⚛ Physics · Electromagnetism 🔥 Day 1

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Electromagnetism — Physics

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Electrostatics

Coulomb's law · Electric field · Potential · Gauss's law · Capacitors

Electric Charge and Coulomb's Law

Electric charge is a fundamental property of matter. Two types: positive (+) and negative (−). Like charges repel; unlike attract. Charge is quantised $(q = ne,\ e = 1.6\times10^{-19}\text{ C})$ and conserved.

Coulomb's law: Force between two point charges q₁ and q₂ separated by distance r in vacuum:

$$ F = \frac{kq_{1}q_{2}}{r^{2}} = \frac{q_{1}q_{2}}{4\pi\epsilon_{0}r^{2}} $$

k = 9×10⁹ N·m²/C²; ε₀ = 8.85×10⁻¹² C²/N·m²
In medium of dielectric constant K: $F = kq_1 \dfrac{q_2}{Kr^2}$ — force decreases
Principle of superposition: net force = vector sum of individual forces

Electric Field

Electric field E at a point is the force per unit positive test charge: $E = \dfrac{F}{q_{0}}$

Source	Electric Field	Direction
Point charge q	$E = \dfrac{kq}{r^{2}}$	Radially outward (q>0) / inward (q<0)
Infinite line charge (λ C/m)	$E = \dfrac{\lambda}{2\pi\epsilon_{0}r}$	Perpendicular to wire, outward
Infinite sheet (σ C/m²)	$E = \dfrac{\sigma}{2\epsilon_{0}}$	Perpendicular to sheet, uniform
Inside conductor	E = 0	—
At conductor surface	$E = \dfrac{\sigma}{\epsilon_{0}}$	Normal to surface

Gauss's Law

The total electric flux through any closed surface (Gaussian surface) equals the net charge enclosed divided by ε₀:

$$ \Phi_E = \oint E\cdot dA = \frac{Q_{enc}}{\epsilon_{0}} $$

Most powerful for symmetric charge distributions (spherical, cylindrical, planar)
Shell theorem: uniformly charged shell — field inside $= 0$, outside $= \dfrac{kQ}{r^2}$ (as if all charge at centre)
Uniformly charged solid sphere: inside $E = \dfrac{kQr}{R^3} \propto r$; outside $E = \dfrac{kQ}{r^2} \propto \dfrac{1}{r^2}$

Electric Potential and Potential Energy

Potential V = Work done by external agent per unit positive charge to bring it from ∞ to that point
Point charge: V = kq/r; Shell/sphere surface: V = kQ/R
Relation: E = −dV/dr (field points from high V to low V)
Potential energy of system of two charges: $U = \dfrac{kq_{1}q_{2}}{r}$
Work to move charge q through potential difference ΔV: W = qΔV
Equipotential surfaces are always perpendicular to field lines

Capacitors and Dielectrics

Configuration	Capacitance C
Parallel plate (vacuum)	$C = \dfrac{\epsilon_{0}A}{d}$
Parallel plate (dielectric K)	$C = K\dfrac{\epsilon_{0}A}{d}$
Spherical (radius R)	$C = 4\pi\epsilon_{0}R$
Series combination	$\dfrac{1}{C_{eq}} = \dfrac{1}{C_{1}} + \dfrac{1}{C_{2}} + \ldots$
Parallel combination	$C_{eq} = C_{1} + C_{2} + \ldots$

Energy stored: $U = \tfrac{1}{2}CV^{2} = \dfrac{Q^{2}}{2C} = \dfrac{QV}{2}$
Energy density in field: $u = \tfrac{1}{2}\epsilon_{0}E^{2}$
On inserting dielectric (battery disconnected): V decreases, C increases, U decreases
On inserting dielectric (battery connected): Q increases, C increases, U increases

💡 JEE Tip: For capacitor problems with battery — voltage is fixed. For problems without battery (isolated) — charge is fixed. These give opposite results for U when dielectric is inserted. Always identify the constraint first.

⚠ Common Error: Electric field inside a conductor is zero — but potential is NOT zero (it equals the surface potential). Don't confuse E=0 with V=0. The conductor is an equipotential body.

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Electrostatics — Formula Sheet

Coulomb's Law & Electric Field

$$ F = \frac{kq_{1}q_{2}}{r^{2}} = \frac{q_{1}q_{2}}{4\pi\epsilon_{0}r^{2}} $$

Coulomb's law; $k = 9\times 10^9$ N·m²/C², $\varepsilon_0 = 8.85\times 10^{-12}$ C²/N·m²

$E = \dfrac{kq}{r^2}$ (point charge)

Electric field magnitude at distance r from point charge q

$E = \dfrac{\lambda}{2\pi\varepsilon_0 r}$ (infinite line)

Field from infinite line charge of linear charge density λ at distance r

$E = \dfrac{\sigma}{2\varepsilon_0}$ (infinite sheet)

Uniform field from infinite plane of surface charge density σ (each side)

Gauss's Law

$$ \oint E\cdot dA = \frac{Q_{enc}}{\epsilon_{0}} $$

Total electric flux through closed surface = enclosed charge / $\varepsilon_0$

$E_{\text{inside sphere}} = \dfrac{kQr}{R^3}$ ($r < R$)

Field inside uniformly charged solid sphere — increases linearly with r

Electric Potential

V = kq/r (point charge)

Potential at distance r from point charge; V = 0 at infinity

E = −dV/dr

Relation between field and potential — field points in direction of decreasing V

$$ U = \frac{kq_{1}q_{2}}{r} $$

Potential energy of two-charge system

$$ W = q(V_{1} - V_{2}) = q\Delta V $$

Work done in moving charge q from potential V₁ to V₂

Capacitors

$C = \dfrac{Q}{V} = \varepsilon_0 \dfrac{A}{d}$ (parallel plate)

Capacitance definition and formula for parallel plate capacitor

$$ U = \tfrac{1}{2}CV^{2} = \frac{Q^{2}}{2C} = \tfrac{1}{2}QV $$

Energy stored in a capacitor — all three forms are equivalent

$u = \tfrac{1}{2}\varepsilon_0 E^2$ (energy density)

Electric field energy stored per unit volume in any region

$C_{series}: \dfrac{1}{C} = \Sigma\, \dfrac{1}{C_i}$ | $C_{parallel}: C = \Sigma C_i$

Capacitor combination rules

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Electrostatics — Solved Examples

Coulomb's Law — Equilibrium of Charges: Three charges $q_1 = +4$ μC, $q_2 = +9$ μC, and $q_3$ are placed on the x-axis. $q_1$ is at $x = 0$, $q_2$ is at $x = 30$ cm. Where should $q_3$ be placed (on the x-axis) so that it experiences zero net force? What is the sign of $q_3$?

For zero net force on q₃, forces from q₁ and q₂ must be equal and opposite. This is possible only between them (for like-sign q₃). Let q₃ be at distance x from q₁ and (30−x) from q₂.
$$ |F_{1}| = |F_{2}|: k|q_{1}||q_{3}|/x^{2} = k|q_{2}||q_{3}|/(30-x)^{2} $$
$$ \frac{4}{x^{2}} = \frac{9}{(30-x)^{2}} \to (30-x)\frac{^{2}}{x^{2}} = \frac{9}{4} \to \frac{30-x}{x} = \frac{3}{2} \to 60-2x = 3x \to x = 12 cm $$
q₃ can be any sign — the net force on q₃ is zero at x=12 cm regardless. But for q₁ and q₂ to also be in equilibrium, q₃ must be negative.

✓ Answer: $q_3$ at $x = 12$ cm from $q_1$ (i.e., 18 cm from $q_2$). For full system equilibrium, $q_3$ must be negative. $q_3 = -(4\times \dfrac{9}{25})$ μC $= -1.44$ μC.

Gauss's Law — Field of Hollow Sphere: A hollow sphere of radius R carries total charge Q uniformly distributed on its surface. Find the electric field at (a) r = R/2 (inside), (b) r = R (surface), (c) r = 2R (outside). Also find the potential at r = R/2.

(a) Inside (r = R/2): Gaussian surface encloses zero charge. By Gauss's law: $E \times 4\pi\frac{R}{2}^{2} = 0 \to$ E = 0
(b) At surface (r = R): Encloses full charge Q. E × 4πR² = Q/ε₀ → $E = \dfrac{kQ}{R^{2}} = \dfrac{Q}{4\pi\epsilon_{0}R^{2}}$
(c) Outside (r = 2R): $E = \dfrac{kQ}{(2R)^{2}} =$ kQ/4R² — behaves like a point charge at centre
Potential at r = R/2 (inside): Since E=0 inside, V is constant and equals the surface potential: V = kQ/R

✓ Key insight: E=0 inside shell but V = kQ/R ≠ 0. The potential is uniform inside because no work is done moving a charge — this doesn't mean V=0!

Capacitor Network with Dielectric: Three capacitors $C_1 = 2$ μF, $C_2 = 3$ μF, $C_3 = 6$ μF are connected in series across a 12 V battery. (a) Find the equivalent capacitance. (b) Find the charge on each capacitor and the voltage across $C_2$. (c) If the battery is then disconnected and a dielectric of $K = 3$ is inserted into $C_1$, find the new voltage across $C_1$.

(a) Series equivalent: $\dfrac{1}{C_{eq}} = \dfrac{1}{2} + \dfrac{1}{3} + \dfrac{1}{6} = \dfrac{3}{6} + \dfrac{2}{6} + \dfrac{1}{6} = \dfrac{6}{6} = 1$ → $C_{eq} =$ 1 μF
(b) Series — same charge on all: $Q = C_{eq} \times V = 1 \times 12 =$ 12 μC on each capacitor
Voltage across $C_2$: $V_2 = \dfrac{Q}{C_2} = \dfrac{12}{3} =$ 4 V. (Check: $V_1 = 6$ V, $V_2 = 4$ V, $V_3 = 2$ V → total $= 12$ V ✓)
(c) Battery disconnected — total charge conserved. New $C'_1 = K C_1 = 3\times 2 = 6$ μF. New series $C'_{eq} = \dfrac{1}{1/6 + 1/3 + 1/6} = \dfrac{1}{1/6 + 2/6 + 1/6} = \dfrac{1}{4/6} = \dfrac{6}{4} = 1.5$ μF
Total charge $Q_{total} = 12$ μC (conserved). New $V_{total} = \dfrac{Q}{C'_{eq}} = \dfrac{12}{1.5} = 8$ V. New $V_1 = \dfrac{Q}{C'_1} = \dfrac{12}{6} =$ 2 V

✓ Answer: (a) 1 μF; (b) $Q = 12$ μC, $V_2 = 4$ V; (c) $V_1 = 2$ V. Inserting dielectric reduces voltage across that capacitor — charge redistributes while total is conserved.

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Electrostatics — Visual Concepts

Click any diagram to zoom

Electric Field Lines — Point Charges — Electric field lines for a positive charge (radiate outward) and negative charge (converge inward). Density of lines indicates field strength. Lines always exit positive charges and enter negative charges. In three dimensions the same lines cross every sphere drawn round the charge, and a sphere's area grows as $4\pi r^2$, so the lines per unit area — the field — fall as $1/r^2$. That is Gauss's law giving Coulomb's law.

Electric Field vs Distance — Uniformly Charged Solid Sphere — E-r graph for uniformly charged solid sphere: rises linearly from 0 to maximum at surface (r=R), then falls as 1/r² outside. For a hollow shell, E=0 for all r<R and kQ/r² outside.

Parallel Plate Capacitor — With and Without Dielectric — Inserting a dielectric reduces the electric field by factor K (dipole alignment opposes external field). Capacitance increases K-fold. If the battery stays connected, charge increases to keep V constant; if disconnected, V decreases.

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Electrostatics — Quick Quiz

Question 1

A point charge +Q is placed at the centre of a hollow spherical shell of radius R carrying charge −2Q. The electric field just outside the shell (at r = R⁺) is:

By Gauss's law, the enclosed charge at $r = R^+$ is $Q + (-2Q) = -Q$. $E \times 4\pi R^2 = -\dfrac{Q}{\varepsilon_0}$ → $E = \dfrac{kQ}{R^2}$ directed inward (toward centre, because net charge is negative). Answer: C ✓

Question 2

Two capacitors of 4 μF and 6 μF are connected in series and then connected to a 100 V supply. The energy stored in the 6 μF capacitor is:

Series: $C_{eq} = \dfrac{4\times 6}{4+6} = 2.4$ μF. $Q = C_{eq}\times V = 2.4\times 100 = 240$ μC (same for both in series). Voltage across 6 μF: $V_6 = \dfrac{Q}{C} = \dfrac{240}{6} = 40$ V. Energy stored: $U = \tfrac{1}{2}CV^2 = \tfrac{1}{2}\times 6\times 10^{-6}\times 1600 =$ 4.8 mJ. Answer: C for the corrected 60 V scenario. JEE problems require careful reading — check the supply voltage carefully.

Question 3

The electric potential at the surface of a charged conducting sphere of radius 10 cm is 300 V. The electric field at the surface is:

For a conducting sphere, $V = \dfrac{kQ}{R}$ and $E = \dfrac{kQ}{R^2}$. Therefore $E = \dfrac{V}{R} = \dfrac{300}{0.10} =$ 3000 V/m. Answer: A ✓. This elegant relation $E = \dfrac{V}{R}$ holds only at the surface of a conducting sphere.

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Current Electricity

Ohm's law · Kirchhoff's laws · Wheatstone bridge · RC circuits

Electric Current and Drift Velocity

Current I = dQ/dt — rate of flow of charge. Conventional current flows from + to −; electrons flow opposite. In a conductor, free electrons drift slowly under E field.

Drift velocity: $v_d = eE\dfrac{\tau}{m} = \dfrac{I}{nAe}$ — extremely small (~10⁻⁴ m/s)
$n$ = free electron density ($\approx \dfrac{10^{28}}{\text{m}^3}$ for copper), $A$ = cross-section area, $\tau$ = relaxation time
Current density: $J = \dfrac{I}{A} = n e v_d = \sigma E$ ($\sigma$ = conductivity $= \dfrac{1}{\rho}$)

Ohm's Law and Resistance

For a conductor at constant temperature: V = IR (Ohm's law — valid for ohmic conductors)

Resistance: $R = \rho L/A$ ($\rho$ = resistivity in Ω·m, $L$ = length, $A$ = area)
Temperature dependence: $\rho(T) = \rho_0[1 + \alpha(T - T_0)]$; for metals $\alpha > 0$ ($\rho$ increases with $T$); for semiconductors $\alpha < 0$
Series: $R_{eq} = R_{1} + R_{2} + \ldots$ (same current, voltages add)
Parallel: $\dfrac{1}{R_{eq}} = \dfrac{1}{R_{1}} + \dfrac{1}{R_{2}} + \ldots$ (same voltage, currents add)

EMF, Internal Resistance and Power

EMF ($\varepsilon$): work done per unit charge by the source = terminal voltage + voltage drop inside
Terminal voltage: $V = \epsilon - Ir$ (during discharge); V = ε + Ir (during charging)
Current: $I = \dfrac{\varepsilon}{R + r}$; short circuit current ($R = 0$): $I_{sc} = \dfrac{\varepsilon}{r}$
Power delivered to external R: $P = I^2 R$; max power when $R = r$ (internal resistance)
Power dissipated: $P = VI = I^{2}R = \dfrac{V^{2}}{R}$
Efficiency of cell: $\eta = \dfrac{R}{R+r}$; max power transfer theorem: $R_{ext} = r_{int}$

Kirchhoff's Laws

Law	Statement	Based on
KCL (Junction rule)	$\Sigma I_{in} = \Sigma I_{out}$ at any junction	Conservation of charge
KVL (Loop rule)	ΣV around any closed loop = 0	Conservation of energy

KVL sign convention: voltage rise (+) when traversing from − to + through a source; voltage drop (−) when traversing through a resistor in current direction
Number of independent equations: junctions give (n−1) KCL; loops give additional equations up to total branches − (n−1)

Wheatstone Bridge and Meter Bridge

Four resistors P, Q, R, S in a diamond configuration with a galvanometer across the middle diagonal:

Balanced condition: no current through galvanometer when P/Q = R/S
Meter bridge: L/(100−L) = R/S where L is the balance length
Potentiometer: null method measures EMF without drawing current — more accurate than voltmeter
Potentiometer comparison of EMFs: $\dfrac{\epsilon_{1}}{\epsilon_{2}} = L_{1}/L_{2}$

RC Circuits (Charging and Discharging)

Charging: $q(t) = Q_0(1 - e^{-t/RC})$; $I(t) = (\dfrac{\varepsilon}{R})e^{-t/RC}$
Discharging: $q(t) = Q_0 e^{-t/RC}$; $I(t) = (\dfrac{Q_0}{RC})e^{-t/RC}$
Time constant $\tau = RC$ — time to charge to 63.2% (or discharge to 36.8%) of maximum
At $t = 5\tau$, capacitor is considered fully charged/discharged (99.3%)
Energy stored at full charge: $U = \tfrac{1}{2}C\varepsilon^2$; energy from battery: $C\varepsilon^2$; heat in $R$: $\tfrac{1}{2}C\varepsilon^2$ (always half, independent of R!)

💡 JEE Tip: In an RC circuit, energy dissipated in resistance = energy stored in capacitor = $\tfrac{1}{2}C\varepsilon^2$ — this is always true regardless of resistance value. The resistance only affects how quickly charging happens, not the final energy split.

⚠ Common Error: For resistors in parallel, the equivalent resistance is always less than the smallest individual resistor. If two equal resistors R are in parallel, $R_{eq} = \dfrac{R}{2}$. Always sanity-check: adding paths must reduce resistance.

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Current Electricity — Formula Sheet

Current, Drift Velocity & Resistance

$I = nAev_d$

Current from drift velocity; $n$ = free electron density, $A$ = area, $e = 1.6\times 10^{-19}$ C

R = ρL/A

Resistance from resistivity; ρ in Ω·m, L in m, A in m²

ρ(T) = ρ₀[1 + α(T − T₀)]

Temperature dependence of resistivity; $\alpha$ = temperature coefficient (K⁻¹)

$R_{series} = R_1 + R_2 + \ldots$ | $\dfrac{1}{R_{parallel}} = \Sigma\, \dfrac{1}{R_i}$

Resistor combination rules

EMF, Power & Cells

$V_{terminal} = \varepsilon - Ir$ (discharge)

Terminal voltage during discharge; $\varepsilon$ = EMF, $r$ = internal resistance, $I$ = current

$$ I = \epsilon / (R + r) $$

Current in simple circuit with internal resistance

$$ P = VI = I^{2}R = \frac{V^{2}}{R} $$

Electrical power — all three forms equivalent via Ohm's law

$P_{max} = \dfrac{\varepsilon^2}{4r}$ (when $R = r$)

Maximum power delivered to external resistance — occurs when $R_{ext} = r_{int}$

Kirchhoff & Wheatstone

ΣI = 0 (KCL) | ΣV = 0 (KVL)

Kirchhoff's current law at junction; voltage law around any closed loop

P/Q = R/S (Wheatstone bridge balanced)

Balance condition — galvanometer reads zero, no current through it

$\dfrac{\varepsilon_1}{\varepsilon_2} = L_1/L_2$ (potentiometer)

Comparison of EMFs using potentiometer — ratio of balance lengths

RC Circuits

τ = RC

Time constant; at t=τ capacitor is 63.2% charged; fully charged at t≈5τ

$q(t) = C\varepsilon(1 - e^{-t/RC})$ (charging)

Charge on capacitor during charging from 0 to Cε

$q(t) = Q_0 e^{-t/RC}$ (discharging)

Charge decays exponentially during discharging

$U_{cap} = \tfrac{1}{2}C\varepsilon^2$ | $H_{resistor} = \tfrac{1}{2}C\varepsilon^2$

Energy stored in capacitor = heat dissipated in resistor — always equal, independent of R

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Current Electricity — Solved Examples

Internal Resistance and Terminal Voltage: A battery of EMF 12 V has internal resistance 2 Ω. It is connected to an external resistance of 10 Ω. Find: (a) the current in the circuit, (b) the terminal voltage, (c) the power delivered to the external resistance, and (d) the efficiency of the cell.

(a) Current: $I = \dfrac{\epsilon}{R + r} = \dfrac{12}{10 + 2} = \dfrac{12}{12} =$ 1 A
(b) Terminal voltage: $V = \epsilon - Ir = 12 - 1\times2 =$ 10 V (also = IR = 1×10 = 10 V ✓)
(c) Power to external R: $P_{ext} = I^{2}R = 1^{2} \times 10 =$ 10 W
(d) Efficiency: $\eta = \dfrac{P_{ext}}{P_{total}} = I^2 \dfrac{R}{I^2(R+r)} = \dfrac{R}{R+r} = \dfrac{10}{12} =$ 83.3%

✓ Summary: $I = 1$ A, $V_{terminal} = 10$ V, $P_{external} = 10$ W, η = 83.3%. Power wasted internally: $I^2 r = 1\times 2 = 2$ W. Total power from battery: 12 W = 10 W (useful) + 2 W (wasted) ✓

Kirchhoff's Laws — Multi-Loop Circuit: In a circuit, battery ε₁ = 6 V (r₁ = 1 Ω) and ε₂ = 4 V (r₂ = 1 Ω) are connected in a loop with R = 2 Ω between them. ε₁ and ε₂ oppose each other (aiding configuration). Find the current in each branch and the terminal voltage of ε₁.

Net EMF in series (opposing batteries): $\epsilon_{net} = \epsilon_{1} - \epsilon_{2} = 6 - 4 = 2 V$ (current driven by ε₁)
Total resistance: $R_{total} = r_1 + r_2 + R = 1 + 1 + 2 = 4$ Ω
Current: $I = \dfrac{\varepsilon_{net}}{R_{total}} = \dfrac{2}{4} =$ 0.5 A (flows in direction of $\varepsilon_1$)
Terminal voltage of ε₁ (discharging): $V_{1} = \epsilon_{1} - Ir_{1} = 6 - 0.5\times1 =$ 5.5 V
Terminal voltage of ε₂ (being charged): $V_{2} = \epsilon_{2} + Ir_{2} = 4 + 0.5\times1 =$ 4.5 V

✓ Verify with KVL: V₁ = Voltage drop across R + V₂ → 5.5 = 0.5×2 + 4.5 = 1 + 4.5 = 5.5 ✓

Wheatstone Bridge + RC Transient: A Wheatstone bridge has P = 100 Ω, Q = 200 Ω, R = 300 Ω, and S is unknown. (a) Find S for bridge balance. (b) If S is replaced by a capacitor C = 2 μF and the galvanometer by a 100 Ω resistor, the battery is 10 V (no internal resistance). Find the steady-state charge on the capacitor, and the time constant of the RC charging.

(a) Balance condition: $\dfrac{P}{Q} = \dfrac{R}{S} \to \dfrac{100}{200} = \dfrac{300}{S} \to S = 300 \times \dfrac{200}{100} =$ 600 Ω
(b) Steady state with C in place of S: At steady state, no current flows through C (fully charged). The galvanometer branch carries current.
In steady state, voltage across C = voltage difference between the two mid-points of the bridge. Upper divider (P, Q in series): $V_{mid1} = 10 \times \dfrac{Q}{P+Q} = 10 \times \dfrac{200}{300} = 6.67$ V
Lower divider (R, galvanometer in series): $V_{mid2} = 10 \times \dfrac{100}{300+100} = 10 \times 0.25 = 2.5$ V
Voltage across C $= V_{mid1} - V_{mid2} = 6.67 - 2.5 = 4.17$ V. Charge: $Q_C = CV = 2\times 10^{-6} \times 4.17 =$ 8.33 μC
Time constant: The capacitor sees the Thevenin resistance of the network. Remove battery (short it): P∥R in series with Q∥G: $R_{th} = (100\,\|\,300) + (200\,\|\,100) = 75 + 66.7 = 141.7$ Ω. $\tau = R_{th} \times C = 141.7 \times 2\times 10^{-6} \approx$ 283 μs

✓ Answers: (a) S = 600 Ω; (b) Q_C ≈ 8.33 μC, τ ≈ 283 μs. The Thevenin method is the cleanest approach for RC transients in complex networks.

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Current Electricity — Visual Concepts

Click any diagram to zoom

V-I Characteristics — Ohmic vs Non-Ohmic — V-I characteristics: ohmic conductors give a straight line (constant R). A filament bulb curves because R increases with temperature. A diode passes current only in one direction — near-zero in reverse bias.

Wheatstone Bridge — Circuit Diagram — Wheatstone bridge circuit. When P/Q = R/S, no current flows through galvanometer G (null deflection). This null method is extremely accurate for resistance measurement because it doesn't depend on the galvanometer calibration.

RC Circuit — Charging and Discharging Curves — RC charging (blue): charge grows as $1-e^{-t/\tau}$, asymptotically approaching Cε. RC discharging (red): charge decays as $e^{-t/\tau}$. Time constant τ = RC — larger R or C means slower charging/discharging.

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Current Electricity — Quick Quiz

Question 1

A wire of resistance 12 Ω is bent into a circle. The resistance between the two ends of a diameter is:

The circle is split into two semicircles at the diameter endpoints. Each semicircle has resistance 12/2 = 6 Ω. These two 6 Ω resistors are in parallel: $R_{eq} = \dfrac{6\times 6}{6+6} =$ 3 Ω. Answer: C ✓

Question 2

A capacitor C = 4 μF is charged to 100 V. The battery is then disconnected and a resistance R = 50 kΩ is connected across it. The time for the voltage to fall to 1/e of its initial value is:

Voltage falls to $\dfrac{V_0}{e}$ at $t = \tau = RC = 50\times 10^3 \times 4\times 10^{-6} = 200\times 10^{-3} =$ 0.2 s. Answer: B ✓. This is the definition of the time constant — the time for voltage to decay to 36.8% (= 1/e) of the initial value.

Question 3

In a potentiometer experiment, the balance length for cell A is 60 cm and for cell B is 40 cm. If the EMF of cell A is 1.5 V, the EMF of cell B is:

$\dfrac{\epsilon_{1}}{\epsilon_{2}} = L_{1}/L_{2} \to \dfrac{1.5}{\epsilon_B} = \dfrac{60}{40} = \dfrac{3}{2} \to \epsilon_B = 1.5 \times \dfrac{2}{3} =$ 1.0 V. Answer: D ✓. The potentiometer is a null deflection method — at balance, the potential drop along the wire equals the cell EMF with no current drawn from the cell.

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Magnetic Effects of Current

Biot-Savart · Ampere's law · Lorentz force · Moving charges · Magnetism in matter

Biot-Savart Law

The magnetic field contribution dB from a current element I·dl at distance r:

$$ dB = \frac{\mu_{0}}{4\pi} \cdot \frac{I\,dl\,\sin\theta}{r^{2}} $$

Configuration	Magnetic Field B	Direction
Infinite straight wire	$\mu_{0}I / (2\pi r)$	Right-hand rule around wire
Centre of circular loop (radius R)	$\mu_{0}I / (2R)$	Along axis, right-hand rule
At axis of circular loop (distance x)	$\mu_0 I R^2 / [2(R^2+x^2)^{3/2}]$	Along axis
Inside solenoid (n turns/m)	$\mu_{0}nI$	Along axis
End of solenoid	$\mu_{0}nI / 2$	Along axis

Ampere's Law

For any closed loop (Amperian loop): $\oint B\cdot dl = \mu_{0} I_{enc}$

Powerful for symmetric configurations — infinite wire, solenoid, toroid
Infinite wire: choose circle of radius r as Amperian loop → $B\cdot 2\pi r = \mu_{0} I$ → $B = \dfrac{\mu_{0}I}{2\pi r}$
Toroid (N turns, mean radius R): inside $B = \dfrac{\mu_{0}NI}{2\pi R}$; outside $B = 0$
Solenoid: $B = \mu_{0}nI$ inside; $B \approx 0$ outside (field completely confined inside)

Lorentz Force and Motion of Charged Particles

Force on charge q moving with velocity v in fields E and B:

$$ F = q(E + v \times B) $$

In uniform B only: particle moves in a circle (F always ⊥ v — no work done by B)
Radius: $r = \dfrac{mv}{qB}$; period: $T = 2\pi \dfrac{m}{qB}$ (independent of speed — cyclotron principle)
In crossed E and B (velocity selector): only particles with $v = \dfrac{E}{B}$ pass undeflected
Helical motion: if v has component along B, that component is unaffected → helix

Force on Current-Carrying Conductors

Force on straight conductor: $F = BIL\sin\theta$ (θ = angle between I and B)
Vector form: $\vec F = I(\vec L \times \vec B)$
Force between two parallel wires (separation d, currents $I_1, I_2$, length L):
$F/L = \mu_{0}I_{1}I_{2} / (2\pi d)$ — attractive if same direction; repulsive if opposite
1 Ampere is defined by $F/L = 2\times 10^{-7}$ N/m when $I_1 = I_2 = 1$ A, $d = 1$ m

Torque on a Current Loop — Galvanometer

Magnetic dipole moment: $m = NIA$ (N turns, current I, area A)
Torque in uniform field: $\vec\tau = \vec m \times \vec B \Rightarrow \tau = NIAB\sin\theta$
Potential energy: $U = -\vec m\cdot\vec B = -mB\cos\theta$; equilibrium when $\vec m \parallel \vec B$ (θ = 0)
Moving-coil galvanometer: $\tau_{magnetic} = \tau_{spring} \to NIAB = k\varphi$ → deflection $\varphi = \dfrac{NIAB}{k}$
Current sensitivity $= \dfrac{\varphi}{I} = \dfrac{NAB}{k}$; voltage sensitivity $= \dfrac{NAB}{kR_g}$

Magnetism in Matter

Type	χ_m	μ_r	Behaviour in B
Diamagnetic	Small negative (−10⁻⁵)	Slightly <1	Weakly repelled
Paramagnetic	Small positive (10⁻⁵ to 10⁻³)	Slightly >1	Weakly attracted
Ferromagnetic	Very large positive (10³)	≫1	Strongly attracted; retains magnetisation

💡 JEE Tip: Cyclotron period $T = 2\pi \dfrac{m}{qB}$ is independent of speed and radius — this is the key principle of the cyclotron. If a proton and deuteron enter the same cyclotron, the deuteron (2× mass, same charge) takes twice the period to complete a revolution.

⚠ Common Error: Magnetic force does NO work on a moving charge ($\vec F \perp \vec v$ always). Speed and kinetic energy do not change. Only the direction changes. Never write $\Delta KE$ due to magnetic force — it's always zero.

📐

Magnetic Effects — Formula Sheet

Biot-Savart & Key Field Results

$$ dB = \frac{\mu_{0}}{4\pi}\,\frac{I\,dl\,\sin\theta}{r^{2}} $$

Biot-Savart law; $\mu_0 = 4\pi\times 10^{-7}$ T·m/A (permeability of free space)

$B = \mu_{0}I / (2\pi r)$ (infinite wire)

Field at perpendicular distance r from a long straight wire

$B = \mu_{0}I / (2R)$ (centre of loop)

Field at centre of circular loop of radius R carrying current I

$B = \mu_{0}nI$ (solenoid interior)

Uniform field inside solenoid; $n = N/L$ = turns per unit length

Ampere's Law

$$ \oint B\cdot dl = \mu_{0} I_{enc} $$

Ampere's circuital law — line integral around any closed path

$B = \mu_{0}NI / (2\pi R)$ (toroid)

Field inside toroid of N total turns and mean radius R; zero outside

Lorentz Force & Particle Motion

$\vec F = q(\vec v \times \vec B)$ | $|F| = qvB\sin\theta$

Magnetic force on moving charge; θ = angle between v and B

$r = mv / (qB)$

Radius of circular motion of charged particle in uniform B field

$T = 2\pi m / (qB)$ (cyclotron period)

Period independent of speed and radius — fundamental to cyclotron operation

$v_{selector} = E / B$ (velocity selector)

Condition for straight-line motion in crossed E and B fields ($qE = qvB$)

Force on Conductors & Torque

$F = BIL\sin\theta$

Force on straight current-carrying conductor of length L in field B

$F/L = \mu_{0}I_{1}I_{2} / (2\pi d)$ (parallel wires)

Force per unit length between parallel wires; attractive for same-direction currents

$\tau = NIAB\sin\theta$ | $m = NIA$

Torque on current loop; m = magnetic dipole moment

🧮

Magnetic Effects — Solved Examples

Field at Centre of Composite Loop: A wire is bent to form a semicircle of radius R = 10 cm, with its two straight ends extended to infinity along opposite directions. A current of 2 A flows through it. Find the net magnetic field at the centre of the semicircle.

Field from the semicircular arc: Half of full loop → $B_{arc} = \tfrac{1}{2} \times \dfrac{\mu_{0}I}{2R} = \dfrac{\mu_{0}I}{4R}$
$B_{arc} = (4\pi\times10^{-7} \times 2) / (4 \times 0.10) = 8\pi\times10^{-7} / 0.40 = 20\pi\times10^{-7} =$ 6.28 μT (into page, by right-hand rule)
Field from the two infinite straight sections: Each extends from the ends of the semicircle to infinity. Both straight sections pass through the centre (their extensions meet at the centre), so the perpendicular distance = 0.
For a straight wire, $B = 0$ at a point on the wire's line of action ($\sin\theta = 0$ for all elements). So $B_{straight} = $ 0 for both sections.

✓ $Net B = \dfrac{\mu_{0}I}{4R} = 6.28 \mu T$ directed into the page. Only the curved part contributes — a very common JEE trap question.

Charged Particle in Uniform Magnetic Field: A proton (m = 1.67×10⁻²⁷ kg, q = 1.6×10⁻¹⁹ C) enters a uniform magnetic field B = 0.5 T perpendicular to the field with speed 3×10⁶ m/s. Find: (a) radius of circular orbit, (b) period of revolution, (c) kinetic energy gained after one full revolution.

(a) Radius: $r = \dfrac{mv}{qB} = (1.67\times10^{-27} \times 3\times10^{6}) / (1.6\times10^{-19} \times 0.5)$
$r = (5.01\times10^{-21}) / (8\times10^{-20}) = 0.0626 m =$ 6.26 cm
(b) Period: $T = 2\pi \dfrac{m}{qB} = (2\pi \times 1.67\times 10^{-27}) / (1.6\times 10^{-19} \times 0.5) = 10.49\times 10^{-27} / (8\times 10^{-20}) =$ $1.31\times 10^{-7}$ s ≈ 131 ns
(c) KE after one revolution: Magnetic force does NO work → speed is unchanged → $\Delta KE = 0$

✓ Answers: r ≈ 6.26 cm; T ≈ 131 ns; $\Delta KE = 0$. The zero KE gain is the most important conceptual point — magnetic force is always perpendicular to velocity, so it cannot do work.

Force Between Wires + Galvanometer Conversion: (a) Two long parallel wires carry currents 8 A and 12 A in the same direction, separated by 4 cm. Find the force per unit length between them and state whether attractive or repulsive. (b) A galvanometer has resistance 30 Ω and full-scale deflection current 2 mA. How do you convert it to: (i) an ammeter reading 0–5 A, (ii) a voltmeter reading 0–10 V?

(a) Force per unit length: $F/L = \dfrac{\mu_{0}I_{1}I_{2}}{2}\pi d = (4\pi\times10^{-7} \times 8 \times 12) / (2\pi \times 0.04)$
$F/L = (4\pi\times10^{-7} \times 96) / (0.08\pi) = (384\pi\times10^{-7}) / (0.08\pi) = 384\dfrac{\times10^{-7}}{0.08} =$ 4.8×10⁻⁴ N/m
Same direction currents → Attractive force.
(b-i) Ammeter (shunt resistor S): $I_g \times R_g = (I - I_g) \times S \to S = I_g \dfrac{R_g}{I - I_g} = \dfrac{2\times10^{-3} \times 30}{5 - 0.002} = \dfrac{0.06}{4.998} \approx$ 0.012 Ω = 12 mΩ in parallel
(b-ii) Voltmeter (series resistor $R_s$): $V = I_g(R_g + R_s) \to R_s = \dfrac{V}{I_g} - R_g = \dfrac{10}{2\times10^{-3}} - 30 = 5000 - 30 =$ 4970 Ω in series

✓ Answers: (a) $4.8\times 10^{-4}$ N/m, attractive; (b-i) shunt S ≈ 12 mΩ; (b-ii) series $R_s$ = 4970 Ω. Low resistance ammeter (≈S), high resistance voltmeter ($\approx R_s + R_g = 5000$ Ω).

📊

Magnetic Effects — Visual Concepts

Click any diagram to zoom

Magnetic Field — Straight Wire and Solenoid — Left: Circular magnetic field lines around a straight wire (right-hand rule — curl fingers in direction of B when thumb points along I). Right: Solenoid produces a uniform field inside (B = μ₀nI), negligible outside. With current out of the page along the top turns (⊙) and into it along the bottom (⊗), the right-hand rule gives B pointing right (→) inside.

Charged Particle Motion in Magnetic Field — With B into the page, F = qv × B on the positive charge at the top of its orbit points down, toward the centre, because the charge is moving left there. So a positive charge circulates anticlockwise (a negative charge would go clockwise), and the particle traces a circular orbit. Period T = 2πm/qB is independent of speed — the principle behind cyclotrons.

Torque on a Rectangular Current Loop in Uniform B — A rectangular current loop in a uniform field B experiences equal and opposite forces on its vertical sides (a couple), producing torque τ = NIABsinθ. The net force on the loop is zero; only rotation occurs. Here the loop lies in the page with B along it: the forces on the vertical sides point into (left) and out of (right) the page, never along B, so the loop turns about the vertical axis until its moment m (into the page, for this clockwise current) lines up with B. This is the basis of the moving-coil galvanometer.

⚡

Magnetic Effects — Quick Quiz

Question 1

A proton and an alpha particle (charge 2e, mass $4m_p$) enter the same uniform magnetic field with the same kinetic energy. The ratio of their radii ($r_{proton} : r_{alpha}$) is:

$r = \dfrac{mv}{qB}$. Same KE → $\tfrac{1}{2}mv^2 = KE \to v = \sqrt{\dfrac{2KE}{m}}$. So $r = m\dfrac{\sqrt{\dfrac{2KE}{m}}}{qB} = \dfrac{\sqrt{2mKE}}{qB}$. $r_p = \dfrac{\sqrt{2m_p KE}}{eB}$; $r_\alpha = \dfrac{\sqrt{2\times 4m_p\times KE}}{2eB} = 2\dfrac{\sqrt{2m_p KE}}{2eB} = \dfrac{\sqrt{2m_p KE}}{eB}$. $r_p : r_\alpha$ = 1 : 1. Answer: B ✓

Question 2

A long straight wire carries current I. A rectangular loop of width a and length b is placed with its near side at distance d from the wire, parallel to it. The net force on the loop is:

The near side (at distance d) is attracted toward the wire; the far side (at $d+a$) is repelled. Net force $= F_{near} - F_{far} = \frac{\mu_{0}I^{2}b}{2\pi}\left[\frac{1}{d} - \frac{1}{d+a}\right] = \frac{\mu_{0}I^{2}b}{2\pi}\cdot\frac{a}{d(d+a)}$. Answer: C ✓. Force is attractive (toward the wire) when loop current in near-side flows parallel to wire current.

Question 3

The magnetic field at the centre of a circular coil of radius R carrying current I is B₀. If the coil is reshaped into a square of the same perimeter carrying the same current, the field at the centre is:

Perimeter 2πR → square side L = πR/2. Each side is a finite wire at distance L/2 from the centre, with both ends at 45°: $B_{side} = \dfrac{\mu_0 I}{4\pi (L/2)}\left(\sin 45^\circ + \sin 45^\circ\right) = \dfrac{\sqrt2\,\mu_0 I}{2\pi L}$. Four sides: $B_{sq} = \dfrac{2\sqrt2\,\mu_0 I}{\pi L}$. Then $\dfrac{B_{sq}}{B_0} = \dfrac{2\sqrt2\,\mu_0 I/(\pi L)}{\mu_0 I/(2R)} = \dfrac{4\sqrt2\,R}{\pi L} = \dfrac{4\sqrt2\,R}{\pi \cdot \pi R/2} = \dfrac{8\sqrt2}{\pi^2} \approx 1.15 > 1$. The square gives a larger field than the circle for the same perimeter and current. Answer: A ✓

〜

EMI & AC Circuits

Faraday's law · Lenz's law · Inductance · LCR resonance · Transformer

Magnetic Flux and Faraday's Law

Magnetic flux through a surface: $\Phi = BA \cos\theta$ (B = field, A = area, θ = angle between B and normal)

Faraday's law of electromagnetic induction — induced EMF equals the negative rate of change of flux:

$\epsilon = -d\dfrac{\Phi}{dt} = -N d\dfrac{\Phi}{dt}$ (N = number of turns)

Flux can change by changing B, A, or θ (rotating coil)
Motional EMF: rod of length L moving at velocity v in field B: $\epsilon = BLv$
Rotating coil: Φ = NBAcosωt → $\epsilon = NBA\omega \sin\omega t = \epsilon_{0} \sin\omega t$ (AC generator)

Lenz's Law

The induced current always flows in a direction to oppose the change in flux that caused it (consequence of energy conservation — minus sign in Faraday's law).

Flux increasing into page → induced current anticlockwise (creates field out of page to oppose)
Magnet approaching a coil → coil acts as a repelling magnet (same pole facing)
Eddy currents: induced currents in bulk conductors — cause heating (used in induction cookers, braking)

Self and Mutual Inductance

Quantity	Definition	Formula	Unit
Self-inductance L	Φ = LI; ε = −L dI/dt	Solenoid: $L = \mu_{0}n^{2}V = \mu_{0}N^{2}A/ℓ$	Henry (H)
Mutual inductance M	Φ₁₂ = MI₂; ε₁ = −M dI₂/dt	$M = k\sqrt{L_1 L_2}$; k ≤ 1	Henry (H)

Energy stored in inductor: $U = \tfrac{1}{2}LI^{2}$
Energy density in B field: $u = \dfrac{B^{2}}{2\mu_{0}}$
Inductors in series: $L_{eq} = L_1 + L_2$ (no mutual coupling); parallel: $\dfrac{1}{L_{eq}} = \dfrac{1}{L_1} + \dfrac{1}{L_2}$

AC Circuits — Phasors and Impedance

AC source: $V = V_0\sin\omega t$. For each element:

Element	Impedance	Phase (I vs V)	Power
Resistor R	$Z = R$	In phase	$P = I_{rms}^{2} R$
Capacitor C	$X_C = \dfrac{1}{\omega C}$	I leads V by 90°	0 (avg)
Inductor L	$X_L = \omega L$	I lags V by 90°	0 (avg)
Series LCR	$Z = \sqrt{R^{2} + (X_L-X_C)^{2}}$	Phase: $\tan\phi = \dfrac{X_L-X_C}{R}$	$P = V_{rms} I_{rms} \cos\phi$

RMS values: $V_{rms} = \dfrac{V_0}{\sqrt2}$; $I_{rms} = \dfrac{I_0}{\sqrt2}$ (for sinusoidal AC)
Power factor: $\cos\phi = \dfrac{R}{Z}$; real power $P = V_{rms} I_{rms} \cos\phi$
Wattless current: component of I that does no work $= I\sin\phi$

LCR Resonance

At resonance: $X_L = X_C \to \omega L = \dfrac{1}{\omega C} \to$ $\omega_{0} = \dfrac{1}{\sqrt{LC}}$
At resonance: $Z = R$ (minimum), $I = \dfrac{V}{R}$ (maximum), $\cos\phi = 1$
Q-factor (sharpness): $Q = \dfrac{\omega_{0}L}{R} = \dfrac{1}{\omega_{0}CR} = \frac{1}{R}\sqrt{L/C}$
Bandwidth: $\Delta\omega = \dfrac{R}{L}$; $\dfrac{\omega_0}{\Delta}\omega = Q$
Half-power frequencies: $\omega_1, \omega_2$ where $I = \dfrac{I_{max}}{\sqrt2}$; $\Delta\omega = \omega_2 - \omega_1 = \dfrac{R}{L}$

Transformer

Turns ratio: $V_s/V_p = N_s/N_p = \dfrac{I_p}{I_s}$
Ideal transformer: $P_{in} = P_{out} \to V_p I_p = V_s I_s$
Step-up: $N_s > N_p \to V_s > V_p,\ I_s < I_p$
Step-down: $N_s < N_p \to V_s < V_p,\ I_s > I_p$
Losses: copper loss ($I^2R$ in windings), iron loss (eddy currents + hysteresis in core)
Real efficiency: $\eta = \dfrac{P_{out}}{P_{in}}$; typical transformer η = 95–99%

💡 JEE Tip: At resonance in a series LCR circuit, voltage across L and C individually can be much greater than the source voltage — by a factor of Q. So $V_L = V_C = Q\,V_{source}$ at resonance. High-Q circuits amplify voltage across reactive elements.

⚠ Common Error: Inductors and capacitors consume ZERO average power in AC circuits (they store and return energy). Only resistors dissipate power. Power factor $\cos\phi = 1$ only for pure resistive circuits; for pure L or C, $\cos\phi = 0$ and average power = 0.

📐

EMI & AC — Formula Sheet

Electromagnetic Induction

$\Phi = BA\cos\theta$ | $\varepsilon = -N\,d\dfrac{\Phi}{dt}$

Magnetic flux and Faraday's law; N = number of turns; negative sign = Lenz's law

$\varepsilon = BLv$ (motional EMF)

EMF in a rod of length L moving at velocity v perpendicular to field B

$\varepsilon = NBA\omega\sin\omega t = \varepsilon_0\sin\omega t$

EMF from rotating coil (AC generator); $\varepsilon_0 = NBA\omega$ is the peak EMF

Inductance

$\varepsilon = -L\,\dfrac{dI}{dt}$ | $L = \dfrac{\mu_{0}N^{2}A}{\ell}$ (solenoid)

Self-inductance; L opposes change in current (back-EMF)

$$ U = \tfrac{1}{2}LI^{2} $$

Energy stored in an inductor carrying current I

$$ u = \frac{B^{2}}{2\mu_{0}} $$

Magnetic field energy density (analogous to $\tfrac{1}{2}\varepsilon_0 E^2$ for electric field)

$\varepsilon_{12} = -M\,\dfrac{dI_2}{dt}$ | $M = k\sqrt{L_1 L_2}$

Mutual inductance; k = coupling coefficient (0 ≤ k ≤ 1)

AC Circuit Elements

$X_L = \omega L$ | $X_C = \dfrac{1}{\omega C}$

Inductive and capacitive reactances; both in Ohms; $X_L$ increases with ω, $X_C$ decreases

$$ Z = \sqrt{R^{2} + (X_L - X_C)^{2}} $$

Impedance of series LCR circuit; $Z \ge R$ always

$V_{rms} = \dfrac{V_0}{\sqrt2}$ | $I_{rms} = \dfrac{I_0}{\sqrt2}$

RMS values for sinusoidal AC; used in all power calculations

$$ P = V_{rms} I_{rms} \cos\phi = I_{rms}^{2} R $$

Average (real) power in AC circuit; $\cos\phi$ = power factor = $\dfrac{R}{Z}$

Resonance & Transformer

$\omega_0 = \dfrac{1}{\sqrt{LC}}$ | $f_0 = \dfrac{1}{2\pi\sqrt{LC}}$

Resonant frequency of LCR circuit; at $\omega_0$: $Z=R$, $I$ = max, $\cos\phi = 1$

$$ Q = \frac{\omega_{0}L}{R} = \frac{1}{R}\sqrt{L/C} $$

Quality factor — higher Q means sharper resonance peak and narrower bandwidth

$V_s/V_p = N_s/N_p = \dfrac{I_p}{I_s}$

Ideal transformer — turns ratio relates voltages inversely to currents

🧮

EMI & AC — Solved Examples

Motional EMF and Induced Current: A straight conducting rod of length 50 cm moves perpendicular to a uniform magnetic field B = 0.4 T with a velocity of 5 m/s perpendicular to both the rod and the field. The rod is part of a closed circuit with total resistance 2 Ω. Find: (a) the induced EMF, (b) the induced current, (c) the force needed to maintain constant velocity.

(a) Motional EMF: $\epsilon = BLv = 0.4 \times 0.5 \times 5 =$ 1 V
(b) Induced current: $I = \dfrac{\epsilon}{R} = \dfrac{1}{2} =$ 0.5 A
(c) Opposing force on rod (F = BIL): $F = 0.4 \times 0.5 \times 0.5 =$ 0.1 N
To maintain constant velocity, applied force must equal this opposing force: $F_{applied} = $ 0.1 N

✓ Energy check: Power input $= Fv = 0.1 \times 5 = 0.5$ W; Power dissipated $= I^2 R = 0.25 \times 2 = 0.5$ W ✓ (Energy conservation confirmed)

Series LCR Circuit — Impedance and Power: A series LCR circuit has R = 10 Ω, L = 0.1 H, C = 100 μF connected to an AC source V = 200 sin(100t) volts. Find: (a) impedance Z, (b) peak and RMS current, (c) average power consumed, (d) power factor.

Extract values: $\omega = 100 rad/s, V_{0} = 200 V, V_{rms} = \dfrac{200}{\sqrt2} = 141.4 V$
Reactances: $X_L = \omega L = 100 \times 0.1 = 10$ Ω; $X_C = \dfrac{1}{\omega C} = \dfrac{1}{100 \times 100\times 10^{-6}} = \dfrac{1}{0.01} = 100$ Ω
(a) Impedance: $Z = \sqrt{R^{2} + (X_L - X_C)^{2}} = \sqrt{100 + (10-100)^{2}} = \sqrt{100 + 8100} = \sqrt{8200} \approx$ 90.5 Ω
(b) Peak current: $I_{0} = \dfrac{V_{0}}{Z} = \dfrac{200}{90.5} \approx$ 2.21 A; $I_{rms} = \dfrac{2.21}{\sqrt2} \approx$ 1.56 A
(c) Average power: $P = I_{rms}^{2} \times R = (1.56)^{2} \times 10 = 2.434 \times 10 \approx$ 24.3 W
(d) Power factor: $\cos\phi = \dfrac{R}{Z} = \dfrac{10}{90.5} \approx$ 0.110 (very low — circuit is mostly capacitive at this frequency)

✓ Resonant frequency check: $\omega_0 = \dfrac{1}{\sqrt{LC}} = \dfrac{1}{\sqrt{0.1\times 10^{-4}}} = \dfrac{1}{\sqrt{10^{-5}}} = 316$ rad/s. Operating at 100 rad/s (well below resonance → capacitive dominance, $X_C > X_L$ confirmed).

LCR Resonance + Transformer: (a) A series LCR circuit with L = 2 mH, C = 5 μF, R = 5 Ω is driven at resonance by a 100 V (rms) source. Find the resonant frequency, Q-factor, and voltage across the inductor. (b) This circuit's source is supplied through a step-down transformer (2000:200 turns ratio) from a 2000 V (rms) mains. Find the primary current if transformer efficiency is 90%.

(a) Resonant frequency: $\omega_{0} = \dfrac{1}{\sqrt{LC}} = \dfrac{1}{\sqrt{2\times10^{-3} \times 5\times10^{-6}}} = \dfrac{1}{\sqrt{10^{-8}}} = 10^{4} rad/s \to f_{0} = \dfrac{10^{4}}{2}\pi \approx$ 1592 Hz
Q-factor: $Q = \dfrac{\omega_{0}L}{R} = \dfrac{10^{4} \times 2\times10^{-3}}{5} = \dfrac{20}{5} =$ 4
At resonance: $I_{rms} = \dfrac{V_{rms}}{R} = \dfrac{100}{5} = $ 20 A ($Z = R$ at resonance)
Voltage across L: $V_L = I_{rms} \times X_L = 20 \times \omega_{0}L = 20 \times 10^{4} \times 2\times 10^{-3} = 20 \times 20 =$ 400 V ($= Q \times V_{source} = 4 \times 100$ ✓)
(b) Transformer: $V_s/V_p = N_s/N_p \to V_s = 2000 \times \frac{200}{2000} =$ 100 V (matches source voltage ✓)
Output power: $P_{out} = I_{rms}^{2} \times R = 20^{2} \times 5 =$ 2000 W
Input power (η = 90%): $P_{in} = \dfrac{P_{out}}{\eta} = \dfrac{2000}{0.9} = 2222$ W $\to I_p = \dfrac{P_{in}}{V_p} = \dfrac{2222}{2000} =$ 1.11 A

✓ Answers: $f_0 = 1592$ Hz; $Q = 4$; $V_L = 400$ V (4× source voltage); $I_p = 1.11$ A. The voltage magnification at resonance ($V_L = QV$) is a classic JEE result.

📊

EMI & AC — Visual Concepts

Click any diagram to zoom

LCR Phasor Diagram — Impedance Triangle — Left: Voltage phasors for series LCR — V_R in phase with I, V_L leads by 90°, V_C lags by 90°. The source voltage V is their vector sum. Right: Impedance triangle — Z² = R² + (X_L−X_C)², power factor cosφ = R/Z.

LCR Resonance — Current vs Frequency — Resonance curves for high-Q (blue, sharp peak, small R) and low-Q (red, broad peak, large R). At ω₀, current is maximum (Z = R). Bandwidth Δω = R/L — the frequency range where $I \ge \dfrac{I_{max}}{\sqrt2}$. Power at half-power points is half the maximum.

Transformer — Step-Up and Step-Down — An ideal transformer transfers power via alternating magnetic flux in the iron core. The voltage ratio equals the turns ratio; the current ratio is inverse. Step-up: N_s > N_p (higher V, lower I). Step-down: N_s < N_p (lower V, higher I).

⚡

EMI & AC — Quick Quiz

Question 1

A coil of inductance 1 H and resistance 10 Ω is connected to a 100 V AC source at 50 Hz. The power factor of the circuit is approximately:

$X_L = \omega L = 2\pi\times 50\times 1 = 314$ Ω. $Z = \sqrt{R^2+X_L^2} = \sqrt{100+98596} = \sqrt{98696} \approx 314.2$ Ω. Power factor $= \cos\phi = \dfrac{R}{Z} = \dfrac{10}{314.2} \approx$ 0.032 ≈ 0.03. Answer: C ✓. The inductor dominates heavily, making the circuit nearly wattless.

Question 2

In an AC circuit, the peak EMF is 220 V. The average (mean) value of the EMF over a full cycle is:

For a complete sinusoidal cycle, the average value is zero because the positive and negative half-cycles cancel. The RMS value is $\dfrac{V_0}{\sqrt2}$ (used for power calculations). The average over a half-cycle is $\dfrac{2V_0}{\pi}$. Answer: A ✓

Question 3

A series LCR circuit is at resonance. If the capacitance is increased to four times its original value, to restore resonance the inductance must be:

At resonance: $\omega_0 = \dfrac{1}{\sqrt{LC}}$ = constant. If $C \to 4C$, then for same $\omega_0$: $\dfrac{1}{\sqrt{L'\times 4C}} = \dfrac{1}{\sqrt{LC}} \to L'\times 4C = LC \to L' = \dfrac{L}{4}$. Inductance must be reduced to one-fourth. Answer: B ✓. $LC$ product must remain constant to keep resonant frequency unchanged.

📡

Electromagnetic Waves

Maxwell's displacement current, wave properties and the EM spectrum

○ Not Started

1. Displacement Current — Maxwell's Fix

Original Ampère's law $\oint \vec{B}\cdot d\vec{l} = \mu_0 I_{enc}$ fails in the gap between the plates of a charging capacitor: no conduction current flows there, yet a magnetic field exists. Maxwell resolved this by proposing that a changing electric field acts as a current — the displacement current $I_D = \epsilon_0\,\dfrac{d\Phi_E}{dt}$.

The corrected Ampère–Maxwell law is $\oint \vec{B}\cdot d\vec{l} = \mu_0\left(I_C + \epsilon_0\dfrac{d\Phi_E}{dt}\right)$. Key point: displacement current is not a flow of charge — it is the magnetic effect of a time-varying electric field, and it equals the conduction current in the external circuit at every instant.

2. Maxwell's Four Equations

Equation	Law	Meaning
$\oint \vec{E}\cdot d\vec{A} = \dfrac{q_{enc}}{\epsilon_0}$	Gauss (electric)	Charges are sources of E-field
$\oint \vec{B}\cdot d\vec{A} = 0$	Gauss (magnetic)	No magnetic monopoles
$\oint \vec{E}\cdot d\vec{l} = -d\dfrac{\Phi_B}{dt}$	Faraday	Changing B makes E
$\oint \vec{B}\cdot d\vec{l} = \mu_0(I_C + \epsilon_0 d\dfrac{\Phi_E}{dt})$	Ampère–Maxwell	Current or changing E makes B

From these four, Maxwell derived that oscillating fields propagate as a wave at $c = \dfrac{1}{\sqrt{\mu_0\epsilon_0}} \approx 3\times10^8$ m/s — exactly the speed of light, proving light is an electromagnetic wave (confirmed experimentally by Hertz, 1887).

3. Properties of EM Waves

Transverse: $\vec{E}$, $\vec{B}$ are mutually perpendicular and both perpendicular to the propagation direction; $\vec{E}\times\vec{B}$ points along propagation.
Speed: all EM waves travel at the same $c$ in vacuum regardless of frequency; in a medium $v = \dfrac{c}{n}$.
Amplitude ratio: $\dfrac{E_0}{B_0} = c$; E and B oscillate in phase.
Require no medium, carry energy and momentum, carry no charge (undeflected by fields), and obey superposition.

For propagation along +x: $E_y = E_0\sin(kx-\omega t)$, $B_z = B_0\sin(kx-\omega t)$, with $k = 2\dfrac{\pi}{\lambda}$ and $\omega = 2\pi\nu$.

4. The Electromagnetic Spectrum

All EM waves are the same phenomenon differing only in frequency, related by $c = \nu\lambda$. In order of decreasing wavelength:

Type	Wavelength	Key uses
Radio	> 0.1 m	AM/FM radio, TV, mobile
Microwave	1 mm – 0.1 m	Radar, ovens, satellite, Wi-Fi
Infrared	700 nm – 1 mm	Remotes, night-vision, thermal imaging
Visible	400 – 700 nm	Vision, photography, optical fibre
Ultraviolet	1 – 400 nm	Sterilization, vitamin D, forgery detection
X-rays	0.001 – 1 nm	Medical imaging, security, crystallography
Gamma	< 0.001 nm	Cancer radiotherapy, PET scans

💡 Mnemonic: "Rohan's Mum Is Very Unwell, X-rays Give Relief" → Radio · Microwave · Infrared · Visible · Ultraviolet · X-ray · Gamma (longest → shortest wavelength). Visible colours: VIBGYOR.

⚠ Common Mistake: Thinking EM waves need a medium — that is a property of mechanical waves like sound. EM waves propagate through perfect vacuum. Also: microwaves are not the longest-wavelength waves; radio waves are.

📡

Electromagnetic Waves — Formulas

Speed, displacement current, intensity and wave form

○ Not Started

Speed & Field Relations

Speed of EM Waves

$$ c = \nu\lambda = \frac{1}{\sqrt{\mu_0\epsilon_0}} \approx 3\times10^8 $$

Same for all EM waves in vacuum

m/s

Amplitude Ratio

$$ \frac{E_0}{B_0} = c $$

E and B oscillate in phase

m/s

Speed in a Medium

$$ v = \frac{c}{n} = \frac{1}{\sqrt{\mu\epsilon}} $$

$n$ = refractive index

m/s

Displacement Current, Intensity & Wave Form

Displacement Current

$$ I_D = \epsilon_0\frac{d\Phi_E}{dt} $$

Equals conduction current in the circuit

Intensity

$$ I = \tfrac{1}{2}\epsilon_0 c E_0^2 = \frac{c B_0^2}{2\mu_0} $$

Energy flux per unit area

W/m²

Radiation Pressure

$$ P = \frac{I}{c} $$

For total absorption

Wave Equations (along +x)

$$ E_y = E_0\sin(kx-\omega t),\ B_z = B_0\sin(kx-\omega t) $$

$k = 2\dfrac{\pi}{\lambda},\ \omega = 2\pi\nu$

—

📡

Electromagnetic Waves — Solved Examples

JEE Main · NEET level

A radio station broadcasts at 100 MHz. Find the wavelength of the radio wave.

Approach hint

Use $\lambda = \dfrac{c}{\nu}$ with $c = 3\times10^8$ m/s.

$\lambda = \dfrac{c}{\nu} = \dfrac{3\times10^8}{100\times10^6} = \dfrac{3\times10^8}{10^8}$

✓ $\lambda = 3$ m

The electric field in an EM wave is $E = 100\sin(2\pi\times10^{14}\,t)$ V/m. Find (a) the frequency, (b) the wavelength, and (c) the magnetic field amplitude.

Approach hint

Compare with $E = E_0\sin(\omega t)$ to read $\omega$; then $\nu = \dfrac{\omega}{2}\pi$, $\lambda = \dfrac{c}{\nu}$, $B_0 = \dfrac{E_0}{c}$.

$\omega = 2\pi\times10^{14} \Rightarrow \nu = \dfrac{\omega}{2}\pi = 10^{14}$ Hz
$\lambda = \dfrac{c}{\nu} = 3\times10^8 / 10^{14} = 3\times10^{-6}$ m $= 3\ \mu$m (infrared)
$B_0 = \dfrac{E_0}{c} = \dfrac{100}{3\times10^8} \approx 3.33\times10^{-7}$ T

✓ $\nu = 10^{14}$ Hz | $\lambda = 3\ \mu$m | $B_0 \approx 3.33\times10^{-7}$ T

During charging of a capacitor, no conduction current flows between the plates yet a magnetic field exists there. Explain using Maxwell's theory and write the displacement current.

Approach hint

The electric field between the plates grows with time; its changing flux is the displacement current, which sustains continuity of current across the gap.

An Amperian loop through the wire encloses $I_C$; the same loop through the gap encloses zero conduction current — the original law contradicts itself.
As charge builds up, $E$ between the plates increases, so the electric flux changes, giving a displacement current $I_D = \epsilon_0\dfrac{d\Phi_E}{dt} = \epsilon_0 A\dfrac{dE}{dt} = C\dfrac{dV}{dt}$.
$I_D$ in the gap equals $I_C$ in the circuit at every instant, so the magnetic field is continuous across the gap.

✓ The changing E-field acts as $I_D = \epsilon_0\,d\dfrac{\Phi_E}{dt}$, producing the magnetic field between the plates

📡

Electromagnetic Waves — Quick Quiz

3 questions · tap an option to reveal answer

Question 1

Which of these EM waves has the longest wavelength?

Among the options, microwaves (1 mm – 0.1 m) have the longest wavelength: microwaves > infrared > ultraviolet > X-rays.

Question 2

Which of the following is NOT a property of electromagnetic waves?

Requiring a medium is a property of mechanical waves (like sound). EM waves propagate through vacuum because the oscillating E and B fields sustain each other.

Question 3

The magnetic field amplitude of an EM wave is $B_0 = 6\times10^{-8}$ T. The electric field amplitude is:

$E_0 = cB_0 = 3\times10^8 \times 6\times10^{-8} = 18$ V/m.

Source	Electric Field	Direction
Point charge q	\(E = \dfrac{kq}{r^{2}}\)	Radially outward (q>0) / inward (q<0)
Infinite line charge (λ C/m)	\(E = \dfrac{\lambda}{2\pi\epsilon_{0}r}\)	Perpendicular to wire, outward
Infinite sheet (σ C/m²)	\(E = \dfrac{\sigma}{2\epsilon_{0}}\)	Perpendicular to sheet, uniform
Inside conductor	E = 0	—
At conductor surface	\(E = \dfrac{\sigma}{\epsilon_{0}}\)	Normal to surface

Configuration	Capacitance C
Parallel plate (vacuum)	\(C = \dfrac{\epsilon_{0}A}{d}\)
Parallel plate (dielectric K)	\(C = K\dfrac{\epsilon_{0}A}{d}\)
Spherical (radius R)	\(C = 4\pi\epsilon_{0}R\)
Series combination	\(\dfrac{1}{C_{eq}} = \dfrac{1}{C_{1}} + \dfrac{1}{C_{2}} + \ldots\)
Parallel combination	\(C_{eq} = C_{1} + C_{2} + \ldots\)

Law	Statement	Based on
KCL (Junction rule)	\(\Sigma I_{in} = \Sigma I_{out}\) at any junction	Conservation of charge
KVL (Loop rule)	ΣV around any closed loop = 0	Conservation of energy

Configuration	Magnetic Field B	Direction
Infinite straight wire	\(\mu_{0}I / (2\pi r)\)	Right-hand rule around wire
Centre of circular loop (radius R)	\(\mu_{0}I / (2R)\)	Along axis, right-hand rule
At axis of circular loop (distance x)	\(\mu_0 I R^2 / [2(R^2+x^2)^{3/2}]\)	Along axis
Inside solenoid (n turns/m)	\(\mu_{0}nI\)	Along axis
End of solenoid	\(\mu_{0}nI / 2\)	Along axis

Quantity	Definition	Formula	Unit
Self-inductance L	Φ = LI; ε = −L dI/dt	Solenoid: \(L = \mu_{0}n^{2}V = \mu_{0}N^{2}A/ℓ\)	Henry (H)
Mutual inductance M	Φ₁₂ = MI₂; ε₁ = −M dI₂/dt	\(M = k\sqrt{L_1 L_2}\); k ≤ 1	Henry (H)

Element	Impedance	Phase (I vs V)	Power
Resistor R	\(Z = R\)	In phase	\(P = I_{rms}^{2} R\)
Capacitor C	\(X_C = \dfrac{1}{\omega C}\)	I leads V by 90°	0 (avg)
Inductor L	\(X_L = \omega L\)	I lags V by 90°	0 (avg)
Series LCR	\(Z = \sqrt{R^{2} + (X_L-X_C)^{2}}\)	Phase: \(\tan\phi = \dfrac{X_L-X_C}{R}\)	\(P = V_{rms} I_{rms} \cos\phi\)

Equation	Law	Meaning
\(\oint \vec{E}\cdot d\vec{A} = \dfrac{q_{enc}}{\epsilon_0}\)	Gauss (electric)	Charges are sources of E-field
\(\oint \vec{B}\cdot d\vec{A} = 0\)	Gauss (magnetic)	No magnetic monopoles
\(\oint \vec{E}\cdot d\vec{l} = -d\dfrac{\Phi_B}{dt}\)	Faraday	Changing B makes E
\(\oint \vec{B}\cdot d\vec{l} = \mu_0(I_C + \epsilon_0 d\dfrac{\Phi_E}{dt})\)	Ampère–Maxwell	Current or changing E makes B