Electromagnetic Induction – Detailed Study Notes (6.1–6.8)

6.1 Introduction

• Electricity and magnetism were historically treated as separate, then found to be interrelated in the early 19th century through experiments by Oersted, Ampere, etc., showing moving charges produce magnetic fields.
• Observations raised questions: Can moving magnets produce electric currents? Does nature permit a relation between electricity and magnetism?
• Answer: Yes. Electromagnetic induction: changing magnetic fields can induce electric currents.
• Faraday and Henry (around 1830) demonstrated that electric currents are induced in closed coils by changing magnetic fields.
• The phenomenon of current generation by changing magnetic fields is called electromagnetic induction.
• Faraday’s public remark on discovery: the use of a new-born baby analogy underscores practical significance beyond theory.
• Practical impact: Faraday and Henry’s work led to generators and transformers; essential for modern civilization.

6.2 The Experiments of Faraday and Henry

• Experiment 6.1 (Fig. 6.1): coil C1 connected to galvanometer G. Pushing the North-pole of a bar magnet toward C1 deflects G; deflection lasts while the magnet moves. If the magnet is pulled away, deflection reverses, indicating current reversal. Deflection larger when the magnet is moved faster.
  • If the magnet is held stationary, no deflection occurs.
  • The polarity of the bar magnet (N/S) affects the direction of deflection; moving the S-pole toward/away yields opposite deflections to the N-pole case.
• Experiment 6.2 (Fig. 6.2): replace the bar magnet with coil C2 connected to a battery. A steady current in C2 produces a steady magnetic field. Moving C2 toward C1 causes deflection in C1; moving C2 away causes deflection in the opposite direction. Deflection lasts as long as C2 moves. If C2 is stationary and C1 moves, similar effects occur: the key point is relative motion between the magnetic field and the circuit.
• Experiment 6.3 (Fig. 6.3): two coils C1 and C2 held stationary. C1 is connected to galvanometer G; C2 is connected to a battery through tapping key K. When K is pressed, current in C2 rises rapidly, increasing the magnetic flux through C1 and inducing emf in C1 (momentary deflection). Holding K steady yields no deflection (constant flux). Releasing K causes a momentary opposite deflection. Insertion of an iron rod along the axis magnifies the deflection.
• Summary of experiments: induction is due to relative motion between magnets and coils, or between coils, and also due to changes in current in a nearby circuit; induction can occur with stationary components if the flux through a coil changes with time.
• Important note: when a coil or loop is used, it is assumed to be conducting with wires coated by insulating material.

6.3 Magnetic Flux

• Faraday's key insight: a simple mathematical relation connects induction to flux change.
• Magnetic flux through a plane of area A in a uniform magnetic field B is defined as
  \Phi_B = \mathbf{B} \cdot \mathbf{A} = B A \cos \theta,
  where \theta is the angle between B and the area vector A.
• For curved surfaces or nonuniform fields, flux is generalized to a sum (integral) over area elements:
  \PhiB = \sumi \mathbf{B}i \cdot \mathbf{A}i \, , \text{or} \ \Phi_B = \int \mathbf{B} \cdot d\mathbf{A},
  where d\mathbf{A} is the area vector of an infinitesimal patch.
• If the field varies across the surface, the total flux is
  \PhiB = \int! Bi \, dAi = \sumi Bi Ai \cos \theta_i.
• SI unit of magnetic flux: Weber (Wb) = ( \text{T} \cdot \text{m}^2 ).
• Magnetic flux is a scalar quantity.

6.4 Faraday’s Law of Induction

• Faraday observed that an emf is induced in a coil when the magnetic flux through the coil changes with time.
• Change in flux can be caused by:
  • Moving a magnet toward/away from a coil (changing B or A or orientation),
  • Moving a current-carrying coil near another coil (changing B),
  • Rotating a coil in a magnetic field to change the angle between B and A, or changing the coil’s shape (area) in a magnetic field.
• Faraday’s law (single loop):
  \mathcal{\varepsilon} = -\frac{d\PhiB}{dt}, where \PhiB is the magnetic flux through the coil.
• For a coil with N turns, the total induced emf is
  \mathcal{\varepsilon} = -N \frac{d\Phi_B}{dt}.
• Direction (sign) is given by Lenz’s law: the induced emf acts to oppose the change in flux that produced it (hence the negative sign).
• Example note: the Earth’s steady magnetic field also threads flux through loops but does not produce emf because the flux is effectively constant over the time interval considered.

6.5 Motional Electromotive Force

• A straight conductor of length l moving with velocity v in a uniform magnetic field B induces emf across its ends.
• Consider a rectangular loop PQRS with the moving side PQ. If PQ moves with velocity v perpendicular to B, the emf across the rod is
  \varepsilon = B l v.
• Derivation via Lorentz force on charge carriers: a charge q moving with velocity v in B experiences a magnetic force F = q( v \times B ).
  • If a conductor of length l moves in the field, the charge separation builds up until the magnetic force is balanced by the electric field; the work per unit charge from moving the charges across the length is
    W/q = \varepsilon = B l v.
• For a stationary conductor in a changing magnetic field, the emf arises from an induced electric field E produced by the time-varying magnetic field, via the relation
  \mathbf{F} = q(\mathbf{E} + \mathbf{v} \times \mathbf{B}) \quad (\text{with } \mathbf{v}=0) \Rightarrow \mathbf{F} = q\mathbf{E}.
• Faraday’s law and the Lorentz-force perspective are consistent: a time-varying B induces an E field, and a moving conductor in a static B experiences a Lorentz force on charges that can drive an emf.

6.6 Lenz’s Law and Conservation of Energy

• Lenz’s law (1834) states: the polarity of the induced emf is such that it tends to oppose the change in magnetic flux that produced it.
• Example reasoning with Fig. 6.1: when the North pole moves toward the coil, the induced current produces a magnetic moment with North polarity facing the approaching magnet, creating a repulsive force that opposes the motion (and thus opposes the flux increase).
• If the open ends of a coil are used (open circuit), an emf is induced across the ends; the direction is determined by Lenz’s law.
• Thought experiments to illustrate energy conservation:
  • If the induced current were opposite to that shown, the magnet would be attracted to the coil and accelerate, which would violate conservation of energy unless energy is supplied from somewhere else—impossible in this setup.
  • With the correct direction, the magnet is repelled as the flux changes; an external agent must do work to move the magnet, and that energy dissipates as Joule heating due to the induced current.

6.7 Inductance

• Inductance connects current to flux linkage. In general, a coil’s flux linkage is proportional to current:
  \PhiB \propto I,\quad \text{or} \quad N\PhiB \propto I.

• If the geometry is fixed, the flux linkage is proportional to current. The proportionality constant is the inductance L, defined by
  N\Phi_B = L I.

• Inductance is a scalar with dimensions [M L^2 T^{-2} A^{-2}] and units of henry (H).

• The self-induced emf in a coil is the back emf that opposes changes in current:
  \varepsilon = -\frac{d}{dt} (L I) = -L \frac{dI}{dt}.

• Energy stored in an inductor (with current I) is
  W = \frac{1}{2} L I^2.

• Self-inductance for a long solenoid of cross-sectional area A, and length l with n turns per unit length:
  B = \mu0 n I \quad (\text{ignoring edge effects}) \text{Total flux linked} = N\PhiB = (n l) (\pi r^2) \mu0 n I = L I, which gives L = \mu0 n^2 A l.

• If a material with relative permeability \mur fills the region, then L = \mur \mu_0 n^2 A l.

• Energy density of the magnetic field in a region is
  uB = \frac{B^2}{2\mu0}.

• Magnetic energy density is analogous to electrostatic energy density in a capacitor, and both scale with the square of the respective field.

• 6.7.1 Mutual Inductance

  • Consider two long coaxial solenoids S1 and S2 with geometry defined (r1, n1 per unit length; r2, n2 per unit length; lengths l). If current I2 in S2 generates flux in S1, the flux linkage is
    N1 \Phi1 = M I_2,
    where M is the mutual inductance between S1 and S2.
  • For these solenoids, the mutual flux linkage is
    \Phi = (\text{field from S2}) \times (\text{area}) = \Big( \mu0 n2 I2 \Big) \pi r1^2,
    so
    M{12} = \mu0 n1 n2 \pi r_1^2 l.
  • By symmetry, current in S1 generates flux in S2, giving
    M{21} = \mu0 n1 n2 \pi r1^2 l, hence (for these geometries) M{12} = M_{21} = M.
  • The general equality M{12} = M{21} holds in broader cases, though explicit calculation may be difficult if geometry is complex.
  • A changing current in one coil induces emf in the other:
    \varepsilon1 = -\frac{d}{dt}(M I2) = -M \frac{dI_2}{dt}.

• 6.7.2 Self-Inductance

  • In a network of coupled coils, total flux through a given coil includes contributions from all coils. For coil 1:
    N1 \Phi1 = M{11} I1 + M{12} I2,
    where M{11} is the self-inductance of coil 1 and M{12} is the mutual inductance with coil 2, etc.
  • The induced emf for coil 1 then is
    \varepsilon1 = -\frac{d}{dt}(M{11} I1 + M{12} I_2).
  • Self-inductance for a long solenoid is again
    L = \mu0 n^2 A l (air). If a material with relative permeability \mur is present, then
    L = \mur \mu0 n^2 A l.

• 6.7.3 Magnetic Energy and Inertia View of Inductance

  • The self-inductance represents the electromagnetic inertia of the circuit against changes in current, analogous to mass in mechanics.
  • The energy stored in the magnetic field when current I flows is
    W = \frac{1}{2} L I^2.
  • The energy stored can be related to the field: magnetic energy density leads to total energy when integrated over the region containing flux.

6.8 AC Generator

• The principle: electromagnetic induction in a rotating coil in a magnetic field converts mechanical energy to electrical energy.
• Basic setup: a coil on a rotor (armature) rotates in a uniform magnetic field B. Ends of the coil connect to an external circuit via slip rings and brushes.
• If the coil has N turns and the angle between the magnetic field and the coil’s area vector is θ(t) = ω t, then the flux is
  \Phi_B(t) = B A \cos(\theta(t)) = B A \cos(\omega t).
• Applying Faraday’s law: the induced emf for the rotating coil is
  \varepsilon = - \frac{d\Phi_B}{dt} = N \frac{d}{dt} (B A \cos(\omega t)) = NBA(-\omega \sin(\omega t)),
  which can be written as
  \varepsilon(t) = NBA \omega \sin(\omega t).
• If angular velocity is constant and angle is defined such that at t = 0 the coil is perpendicular to B, then the instantaneous emf is
  \varepsilon(t) = \varepsilon0 \sin(\omega t), \text{with} \ \varepsilon0 = NBA \omega.
• Since sin(ωt) varies in time, the polarity of the emf reverses periodically; the current is alternating (AC).
• Frequency relation: ω = 2π f, so
  \varepsilon(t) = NBA \omega \sin(\omega t) = NBA (2\pi f) \sin(2\pi f t).
• Practical power generation: hydroelectric, thermal, or nuclear power, with generators producing up to hundreds of megawatts.
• Example: 6.10 (Kamla’s bike generator) shows a practical calculation of the maximum emf from rotating a coil on a bicycle:
  • N = 100 turns, A = 0.10 m^2, B = 0.01 T, rotation frequency f = 0.5 Hz (half a revolution per second).
  • Maximum emf: \varepsilon_{max} = NBA \omega = N B A (2\pi f) = 100 \times 0.01 \times 0.10 \times 2\pi \times 0.5 = 0.314\,\text{V}.

6.9 Example Highlights and Calculations

• Example 6.1: Induced emf magnitude equals time rate of change of magnetic flux through the circuit. For a single loop, \mathcal{\varepsilon} = -\frac{d\PhiB}{dt}. For N turns, \mathcal{\varepsilon} = -N \frac{d\PhiB}{dt}.
• Example 6.2: A square loop (side 0.10 m, R = 0.5 Ω) in a uniform B = 0.10 T, with B reduced to zero in 0.70 s. Initial flux is BA cos θ with θ = 45°, so initial flux is roughly 0.1 × 0.01 × cos 45°; the emf magnitude ~ 1.0 mV, giving a current ~ 2 mA. (Earth’s field not contributing a changing flux during the interval.)
• Example 6.3: Circular coil, radius 0.10 m, 500 turns, R = 2 Ω, horizontal Earth field component BH = 3.0×10^{-5} T, rotated by 180° in 0.25 s. Initial flux: ΦB,i = B A cos 0°; final flux: Φ_B,f = B A cos 180°. Induced emf magnitude ≈ 3.8×10^{-3} V; current ≈ 1.9×10^{-3} A.
• Example 6.6: Metallic rod of length 1 m rotated in a uniform B = 1 T; rotation frequency such that v = ω r; emf between centre and rim is about 157 V (calculation via ∫ B v r dr or using equivalent sector method).
• Example 6.7: Wheel with 10 spokes, each of length 0.5 m, rotating at 120 rev/min in a horizontal earth-field component HE = 0.4 G. Emf across axle-to-rim is ~6.28×10^{-5} V; the spokes’ emfs are in parallel and thus do not add.
• Example 6.8–6.9 discuss mutual and self inductance with specific configurations (concentric coils; air vs. medium with permeability), deriving M12, M21, and L for common geometries.
• Example 6.10 (Kamla’s bicycle generator): maximum emf for a 100-turn coil, area 0.10 m^2, B = 0.01 T, f = 0.5 Hz is 0.314 V.

6.10 Summary of Key Formulas and Concepts

• Magnetic flux through a surface of area A in a uniform field B:
  \Phi_B = B A \cos \theta.
• For general surfaces and nonuniform fields:
  \PhiB = \int \mathbf{B} \cdot d\mathbf{A} = \sumi \mathbf{B}i \cdot \mathbf{A}i.
• Faraday’s law (induction in a closed loop):
  \mathcal{\varepsilon} = -\frac{d\PhiB}{dt}, and for N turns: \mathcal{\varepsilon} = -N \frac{d\PhiB}{dt}.
• Lenz’s law: the emf polarity is such that the induced current opposes the flux change that produced it (negative sign in Faraday’s law).
• Motional emf in a conductor of length l moving with velocity v in a magnetic field B:
  \varepsilon = B l v.
• Inductance and flux linkage:
  • Flux linkage: N \Phi_B = L I.
  • Self-inductance: \varepsilon = -L \frac{dI}{dt}.
  • Energy stored in an inductor: W = \frac{1}{2} L I^2.
• Mutual inductance: current change in coil 2 induces emf in coil 1:
  \varepsilon1 = -\frac{d}{dt} (M I2) = -M \frac{d I2}{dt}, with M{12} = M{21} = M = \mu0 n1 n2 \pi r_1^2 l for long coaxial solenoids (example geometry).
• Energy density of magnetic field: uB = \frac{B^2}{2\mu0}.
• AC generator emf for rotating coil (N turns, area A, angular speed ω):
  \varepsilon(t) = -\frac{d}{dt}(N B A \cos(\omega t)) = NBA \omega \sin(\omega t),
  with maximum value \varepsilon_{\max} = NBA \omega and frequency relation \omega = 2\pi f.
• For practical generator power, the frequency is typically 50 Hz (India) or 60 Hz (USA) depending on the grid.

POINTS TO PONDER

• Electricity and magnetism are intimately connected: moving charges create magnetic fields; changing magnetic fields create electric currents.
• In a closed circuit, induced currents oppose changes in magnetic flux (conservation of energy). In an open circuit, an emf can exist across the ends.
• The motional emf perspective (Lorentz force) and Faraday’s law perspective are complementary: moving charges in a static field vs. time-varying fields generate electric fields.
• This interplay hints at the relativity of electromagnetism and the broader unity of physical laws.

EXERCISES (selected from Transcript)

• 6.1 Predict the direction of induced current in scenarios using Fig. 6.15(a)–(f).
• 6.2 Use Lenz’s law for Fig. 6.16(a)–(b) to determine induced current directions when a wire is deformed.
• 6.3 A long solenoid with 15 turns per cm has a small loop of area 2.0 cm^2 inside, normal to its axis. If current through the solenoid changes from 2.0 A to 4.0 A in 0.1 s, find the induced emf in the loop.
• 6.4 A rectangular loop 8 cm × 2 cm with a small cut moves out of a region with a uniform magnetic field B = 0.3 T directed normal to the loop. If velocity is 1 cm s^{-1} perpendicular to the longer side, compute the emf across the cut and its duration. Repeat for motion perpendicular to the shorter side.
• 6.5 A 1.0 m long metallic rod is rotated with angular frequency 400 rad s^{-1} about an axis through one end, with a uniform B = 0.5 T parallel to the axis. Compute the emf between the center and the rim.
• 6.6 A 10 m long straight wire falls with speed 5.0 m s^{-1} perpendicular to the horizontal component of Earth’s field HE = 0.30 × 10^{-4} Wb m^{-2}. (a) What is the instantaneous emf? (b) Direction of emf? (c) Which end is at higher potential?
• 6.7 A circuit current falls from 5.0 A to 0.0 A in 0.1 s. If the average emf is 200 V, estimate the self-inductance of the circuit.
• 6.8 A pair of adjacent coils has mutual inductance M = 1.5 H. If the current in one coil changes from 0 to 20 A in 0.5 s, what is the change in flux linkage with the other coil? Reprint 2025-26