Electrostatics studies charges at rest, the forces between them and the electric field and potential they produce. Gauss's law relates the electric flux through a closed surface to the charge inside it. It makes it easy to find fields of symmetric charge distributions such as a line, a sheet or a sphere.
महत्त्वाचे मुद्दे
Coulomb's law: force between two point charges is proportional to the product of charges and inversely proportional to the square of the distance.
Electric field E = F/q₀ (force per unit positive test charge); unit N/C or V/m.
Electric potential V = work done per unit charge in bringing a test charge from infinity; E = −∇V.
Field lines start on positive and end on negative charges; they never cross.
Electric flux Φ = ∮E·dA through a surface.
Gauss's law: total flux through any closed surface = (charge enclosed)/ε₀. The surface chosen is called a Gaussian surface.
Gauss's law is true for any shape of surface, but it is useful only where symmetry makes E constant on the surface.
Inside a conductor in equilibrium, E = 0; any excess charge lies on its outer surface.
A charged spherical shell acts like a point charge outside it; inside it the field is zero.
Differential form: ∇·E = ρ/ε₀. With E = −∇V this gives Poisson's equation ∇²V = −ρ/ε₀, and in charge-free space Laplace's equation ∇²V = 0.
Electric dipole: equal and opposite charges ±q separated by 2a; dipole moment p = q × 2a, directed from −q to +q.
महत्त्वाच्या संज्ञा
Permittivity of free space ε₀ = 8.85 × 10⁻¹² C² N⁻¹ m⁻²; 1/(4πε₀) = 9 × 10⁹ N m² C⁻².
Surface charge density σ – charge per unit area; linear density λ – charge per unit length.
Equipotential surface – surface on which V is the same everywhere (समविभव पृष्ठ).
Gaussian surface – imaginary closed surface used to apply Gauss's law.
सूत्रे
Coulomb: F = (1/4πε₀) q₁q₂/r²
Point charge: E = q/(4πε₀r²), V = q/(4πε₀r)
Gauss's law: ∮E·dA = q_enc/ε₀
Infinite line charge: E = λ/(2πε₀r)
Infinite plane sheet: E = σ/(2ε₀); just outside a charged conductor: E = σ/ε₀
Spherical shell (radius R): E = q/(4πε₀r²) for r > R; E = 0 for r < R
Uniformly charged solid sphere, inside: E = qr/(4πε₀R³)
Dipole on axis: E = 2p/(4πε₀r³); on equator: E = p/(4πε₀r³) (for r ≫ a)
५ महत्त्वाचे प्रश्न
Q1. State Gauss's law in electrostatics.
The total electric flux through any closed surface equals 1/ε₀ times the total charge enclosed by that surface: ∮E·dA = q_enc/ε₀.
Q2. Two charges of 2 μC and 3 μC are 0.3 m apart in air. Find the force between them.
Q3. Using Gauss's law, find the electric field due to an infinite line charge.
Line with charge per unit length λ. Choose a coaxial cylinder of radius r and length l as Gaussian surface.
By symmetry E is radial and has the same magnitude on the curved surface.
Flux through the flat ends = 0 (E parallel to the ends).
Flux through curved surface = E × 2πrl.
Charge enclosed = λl. Gauss: E × 2πrl = λl/ε₀.
E = λ/(2πε₀r), so E ∝ 1/r.
Q4. Using Gauss's law, find the field of an infinite plane sheet of charge.
Sheet with surface charge density σ. Choose a cylinder (pill-box) crossing the sheet with end area A.
E is perpendicular to the sheet, pointing away on both sides; no flux through the curved surface.
Total flux = EA + EA = 2EA.
Charge enclosed = σA.
2EA = σA/ε₀ → E = σ/(2ε₀), independent of distance.
Q5. A point charge of 1 μC is at the centre of a cube of side 10 cm. Find the flux through the whole cube and through one face.
By Gauss's law, total flux = q/ε₀ = 10⁻⁶/(8.85 × 10⁻¹²).
Φ ≈ 1.13 × 10⁵ N m² C⁻¹ (the cube size does not matter).
By symmetry, each of the 6 faces gets one-sixth.
Flux through one face ≈ 1.13 × 10⁵/6 ≈ 1.88 × 10⁴ N m² C⁻¹.
परीक्षेसाठी टीप
In every Gauss's law derivation, draw the Gaussian surface, state the symmetry, split the flux into parts, and only then apply the law. Remember sheet σ/2ε₀ but conductor surface σ/ε₀.
Dielectrics are insulators that become polarised in an electric field. Placing a dielectric between the plates of a capacitor raises its capacitance. This topic covers polar and non-polar molecules, polarisation, the three electric vectors E, P and D, Gauss's law in a dielectric, dielectric constant and susceptibility, the capacitor with a dielectric slab, and the Clausius–Mossotti relation.
महत्त्वाचे मुद्दे
A dielectric has no free charges, but its charges shift slightly in an electric field, forming tiny dipoles.
Non-polar molecules (H₂, O₂, CO₂) have no permanent dipole; the field induces one. Polar molecules (H₂O, HCl, NH₃) have permanent dipoles that line up with the field.
Polarisation P = dipole moment per unit volume (C m⁻²). It produces bound surface charge σ_b = P·n̂ and bound volume charge ρ_b = −∇·P.
The field inside a dielectric is reduced: E = E₀/K, where K (or ε_r) is the dielectric constant.
Electric displacement D = ε₀E + P. Gauss's law in a dielectric uses only free charge: ∮D·dS = q_free, or ∇·D = ρ_free.
For a linear dielectric P = ε₀χE, so D = ε₀(1 + χ)E = εE, and K = 1 + χ.
Filling a capacitor completely with a dielectric multiplies its capacitance by K.
Clausius–Mossotti relation links the microscopic polarisability α with the measurable K.
Types of polarisation: electronic, ionic, orientational (only polar molecules; falls with temperature) and space-charge.
Dielectric strength: the largest field a dielectric can take before breakdown (air ≈ 3 × 10⁶ V m⁻¹).
महत्त्वाच्या संज्ञा
Electric susceptibility (χ) – ratio P/(ε₀E); χ = K − 1.
Polarisability (α) – induced dipole moment per unit local field, p = αE_loc.
Bound charge – charge that appears on a dielectric's surface because of polarisation; it cannot move freely.
Local field – actual field acting on a molecule, E_loc = E + P/(3ε₀) for cubic symmetry.
Dielectric loss – energy turned into heat in an alternating field.
सूत्रे
D = ε₀E + P = ε₀KE; P = ε₀χE = ε₀(K − 1)E
Bound charge density on the surface σ_b = P = σ_f(1 − 1/K)
Parallel-plate capacitor: C = Kε₀A/d
Partly filled (slab of thickness t): C = ε₀A/[d − t(1 − 1/K)]
Energy density u = ½ E·D = ½ε₀KE²
Clausius–Mossotti: (K − 1)/(K + 2) = Nα/(3ε₀)
५ महत्त्वाचे प्रश्न
Q1. Distinguish between polar and non-polar dielectrics.
Non-polar: the centres of positive and negative charge coincide, so there is no permanent dipole moment; dipoles are only induced by a field (O₂, N₂, benzene). Polar: the centres do not coincide, so molecules have permanent dipoles that are randomly oriented without a field and partly line up in a field (H₂O, NH₃). Polar materials usually have larger K and their K decreases with rising temperature.
Q2. A parallel-plate capacitor has A = 0.02 m² and d = 2 mm. Find C with air and with mica (K = 6) filling the gap.
C₀ = ε₀A/d = 8.85 × 10⁻¹² × 0.02/(2 × 10⁻³) = 8.85 × 10⁻¹¹ F (88.5 pF). With mica: C = KC₀ = 6 × 88.5 = 531 pF.
Q3. Derive the capacitance of a parallel-plate capacitor with a dielectric slab of thickness t.
Free charge density σ on the plates. In air, E₀ = σ/ε₀; inside the slab, E = σ/(Kε₀).
Potential difference V = E₀(d − t) + Et = (σ/ε₀)[(d − t) + t/K].
C = Q/V = σA/V = ε₀A/[d − t + t/K] = ε₀A/[d − t(1 − 1/K)].
Check: t = 0 gives ε₀A/d; t = d gives Kε₀A/d.
Q4. Derive the Clausius–Mossotti relation (outline).
Each molecule feels the local field E_loc = E + P/(3ε₀) (Lorentz field inside a spherical cavity).
With N molecules per unit volume, each of dipole moment αE_loc: P = Nα E_loc = Nα[E + P/(3ε₀)].
Network theorems simplify the analysis of electrical circuits containing many sources and resistors. This topic covers Ohm's law, Kirchhoff's laws, the superposition theorem, Thevenin's and Norton's theorems, the maximum power transfer theorem and star–delta conversion.
महत्त्वाचे मुद्दे
Ohm's law: V = IR at constant temperature.
Kirchhoff's current law (KCL): the algebraic sum of currents at a junction is zero (conservation of charge).
Kirchhoff's voltage law (KVL): the algebraic sum of EMFs and potential drops round any closed loop is zero (conservation of energy).
Ideal voltage source: zero internal resistance. Ideal current source: infinite internal resistance.
Superposition theorem: in a linear network with several sources, the current in any branch is the algebraic sum of the currents due to each source acting alone (other voltage sources shorted, current sources opened, internal resistances kept).
Thevenin's theorem: any linear two-terminal network can be replaced by a voltage source V_th in series with a resistance R_th.
V_th = open-circuit voltage across the terminals. R_th = resistance seen from the terminals with sources replaced by their internal resistances.
Norton's theorem: the same network can be replaced by a current source I_N in parallel with R_N (= R_th). I_N = short-circuit current.
Thevenin and Norton forms are related by V_th = I_N R_th.
Maximum power transfer theorem: a load gets maximum power when its resistance equals the Thevenin resistance of the source network. Efficiency at that point is only 50%.
Star–delta transformation simplifies networks that are neither series nor parallel.
Reciprocity theorem: in a linear passive network, interchanging the positions of a source and an ammeter does not change the ammeter reading.
महत्त्वाच्या संज्ञा
Linear network – parameters (R, L, C) do not change with voltage or current.
Node (संधी बिंदू) – point where two or more elements meet.
Mesh / loop – closed path in a circuit.
Load – the element that receives power from the network.
सूत्रे
V = IR; series R = R₁ + R₂ + …; parallel 1/R = 1/R₁ + 1/R₂ + …
Star to delta: R_AB = R_A + R_B + R_A R_B/R_C (and similarly)
५ महत्त्वाचे प्रश्न
Q1. State Thevenin's theorem.
Any linear two-terminal network of sources and resistances can be replaced by a single voltage source V_th (the open-circuit voltage across the terminals) in series with a resistance R_th (the resistance seen from the terminals with all sources replaced by their internal resistances).
Q2. State the maximum power transfer theorem.
A resistive load draws maximum power from a linear network when the load resistance equals the Thevenin (internal) resistance of the network: R_L = R_th. Then P_max = V_th²/(4R_th).
Q3. A 12 V battery (negligible internal resistance) is in series with 4 Ω, and a 12 Ω resistor is connected across the output terminals A–B. Find the Thevenin equivalent and the current in a 6 Ω load across A–B.
Check by direct method: total R = 4 + (12 ∥ 6) = 4 + 4 = 8 Ω; I = 1.5 A; voltage across A–B = 1.5 × 4 = 6 V; load current = 6/6 = 1 A ✓.
Q4. For the circuit of Q3, find the Norton equivalent and the maximum power that can be given to a load.
Short A–B: short-circuit current I_N = 12/4 = 3 A (12 Ω is shorted).
R_N = R_th = 3 Ω. Check: I_N R_N = 9 V = V_th ✓.
Maximum power when R_L = 3 Ω.
P_max = V_th²/(4R_th) = 81/12 = 6.75 W.
Q5. Explain the superposition theorem with the steps for applying it.
Applies only to linear, bilateral networks with more than one source.
Keep one source; replace other voltage sources by short circuits and current sources by open circuits (keeping internal resistances).
Find the current in the required branch due to this one source.
Repeat for each source in turn.
Add all the currents algebraically (with directions) to get the actual current.
It cannot be used directly for power, since power depends on I².
परीक्षेसाठी टीप
For Thevenin problems, draw three small diagrams: one for V_th (load removed), one for R_th (sources killed), and the final equivalent with the load. Always check your answer with Norton's form or a direct method.
Alternating current (AC) changes its direction periodically, usually as a sine wave. This topic covers mean and rms values, the behaviour of R, L and C in AC circuits, reactance and impedance, phasor diagrams, the series LCR circuit and its resonance, the quality factor, the parallel resonant circuit, and power and power factor. The j-operator (complex number) method is also used.
महत्त्वाचे मुद्दे
AC voltage: V = V₀ sin ωt, with ω = 2πf. In India the mains is 230 V (rms), 50 Hz.
rms value I_rms = I₀/√2 is the steady DC that would give the same heating. Mean value over a half cycle = 2I₀/π.
Pure R: current and voltage in phase.
Pure L: current lags voltage by 90°; inductive reactance X_L = ωL rises with frequency.
Pure C: current leads voltage by 90°; capacitive reactance X_C = 1/(ωC) falls with frequency.
Series LCR: impedance Z = √[R² + (X_L − X_C)²]; phase angle tan φ = (X_L − X_C)/R.
Series resonance at ω₀ = 1/√(LC): X_L = X_C, Z = R (minimum), current maximum. Used in radio tuning ("acceptor circuit").
Parallel resonance (LC tank): impedance is maximum and line current minimum ("rejector circuit"); dynamic resistance L/(CR).
Quality factor Q = ω₀L/R = (1/R)√(L/C): sharpness of resonance; the voltage across L or C at resonance is Q times the supply voltage.
Average power P = V_rms I_rms cos φ. cos φ is the power factor. A pure L or C draws no average power (wattless current).
In j-notation: Z = R + j(ωL − 1/ωC), where j = √(−1).
महत्त्वाच्या संज्ञा
Reactance – opposition to AC by an inductor or capacitor (ohm); stores energy instead of using it.
Impedance (Z) – total opposition to AC in a circuit.
Admittance (Y) – 1/Z.
Bandwidth – range of frequencies between the half-power points, Δf = f₀/Q.
Phasor – rotating vector representing a sinusoidal quantity.
In a circuit with only L or only C, the phase difference between voltage and current is 90°, so cos φ = 0 and average power P = V_rms I_rms cos φ = 0. The current flows but no energy is used on average; energy is stored in one quarter-cycle and returned in the next. Such current is called wattless current.
Q2. Find the rms and peak values for the Indian mains supply.
V_rms = 230 V, so V₀ = √2 × 230 ≈ 325 V. Mean value over a half cycle = 2V₀/π = 2 × 325/3.1416 ≈ 207 V.
Q3. Derive the impedance and the resonance condition of a series LCR circuit using phasors.
Same current I in all elements. Voltages: V_R = IR in phase with I; V_L = IX_L leading by 90°; V_C = IX_C lagging by 90°.
V_L and V_C are opposite, so the net reactive voltage is I(X_L − X_C), perpendicular to V_R.
V = √[V_R² + (V_L − V_C)²] = I√[R² + (X_L − X_C)²], so Z = √[R² + (X_L − X_C)²] and tan φ = (X_L − X_C)/R.
Z is least when X_L = X_C, i.e. ωL = 1/ωC → ω₀ = 1/√(LC). At resonance Z = R, φ = 0 and I = V/R is maximum.
Q4. A series circuit has R = 10 Ω, L = 0.1 H and C = 10 μF, connected to 100 V (rms). Find f₀, the current at resonance, Q and the voltage across L at resonance.
f₀ = 1/(2π√(0.1 × 10⁻⁵)) = 1/(2π × 10⁻³) ≈ 159 Hz
I₀ = V/R = 100/10 = 10 A (rms)
ω₀ = 1000 rad s⁻¹, Q = ω₀L/R = 1000 × 0.1/10 = 10
V_L = I X_L = 10 × 100 = 1000 V, i.e. Q times the supply. Such high voltages make resonance dangerous in power circuits.
Q5. A coil of resistance 30 Ω and inductance 0.127 H is connected to 230 V, 50 Hz mains. Find the impedance, current, power factor and power.
X_L = 2π × 50 × 0.127 ≈ 39.9 Ω ≈ 40 Ω
Z = √(30² + 40²) = 50 Ω
I = 230/50 = 4.6 A
cos φ = R/Z = 30/50 = 0.6 (current lags by 53°)
P = V I cos φ = 230 × 4.6 × 0.6 ≈ 635 W (check: I²R = 4.6² × 30 = 634.8 W ✔)
परीक्षेसाठी टीप
Draw the phasor diagram for every LCR question; it gives Z and φ in one step. Remember the series circuit is an acceptor (minimum Z) and the parallel circuit a rejector (maximum Z).
Magnetostatics deals with magnetic fields produced by steady currents. The Biot–Savart law gives the field of a small current element, and Ampère's circuital law gives fields of symmetric current distributions easily. This topic also covers the force on currents and charges in a magnetic field.
महत्त्वाचे मुद्दे
A steady current produces a magnetic field around it (Oersted, 1820).
Biot–Savart law: the field dB due to a current element I dl at distance r is proportional to I dl sin θ/r².
Direction of B round a straight wire: right-hand thumb rule.
Ampère's circuital law: the line integral of B round any closed path equals μ₀ times the current enclosed.
Ampère's law is useful for long straight wires, solenoids and toroids, where B is constant along the chosen path.
Inside a long solenoid the field is uniform and along its axis; outside it is nearly zero.
Lorentz force: a charge q moving with velocity v in fields E and B feels F = q(E + v × B). The magnetic part does no work.
A current-carrying conductor in a magnetic field feels F = I(l × B).
Two parallel wires with currents in the same direction attract; in opposite directions they repel. This defines the ampere.
Differential forms: ∇·B = 0 (no magnetic monopoles); ∇ × B = μ₀J.
Magnetic vector potential A: B = ∇ × A.
A current loop behaves like a magnetic dipole with moment m = NIA.
महत्त्वाच्या संज्ञा
Permeability of free space μ₀ = 4π × 10⁻⁷ T m A⁻¹ (H/m).
Magnetic field (चुंबकीय क्षेत्र) B – unit tesla (T); 1 T = 10⁴ gauss.
Solenoid – long coil of many turns; toroid – solenoid bent into a ring.
Magnetic dipole moment – NIA, unit A m².
सूत्रे
Biot–Savart: dB = (μ₀/4π) I dl sin θ/r²
Long straight wire: B = μ₀I/(2πr)
Centre of circular coil (N turns, radius R): B = μ₀NI/(2R)
On the axis of a coil: B = μ₀NIR²/[2(R² + x²)^(3/2)]
Ampère: ∮B·dl = μ₀I_enc
Long solenoid: B = μ₀nI (n = turns per metre); toroid: B = μ₀NI/(2πr)
Force per unit length between parallel wires: F/l = μ₀I₁I₂/(2πd)
Force on a moving charge: F = qvB sin θ; radius of circular path r = mv/(qB)
५ महत्त्वाचे प्रश्न
Q1. State Ampère's circuital law.
The line integral of the magnetic field B around any closed path equals μ₀ times the net current passing through the area bounded by the path: ∮B·dl = μ₀I_enc.
Q2. Find the magnetic field 5 cm from a long straight wire carrying 10 A.
Learn the four standard fields (straight wire, coil centre, coil axis, solenoid) with their figures. In Ampère's law derivations, always show which parts of the loop contribute zero and why.
A changing magnetic flux through a circuit produces an emf in it. This is electromagnetic induction, discovered by Faraday in 1831. This topic covers Faraday's and Lenz's laws, motional emf, the integral and differential forms of Faraday's law, self and mutual inductance, energy stored in a magnetic field, eddy currents, and growth and decay of current in an LR circuit.
महत्त्वाचे मुद्दे
Magnetic flux Φ = ∫B·dS (weber, Wb). Flux changes if B, the area or the angle changes.
Lenz's law: the induced current flows so as to oppose the change that causes it. The minus sign in Faraday's law expresses this; it follows from conservation of energy.
Motional emf: a rod of length l moving with speed v perpendicular to B has emf e = Blv.
Integral form: ∮E·dl = −(d/dt)∫B·dS. Using Stokes' theorem, differential form: ∇×E = −∂B/∂t. A changing B creates a non-conservative (curly) electric field.
Self-inductance L: a changing current in a coil induces emf in the same coil, e = −L dI/dt. Unit: henry (H).
Mutual inductance M: a changing current in one coil induces emf in a nearby coil, e₂ = −M dI₁/dt. M₁₂ = M₂₁ (reciprocity).
Energy stored in an inductor U = ½LI²; energy density in a magnetic field u = B²/(2μ₀).
Eddy currents are loops of current induced in bulk conductors. Uses: induction furnace, electromagnetic braking, damping in galvanometers, speedometers. Losses are cut by laminating transformer cores.
In an LR circuit, current grows as I = I₀(1 − e^(−t/τ)) and decays as I = I₀e^(−t/τ), with time constant τ = L/R.
महत्त्वाच्या संज्ञा
Henry – inductance of a coil in which 1 A s⁻¹ change of current induces 1 V.
Coefficient of coupling (k) – M = k√(L₁L₂), 0 ≤ k ≤ 1.
Time constant (τ = L/R) – time for current to reach 63% of its final value while growing.
Back emf – self-induced emf opposing the change of current.
सूत्रे
e = −N dΦ/dt; motional emf e = Blv
∮E·dl = −dΦ/dt; ∇×E = −∂B/∂t
Long solenoid: L = μ₀N²A/l = μ₀n²Al
Two coaxial solenoids: M = μ₀N₁N₂A/l
Energy U = ½LI²; u = B²/2μ₀
Series inductors with coupling: L = L₁ + L₂ ± 2M
LR growth: I = (E/R)(1 − e^(−Rt/L)); decay: I = I₀e^(−Rt/L)
५ महत्त्वाचे प्रश्न
Q1. State Faraday's laws and Lenz's law.
Faraday: (1) whenever the magnetic flux linked with a circuit changes, an emf is induced in it; (2) the emf equals the rate of change of flux linkage, e = −N dΦ/dt. Lenz: the induced current always opposes the change of flux that produces it. If a magnet's N-pole approaches a coil, the near face of the coil becomes an N-pole and repels it, so work must be done — this work appears as electrical energy.
Q2. A rod 0.5 m long moves at 4 m s⁻¹ perpendicular to a field of 0.2 T. Find the emf. If the circuit resistance is 2 Ω, find the current and the force needed to keep it moving.
e = Blv = 0.2 × 0.5 × 4 = 0.4 V; I = 0.4/2 = 0.2 A; F = BIl = 0.2 × 0.2 × 0.5 = 0.02 N. Check: mechanical power Fv = 0.08 W = electrical power I²R = 0.04 × 2 = 0.08 W ✔.
Q3. Derive the self-inductance of a long solenoid and the energy stored in it.
Field inside: B = μ₀nI, where n = N/l.
Flux linkage: NΦ = N × B × A = N μ₀(N/l)I A.
L = NΦ/I = μ₀N²A/l.
Work done against the back emf while the current rises from 0 to I: W = ∫ L I dI = ½LI².
Put L and I = Bl/(μ₀N): U = ½(μ₀N²A/l)(Bl/μ₀N)² = (B²/2μ₀)(Al). Dividing by volume Al gives u = B²/(2μ₀).
Q4. A solenoid has 1000 turns, length 50 cm and cross-section 10 cm². Find L and the energy stored when it carries 2 A.
L = μ₀N²A/l = 4π × 10⁻⁷ × 10⁶ × 10⁻³/0.5 = 4π × 10⁻⁴/0.5 ≈ 2.51 × 10⁻³ H (2.51 mH). U = ½LI² = ½ × 2.51 × 10⁻³ × 4 ≈ 5.0 × 10⁻³ J.
Q5. Derive the differential form of Faraday's law and explain its meaning.
Integral form: ∮_C E·dl = −(d/dt)∫_S B·dS.
Stokes' theorem on the left: ∮E·dl = ∫(∇×E)·dS.
For a fixed surface, the time derivative moves inside: ∫(∇×E)·dS = −∫(∂B/∂t)·dS.
This holds for every surface, so ∇×E = −∂B/∂t.
Meaning: a time-varying magnetic field produces an electric field whose lines form closed loops. Unlike the electrostatic field, this field has non-zero curl and cannot be written as −∇V alone. It is one of Maxwell's four equations.
परीक्षेसाठी टीप
Lenz's law questions need a clear statement of which way the induced current flows and why (opposing the change, not the field). In numericals, check your answer with energy conservation where possible.
Maxwell brought together the laws of electricity and magnetism into four equations and added the displacement current. From these equations he predicted electromagnetic (EM) waves travelling at the speed of light, showing that light itself is an EM wave. This topic covers displacement current, the four Maxwell equations in differential and integral form, the wave equation, properties of EM waves, the Poynting vector and the EM spectrum.
महत्त्वाचे मुद्दे
Ampere's law ∇×B = μ₀J fails for changing fields: taking the divergence gives ∇·J = 0, which contradicts the continuity equation when charge density changes (for example in a charging capacitor).
Maxwell added the displacement current density J_d = ε₀ ∂E/∂t. It is a changing electric field, not a flow of charge, but it produces a magnetic field just like a real current.
∇·E = ρ/ε₀ (Gauss's law)
∇·B = 0 (no magnetic monopoles)
∇×E = −∂B/∂t (Faraday's law)
∇×B = μ₀J + μ₀ε₀ ∂E/∂t (Ampere–Maxwell law)
The four Maxwell equations (free space with charges and currents):
In empty space (ρ = 0, J = 0) the equations give wave equations for E and B with speed c = 1/√(μ₀ε₀) = 3 × 10⁸ m s⁻¹.
EM waves are transverse: E and B are perpendicular to each other and to the direction of travel.
E and B are in phase, and E₀ = cB₀.
EM waves need no medium and carry energy and momentum. The energy flow per unit area per second is given by the Poynting vector S = (1/μ₀)(E × B).
The EM spectrum, in order of increasing frequency: radio, microwaves, infrared, visible, ultraviolet, X-rays, gamma rays. All travel at c in vacuum.
Hertz produced and detected EM waves in 1887, confirming Maxwell's prediction.
महत्त्वाच्या संज्ञा
Displacement current – term ε₀ dΦ_E/dt that completes Ampere's law.
Poynting vector – vector giving the direction and rate of energy flow in an EM wave (W m⁻²).
Intensity – time-average of |S|, I = ½cε₀E₀².
Radiation pressure – pressure exerted by EM waves, I/c for full absorption.
Impedance of free space – E₀/H₀ = √(μ₀/ε₀) ≈ 377 Ω.
Q1. What is displacement current? Why did Maxwell introduce it?
It is the term ε₀ dΦ_E/dt (or density ε₀∂E/∂t) added to the conduction current in Ampere's law. Between the plates of a charging capacitor there is no conduction current, but a magnetic field is still observed; the changing electric field there acts as a current. Without it, Ampere's law would contradict the conservation of charge (continuity equation).
Q2. Find the speed of EM waves in vacuum from μ₀ and ε₀.
c = 1/√(μ₀ε₀) = 1/√(4π × 10⁻⁷ × 8.854 × 10⁻¹²) = 1/√(1.113 × 10⁻¹⁷) = 1/(3.336 × 10⁻⁹) ≈ 3.0 × 10⁸ m s⁻¹ — exactly the measured speed of light.
Q3. Derive the wave equation for E in free space from Maxwell's equations.
Take the curl of ∇×E = −∂B/∂t: ∇×(∇×E) = −∂(∇×B)/∂t.
With J = 0: ∇×B = μ₀ε₀ ∂E/∂t, so the right side is −μ₀ε₀ ∂²E/∂t².
Use the identity ∇×(∇×E) = ∇(∇·E) − ∇²E, and ∇·E = 0 in free space.
So −∇²E = −μ₀ε₀ ∂²E/∂t², i.e. ∇²E = μ₀ε₀ ∂²E/∂t².
Comparing with the standard wave equation ∇²ψ = (1/v²)∂²ψ/∂t² gives v = 1/√(μ₀ε₀) = c. The same steps give the equation for B.
Q4. Show that EM waves are transverse.
For a plane wave E = E₀ e^(i(kx − ωt)) travelling along x, ∇·E = 0 gives ik E₀x = 0, so E₀x = 0: E has no component along the direction of travel. Similarly ∇·B = 0 gives B₀x = 0. From ∇×E = −∂B/∂t, k × E₀ = ωB₀, so B is perpendicular to both k and E, with B₀ = E₀/c.
Q5. Sunlight reaching the Earth has intensity 1400 W m⁻². Find the peak electric and magnetic fields.
Write all four Maxwell equations with their names in both forms — it is the most common 5-mark question. Mention that the wave speed comes out equal to c: this was the evidence that light is an EM wave.
All materials respond to a magnetic field, but in different ways. Based on this response they are classified as diamagnetic, paramagnetic and ferromagnetic (with antiferromagnetic and ferrimagnetic as special cases). This topic covers B, H and M, susceptibility and permeability, Curie's law, domain theory and the hysteresis loop.
महत्त्वाचे मुद्दे
Magnetising field H: field due to free (conduction) currents, unit A/m.
Magnetisation M: magnetic moment per unit volume of the material, unit A/m.
Diamagnetic (bismuth, copper, water, gold): atoms have no permanent moment; induced moment opposes the field; χ small and negative; independent of temperature; weakly repelled. A superconductor is a perfect diamagnet (χ = −1).
Paramagnetic (aluminium, platinum, oxygen): atoms have permanent moments, randomly oriented; they partly line up in a field; χ small and positive; weakly attracted.
Curie's law for paramagnets: χ = C/T.
Ferromagnetic (iron, cobalt, nickel): χ large and positive; strongly attracted; shows hysteresis.
Domain theory: a ferromagnet contains small regions (domains) that are fully magnetised. In an unmagnetised sample domains point randomly. A field grows favourable domains (wall movement) and rotates them.
Above the Curie temperature a ferromagnet becomes paramagnetic, obeying the Curie–Weiss law χ = C/(T − θ). For iron the Curie temperature is about 770 °C.
Antiferromagnetic (MnO): neighbouring moments are equal and opposite. Ferrimagnetic (ferrites): opposite moments of unequal size give net magnetisation.
Hysteresis: B lags behind H when a ferromagnet is taken through a cycle. Area of the loop = energy lost per cycle per unit volume.
Soft iron: narrow loop, low coercivity, low loss — used in transformer cores and electromagnets. Steel/alnico: wide loop, high retentivity and coercivity — used for permanent magnets.
महत्त्वाच्या संज्ञा
Retentivity (remanence) – B left when H is reduced to zero.
Coercivity – reverse H needed to make B zero.
Magnetic domain (चुंबकीय क्षेत्रक) – small region with all atomic moments aligned.
Curie temperature – temperature above which ferromagnetism disappears.
सूत्रे
B = μ₀(H + M) = μ₀μᵣH
χ = M/H; μᵣ = 1 + χ
Curie's law: χ = C/T
Curie–Weiss law: χ = C/(T − θ), for T > θ
Energy loss per cycle = (area of B–H loop) per unit volume
५ महत्त्वाचे प्रश्न
Q1. Distinguish between diamagnetic and paramagnetic substances.
Diamagnetic substances have no permanent atomic moment, small negative χ independent of temperature, and are weakly repelled. Paramagnetic substances have permanent atomic moments, small positive χ that decreases with temperature (χ = C/T), and are weakly attracted.
Q2. What is the Curie temperature?
It is the temperature above which a ferromagnetic material loses its ferromagnetism and becomes paramagnetic, because thermal agitation destroys the alignment within domains. For iron it is about 770 °C.
Q3. Explain the hysteresis loop of a ferromagnetic material.
Start with an unmagnetised sample; increasing H raises B along the magnetisation curve to saturation.
Reduce H to zero: B does not fall to zero — the remaining value is the retentivity.
Apply reverse H: B becomes zero at the coercivity value.
Further reverse H gives reverse saturation; bringing H back completes a closed loop.
The area of the loop gives energy lost as heat per cycle per unit volume.
Soft iron gives a thin loop (cores); steel gives a fat loop (permanent magnets).
Q4. An iron rod with μᵣ = 500 is placed in a field H = 1000 A/m. Find B, M and χ.
Q5. The susceptibility of a paramagnetic salt is 2.8 × 10⁻⁴ at 300 K. Find it at 350 K.
Curie's law: χ ∝ 1/T, so χ₁T₁ = χ₂T₂.
χ₂ = χ₁T₁/T₂ = 2.8 × 10⁻⁴ × 300/350.
χ₂ = 2.8 × 10⁻⁴ × 0.857 ≈ 2.4 × 10⁻⁴.
Susceptibility falls as temperature rises.
परीक्षेसाठी टीप
Make a comparison table of dia-, para- and ferromagnetism (sign of χ, size, temperature dependence, examples) — it fits many short and long questions. Draw the hysteresis loop with retentivity and coercivity labelled.
A p-n junction diode allows current mainly in one direction. This one-way property is used to convert AC into DC (rectification). This topic covers the formation of a p-n junction, its characteristics, the Zener diode as a voltage regulator, half-wave and full-wave rectifiers, and filter circuits.
महत्त्वाचे मुद्दे
When p-type and n-type regions are joined, electrons and holes diffuse across and recombine, leaving a depletion layer of fixed ions with an internal field.
This creates a barrier potential: about 0.7 V for silicon and 0.3 V for germanium.
Forward bias (p to +, n to −) lowers the barrier; current rises sharply after the knee (cut-in) voltage.
Reverse bias widens the depletion layer; only a tiny reverse saturation current flows due to minority carriers.
At a large reverse voltage, breakdown occurs: Zener breakdown (heavily doped, strong field pulls electrons from bonds, low voltage) or avalanche breakdown (lightly doped, carriers ionise atoms by collision, higher voltage).
A Zener diode works in reverse breakdown, keeping its voltage nearly constant — used as a voltage regulator.
Half-wave rectifier: one diode; conducts only in one half-cycle.
Full-wave centre-tap rectifier: two diodes and a centre-tapped transformer; both halves used.
Bridge rectifier: four diodes, no centre tap needed; lower PIV per diode than centre-tap.
Ripple: AC part left in the rectified output. Filters (capacitor, choke, π-filter) smooth it.
Special diodes: LED (light emitting), photodiode (light sensing, reverse biased), solar cell, varactor.
महत्त्वाच्या संज्ञा
Knee voltage – forward voltage above which current increases rapidly.
PIV (peak inverse voltage) – maximum reverse voltage a diode faces in a circuit.
Ripple factor (γ) – rms value of AC component ÷ DC value.
Rectifier (दिष्टकारी) – circuit converting AC to DC.
सूत्रे
Diode equation: I = I₀(e^(V/ηV_T) − 1), V_T = kT/e ≈ 26 mV at room temperature
It is the region near the p-n junction that has no free charge carriers, only fixed positive ions (n side) and negative ions (p side). It forms because carriers diffuse across and recombine, and it creates the barrier potential.
Q2. Distinguish between Zener and avalanche breakdown.
Zener breakdown occurs in heavily doped junctions at low reverse voltage (below about 5–6 V); the strong field breaks covalent bonds directly. Avalanche breakdown occurs in lightly doped junctions at higher voltage, where accelerated carriers knock out more carriers by collision.
Q3. Derive the efficiency of a half-wave rectifier.
Output current i = I_m sin ωt for 0 to π, zero for π to 2π.
I_dc = (1/2π) ∫₀^π I_m sin θ dθ = I_m/π.
I_rms = √[(1/2π) ∫₀^π I_m² sin²θ dθ] = I_m/2.
DC output power = I_dc² R_L = I_m²R_L/π². AC input power = I_rms²(R_L + r_f) = (I_m²/4)(R_L + r_f).
Q5. A Zener regulator has V_in = 15 V, V_z = 10 V, R_s = 100 Ω and R_L = 1 kΩ. Find the currents.
Current through R_s: I_s = (15 − 10)/100 = 0.05 A = 50 mA.
Load current: I_L = 10/1000 = 0.01 A = 10 mA.
Zener current: I_z = I_s − I_L = 50 − 10 = 40 mA.
If the load changes, I_z adjusts so the output stays at 10 V.
परीक्षेसाठी टीप
Draw rectifier circuits with input and output waveforms; examiners give most marks for neat diagrams. Memorise the comparison table: V_dc, efficiency, ripple factor and PIV for half-wave, centre-tap and bridge.
Optoelectronic and Special Diodes: LED, Photodiode, Solar Cell, Varactor
थोडक्यात सारांश
Besides ordinary rectifier diodes, many p-n junction devices are designed for special jobs: giving out light, sensing light, turning sunlight into electricity, or acting as a voltage-controlled capacitor. This topic covers the LED, photodiode, solar cell, varactor diode and tunnel diode — their principle, construction, characteristics and uses. (The Zener diode and rectifiers are covered in the previous topic.)
महत्त्वाचे मुद्दे
LED (light emitting diode): a forward-biased p-n junction of a direct band-gap semiconductor (GaAs, GaAsP, GaN). Electrons and holes recombine at the junction and the energy comes out as light (electroluminescence).
The colour of an LED depends on the band gap: λ = hc/Eg. GaAs (1.43 eV) gives infrared; GaAsP gives red to yellow; GaN gives blue.
Silicon and germanium have an indirect band gap, so they give heat, not light, on recombination.
Photodiode: works in reverse bias. Light falling on the junction creates electron–hole pairs, increasing the reverse current in proportion to light intensity. Fast response; used in light meters, optical-fibre receivers, smoke detectors.
Solar cell: an unbiased p-n junction with a large area. Light creates electron–hole pairs; the junction field separates them, giving an open-circuit voltage (about 0.5–0.6 V for Si) and a short-circuit current. It works in the fourth quadrant of the I–V graph.
Solar cell figures of merit: fill factor FF = (V_m I_m)/(V_oc I_sc) and efficiency η = (V_m I_m)/(P_in).
Varactor (varicap): a reverse-biased diode used as a variable capacitor. Junction capacitance falls as reverse voltage rises because the depletion layer widens. Used in TV/radio tuners and FM modulators.
Tunnel diode: very heavily doped junction with a very thin depletion layer. Electrons tunnel through it, giving a negative resistance region. Used in high-frequency oscillators.
महत्त्वाच्या संज्ञा
Electroluminescence – emission of light when current passes through a material.
Direct band gap – conduction band minimum and valence band maximum at the same momentum; good for light emission.
Dark current – small reverse current of a photodiode with no light.
Open-circuit voltage (V_oc) – voltage across a solar cell when no current is drawn.
Negative resistance – region where current falls as voltage rises.
सूत्रे
LED wavelength λ = hc/Eg; λ (nm) ≈ 1240/Eg (eV)
Photodiode current I = I_dark + I_photo, with I_photo ∝ light intensity
Solar cell: FF = V_m I_m/(V_oc I_sc); η = V_m I_m/P_in × 100%
Junction (depletion) capacitance C = εA/W; for an abrupt junction C ∝ 1/√(V_R + V_bi)
५ महत्त्वाचे प्रश्न
Q1. Why is a photodiode operated in reverse bias?
In reverse bias the dark current is very small, so the extra current caused by light is easy to measure. The reverse current is almost proportional to light intensity, and the wide depletion layer collects the light-generated carriers quickly. In forward bias the large diffusion current would swamp the small photocurrent.
Q2. An LED is made of a material with band gap 1.9 eV. Find the wavelength and colour of the light.
λ = 1240/Eg = 1240/1.9 ≈ 653 nm, which is red light.
Q3. Explain the construction and working of a solar cell with its I–V characteristic.
Construction: a thin n-layer on a p-type silicon wafer, with a large junction area facing the light; metal fingers on top and a full metal contact at the back; an anti-reflection coating.
Working: photons with energy greater than Eg create electron–hole pairs near the junction. The built-in field pushes electrons to the n-side and holes to the p-side, so the n-side becomes negative and the p-side positive: an emf appears.
Characteristic: drawn in the fourth quadrant, current I_sc at V = 0 and voltage V_oc at I = 0. The maximum power point (V_m, I_m) is the corner of the largest rectangle under the curve.
Uses: satellites, solar panels for homes, calculators, street lights.
Q4. A solar cell has V_oc = 0.6 V, I_sc = 3 A, V_m = 0.5 V and I_m = 2.8 A. Light of 1000 W m⁻² falls on its area of 100 cm². Find the fill factor and efficiency.
FF = (0.5 × 2.8)/(0.6 × 3) = 1.4/1.8 ≈ 0.78
P_in = 1000 × 100 × 10⁻⁴ = 10 W
η = 1.4/10 × 100 = 14%
Q5. What is a varactor diode? Explain how its capacitance changes.
A varactor is a p-n junction diode designed to be used as a voltage-controlled capacitor in reverse bias. The depletion layer acts as the dielectric and the p and n regions as plates. Increasing the reverse voltage widens the depletion layer W, so C = εA/W falls. For an abrupt junction, C ∝ (V_R + V_bi)^(−½). Uses: electronic tuning in radios and TVs, frequency multipliers, automatic frequency control.
परीक्षेसाठी टीप
Learn λ (nm) = 1240/Eg (eV) — it makes LED numericals one-liners. Draw the I–V graphs: LED in the first quadrant, photodiode in the third, solar cell in the fourth.