Electromagnetic Induction & Alternating Current

Magnetic flux through a surface is $\Phi$ = BA cos $\theta$, where B is the magnetic field (T), A is the area ($m^{2}$), and $\theta$ is the angle between B and the area normal; [$\Phi$] = [M $L^{2} \; T^{-2} \; A^{-1}$], SI unit: weber (Wb).

Faraday's first law states that a changing magnetic flux through a circuit induces an EMF. Faraday's second law quantifies this: EMF = -dPhi/dt; [EMF] = [M $L^{2} \; T^{-3} \; A^{-1}$], SI unit: volt (V). The negative sign embodies Lenz's law: the induced current opposes the change in flux that produced it, which is a consequence of conservation of energy. If flux through a loop increases, the induced current creates a magnetic field opposing the increase; if flux decreases, the induced current supports the original field.

Motional EMF: When a conducting rod of length l moves with velocity v perpendicular to a uniform magnetic field B, an EMF $\epsilon$ = Bvl is induced; [$\epsilon$] = [M $L^{2} \; T^{-3} \; A^{-1}$], SI unit: V. This arises from the Lorentz force F = qvB on free charges in the moving rod. To determine polarity: use F = qv x B (right-hand rule) — thumb along v, fingers along l, the palm pushes positive charges, identifying the higher-potential end.

For a rotating coil in a magnetic field: $\epsilon$ = NBAw sin(wt) = $\epsilon_{0}$ sin(wt), where $\epsilon_{0}$ = NBAw is the peak EMF, N = number of turns, A = area, w = angular frequency; [$\epsilon_{0}$] = [M $L^{2} \; T^{-3} \; A^{-1}$], SI unit: V.

Self-inductance L opposes changes in current through a coil: $\epsilon$ = -L(dI/dt); [L] = [M $L^{2} \; T^{-2} \; A^{-2}$], SI unit: henry (H).
For a solenoid: L = $\mu_{0} \; n^{2}$ Al = $\mu_{0} \; N^{2}$ A/l, where n = N/l.
Mutual inductance M couples two coils: $\epsilon_{2}$ = -M(dI_1/dt).
For coaxial solenoids: M = $\mu_{0} \; n_{1} \; n_{2}$ Al.
Energy stored in an inductor: U = ($\frac{1}{2}$)$LI^{2}$; [U] = [M $L^{2} \; T^{-2}$], SI unit: J.

Eddy currents are induced in bulk conductors by changing flux, causing heating. Applications include induction furnaces, electromagnetic braking, and speedometers. They are minimized by lamination (thin insulated sheets).

AC fundamentals: An alternating voltage v = $V_{0}$ sin(wt) has peak value $V_{0}$, RMS value $V_{rms}$ = $\frac{V_{0}}{\sqrt{2}}$, and mean value (half cycle) $V_{mean}$ = $\frac{2V_{0}}{\pi}$. Note: RMS and mean values are different — $V_{rms}$ is used for power calculations.

AC through pure R: V and I are in phase; average power P = $V_{rms} \; I_{rms}$.
Through pure L: current lags voltage by $\frac{\pi}{2}$; inductive reactance $X_{L}$ = wL = 2 $\pi$ fL; [$X_{L}$] = [M $L^{2} \; T^{-3} \; A^{-2}$], SI unit: ohm; average power = 0 (wattless).
Through pure C: current leads voltage by $\frac{\pi}{2}$; capacitive reactance $X_{C}$ = 1/(wC); [$X_{C}$] = [M $L^{2} \; T^{-3} \; A^{-2}$], SI unit: ohm; average power = 0.

LCR series circuit: impedance Z = $\sqrt{R^{2} + ({X}_{L} - {X}_{C}}$²); [Z] = [M $L^{2} \; T^{-3} \; A^{-2}$], SI unit: ohm.
Phase angle: tan $\phi$ = ($X_{L}$ - $X_{C}$)/R. If $X_{L}$ > $X_{C}$, circuit is inductive (I lags V); if $X_{C}$ > $X_{L}$, capacitive (I leads V).

At resonance: $X_{L}$ = $X_{C}$, so $f_{0}$ = 1/(2 $\pi \; \sqrt{LC}$).
At resonance, Z = R (minimum, NOT zero), current is maximum, and phase angle is zero. The circuit behaves as purely resistive.

Power in AC: P = $V_{rms} \; I_{rms}$ cos $\phi$; power factor cos $\phi$ = R/Z; wattless current component = $I_{rms}$ sin $\phi$. Only the resistive component dissipates power.

Transformer: $\frac{V_{s}}{V_{p}}$ = $\frac{N_{s}}{N_{p}}$ = $\frac{I_{p}}{I_{s}}$ (note: current ratio is inverse of turns ratio). Step-up: $N_{s}$ > $N_{p}$ (voltage increases, current decreases). Step-down: $N_{s}$ < $N_{p}$.
Ideal transformer: $P_{input}$ = $P_{output}$. Energy losses: copper loss ($I^{2}$R in windings), iron/eddy current loss, flux leakage, hysteresis loss.

The key testable concept is Faraday's law with Lenz's law direction, motional EMF calculations, and LCR series circuit impedance and resonance conditions.

Solved Numericals

N₁. A conducting rod of length 50 cm moves with velocity 4 m/s perpendicular to a uniform magnetic field of 0.5 T. Find the motional EMF. If the rod is part of a closed circuit of resistance 2 ohm, find the induced current and force needed to maintain constant velocity.

Given: l = 50 cm = 0.50 m, v = 4 m/s, B = 0.5 T, R = 2 ohm.

Motional EMF: $\epsilon$ = Bvl = 0.5 T x 4 m/s x 0.50 m = 1.0 V.

Induced current: I = $\epsilon$/R = 1.0 V / 2 ohm = 0.5 A.

Force on current-carrying rod in field: F = BIl = 0.5 T x 0.5 A x 0.50 m = 0.125 N.

By Lenz's law, this force opposes the motion. To maintain constant velocity, an external force of 0.125 N must be applied in the direction of motion.

Power check: $P_{external}$ = Fv = 0.125 N x 4 m/s = 0.5 W. $P_{dissipated}$ = $I^{2}$ R = 0.25 x 2 = 0.5 W. Energy is conserved.

N₂. An LCR series circuit has R = 100 ohm, L = 0.5 H, C = 10 uF connected to 200 V, 50 Hz AC supply.

Given: R = 100 ohm, L = 0.5 H, C = 10 uF = 10 x $10^{-6}$ F, $V_{rms}$ = 200 V, f = 50 Hz.

Inductive reactance: $X_{L}$ = 2 $\pi$ fL = 2 $\pi$ x 50 x 0.5 = 50 $\pi$ = 157.08 ohm.

Capacitive reactance: $X_{C}$ = 1/(2 $\pi$ fC) = 1/(2 $\pi$ x 50 x 10 x $10^{-6}$) = 1/($\pi$ x $10^{-3}$) = 1000/$\pi$ = 318.31 ohm.

Impedance: Z = $\sqrt{R^{2} + ({X}_{L} - {X}_{C}}$²) = $\sqrt{100^{2} + (157.08 - 318.31}$²) = $\sqrt{10000 + (-161.23}$²) = $\sqrt{10000 + 25995}$ = $\sqrt{35995}$ = 189.72 ohm.

RMS current: $I_{rms}$ = $V_{rms}$/Z = $\frac{200}{189}$.72 = 1.054 A.

Phase angle: tan $\phi$ = ($X_{L}$ - $X_{C}$)/R = (157.08 - 318.31)/100 = -161.$\frac{23}{100}$ = -1.6123. $\phi$ = arctan(-1.6123) = -58.2 deg.

Since $\phi$ is negative ($X_{C}$ > $X_{L}$), the circuit is capacitive (I leads V).

Power factor: cos $\phi$ = R/Z = $\frac{100}{189}$.72 = 0.527.

Average power: P = $V_{rms} \; I_{rms}$ cos $\phi$ = 200 x 1.054 x 0.527 = 111.1 W.

Alternatively: P = $I_{rms}^{2}$ x R = (1.054)² x 100 = 111.1 W. Verified.

N₃. A step-down transformer converts 2200 V to 220 V. The secondary coil has 100 turns. Find the primary turns. If the secondary current is 10 A and efficiency is 90%, find the primary current.

Turns ratio: $\frac{V_{s}}{V_{p}}$ = $\frac{N_{s}}{N_{p}}$ $\frac{220}{2200}$ = 100/$N_{p}$ $N_{p}$ = 100 x $\frac{2200}{220}$ = 1000 turns.

Output power: $P_{s}$ = $V_{s}$ x $I_{s}$ = 220 V x 10 A = 2200 W.

Efficiency: $\eta$ = $\frac{P_{s}}{P_{p}}$ = 0.90. Input power: $P_{p}$ = $\frac{P_{s}}{\eta}$ = $\frac{2200}{0}$.90 = 2444.4 W.

Primary current: $I_{p}$ = $\frac{P_{p}}{V_{p}}$ = 2444.$\frac{4}{2200}$ = 1.11 A.

For an ideal transformer (100% efficiency): $I_{p}$ = $I_{s}$ x $\frac{N_{s}}{N_{p}}$ = 10 x $\frac{100}{1000}$ = 1.0 A. The actual primary current (1.11 A) is higher due to 10% energy losses.