Class 12 – Physics
Master 30 advanced topics covering electrostatics, current electricity, magnetism, electromagnetic induction, alternating current, optics, modern physics and semiconductor electronics. Every topic now includes detailed theory, key formulas, a fully worked numerical, common mistakes and practice questions.
📘 How to use this page
Read the theory first, then study the formula box, and only then attempt the worked example on your own before looking at the solution. Use the “Common mistake” boxes to avoid typical board-exam errors. Finish each topic by solving the practice questions without notes. Physics is learned by solving problems, so aim to do at least five numericals per topic.
⚡ Electrostatics and Capacitance
Electric Charge Distribution and Quantisation
Electric charge is a fundamental property of matter that comes in two kinds, positive and negative. Like charges repel and unlike charges attract. Three properties define how charge behaves: it is additive (total charge is the algebraic sum), conserved (it can be transferred but never created or destroyed in an isolated system), and quantised (any observable charge is an integral multiple of the elementary charge e = 1.6 × 10⁻¹⁹ C, so q = ne).
At the macroscopic scale, the charge on an object is so large compared with e that quantisation goes unnoticed and charge appears continuous. This is why we can describe a charged body using smooth charge densities. When charge is spread along a line, over a surface, or through a volume, we use linear (λ), surface (σ) and volume (ρ) charge density respectively. For non-uniform distributions the density varies with position and charge is found by integration, for example Q = ∫λ dl.
Ways of charging
- Friction: electrons transfer from one body to another (glass rubbed with silk becomes positive).
- Conduction: contact with a charged body shares charge; both end up with the same sign.
- Induction: a charged body brought near a conductor redistributes its charge; earthing then leaves the conductor with a charge of opposite sign.
q = ±ne | λ = Q/L (C m⁻¹) | σ = Q/A (C m⁻²) | ρ = Q/V (C m⁻³)
λ = Q/L = 5 × 10⁻⁶ / 2 = 2.5 × 10⁻⁶ C m⁻¹. Number of electrons n = Q/e = 5 × 10⁻⁶ / 1.6 × 10⁻¹⁹ ≈ 3.1 × 10¹³.
- Compare linear, surface and volume charge density and write their SI units.
- A metal sphere of radius 10 cm carries 4 μC uniformly on its surface. Find σ.
- How many electrons together carry a charge of 1 C?
Coulomb’s Law in Multi-Charge Systems
Coulomb’s law states that the electrostatic force between two point charges is directly proportional to the product of the charges and inversely proportional to the square of the distance between them. The force acts along the line joining the charges, is attractive for unlike charges and repulsive for like charges, and obeys Newton’s third law.
When more than two charges are present, the principle of superposition applies: the force on any charge is the vector sum of the forces due to each of the other charges taken one at a time. The presence of other charges does not change the pairwise force. Vector addition therefore becomes the key skill: resolve each force into components, add components, then find magnitude and direction.
In a medium of relative permittivity (dielectric constant) K, the force reduces by the factor K. Compared with gravity, the electrostatic force between an electron and a proton is about 10³⁹ times stronger, which is why electrical forces dominate atomic and molecular structure.
F = kq₁q₂/r² = q₁q₂ / (4πε₀r²), k = 9 × 10⁹ N m² C⁻² | In a medium: F = F₀/K | Vector form: F₁₂ = k q₁q₂ r̂₁₂ / r²
Two adjacent charges each give kq²/a², at right angles; their resultant is √2 kq²/a² along the diagonal. The opposite corner (distance a√2) gives kq²/(2a²) in the same direction. Net force = (√2 + ½) kq²/a² ≈ 1.91 kq²/a², directed away from the centre.
- Two 2 μC charges are 3 cm apart in air. Find the force between them (answer: 40 N).
- Draw force vectors for three charges arranged at the vertices of a triangle.
- Three charges +q, −2q and +q lie on a line with equal spacing d. Find the net force on the middle charge.
Electric Field Due to Continuous Charge Distributions
The electric field at a point is the force per unit positive test charge placed there, E = F/q₀. For a point charge, E = kq/r² directed radially outward for positive charge and inward for negative. Field lines begin on positive charges, end on negative charges, never intersect, and are denser where the field is stronger.
For a continuous distribution, divide the body into small elements dq, write the field dE of each element, and integrate. The skill lies in choosing dq well and exploiting symmetry so that components cancel. For a rod use dq = λ dx, for a ring use dq = λ R dθ, and for a disc use thin concentric rings of radius r and width dr, with dq = σ 2πr dr.
Ring on axis: E = kQx / (x² + R²)3/2 (maximum at x = R/√2; zero at centre)
Disc on axis: E = (σ/2ε₀) [1 − x/√(x² + R²)]
Infinite line: E = λ / (2πε₀ r) | Infinite sheet: E = σ/2ε₀ | Two opposite sheets: E = σ/ε₀ between, 0 outside
x² + R² = 25 cm², so (x² + R²)3/2 = (0.05)³ = 1.25 × 10⁻⁴ m³. E = (9 × 10⁹)(10⁻⁸)(0.04) / 1.25 × 10⁻⁴ = 2.88 × 10⁴ N C⁻¹, directed along the axis away from the ring.
- Identify a suitable differential charge element dq for a charged rod of length L.
- Show that the field at the centre of a uniformly charged ring is zero.
- Find the field of a disc at a large distance and compare with a point charge.
Electric Dipoles in External Electric Fields
An electric dipole is a pair of equal and opposite charges ±q separated by a small distance 2a. Its strength is the dipole moment p = q × 2a, a vector pointing from the negative to the positive charge. Many molecules such as H₂O and HCl are natural dipoles.
In a uniform field, the net force on a dipole is zero because the forces on the two charges are equal and opposite, but a torque acts that tends to rotate it into alignment with the field. In a non-uniform field there is also a net force, which draws the dipole toward the region of stronger field. This explains why small pieces of paper are attracted to a charged comb.
τ = p × E, τ = pE sinθ | U = −p·E = −pE cosθ
Work done in rotating from θ₁ to θ₂: W = pE (cosθ₁ − cosθ₂)
Axial point (far): E = 2kp/r³ | Equatorial point (far): E = kp/r³, directed opposite to p
Potential: V = kp cosθ / r² (zero on the equatorial line)
τ = pE sinθ = 4 × 10⁻⁹ × 2 × 10⁴ × 0.5 = 4 × 10⁻⁵ N m. Work = pE (cos30° − cos90°) = 8 × 10⁻⁵ × 0.866 ≈ 6.9 × 10⁻⁵ J.
- Determine the orientations corresponding to maximum and zero torque.
- Which orientation is stable equilibrium and what is the potential energy there?
- Why does a dipole in a uniform field experience no net force?
Electric Flux and Applications of Gauss’s Law
Electric flux measures how much electric field passes through a surface: Φ = ∮E·dA. It is a scalar, with sign taken positive for field lines leaving a closed surface and negative for entering. Gauss’s law states that the total flux through any closed surface equals the net enclosed charge divided by ε₀, regardless of the surface’s shape or where inside the charge sits.
The law is always true, but it is useful for finding E only when symmetry (spherical, cylindrical or planar) lets us pull E out of the integral. The method: choose a Gaussian surface on which E is constant and perpendicular to the surface (or parallel, contributing zero), calculate the flux, and equate to qenc/ε₀. Only enclosed charge matters for the total flux, but E itself at the surface is due to all charges, inside and outside.
Φ = ∮E·dA = qenc/ε₀
Uniformly charged thin shell: E = kQ/r² (r ≥ R), E = 0 (r < R)
Solid non-conducting sphere: E = kQr/R³ (inside), kQ/r² (outside)
Infinite wire: E = λ/2πε₀r | Infinite plane sheet: E = σ/2ε₀ | Conductor surface: E = σ/ε₀
Φ = q/ε₀ = 8.85 × 10⁻⁹ / 8.85 × 10⁻¹² = 1000 N m² C⁻¹. By symmetry each face receives 1000/6 ≈ 167 N m² C⁻¹.
- Choose a suitable Gaussian surface for an infinitely long charged wire and derive E.
- Why is the field inside a hollow charged conductor zero?
- Find the flux through a hemispherical surface of radius R placed in a uniform field E with its axis along E.
Electric Potential and Equipotential Surfaces
Electric potential at a point is the work done per unit positive charge in bringing it from infinity to that point, V = W/q. Potential difference between two points is the work per unit charge moving between them. Potential is a scalar, so contributions from several charges simply add algebraically, which is often much easier than adding vector fields.
Field and potential are linked by E = −dV/dr: the field points in the direction of fastest decrease of potential and its magnitude equals the potential gradient. An equipotential surface is one on which V is constant. No work is done along it, so the field is always perpendicular to equipotential surfaces. The surface of a conductor in electrostatic equilibrium is an equipotential, and so is the entire interior.
V = kq/r | V = Σ kqᵢ/rᵢ | WAB = q (VB − VA) | E = −dV/dr
Uniform field: E = V/d | Charged shell: V = kQ/R inside and on surface
V = 9 × 10⁹ × 2 × 10⁻⁹ / 0.3 = 60 V. W = qV = 5 × 10⁻⁹ × 60 = 3 × 10⁻⁷ J.
- Sketch field lines and equipotential surfaces around a positive charge.
- Why can two equipotential surfaces never intersect?
- A uniform field of 500 V m⁻¹ exists. What is the potential difference between two points 4 cm apart along the field?
Electrostatic Potential Energy of Charge Systems
Electrostatic potential energy is the work required to assemble a configuration of charges by bringing them in from infinity. It is stored in the arrangement of charges and is independent of the path taken to build it. For two charges the energy is positive if they are alike (work must be done against repulsion) and negative if unlike (the system is bound).
For three or more charges, add the interaction energy of every distinct pair exactly once. A charge q in an external potential V has energy U = qV. The work done by an external agent in moving a charge slowly equals the change in potential energy, while the work done by the field is its negative.
U = kq₁q₂/r | Utotal = k(q₁q₂/r₁₂ + q₂q₃/r₂₃ + q₁q₃/r₁₃) | U = qV | 1 eV = 1.6 × 10⁻¹⁹ J
Energy conservation: ½mv² + qV = constant
KE = eV = 100 eV = 1.6 × 10⁻¹⁷ J. v = √(2eV/m) = √(2 × 1.6 × 10⁻¹⁷ / 9.1 × 10⁻³¹) ≈ 5.9 × 10⁶ m s⁻¹.
- Predict the sign of potential energy for like and unlike charges.
- Find the energy to assemble three equal charges q at the vertices of an equilateral triangle of side a (answer: 3kq²/a).
- A proton and an alpha particle are accelerated through the same potential difference. Compare their kinetic energies.
Capacitors and Dielectric Materials
A capacitor consists of two conductors separated by an insulator and stores energy in the electric field between them. Its capacitance C = Q/V depends only on the geometry of the conductors and the medium between them, never on the charge or voltage. The SI unit is the farad (F); practical values are in μF, nF and pF.
Inserting a dielectric places polarisable molecules in the field. They align (or are distorted) so that induced surface charges produce an opposing field, reducing the net field within the dielectric to E₀/K. For the same charge, the potential difference falls by K and capacitance rises by K. A conducting slab behaves like K → ∞.
Parallel plate: C = ε₀A/d; with dielectric C = Kε₀A/d
Partly filled (slab thickness t): C = ε₀A / (d − t + t/K); conducting slab: C = ε₀A/(d − t)
Isolated sphere: C = 4πε₀R | Spherical capacitor: C = 4πε₀ ab/(b − a)
C₀ = 8.85 × 10⁻¹² × 0.01 / 10⁻³ = 8.85 × 10⁻¹¹ F ≈ 88.5 pF. With dielectric C = 5 × 88.5 = 442.5 pF.
Battery connected vs disconnected
| Quantity | Battery stays connected (V constant) | Battery removed (Q constant) |
|---|---|---|
| Charge Q | Increases K times | Unchanged |
| Voltage V | Unchanged | Decreases K times |
| Electric field E | Unchanged | Decreases K times |
| Energy U | Increases K times | Decreases K times |
- Explain how capacitance changes when plate area increases.
- A 4 μF capacitor charged to 50 V is isolated and a K = 4 dielectric is inserted. Find the new voltage.
- Why is air breakdown (about 3 × 10⁶ V m⁻¹) a limit on capacitor voltage?
Combination and Energy Storage of Capacitors
In a series combination the same charge flows onto every capacitor and the voltages add, so reciprocals of capacitance add. The equivalent capacitance is smaller than the smallest individual one. In a parallel combination every capacitor has the same potential difference and charges add, so capacitances add directly.
The energy stored equals the work done in charging, which is the area under the V–q graph. It resides in the electric field itself, with energy density u = ½ε₀E² (or ½Kε₀E² in a dielectric). When a charged capacitor is connected to an uncharged one, charge redistributes until potentials are equal, and some energy is always lost as heat and radiation even though charge is conserved.
Series: 1/C = 1/C₁ + 1/C₂ + … | Parallel: C = C₁ + C₂ + …
U = ½CV² = ½QV = Q²/2C | u = ½ε₀E²
Common potential after sharing: V = (C₁V₁ + C₂V₂)/(C₁ + C₂); energy loss = C₁C₂(V₁ − V₂)² / 2(C₁ + C₂)
C = (2 × 3)/(2 + 3) = 1.2 μF. Q = CV = 120 μC on each. U = ½CV² = ½ × 1.2 × 10⁻⁶ × 10⁴ = 6 mJ. The voltages are 60 V across 2 μF and 40 V across 3 μF.
- Compare two identical capacitors C connected in series and in parallel.
- A 6 μF capacitor charged to 10 V is connected to an uncharged 3 μF capacitor. Find the common voltage and energy lost.
- Find the energy density at a point where E = 10⁴ V m⁻¹ in air.
🔋 Current Electricity and Circuit Analysis
Drift Velocity and Microscopic Theory of Current
Inside a metal, free electrons move randomly at high thermal speeds (about 10⁵ m s⁻¹) but with zero net displacement. When an electric field is applied, each electron is accelerated between collisions with lattice ions, giving a very small average drift superimposed on the random motion. The mean time between collisions is the relaxation time τ.
The drift speed is surprisingly small (a fraction of a millimetre per second) yet a bulb lights almost instantly because the electric field is established throughout the wire at nearly the speed of light, setting all electrons in motion together. Mobility μ = vd/E measures how easily carriers drift, and microscopic Ohm’s law links current density to field: J = σE.
I = nAevd | vd = eEτ/m | μ = vd/E = eτ/m | J = I/A = nevd | J = σE
vd = I/(nAe) = 1 / (8.5 × 10²⁸ × 10⁻⁶ × 1.6 × 10⁻¹⁹) ≈ 7.4 × 10⁻⁵ m s⁻¹, or about 0.07 mm s⁻¹.
- Explain why conventional current is opposite to electron drift in a metal.
- If the temperature of a conductor rises, what happens to τ and vd at fixed E?
- Two wires of the same material have radii in the ratio 1:2 and carry equal current. Compare drift speeds.
Electrical Resistivity and Temperature Dependence
Ohm’s law states that for a conductor at constant temperature V = IR. Resistance R depends on length, area and material: R = ρL/A. The property ρ is the resistivity, a characteristic of the material. Microscopically, ρ = m/(ne²τ), so it increases when collisions become more frequent (smaller τ).
In metals, heating increases lattice vibrations, reducing τ, so resistivity rises roughly linearly with temperature. In semiconductors and insulators, heating releases many more carriers, so resistivity decreases (negative temperature coefficient). Alloys like manganin and constantan have very small temperature coefficients and are used in standard resistors. Some materials become superconducting, with zero resistance below a critical temperature.
R = ρL/A | ρ = m/(ne²τ) | RT = R₀[1 + α(T − T₀)]
Series: R = ΣRᵢ | Parallel: 1/R = Σ1/Rᵢ | P = VI = I²R = V²/R
Volume is constant, so area halves. R′ = ρ(2L)/(A/2) = 4ρL/A = 40 Ω. In general, stretching to n times the length multiplies resistance by n².
- Predict the effect of doubling wire length while keeping area fixed.
- A copper coil has 50 Ω at 20 °C. Find its resistance at 100 °C if α = 4 × 10⁻³ °C⁻¹.
- Why are nichrome and manganin used for heaters and standard resistors respectively?
Kirchhoff’s Rules and Complex Electrical Circuits
Some circuits cannot be simplified using series and parallel rules. Kirchhoff’s two rules solve any network. The junction rule says that total current entering a junction equals total leaving it (conservation of charge). The loop rule says that the algebraic sum of potential changes around any closed loop is zero (conservation of energy).
Procedure: assign current directions arbitrarily, mark each element’s polarity, pick loops, and write the loop rule with a consistent sign convention. Crossing a resistor in the direction of current is a drop (−IR); against the current is a rise (+IR). Crossing a cell from − to + is a rise (+ε). Write as many independent equations as there are unknowns. Remember also that a real cell has internal resistance r, so its terminal voltage is V = ε − Ir while discharging.
ΣIin = ΣIout | ΣΔV = 0 around any loop
Terminal voltage: V = ε − Ir (discharging), V = ε + Ir (charging)
n cells in series: εeq = nε, req = nr | m identical cells in parallel: εeq = ε, req = r/m
I = ε/(R + r) = 12/6 = 2 A. V = ε − Ir = 12 − 2 = 10 V (also = IR = 10 V). In a second case, 10 V and 4 V cells opposing in series through 3 Ω give I = (10 − 4)/3 = 2 A in the direction set by the larger cell.
- Write independent equations for a two-loop circuit with two cells and three resistors.
- Find the current through a cell when it is short-circuited.
- When is maximum power delivered to an external load? (Answer: R = r.)
Wheatstone Bridge and Electrical Measurement
A Wheatstone bridge has four resistors P, Q, R, S in a diamond with a galvanometer across one diagonal and a cell across the other. When the bridge is balanced, the two junctions connected to the galvanometer are at the same potential, so no current flows through it. This gives P/Q = R/S, which lets an unknown resistance be found without needing the cell’s emf or internal resistance, a strong advantage for accurate measurement.
The metre bridge is a practical form: a uniform wire of length 100 cm acts as the ratio arm. The potentiometer is another null-method device. It compares emfs (ε₁/ε₂ = l₁/l₂) and measures internal resistance without drawing current from the cell at balance, making it more accurate than a voltmeter.
Balance: P/Q = R/S | Metre bridge: X = R(100 − l)/l
Potentiometer: ε₁/ε₂ = l₁/l₂ | Internal resistance: r = R(l₁ − l₂)/l₂
X = R(100 − l)/l = 10 × 60/40 = 15 Ω.
- Explain why galvanometer current becomes zero at bridge balance.
- Why is a potentiometer preferred over a voltmeter for measuring emf?
- Why are copper connecting strips used at the metre-bridge ends?
🧲 Magnetism and Electromagnetic Induction
Motion of Charged Particles in Magnetic Fields
A magnetic field exerts the Lorentz force F = qv × B on a moving charge. Because this force is always perpendicular to velocity, it changes the direction but never the speed or kinetic energy of the particle; the field does no work. A charge at rest feels no magnetic force.
If v is perpendicular to B, the particle moves in a circle because the magnetic force supplies the centripetal force. If v makes an angle with B, it follows a helix: circular motion due to the perpendicular component and uniform motion along the field due to the parallel component. If v is parallel to B, the particle goes undeflected. The time period of circular motion is independent of speed, the principle behind the cyclotron. When both fields exist, F = q(E + v × B), and a velocity selector passes only particles with v = E/B.
F = qvB sinθ | r = mv/(qB) = p/(qB) | T = 2πm/(qB) | f = qB/(2πm)
Helix pitch: p = v cosθ · T | Velocity selector: v = E/B
r = mv/(eB) = (1.67 × 10⁻²⁷ × 2 × 10⁶)/(1.6 × 10⁻¹⁹ × 0.1) ≈ 0.21 m. T = 2πm/(eB) ≈ 6.6 × 10⁻⁷ s.
- Use the right-hand rule to predict force direction on a positive charge moving east in a field pointing north.
- An electron and a proton enter a field with equal momentum. Compare their radii.
- Why does the cyclotron frequency not change as the particle speeds up (non-relativistic)?
Magnetic Force on Current-Carrying Conductors
A conductor carrying current in a magnetic field feels a force because the field acts on each moving charge carrier and the force is transmitted to the lattice. For a straight wire, F = IL × B. Two parallel conductors exert forces on each other: parallel currents attract and antiparallel currents repel. This effect defines the SI unit of current, the ampere.
A current-carrying loop in a uniform field experiences zero net force but a torque τ = m × B with magnetic moment m = NIA. This is the working principle of the moving-coil galvanometer, where a restoring spring torque balances magnetic torque so that deflection is proportional to current. A galvanometer becomes an ammeter when a small shunt resistance is placed in parallel, and a voltmeter when a large resistance is placed in series.
Fields produced by currents
Long straight wire: B = μ₀I/2πr. Centre of a circular loop: B = μ₀I/2R (for N turns, multiply by N). Solenoid: B = μ₀nI inside, where n is turns per unit length. Here μ₀ = 4π × 10⁻⁷ T m A⁻¹.
F = BIL sinθ | τ = NIAB sinθ | F/L = μ₀I₁I₂/2πd = 2 × 10⁻⁷ I₁I₂/d
Shunt: S = IgG/(I − Ig) | Series resistance: R = V/Ig − G
F = BIL = 0.2 × 4 × 0.5 = 0.4 N. Between the wires: F/L = 2 × 10⁻⁷ × 1 × 1 / 1 = 2 × 10⁻⁷ N m⁻¹.
- Find the orientation that gives maximum magnetic force on a straight wire.
- A galvanometer of resistance 50 Ω gives full-scale deflection at 2 mA. Convert it into a 0–1 A ammeter.
- Why is a radial magnetic field used in a moving-coil galvanometer?
Magnetic Dipoles and Properties of Magnetic Materials
A current loop is equivalent to a magnetic dipole with moment m = NIA, directed by the right-hand rule. A bar magnet behaves similarly, and magnetic field lines form continuous closed loops, so no isolated magnetic monopole exists and the net magnetic flux through a closed surface is always zero (Gauss’s law for magnetism). In a uniform field a dipole has torque τ = mB sinθ and energy U = −m·B.
Materials respond to an applied field through their magnetisation M and magnetic susceptibility χ. Diamagnetic substances have small negative χ and are slightly repelled (bismuth, copper, water). Paramagnetic substances have small positive χ and are weakly attracted (aluminium, sodium); their susceptibility follows Curie’s law χ ∝ 1/T. Ferromagnetic substances (iron, cobalt, nickel) have very large positive χ due to domains, and become paramagnetic above the Curie temperature. Their magnetisation traces a hysteresis loop, which defines retentivity and coercivity.
m = NIA | Axial field of bar magnet: B = μ₀ 2m / 4πr³ | Equatorial: B = μ₀ m / 4πr³
B = μ₀(H + M) | M = χH | μr = 1 + χ
| Property | Diamagnetic | Paramagnetic | Ferromagnetic |
|---|---|---|---|
| Susceptibility χ | Small, negative | Small, positive | Large, positive |
| Relative permeability | Slightly < 1 | Slightly > 1 | ≫ 1 |
| In a field | Weakly repelled | Weakly attracted | Strongly attracted |
m = NIA = 100 × 2 × 0.02 = 4 A m². τmax = mB = 4 × 0.5 = 2 N m.
- Compare diamagnetic, paramagnetic and ferromagnetic behaviour.
- Why do magnetic field lines form closed loops?
- What does the area of a hysteresis loop represent?
Electromagnetic Induction and Faraday’s Law
Faraday discovered that a changing magnetic flux through a circuit induces an emf. Magnetic flux is Φ = B·A = BA cosθ, with unit weber (Wb). Flux can change in three ways: by changing B, by changing the area A, or by changing the orientation angle θ. A steady flux, however large, induces nothing.
Faraday’s law states that the induced emf equals the rate of change of flux linkage: ε = −N dΦ/dt. A conductor of length l moving with speed v perpendicular to a uniform field has motional emf ε = Blv. The induced current through a closed circuit of resistance R is I = ε/R, and the charge that flows depends only on the total flux change, q = ΔΦ/R, not on how quickly it happens. Practical applications include AC generators, transformers, induction cooktops and magnetic braking by eddy currents.
Φ = BA cosθ | ε = −N dΦ/dt | ε = Blv | q = NΔΦ/R
Rotating coil: ε = NBAω sin ωt, with ε₀ = NBAω
ε = NAΔB/Δt = 200 × 0.05 × 0.4 / 0.2 = 20 V. For a 0.5 m rod moving at 4 m s⁻¹ across a 0.3 T field, ε = Blv = 0.3 × 0.5 × 4 = 0.6 V.
- List three different ways of changing magnetic flux through a loop.
- A 20 cm square loop of resistance 2 Ω is pulled out of a field of 0.5 T in 0.1 s. Find the induced current.
- Explain how eddy currents are reduced by laminating a core.
Lenz’s Law and Induced Electric Fields
Lenz’s law gives the direction of the induced current: it flows so as to oppose the change in magnetic flux that caused it. This is not an extra rule but a consequence of conservation of energy. If the induced current aided the change, flux would grow without limit and energy would be created from nothing. The minus sign in Faraday’s law expresses exactly this opposition.
To use it: (1) decide whether flux through the loop is increasing or decreasing, (2) state the direction of the induced field that would oppose that change, (3) use the right-hand rule to find the current producing that field. Mechanical work done against the opposing force is converted into electrical energy and then heat. When flux changes in a region without a circuit, it induces a non-conservative electric field with ∮E·dl = −dΦ/dt, whose field lines are closed loops.
ε = Blv | I = Blv/R | Magnetic drag F = B²l²v/R | Power = Fv = B²l²v²/R = I²R
ε = 0.5 × 0.4 × 5 = 1 V; I = 0.5 A; F = BIl = 0.1 N; mechanical power = Fv = 0.5 W, equal to the heating I²R = 0.25 × 2 = 0.5 W. Energy is conserved.
- Predict induced current direction when a north pole is withdrawn from a coil.
- Explain why a magnet falls slowly through a copper pipe.
- Why is the induced electric field non-conservative?
Self-Inductance and Mutual Inductance
When current in a coil changes, its own flux changes and induces an emf opposing that change. This property is self-inductance L, defined by Φ = LI, with unit henry (H). It plays the role of electrical “inertia”. Two nearby coils show mutual inductance M: a changing current in one induces an emf in the other, ε₂ = −M dI₁/dt.
An inductor stores energy in its magnetic field. In an LR circuit, the current does not jump instantly but grows exponentially with time constant τ = L/R, reaching about 63% of its final value after one time constant. On breaking such a circuit, a large back emf can produce sparks across the switch.
ε = −L dI/dt | Solenoid: L = μ₀N²A/l = μ₀n²Al | U = ½LI²
Mutual (coaxial solenoids): M = μ₀N₁N₂A/l | Growth: I = I₀(1 − e−t/τ), τ = L/R
L = μ₀N²A/l = 4π × 10⁻⁷ × (500)² × 4 × 10⁻⁴ / 0.5 ≈ 2.5 × 10⁻⁴ H (0.25 mH). U = ½LI² = ½ × 2.5 × 10⁻⁴ × 4 = 5 × 10⁻⁴ J.
- Explain why an inductor opposes a sudden change in current.
- How does inserting an iron core affect L, and why?
- An L = 2 H, R = 10 Ω circuit is connected to a 20 V cell. Find the time constant and final current.
〰️ Alternating Current and Electromagnetic Waves
Alternating Current and Phasor Analysis
Alternating current (AC) varies sinusoidally with time, i = I₀ sin ωt, where ω = 2πf. Its average over a full cycle is zero, so the useful measure is the RMS value, defined so that it produces the same heating as an equal DC value. Household mains of 220 V is an RMS value; its peak is about 311 V.
Different components shift the phase between voltage and current. In a resistor, they are in phase. In an inductor, voltage leads current by 90° and the opposition is inductive reactance XL = ωL. In a capacitor, current leads voltage by 90° and the opposition is XC = 1/ωC. In a series LCR circuit a phasor diagram gives the impedance Z and phase angle φ. Average power depends on the power factor cosφ, so a pure inductor or capacitor consumes no power (wattless current).
Irms = I₀/√2 ≈ 0.707 I₀ | Vrms = V₀/√2 | XL = ωL | XC = 1/ωC
Z = √[R² + (XL − XC)²] | tanφ = (XL − XC)/R | P = VrmsIrms cosφ
Z = √(30² + 40²) = 50 Ω; Irms = 220/50 = 4.4 A; cosφ = R/Z = 0.6; P = 220 × 4.4 × 0.6 ≈ 581 W, which equals I²R = 19.36 × 30 = 581 W.
- Compare phase relations in pure R, L and C circuits.
- Find XL of a 0.1 H inductor at 50 Hz.
- Why is the average power in a pure capacitor zero?
LCR Circuits and Electrical Resonance
In a series LCR circuit, XL increases with frequency while XC decreases. At one special frequency they are equal and cancel, so impedance becomes purely resistive and minimal (Z = R), and current is at its maximum. This is resonance, occurring at ω₀ = 1/√(LC). Voltage and current are then in phase and the power factor is 1.
The sharpness of resonance is described by the quality factor Q. A high Q (low resistance) gives a tall, narrow current peak and good frequency selectivity, while a low Q gives a broad, flat response. Radio tuning uses this: varying a capacitor changes the resonant frequency to select one station out of many. Note that individual voltages across L and C at resonance can far exceed the supply voltage, each being Q times larger.
ω₀ = 1/√(LC) | f₀ = 1/(2π√(LC)) | Imax = V/R
Q = ω₀L/R = 1/(ω₀RC) = (1/R)√(L/C) | Bandwidth Δω = R/L | Q = ω₀/Δω
LC = 10⁻⁹ s², so ω₀ = 1/√(10⁻⁹) ≈ 3.16 × 10⁴ rad s⁻¹ and f₀ ≈ 5.03 kHz. Q = ω₀L/R = 3.16 × 10⁴ × 2 × 10⁻³ / 10 ≈ 6.3.
- Explain why resonance is useful in radio tuning.
- What happens to resonant frequency if C is made four times larger?
- At resonance, what is the phase difference between supply voltage and current?
Transformers and AC Power Transmission
A transformer consists of two coils wound on a common soft-iron core. An alternating current in the primary produces a changing flux in the core that links the secondary and induces an emf by mutual induction. With turns Np and Ns, the voltage ratio equals the turns ratio. A step-up transformer (Ns > Np) raises voltage and lowers current; a step-down transformer does the reverse. Energy is conserved, so an ideal transformer does not increase power.
Real transformers lose energy by copper loss (I²R heating of windings), eddy currents (reduced by laminating the core), hysteresis (reduced by using soft iron), and flux leakage. Efficiencies of 95–99% are common. Power is transmitted at very high voltage because for fixed power P = VI, higher V means lower current, and the line loss I²R falls as the square of the current.
Vs/Vp = Ns/Np = Ip/Is (ideal) | Efficiency η = (Pout/Pin) × 100%
Line loss: Ploss = I²Rline = (P/V)²Rline
Ns = 1000 × 220/2200 = 100. Ip = Is × Ns/Np = 1 A. Line loss: 100 kW sent through 5 Ω at 2000 V has I = 50 A and loss = 12.5 kW; at 20,000 V, I = 5 A and loss = 125 W, a hundred times smaller.
- Explain why electrical power is transmitted at very high voltage.
- List four energy losses in a transformer and how each is reduced.
- A 90%-efficient transformer takes 2 kW. How much power reaches the load?
Electromagnetic Waves and Their Applications
Maxwell realised that Ampere’s law was incomplete: a changing electric field also produces a magnetic field, as though it were a current. This displacement current Id = ε₀ dΦE/dt completes the four Maxwell equations and predicts self-sustaining electromagnetic waves. In these waves E and B oscillate in phase, perpendicular to each other and to the direction of propagation (transverse), and need no medium. In vacuum all travel at c = 1/√(μ₀ε₀) ≈ 3 × 10⁸ m s⁻¹, and E₀/B₀ = c.
The waves carry energy and momentum. They are produced by accelerating charges, with radio waves generated by oscillating circuits. The spectrum is ordered by frequency, all being the same kind of wave.
| Region | Approx. wavelength | Typical use |
|---|---|---|
| Radio waves | > 0.1 m | Broadcasting, communication |
| Microwaves | 1 mm – 0.1 m | Radar, ovens, satellite links |
| Infrared | 700 nm – 1 mm | Remote controls, thermal imaging |
| Visible | 400 – 700 nm | Vision, photography |
| Ultraviolet | 1 – 400 nm | Sterilisation, forensics |
| X-rays | 10⁻³ – 1 nm | Medical imaging, crystallography |
| Gamma rays | < 10⁻³ nm | Cancer therapy, nuclear studies |
c = fλ = E₀/B₀ = 1/√(μ₀ε₀) | Id = ε₀ dΦE/dt
Energy densities: uE = ½ε₀E², uB = B²/2μ₀ (equal in a wave) | Intensity I = ½ε₀cE₀²
B₀ = E₀/c = 60/(3 × 10⁸) = 2 × 10⁻⁷ T. λ = c/f = 3 × 10⁸ / 10⁸ = 3 m.
- Arrange the main electromagnetic regions in increasing frequency.
- Why can electromagnetic waves travel through vacuum while sound cannot?
- Why do ozone layers matter for ultraviolet radiation?
🔭 Optics
Ray Optics and Advanced Optical Instruments
Ray optics treats light as straight rays that obey the laws of reflection and refraction (Snell’s law n₁ sinθ₁ = n₂ sinθ₂). Use the Cartesian sign convention: distances are measured from the pole or optical centre, positive along the direction of incident light. When light travels from a denser to a rarer medium at an angle above the critical angle (sin ic = 1/n), it is totally reflected. This effect drives optical fibres, prisms in binoculars, and the sparkle of diamonds.
Thin lenses form images by refraction, with focal length given by the lens maker’s formula. The power of a lens is the reciprocal of focal length in metres, measured in dioptres. In a compound microscope, a short-focus objective forms a magnified real image that the eyepiece magnifies further. In an astronomical telescope, a large-aperture objective collects light from a distant object and the eyepiece magnifies the image. Eye defects are corrected with lenses: myopia with a concave lens, hypermetropia with a convex lens.
Mirror: 1/v + 1/u = 1/f | Lens: 1/v − 1/u = 1/f | m = v/u (lens), m = −v/u (mirror)
Lens maker: 1/f = (n − 1)(1/R₁ − 1/R₂) | P = 1/f (m) | Combination: P = P₁ + P₂
Prism: n = sin[(A + δm)/2] / sin(A/2) | Microscope (image at near point): M = (L/fo)(1 + D/fe)
Telescope (normal adjustment): M = fo/fe, length = fo + fe
u = −30, f = +20: 1/v = 1/f + 1/u = 1/20 − 1/30 = 1/60, so v = +60 cm (real image beyond the lens). m = v/u = −2, so it is inverted and twice the size. For a telescope with fo = 100 cm and fe = 5 cm, M = 20 and tube length = 105 cm.
- Draw a labelled ray diagram for an astronomical telescope.
- Find the critical angle for glass of refractive index 1.5.
- A myopic person has a far point of 80 cm. Find the power of the corrective lens.
Wave Optics, Interference and Diffraction
Huygens’ principle treats every point on a wavefront as a source of secondary wavelets; the new wavefront is their common tangent. It explains reflection and refraction and shows that light slows down in denser media. Interference occurs when two coherent waves (constant phase difference) superpose, giving bright fringes where path difference is nλ and dark fringes where it is (n + ½)λ.
In Young’s double-slit experiment, fringes of equal width β = λD/d appear on a screen. Diffraction is the bending of light around obstacles or through narrow openings of size comparable to the wavelength. In single-slit diffraction a broad central maximum is flanked by weaker, narrower maxima, with minima at a sinθ = nλ. Unlike interference fringes, diffraction fringes have unequal widths and intensities. Polarisation demonstrates that light is transverse: unpolarised light passing through a polariser emerges with half its intensity, and Malus’s law governs a second polariser.
Fringe width β = λD/d | Position of nth bright fringe: y = nλD/d
I = 4I₀ cos²(φ/2) | Imax/Imin = (a₁ + a₂)²/(a₁ − a₂)²
Single slit minima: a sinθ = nλ; central maximum width = 2λD/a
Malus: I = I₀ cos²θ | Brewster: tanθB = n
β = λD/d = (600 × 10⁻⁹ × 1.5)/(3 × 10⁻⁴) = 3 × 10⁻³ m = 3 mm. If the whole setup is immersed in water (n = 1.33), the fringes shrink to β/n ≈ 2.25 mm.
- State the path-difference conditions for constructive and destructive interference.
- What happens to fringe width if slit separation is doubled?
- Light of intensity I₀ passes through two polaroids with axes at 60°. Find the final intensity (assume I₀ is polarised light).
⚛️ Modern Physics and Electronics
Photoelectric Effect and Photon Behaviour
When light of sufficiently high frequency strikes a metal surface, electrons are ejected. Experiments show that (1) there is a threshold frequency below which no emission occurs, however intense the light, (2) the maximum kinetic energy of electrons depends on frequency, not intensity, (3) the photocurrent is proportional to intensity, and (4) emission is essentially instantaneous. The wave theory cannot explain these results.
Einstein explained them by proposing that light consists of quanta, called photons, each of energy E = hν. One photon gives all its energy to one electron. Part of this, the work function φ, is needed to escape; the remainder becomes kinetic energy. Increasing intensity increases the number of photons, hence the number of electrons, while increasing frequency increases the energy per photon. The stopping potential V₀ is the retarding voltage needed to stop even the fastest electrons.
E = hν = hc/λ | hν = φ + Kmax | Kmax = eV₀ | ν₀ = φ/h
V₀–ν graph: slope = h/e | Photon momentum p = h/λ | hc ≈ 1240 eV nm
E = 1240/400 = 3.1 eV. Kmax = 3.1 − 2.0 = 1.1 eV, so V₀ = 1.1 V. λ₀ = 1240/2.0 = 620 nm; light longer than 620 nm will not eject electrons.
- Distinguish the effects of increasing intensity and increasing frequency.
- Why does the wave theory fail to explain the absence of time lag?
- What quantity does the slope of a V₀–ν graph give, and what does the intercept on the ν axis give?
Matter Waves and Wave-Particle Duality
De Broglie extended the dual nature of light to matter: every moving particle has an associated wave with wavelength λ = h/p. For everyday objects this wavelength is far too small to observe, but for electrons, neutrons and atoms it is comparable to atomic spacings, giving measurable diffraction. Davisson and Germer confirmed this by scattering electrons from a nickel crystal, observing diffraction peaks that matched λ calculated from de Broglie’s relation.
Wave-particle duality means that radiation and matter show wave-like properties in propagation (interference, diffraction) and particle-like properties in interaction (photoelectric effect, collisions). De Broglie’s idea also explains Bohr’s quantisation: a stable orbit must hold a whole number of wavelengths, 2πr = nλ. Matter waves are not mechanical or electromagnetic waves; their intensity gives the probability of finding the particle.
λ = h/p = h/mv = h/√(2mK) | Electron through V volts: λ = 1.227/√V nm
Charged particle: λ = h/√(2mqV) | h = 6.63 × 10⁻³⁴ J s
Electron: λ = 1.227/√150 ≈ 0.100 nm, comparable to atomic spacing. Ball: λ = 6.63 × 10⁻³⁴ / (0.15 × 30) ≈ 1.5 × 10⁻³³ m, undetectably small.
- Compare the wavelengths of two particles with different momenta.
- An electron and a proton have equal kinetic energy. Which has the longer wavelength?
- Why do we not observe wave behaviour for a moving car?
Atomic Energy Levels and Spectral Transitions
Rutherford’s alpha-scattering experiment showed that an atom has a tiny, dense, positive nucleus with electrons around it. Classical physics predicts such an atom should collapse and emit a continuous spectrum, neither of which happens. Bohr’s model for hydrogen-like atoms fixes this with postulates: electrons move in certain stationary orbits without radiating, with angular momentum quantised as mvr = nh/2π, and radiation is emitted or absorbed only when an electron jumps between orbits, with photon energy equal to the energy difference.
The energy of level n is negative, indicating a bound state; the lowest level (n = 1) is the ground state. Spectral lines fall in series according to the lower level: Lyman (to n = 1, ultraviolet), Balmer (n = 2, visible), Paschen (n = 3, infrared), then Brackett and Pfund. Bohr’s model works well for hydrogen but cannot explain multi-electron atoms, fine structure or relative intensities, which require quantum mechanics.
En = −13.6/n² eV | rn = 0.529 n² Å | vn ≈ 2.19 × 10⁶/n m s⁻¹
hν = Ei − Ef | 1/λ = R(1/n₁² − 1/n₂²), R = 1.097 × 10⁷ m⁻¹
For hydrogen-like ions: En = −13.6 Z²/n² eV
ΔE = 13.6(1/4 − 1/9) = 13.6 × 5/36 ≈ 1.89 eV. λ = 1240/1.89 ≈ 656 nm, a red line in the visible region. The ionisation energy of hydrogen from the ground state is 13.6 eV.
- Explain why atomic spectra contain specific lines rather than every wavelength.
- Find the shortest wavelength in the Lyman series.
- How many different spectral lines can arise when electrons fall from n = 4 to lower levels?
Nuclear Binding Energy and Radioactive Processes
The nucleus consists of protons and neutrons (nucleons) bound by the short-range, charge-independent strong nuclear force. Nuclear radius follows R = R₀A1/3 with R₀ ≈ 1.2 fm, giving a density of about 2.3 × 10¹⁷ kg m⁻³ that is nearly the same for all nuclei. The mass of a nucleus is less than the sum of its nucleons; this mass defect corresponds to the binding energy, via E = mc². The binding energy per nucleon peaks near 8.75 MeV around iron (A ≈ 56), so energy is released both when heavy nuclei split (fission) and when light nuclei combine (fusion).
Unstable nuclei decay spontaneously. In alpha decay the nucleus loses ⁴He (A − 4, Z − 2). In beta-minus decay a neutron converts to a proton (Z + 1, A unchanged), emitting an electron and an antineutrino. Gamma decay releases a photon with no change in A or Z. Decay is random for individual nuclei but follows an exponential law for large numbers.
Δm = Zmp + (A − Z)mn − M | BE = Δm × 931.5 MeV/u
N = N₀e−λt = N₀(½)t/T | T½ = 0.693/λ | τ = 1/λ | Activity A = λN
BE = 0.0304 × 931.5 ≈ 28.3 MeV, so BE/A ≈ 7.07 MeV per nucleon. Time = 3 half-lives, so remaining = 80 × (½)³ = 10 g.
- Compare changes in mass number and atomic number during alpha and beta decay.
- ²³⁸U emits one alpha particle. Write the daughter nucleus.
- Explain, using the binding energy curve, why both fission and fusion release energy.
Semiconductor Devices and Modern Electronic Circuits
In solids, atomic levels merge into energy bands. In metals the valence and conduction bands overlap; in insulators a wide gap (> 3 eV) separates them; in semiconductors the gap is small (Si ≈ 1.1 eV, Ge ≈ 0.7 eV), so thermal energy promotes some electrons. An intrinsic semiconductor has equal electron and hole concentrations. Doping with pentavalent atoms (P, As) adds donor levels and makes n-type material; trivalent atoms (B, Al, In) create acceptor levels and make p-type material. In equilibrium nenh = ni².
Joining p and n regions creates a p-n junction with a depletion layer and a barrier potential (about 0.7 V in silicon, 0.3 V in germanium). In forward bias the barrier shrinks and current flows readily; in reverse bias the barrier grows and only a tiny leakage current flows, so a diode conducts in one direction and works as a rectifier. Special diodes include the Zener (voltage regulation in reverse breakdown), LED (emits light in forward bias), photodiode (light detector in reverse bias) and solar cell (converts light to electrical energy). Logic gates (AND, OR, NOT, NAND, NOR) form the basis of digital electronics; NAND and NOR are universal gates.
nenh = ni² | Half-wave rectifier ripple frequency = fin | Full-wave ripple frequency = 2fin
Logic: AND (Y = A·B), OR (Y = A + B), NOT (Y = Ā), NAND (Y = (A·B)‾), NOR (Y = (A + B)‾)
nh = ni²/ne = (2.25 × 10³²)/(5 × 10²²) = 4.5 × 10⁹ m⁻³, far smaller than ne, so electrons are the majority carriers. For a 50 Hz input, a full-wave rectifier gives 100 Hz output ripple.
- Compare intrinsic, n-type and p-type semiconductors.
- Explain how a p-n junction diode acts as a rectifier with a neat circuit diagram.
- Draw the truth table of a NAND gate and show it can be used as a NOT gate.
📋 Constants and Exam Tips
Important Constants
| Quantity | Symbol | Value |
|---|---|---|
| Speed of light | c | 3 × 10⁸ m s⁻¹ |
| Elementary charge | e | 1.6 × 10⁻¹⁹ C |
| Coulomb constant | k = 1/4πε₀ | 9 × 10⁹ N m² C⁻² |
| Permittivity of free space | ε₀ | 8.85 × 10⁻¹² C² N⁻¹ m⁻² |
| Permeability of free space | μ₀ | 4π × 10⁻⁷ T m A⁻¹ |
| Planck’s constant | h | 6.63 × 10⁻³⁴ J s |
| Electron mass | mₑ | 9.1 × 10⁻³¹ kg |
| Proton mass | mₚ | 1.67 × 10⁻²⁷ kg |
| Atomic mass unit | 1 u | 931.5 MeV/c² = 1.66 × 10⁻²⁷ kg |
| Rydberg constant | R | 1.097 × 10⁷ m⁻¹ |
| Electron volt | 1 eV | 1.6 × 10⁻¹⁹ J |
Exam strategy
- Start with a diagram. Ray diagrams, circuit diagrams and force vectors earn marks and prevent sign errors.
- Write the formula first, then substitute. Marks are awarded for the correct formula even when arithmetic slips.
- Check units and powers of ten. Convert μC, cm, mm and nF to SI before calculating.
- Use limiting cases to test your answer. A formula that gives nonsense at x → 0 or x → ∞ is likely wrong.
- Learn derivations with their diagrams. Typical long-answer questions include Gauss’s law applications, capacitor with dielectric, Biot–Savart results, resonance in LCR, Young’s experiment and Bohr’s model.
- Revise definitions precisely. Terms such as “work function”, “mass defect” and “drift velocity” need exact wording.