⚛ 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.

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:

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

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.

Charged Particle Motion in Magnetic Field — A positive charge moving right in a field directed into the page experiences an upward-to-downward (centripetal) force. 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. 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. Field at centre of square: $B_{sq} = 4 \times (\mu_0 \dfrac{I}{4}\pi)\dfrac{\dfrac{2L}{2}}{(\dfrac{L}{2})^2}$ ... using result $B_{sq} = 4\sqrt2\,\mu_0 \dfrac{I}{\pi} L$. Numerically $\dfrac{B_{sq}}{B_0} = \dfrac{4\sqrt2\mu_0 \dfrac{I}{\pi} L}{\mu_0 \dfrac{I}{2R}} = 8\sqrt2 \dfrac{R}{\pi} L = 8\sqrt2 \dfrac{R}{\pi^2 R/2} = 16\dfrac{\sqrt2}{\pi^2} \approx 2.28 > 1$. 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.

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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)

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

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

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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

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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