1. Chapter at a Glance
Electromagnetic induction (EMI) is the production of an EMF by a changing magnetic flux. Faraday discovered it, and it is the principle behind generators, transformers and induction cooktops. This chapter covers magnetic flux, Faraday's and Lenz's laws, motional EMF, eddy currents, self- and mutual inductance, energy stored in an inductor, and the AC generator. NEET asks about one question from here every year.
- Induced EMF from a changing flux (Φ given as a function of time).
- Direction of the induced current by Lenz's law.
- Motional EMF of a sliding or rotating rod.
- Self-inductance of a solenoid, mutual inductance and energy ½LI².
- Peak EMF of an AC generator.
2. Magnetic Flux and Faraday's Law
- An EMF is induced whenever the flux through a circuit changes. The flux can change because B changes, the area changes, or the angle θ changes.
- The induced current I = ε/R; the charge that flows is q = NΔΦ/R, which does not depend on how fast the flux changes.
3. Lenz's Law
- The induced current flows in a direction that opposes the change in flux that produces it. This is the meaning of the minus sign in Faraday's law.
- Lenz's law follows from conservation of energy: if the induced current helped the change, energy would be created from nothing.
- When a north pole approaches a coil, the near face of the coil becomes a north pole (repelling it). When the north pole moves away, the near face becomes a south pole (attracting it).
4. Motional EMF and Eddy Currents
- The magnetic force on the free charges in a moving rod pushes them to one end, creating the EMF.
- If the rod slides on rails joined by a resistor R: I = Blv/R, the retarding force is F = B²l²v/R, and the mechanical power Fv equals the electrical power I²R.
- Eddy currents are induced currents that circulate inside bulk conductors. They cause heating and damping.
- They are reduced by making cores from thin laminated sheets.
- They are useful in electromagnetic brakes, induction furnaces, speedometers and the damping of galvanometers.
5. Self- and Mutual Inductance
| Quantity | Definition | Formula | NEET Note |
|---|---|---|---|
| Self-inductance L | NΦ = LI | ε = −L dI/dt | Opposes change of its own current |
| Long solenoid | N turns, length l, area A | L = μ₀N²A/l = μ₀n²Al | L ∝ N²; with a core, × μr |
| Mutual inductance M | N₂Φ₂ = MI₁ | ε₂ = −M dI₁/dt | M₁₂ = M₂₁ |
| Coaxial solenoids | inner radius r₁ | M = μ₀n₁n₂πr₁²l | Depends on the smaller area |
| Energy in an inductor | Stored in the magnetic field | U = ½LI² | Like ½mv² (L acts like inertia) |
The unit of inductance is the henry (H). Inductance behaves like electrical inertia: it opposes any change in current, just as mass opposes a change in velocity.
6. AC Generator
- A coil of N turns and area A rotates with angular speed ω in a field B. The flux is Φ = NBA cos ωt.
- Induced EMF: ε = NBAω sin ωt, with peak value ε₀ = NBAω.
- The EMF is maximum when the plane of the coil is parallel to B (flux zero but changing fastest), and zero when the plane is perpendicular to B (flux maximum).
- Mechanical energy is converted into electrical energy.
7. Formula Master Table
| Quantity | Formula |
|---|---|
| Magnetic flux | Φ = BA cosθ |
| Faraday's law | ε = −N dΦ/dt |
| Charge flowed | q = NΔΦ/R |
| Sliding rod | ε = Blv |
| Rotating rod | ε = ½Bωl² |
| Retarding force on rod | F = B²l²v/R |
| Self-induced EMF | ε = −L dI/dt |
| Solenoid inductance | L = μ₀N²A/l |
| Mutual inductance | M = μ₀n₁n₂πr₁²l |
| Energy in an inductor | U = ½LI² |
| Generator EMF | ε = NBAω sin ωt |
8. Master Summary Table
| Concept | Key idea | NEET Note |
|---|---|---|
| Faraday's law | EMF = rate of change of flux | Flux can change via B, A or θ |
| Lenz's law | Opposes the change | Energy conservation |
| Motional EMF | Blv | Rod must cut field lines |
| Eddy currents | Loops in bulk metal | Reduced by laminations |
| Self-inductance | L ∝ N² | Electrical inertia |
| Mutual inductance | M₁₂ = M₂₁ | Principle of the transformer |
| Generator | ε₀ = NBAω | Max when plane ∥ B |
9. Mind Map
- Flux
- Faraday
- Lenz
- Blv
- ½Bωl²
- Rails and resistor
- Self L
- Mutual M
- ½LI²
- Generator
- Eddy currents
- Brakes, furnaces
10. Tips and Tricks
- If Φ is given as a function of t, differentiate it and put in the time.
- A loop moving through a uniform field with its whole area inside has no induced EMF (flux is constant).
- A falling magnet through a copper tube falls slowly because eddy currents oppose its motion.
- L ∝ N²: doubling the turns (same length and area) makes L four times.
- Thinking a large flux means a large EMF (only the rate of change matters).
- Getting the generator EMF maximum at maximum flux (it is zero there).
- Forgetting the ½ in the rotating-rod EMF.
- Using Lenz's law to oppose the flux itself rather than the change in flux.
11. Important One-liners
- The SI unit of magnetic flux is the weber; 1 Wb = 1 T m².
- Lenz's law is a consequence of the law of conservation of energy.
- The charge induced in a coil depends on the change in flux, not on the time taken.
- Transformer cores are laminated to reduce eddy-current losses.
- The self-inductance of a coil increases if a soft-iron core is inserted.
- Energy in an inductor is stored in its magnetic field.
- An AC generator changes mechanical energy into electrical energy.
12. Solved Numericals: Average to Advanced
Five average and five advanced numericals. Take μ₀ = 4π × 10⁻⁷ T m A⁻¹. Click to reveal each answer and working.
Average level
A coil of 50 turns and area 0.04 m² is held perpendicular to a field of 0.3 T. The field falls to zero in 0.05 s. Find the average induced EMF.
Show answer & solution
Concept: ε = NΔΦ/Δt.
- ΔΦ (one turn) = 0.3 × 0.04 = 0.012 Wb.
- ε = 50 × 0.012/0.05 = 12 V.
Answer: 12 V
A 0.5 m rod moves at 4 m s⁻¹ at right angles to a field of 0.3 T. Find the EMF across its ends.
Show answer & solution
Concept: ε = Blv.
- ε = 0.3 × 0.5 × 4.
- ε = 0.6 V.
Answer: 0.6 V
The current in a 0.2 H coil falls steadily from 5 A to 1 A in 0.02 s. Find the self-induced EMF.
Show answer & solution
Concept: |ε| = L ΔI/Δt.
- ΔI = 4 A.
- ε = 0.2 × 4/0.02 = 40 V (it acts to keep the current flowing).
Answer: 40 V
Find the energy stored in a 50 mH inductor carrying 4 A.
Show answer & solution
Concept: U = ½LI².
- U = ½ × 0.05 × 16.
- U = 0.4 J.
Answer: 0.4 J
The flux through a loop varies as Φ = (3t² + 2t) mWb. Find the induced EMF at t = 2 s.
Show answer & solution
Concept: |ε| = dΦ/dt.
- dΦ/dt = 6t + 2 mWb s⁻¹.
- At t = 2 s: 6 × 2 + 2 = 14 mV.
Answer: 14 mV
Advanced level
A 1 m metal rod rotates at 20 rad s⁻¹ about one of its ends, in a plane perpendicular to a uniform field of 0.5 T. Find the EMF between its ends.
Show answer & solution
Concept: each element moves at speed ωr; integrating gives ε = ½Bωl².
- ε = ½ × 0.5 × 20 × (1)².
- ε = 5 V.
Answer: 5 V
A generator coil has 200 turns of area 0.025 m² and rotates at 25 revolutions per second in a field of 0.3 T. Find the peak and rms EMF.
Show answer & solution
Concept: ε₀ = NBAω; ω = 2πf.
- ω = 2π × 25 = 50π rad s⁻¹.
- ε₀ = 200 × 0.3 × 0.025 × 50π = 1.5 × 50π ≈ 235.6 V.
- εrms = ε₀/√2 ≈ 166.6 V.
Answer: about 236 V peak; about 167 V rms
A 25 cm rod slides at a steady 6 m s⁻¹ on frictionless rails in a 0.8 T field. The rails are joined by a 3 Ω resistor. Find the EMF, current, the force needed to keep the rod moving, and the power dissipated.
Show answer & solution
Concept: ε = Blv; F = BIl; power balance Fv = I²R.
- ε = 0.8 × 0.25 × 6 = 1.2 V; I = 1.2/3 = 0.4 A.
- Magnetic (retarding) force = BIl = 0.8 × 0.4 × 0.25 = 0.08 N, so an equal applied force is needed.
- Power = Fv = 0.08 × 6 = 0.48 W = I²R = 0.16 × 3 ✓.
Answer: 1.2 V; 0.4 A; 0.08 N; 0.48 W
A long outer solenoid has 2000 turns per metre. A small coil of 100 turns and radius 2 cm is placed coaxially inside it. Find their mutual inductance.
Show answer & solution
Concept: field of outer solenoid B = μ₀n₁I; flux linked with inner coil = N₂BA₂ = MI.
- M = μ₀n₁N₂πr₂².
- = 4π × 10⁻⁷ × 2000 × 100 × π × (0.02)².
- M ≈ 3.16 × 10⁻⁴ H ≈ 0.32 mH.
Answer: about 0.32 mH
An air-cored solenoid has 500 turns, length 0.5 m and cross-section 4 cm². Find its self-inductance and the energy stored when it carries 2 A. What would L become if the number of turns were doubled (same length and area)?
Show answer & solution
Concept: L = μ₀N²A/l; U = ½LI².
- L = 4π × 10⁻⁷ × (500)² × 4 × 10⁻⁴/0.5 ≈ 2.51 × 10⁻⁴ H.
- U = ½ × 2.51 × 10⁻⁴ × 4 ≈ 5.0 × 10⁻⁴ J.
- L ∝ N², so doubling N makes L four times: about 1.0 × 10⁻³ H.
Answer: about 0.25 mH; 0.5 mJ; about 1.0 mH
13. NEET-pattern Questions with Detailed Answers
These questions are original and based on concepts that NEET UG asks again and again.
Lenz's law is a consequence of the conservation of:
- Charge
- Energy
- Momentum
- Mass
Show solution
The induced current opposes the change, so work must be done to produce electrical energy. If it did not oppose the change, energy would appear from nothing.
Answer: (B)
The SI unit of self-inductance is:
- Weber
- Tesla
- Henry
- Farad
Show solution
Inductance is measured in henry (1 H = 1 Wb A⁻¹). The weber is the unit of flux, the tesla of field and the farad of capacitance.
Answer: (C)
The core of a transformer is laminated to:
- Increase the flux
- Reduce eddy-current losses
- Reduce hysteresis loss
- Increase the EMF
Show solution
Thin insulated sheets break up the eddy-current loops and so reduce heating. Hysteresis loss is reduced by using soft iron, not by laminating.
Answer: (B)
In an AC generator, the induced EMF is maximum when the plane of the coil is:
- Perpendicular to B
- Parallel to B
- At 45° to B
- At any position
Show solution
ε = NBAω sin ωt is maximum when the flux is zero and changing fastest. That happens when the plane of the coil is parallel to B. In the perpendicular position the flux is maximum and the EMF is zero.
Answer: (B)
A magnet is dropped through a long vertical copper tube. Compared with free fall, it:
- Falls faster
- Falls with acceleration g
- Falls more slowly
- Stops at the top
Show solution
Eddy currents induced in the copper tube oppose the motion of the magnet (Lenz's law). So it falls with an acceleration less than g, and can even reach a terminal speed.
Answer: (C)
If the number of turns of a solenoid is doubled while its length and area are unchanged, its self-inductance becomes:
- Double
- Half
- Four times
- Unchanged
Show solution
L = μ₀N²A/l ∝ N². Doubling N makes L four times as large.
Answer: (C)
A coil of resistance 10 Ω and 100 turns has its flux changed from 5 mWb to 1 mWb per turn. The charge that flows through it is:
- 0.04 C
- 0.4 mC
- 4 C
- Depends on the time taken
Show solution
q = NΔΦ/R = 100 × 4 × 10⁻³/10 = 0.04 C. The charge does not depend on the time taken, so (D) is wrong. (B) forgets the factor of N.
Answer: (A)
A square loop moves at constant velocity through a region of uniform magnetic field. An EMF is induced:
- Throughout the motion
- Only while it enters and leaves the field region
- Only while it is fully inside
- Never
Show solution
An EMF needs a change in flux. The flux changes only while part of the loop is entering or leaving the field. When the loop is fully inside, the flux is constant and the EMF is zero.
Answer: (B)
14. Quick Revision Summary
- Φ = BA cosθ; ε = −N dΦ/dt; q = NΔΦ/R.
- Lenz: oppose the change (energy conservation).
- ε = Blv (sliding), ½Bωl² (rotating).
- L = μ₀N²A/l; U = ½LI²; M₁₂ = M₂₁.
- Generator ε₀ = NBAω; eddy currents reduced by laminations.