. It reduces the area of each available loop, so the induced EMF round it is smaller; and it forces the current through a much narrower cross-section, so the resistance of each path is much larger. Both effects cut the eddy current, so the wasteful heating drops sharply while the useful mutual induction between the windings is left untouched.
Marking note: the marks are for reduced loop area and increased resistance of the path. Saying only “to reduce eddy currents” restates the question and earns little.
Eddy currents are not a separate law
There is no “law of eddy currents” to learn. They are Faraday plus Lenz applied to a continuous conductor. If a question about them puzzles you, go back to the routine: which knob is turning, is the flux growing or shrinking, and which way must the induced current swirl to fight that. The answer will be there.
Electromagnetic Induction Class 12 Physics Numericals With Solutions
Mixed exam-style problems, deliberately shuffled so you cannot coast on the section you just read. For each one, name the knob before you touch your calculator. Every answer below has been computed and cross-checked.
Example 23 — circular coil in a decaying field
A circular coil of radius 8.0 cm has 100 turns and a total circuit resistance of 5.0 Ω. The magnetic field perpendicular to its plane is decreasing at a steady 0.10 T s−1. Find the induced EMF and current.
Knob: B.
A = πr² = π(0.080)² = 2.0106 × 10−2 m²
|ε| = NA |dB/dt| = 100 × 2.0106 × 10−2 × 0.10 = 0.201 V
I = 0.2011/5.0 = 4.02 × 10−2 A = 40.2 mA
Direction: the flux is shrinking, so the induced current flows so as to maintain the original field — i.e. in the same rotational sense as the original current that would have produced that field.
Knob: B.
A = πr² = π(0.080)² = 2.0106 × 10−2 m²
|ε| = NA |dB/dt| = 100 × 2.0106 × 10−2 × 0.10 = 0.201 V
I = 0.2011/5.0 = 4.02 × 10−2 A = 40.2 mA
Direction: the flux is shrinking, so the induced current flows so as to maintain the original field — i.e. in the same rotational sense as the original current that would have produced that field.
Example 24 — coil flipped through 180° — charge through the galvanometer
A coil of 200 turns and area 2.0 × 10−3 m² sits with its plane perpendicular to a field of 0.50 T. It is turned through 180°. The total circuit resistance is 20 Ω. Find the charge that circulates.
Knob: θ. Flipping through 180° changes cosθ from +1 to −1, so the flux reverses — the change is twice the original value, not zero and not once.
Δ(NΦ) = NA[B − (−B)] = NA(2B) = 200 × 2.0 × 10−3 × 1.00 = 0.40 Wb-turns
q = Δ(NΦ)/R = 0.40/20 = 2.0 × 10−2 C = 20 mC
The trap: writing NAB = 0.20 Wb-turns and getting half the right answer. A reversal is a change of 2B. Also note the time was never given — and never needed, because charge does not depend on it.
Knob: θ. Flipping through 180° changes cosθ from +1 to −1, so the flux reverses — the change is twice the original value, not zero and not once.
Δ(NΦ) = NA[B − (−B)] = NA(2B) = 200 × 2.0 × 10−3 × 1.00 = 0.40 Wb-turns
q = Δ(NΦ)/R = 0.40/20 = 2.0 × 10−2 C = 20 mC
The trap: writing NAB = 0.20 Wb-turns and getting half the right answer. A reversal is a change of 2B. Also note the time was never given — and never needed, because charge does not depend on it.
Example 25 — rod on rails — find the mechanical power
A rod of length 0.40 m is pulled at 3.0 m s−1 along rails in a perpendicular field of 0.50 T. The total circuit resistance is 0.20 Ω. Find the current, the force needed and the mechanical power supplied.
Knob: A.
ε = Blv = 0.50 × 0.40 × 3.0 = 0.60 V
I = 0.60/0.20 = 3.0 A
F = BIl = 0.50 × 3.0 × 0.40 = 0.60 N
P = Fv = 0.60 × 3.0 = 1.8 W
Check: εI = 0.60 × 3.0 = 1.8 W ✓ and I²R = 9.0 × 0.20 = 1.8 W ✓
Knob: A.
ε = Blv = 0.50 × 0.40 × 3.0 = 0.60 V
I = 0.60/0.20 = 3.0 A
F = BIl = 0.50 × 3.0 × 0.40 = 0.60 N
P = Fv = 0.60 × 3.0 = 1.8 W
Check: εI = 0.60 × 3.0 = 1.8 W ✓ and I²R = 9.0 × 0.20 = 1.8 W ✓
Example 26 — self-inductance and energy from a stated EMF
An EMF of 8.0 V is induced in a coil when the current through it changes at 32 A s−1. Find the self-inductance, and the energy stored when the coil carries 4.0 A.
L = |ε| / (di/dt) = 8.0/32 = 0.25 H
U = ½LI² = ½ × 0.25 × 16 = 2.0 J
L = |ε| / (di/dt) = 8.0/32 = 0.25 H
U = ½LI² = ½ × 0.25 × 16 = 2.0 J
Example 27 — flux from a non-linear field — use calculus
A square loop of side 10 cm and resistance 0.50 Ω lies with its plane perpendicular to a field that varies as B = (0.50t² + 0.20t) T. Find the induced current at t = 1.8 s and the rate of heat production at that instant.
A = (0.10)² = 1.0 × 10−2 m²
Φ = AB = 10−2(0.50t² + 0.20t)
ε = −dΦ/dt = −10−2(1.0t + 0.20)
At t = 1.8 s: |ε| = 10−2(1.8 + 0.20) = 10−2 × 2.00 = 0.020 V = 20 mV
I = 0.020/0.50 = 0.040 A = 40 mA
P = I²R = (0.040)² × 0.50 = 8.0 × 10−4 W = 0.80 mW
Method note: when B is not linear in t, you cannot use ΔB/Δt. Differentiate. And differentiate the flux, not the field, so the area is never left behind.
A = (0.10)² = 1.0 × 10−2 m²
Φ = AB = 10−2(0.50t² + 0.20t)
ε = −dΦ/dt = −10−2(1.0t + 0.20)
At t = 1.8 s: |ε| = 10−2(1.8 + 0.20) = 10−2 × 2.00 = 0.020 V = 20 mV
I = 0.020/0.50 = 0.040 A = 40 mA
P = I²R = (0.040)² × 0.50 = 8.0 × 10−4 W = 0.80 mW
Method note: when B is not linear in t, you cannot use ΔB/Δt. Differentiate. And differentiate the flux, not the field, so the area is never left behind.
Example 28 — coil pulled out of a field
A 500-turn coil of area 4.0 cm² lies with its plane perpendicular to a uniform field of 0.60 T. It is pulled completely clear of the field in 0.20 s. The total circuit resistance is 10 Ω. Find the average EMF, the average current and the charge that flows.
Knob: A — the area inside the field falls to zero.
A = 4.0 × 10−4 m²
Δ(NΦ) = 500 × 4.0 × 10−4 × 0.60 = 0.12 Wb-turns
εav = 0.12/0.20 = 0.60 V
Iav = 0.60/10 = 0.060 A = 60 mA
q = IavΔt = 0.060 × 0.20 = 1.2 × 10−2 C; cross-check q = Δ(NΦ)/R = 0.12/10 = 1.2 × 10−2 C ✓
Knob: A — the area inside the field falls to zero.
A = 4.0 × 10−4 m²
Δ(NΦ) = 500 × 4.0 × 10−4 × 0.60 = 0.12 Wb-turns
εav = 0.12/0.20 = 0.60 V
Iav = 0.60/10 = 0.060 A = 60 mA
q = IavΔt = 0.060 × 0.20 = 1.2 × 10−2 C; cross-check q = Δ(NΦ)/R = 0.12/10 = 1.2 × 10−2 C ✓
Example 29 — energy in a solenoid, computed two independent ways
A solenoid of length 0.50 m and cross-sectional area 10 cm² has 2000 turns per metre and carries 3.0 A. Find the stored magnetic energy, using both the energy-density route and the ½LI² route.
A = 1.0 × 10−3 m², n = 2000 m−1
Route 1 — energy density.
B = μ0nI = 1.2566 × 10−6 × 2000 × 3.0 = 7.540 × 10−3 T
uB = B²/(2μ0) = (7.540 × 10−3)²/(2.5133 × 10−6) = 22.62 J m−3
Volume = Al = 1.0 × 10−3 × 0.50 = 5.0 × 10−4 m³
U = 22.62 × 5.0 × 10−4 = 1.13 × 10−2 J
Route 2 — ½LI².
L = μ0n²Al = 1.2566 × 10−6 × 4.0 × 106 × 1.0 × 10−3 × 0.50 = 2.513 × 10−3 H
U = ½ × 2.513 × 10−3 × (3.0)² = 1.13 × 10−2 J ✓
Why they agree: ½LI² and B²/2μ0 are the same statement seen from two ends — one counts the work done by the circuit, the other counts the energy sitting in the field. If a numerical gives you both routes, use the second as a free check.
A = 1.0 × 10−3 m², n = 2000 m−1
Route 1 — energy density.
B = μ0nI = 1.2566 × 10−6 × 2000 × 3.0 = 7.540 × 10−3 T
uB = B²/(2μ0) = (7.540 × 10−3)²/(2.5133 × 10−6) = 22.62 J m−3
Volume = Al = 1.0 × 10−3 × 0.50 = 5.0 × 10−4 m³
U = 22.62 × 5.0 × 10−4 = 1.13 × 10−2 J
Route 2 — ½LI².
L = μ0n²Al = 1.2566 × 10−6 × 4.0 × 106 × 1.0 × 10−3 × 0.50 = 2.513 × 10−3 H
U = ½ × 2.513 × 10−3 × (3.0)² = 1.13 × 10−2 J ✓
Why they agree: ½LI² and B²/2μ0 are the same statement seen from two ends — one counts the work done by the circuit, the other counts the energy sitting in the field. If a numerical gives you both routes, use the second as a free check.
Formula List and Quick Revision
Everything in the chapter, on one screen. If you can reconstruct the third column from the second, you understand the chapter; if you can only recite the second, you are still memorising.
| Quantity | Formula | What it is really saying |
|---|---|---|
| Magnetic flux | Φ = BA cosθ (Wb) | How many field lines get through. θ is measured from the normal. |
| Flux linkage | NΦ (Wb-turns) | N turns each catch the same flux, so the effect multiplies. |
| Faraday’s law | ε = −N dΦ/dt | EMF is the rate of change of flux linkage. The minus sign is energy conservation. |
| Average EMF | εav = −N ΔΦ/Δt | Use only when the change is uniform. Otherwise differentiate. |
| Induced current | I = ε/R = −(N/R) dΦ/dt | Ohm’s law applied to the induced EMF. |
| Induced charge | q = N ΔΦ/R | Independent of time. Fast or slow, the same charge circulates. |
| Motional EMF (sliding rod) | ε = Blv | B, l and v mutually perpendicular. This is the ‘area’ knob. |
| Motional EMF (rotating rod) | ε = ½BωL² | Bl × the average speed of the rod — hence the ½. |
| Force to keep the rod moving | F = BIl = B²l²v/R | Always a brake. Substituting I = Blv/R gives the second form. |
| Power balance | Fv = εI = I²R | Mechanical work in = electrical energy out = heat. Use it as a free check. |
| Self-inductance | L = NΦ/i (H) | A fixed ratio set by geometry and the core. |
| Back EMF | ε = −L di/dt | The coil objecting to a change in its own current. |
| L of a long solenoid | L = μ0N²A/l = μ0n²Al | Note the N²: double the turns, quadruple the inductance. |
| Mutual inductance | M = N2Φ2/i1 (H) | One number for the pair: M12 = M21. |
| Mutually induced EMF | ε2 = −M di1/dt | The second coil objecting to a change in the first coil’s current. |
| M of two coaxial solenoids | M = μ0N1N2A/l | Same geometry as L, with the two turn-counts sharing the job. |
| Coefficient of coupling | M = k√(L1L2), 0 ≤ k ≤ 1 | k = 1 means no leakage flux at all. |
| Energy in an inductor | U = ½LI² (J) | The work you did against the back EMF, now sitting in the field. |
| Magnetic energy density | uB = B²/(2μ0) (J m−3) | The magnetic twin of ½ε0E². |
| Inductors in series (no coupling) | L = L1 + L2 | Same shape as resistors in series. |
| Inductors in parallel (no coupling) | 1/L = 1/L1 + 1/L2 | Same shape as resistors in parallel. |
The three-line summary
1. Φ = BA cosθ — three knobs, and only three: Bend it, Area it, Angle it.
2. Faraday gives the size: EMF = rate of change of flux linkage.
3. Lenz gives the direction: oppose the CHANGE, not the FIELD — and the mechanical effect is always a brake, because energy has to be paid for.
2. Faraday gives the size: EMF = rate of change of flux linkage.
3. Lenz gives the direction: oppose the CHANGE, not the FIELD — and the mechanical effect is always a brake, because energy has to be paid for.
The 60-second pre-exam scan
Read only these five lines the morning of the paper: Φ = BA cosθ with θ from the normal; ε = −N dΦ/dt; q = NΔΦ/R is time-independent; ε = Blv and ½BωL²; L = μ0N²A/l and U = ½LI². Then remind yourself once: oppose the change, not the field.
One last thread to pull: the symmetry you have just learnt — a changing magnetic field creating an electric field — is exactly what led Maxwell to ask whether the reverse also happens. It does, and that question grows into Electromagnetic Waves, the next chapter in Unit V. Finish this chapter first; then that one will feel inevitable rather than new.
Practice Worksheet
Ten questions covering the whole chapter, in roughly increasing difficulty. Attempt each one on paper first, name the knob, then open the answer. If you get one wrong, do not re-read the answer — go back to the section it came from.
Q1. A coil of 250 turns and radius 5.0 cm sits in a uniform field of 0.30 T. The field makes an angle of 30° with the plane of the coil. Find the total flux linkage.
Watch the word “plane”. The angle with the normal is θ = 90° − 30° = 60°, so cosθ = 0.50.
A = π(0.050)² = 7.854 × 10−3 m²
NΦ = 250 × 0.30 × 7.854 × 10−3 × 0.50 = 0.295 Wb-turns (to 3 s.f.)
Using cos 30° = 0.866 by mistake would have given 0.510 Wb-turns — the single most common error in this chapter.
A = π(0.050)² = 7.854 × 10−3 m²
NΦ = 250 × 0.30 × 7.854 × 10−3 × 0.50 = 0.295 Wb-turns (to 3 s.f.)
Using cos 30° = 0.866 by mistake would have given 0.510 Wb-turns — the single most common error in this chapter.
Q2. The flux through a single-turn loop is Φ = (4t³ + 5t + 2) × 10−3 Wb. Find the magnitude of the induced EMF at t = 1.5 s. What happened to the constant term?
ε = −dΦ/dt = −(12t² + 5) × 10−3 V
At t = 1.5 s: 12(1.5)² = 12 × 2.25 = 27; 27 + 5 = 32
|ε| = 32 × 10−3 V = 32 mV
The constant 2 mWb differentiated to zero. Standing flux contributes nothing to the EMF — only its rate of change matters.
At t = 1.5 s: 12(1.5)² = 12 × 2.25 = 27; 27 + 5 = 32
|ε| = 32 × 10−3 V = 32 mV
The constant 2 mWb differentiated to zero. Standing flux contributes nothing to the EMF — only its rate of change matters.
Q3. The north pole of a magnet is pulled away from a coil. State (a) the direction of the induced current as seen from the magnet’s side, (b) the effective polarity of the coil’s near face, and (c) the force on the magnet.
Step 1: field through the coil points away from the magnet.
Step 2: the magnet is receding, so that flux is shrinking.
Step 3: the induced field inside the coil must therefore point away from the magnet, in the same direction as the original, to prop the flux up.
Step 4: right thumb pointing away from the viewer ⇒ fingers curl clockwise as seen from the magnet’s side.
(b) The near face becomes a south pole. (c) South facing the magnet’s north ⇒ the coil attracts the magnet, resisting your pull.
Sanity check: it is a brake in both cases — repulsion when you push in, attraction when you pull out. Never a helpful push.
Step 2: the magnet is receding, so that flux is shrinking.
Step 3: the induced field inside the coil must therefore point away from the magnet, in the same direction as the original, to prop the flux up.
Step 4: right thumb pointing away from the viewer ⇒ fingers curl clockwise as seen from the magnet’s side.
(b) The near face becomes a south pole. (c) South facing the magnet’s north ⇒ the coil attracts the magnet, resisting your pull.
Sanity check: it is a brake in both cases — repulsion when you push in, attraction when you pull out. Never a helpful push.
Q4. A metal rod of length 75 cm rotates about one end in a horizontal plane at 300 rev min−1. A uniform field of 0.40 T is perpendicular to the plane of rotation. Find the EMF between the ends of the rod.
First convert the rate: ω = 300 × 2π/60 = 10π = 31.416 rad s−1
ε = ½BωL² = ½ × 0.40 × 31.416 × (0.75)²
= 0.20 × 31.416 × 0.5625 = 3.53 V (to 3 s.f.)
Forgetting the rev min−1 → rad s−1 conversion is the usual slip here; it would give an answer 2π/60 times too large.
ε = ½BωL² = ½ × 0.40 × 31.416 × (0.75)²
= 0.20 × 31.416 × 0.5625 = 3.53 V (to 3 s.f.)
Forgetting the rev min−1 → rad s−1 conversion is the usual slip here; it would give an answer 2π/60 times too large.
Q5. A solenoid 0.40 m long has 1000 turns and a circular cross-section of radius 2.0 cm. Find its self-inductance. Then state what happens to L if the number of turns is doubled with the length unchanged.
A = π(0.020)² = 1.2566 × 10−3 m²
L = μ0N²A/l = (1.2566 × 10−6) × 106 × 1.2566 × 10−3 / 0.40
= 1.5791 × 10−3/0.40 = 3.95 × 10−3 H = 3.95 mH
Doubling N with l fixed multiplies L by four, because L ∝ N². New value ≈ 15.8 mH.
L = μ0N²A/l = (1.2566 × 10−6) × 106 × 1.2566 × 10−3 / 0.40
= 1.5791 × 10−3/0.40 = 3.95 × 10−3 H = 3.95 mH
Doubling N with l fixed multiplies L by four, because L ∝ N². New value ≈ 15.8 mH.
Q6. An EMF of 0.60 V is induced in coil 2 when the current in coil 1 changes at 12 A s−1. (a) Find M. (b) Find the EMF induced in coil 1 if instead the current in coil 2 changes at 20 A s−1. State the principle you used.
(a) M = |ε2| / (di1/dt) = 0.60/12 = 0.050 H = 50 mH
(b) By the reciprocity theorem, M12 = M21 = M, so the same 50 mH applies in the reverse direction:
|ε1| = M × di2/dt = 0.050 × 20 = 1.0 V
A pair of coils has only one mutual inductance, whichever coil you drive.
(b) By the reciprocity theorem, M12 = M21 = M, so the same 50 mH applies in the reverse direction:
|ε1| = M × di2/dt = 0.050 × 20 = 1.0 V
A pair of coils has only one mutual inductance, whichever coil you drive.
Q7. The current in a 3.0 H inductor is reduced from 6.0 A to 2.0 A. How much energy is released from the magnetic field? Why is the answer not ½L(ΔI)²?
Ui = ½ × 3.0 × (6.0)² = 54 J
Uf = ½ × 3.0 × (2.0)² = 6.0 J
Energy released = 54 − 6.0 = 48 J
It is not ½L(ΔI)² = ½(3.0)(4.0)² = 24 J, because energy goes as I² and squaring is not linear: Ii² − If² ≠ (Ii − If)². Always compute the two energies and subtract.
Uf = ½ × 3.0 × (2.0)² = 6.0 J
Energy released = 54 − 6.0 = 48 J
It is not ½L(ΔI)² = ½(3.0)(4.0)² = 24 J, because energy goes as I² and squaring is not linear: Ii² − If² ≠ (Ii − If)². Always compute the two energies and subtract.
Q8. Show that the charge circulating in a circuit during a flux change is independent of the time taken. Then apply it: a 100-turn coil of area 25 cm² is in a field of 0.80 T perpendicular to its plane; the field is switched off completely. Circuit resistance is 4.0 Ω. Find the charge.
Proof: I = ε/R = (N/R) |dΦ/dt|, so
q = ∫I dt = ∫(N/R)(dΦ/dt) dt = (N/R)∫dΦ = N ΔΦ/R. The dt cancels, so q depends only on the total flux change and the resistance.
Application: A = 25 cm² = 2.5 × 10−3 m²
NΔΦ = 100 × 2.5 × 10−3 × 0.80 = 0.20 Wb-turns
q = 0.20/4.0 = 0.050 C = 50 mC
Notice no time was given, and none was needed.
q = ∫I dt = ∫(N/R)(dΦ/dt) dt = (N/R)∫dΦ = N ΔΦ/R. The dt cancels, so q depends only on the total flux change and the resistance.
Application: A = 25 cm² = 2.5 × 10−3 m²
NΔΦ = 100 × 2.5 × 10−3 × 0.80 = 0.20 Wb-turns
q = 0.20/4.0 = 0.050 C = 50 mC
Notice no time was given, and none was needed.
Q9. Two coils have self-inductances 0.050 H and 0.20 H, with a coefficient of coupling of 0.60. (a) Find M. (b) What is the largest M these coils could have, and what physical change would get you there?
(a) √(L1L2) = √(0.050 × 0.20) = √0.010 = 0.10 H
M = k√(L1L2) = 0.60 × 0.10 = 0.060 H = 60 mH
(b) The maximum is at k = 1, giving 0.10 H = 100 mH. You would get there by eliminating leakage flux — winding both coils tightly on the same closed soft-iron core, one directly over the other, so that every field line from one threads the other.
M = k√(L1L2) = 0.60 × 0.10 = 0.060 H = 60 mH
(b) The maximum is at k = 1, giving 0.10 H = 100 mH. You would get there by eliminating leakage flux — winding both coils tightly on the same closed soft-iron core, one directly over the other, so that every field line from one threads the other.
Q10. (a) Find the magnetic energy density in a transformer core where B = 1.2 T. (b) A student says “eddy currents are wasteful, so we should use a solid core to keep them in one place.” Correct the student. (c) State one condition under which a loop moving through a magnetic field has no induced EMF.
(a) uB = B²/(2μ0) = (1.2)²/(2 × 1.2566 × 10−6) = 1.44/2.5133 × 10−6 = 5.73 × 105 J m−3
Compare with 159 J m−3 at 0.02 T — the square makes an enormous difference, which is exactly why cores are used.
(b) A solid core makes the problem worse, not better. Wide continuous conducting paths mean large enclosed area (large induced EMF) and large cross-section (low resistance), so the eddy currents and the I²R heat loss are both large. Laminating the core into thin insulated sheets cuts the loop area and raises the path resistance, reducing the loss while leaving the useful mutual induction untouched.
(c) Any situation where the flux does not change: for example a loop moving entirely within a uniform field (constant area, constant B, constant angle), or a rod sliding parallel to the field lines so that no area is swept across them. Motion by itself induces nothing — only a change in flux does.
Compare with 159 J m−3 at 0.02 T — the square makes an enormous difference, which is exactly why cores are used.
(b) A solid core makes the problem worse, not better. Wide continuous conducting paths mean large enclosed area (large induced EMF) and large cross-section (low resistance), so the eddy currents and the I²R heat loss are both large. Laminating the core into thin insulated sheets cuts the loop area and raises the path resistance, reducing the loss while leaving the useful mutual induction untouched.
(c) Any situation where the flux does not change: for example a loop moving entirely within a uniform field (constant area, constant B, constant angle), or a rod sliding parallel to the field lines so that no area is swept across them. Motion by itself induces nothing — only a change in flux does.
Kaizen: one small improvement, repeated. Tonight, do not re-read this chapter — just draw the B–A–θ knob-box from memory and write one line under it: oppose the change, not the field. Tomorrow, add the four Lenz steps. By the end of the week you will have built the whole chapter out of two sketches, and the numericals will start feeling like arithmetic.