Induced current $I(t) = \frac{|E|}{R} = \frac{1}{R} \left| \frac{d\Phi}{dt} \right|$. Total charge $Q = \int_{0}^{\infty} I(t) dt = \frac{1}{R} \int_{\Phi(0)}^{\Phi(\infty)} |d\Phi|$.
$\Phi(0) = B_0 A$. $\Phi(\infty) = B_0 A e^{-\infty} = 0$.
$Q = \frac{|\Phi(\infty) - \Phi(0)|}{R} = \frac{|0 - B_0 A|}{R} = \mathbf{\frac{B_0 A}{R}}$. Independent of $\tau$.
From Faraday's Law in integral form: $\oint \vec{E} \cdot d\vec{l} = -\frac{d\Phi_B}{dt}$. Due to symmetry, $E$ is tangential.
For $r < R$: $E(2\pi r) = -\frac{d}{dt}(B \pi r^2) = -\pi r^2 \alpha \implies E_{in} = \mathbf{-\frac{\alpha r}{2}}$. (Linear dependence).
For $r > R$: $E(2\pi r) = -\frac{d}{dt}(B \pi R^2) = -\pi R^2 \alpha \implies E_{out} = \mathbf{-\frac{\alpha R^2}{2r}}$. (Hyperbolic decay).
Radius $r(t) = r_0 - vt$. Area $A(t) = \pi (r_0 - vt)^2$.
Flux $\Phi(t) = B_0 \pi (r_0 - vt)^2$.
$E = -\frac{d\Phi}{dt} = -B_0 \pi \cdot 2(r_0 - vt)(-v) = \mathbf{2\pi B_0 v (r_0 - vt)}$. EMF decreases linearly to 0.
Assume the leading edge is at $x+a$ and trailing edge at $x$.
Motional EMF on leading edge $E_1 = B(x+a) a v = B_0 \left(\frac{x+a}{a}\right) a v = B_0 v (x+a)$.
Motional EMF on trailing edge $E_2 = B(x) a v = B_0 \left(\frac{x}{a}\right) a v = B_0 v x$.
Net EMF in loop $E_{net} = E_1 - E_2 = B_0 v a$. Current $I = \frac{E_{net}}{R} = \mathbf{\frac{B_0 v a}{R}}$.
Induced E-field at boundary $r=R$ is $E = \frac{R}{2} \left| \frac{dB}{dt} \right| = \frac{R\alpha}{2}$.
Tangential electrical force on charge $dq$ is $dF = E dq$. Torque $d\tau = R dF = E R dq$.
Total torque $\tau = E R Q_{total} = \left(\frac{R\alpha}{2}\right) R (2\pi R \lambda) = \pi \lambda \alpha R^3$.
Moment of inertia $I = MR^2$. Angular acceleration $\alpha_{ang} = \frac{\tau}{I} = \frac{\pi \lambda \alpha R^3}{MR^2} = \mathbf{\frac{\pi \lambda \alpha R}{M}}$.
$E = -\frac{d\Phi}{dt} = -\frac{d}{dt}(\alpha t^2 - \beta t) = -(2\alpha t - \beta)$.
Current reverses direction when EMF crosses zero.
$2\alpha t - \beta = 0 \implies \mathbf{t = \frac{\beta}{2\alpha}}$.
Total flux $\Phi_{tot} = \Phi_{ext} + \Phi_{self} = B_{ext}A + LI$.
From Faraday's law, $E = -\frac{d\Phi_{tot}}{dt}$. For a superconductor, $R=0$, so voltage drop $IR = 0$.
Therefore, $E = 0 \implies \frac{d\Phi_{tot}}{dt} = 0 \implies \mathbf{\Phi_{tot} = constant}$. The induced current flawlessly cancels any external flux changes.
Graph consists of three distinct phases:
1. Entering (width $w$): Constant EMF $E = Bwv$ (positive pulse for duration $w/v$).
2. Fully inside (moves $2w$ distance): EMF $= 0$ (zero pulse for duration $2w/v$).
3. Exiting (width $w$): Constant EMF $E = -Bwv$ (negative pulse for duration $w/v$).
Initial flux $\Phi_i = 0$. Final $\Phi_f = NAB$. EMF $E(t) = NAB\omega \sin(\omega t)$.
Heat $H = \int \frac{E^2}{R} dt = \frac{(NAB\omega)^2}{R} \int_0^{\pi/2\omega} \sin^2(\omega t) dt$.
Integral of $\sin^2$ over quarter cycle is $\frac{1}{2} \cdot \frac{\pi}{2\omega} = \frac{\pi}{4\omega}$.
$H = \frac{N^2 A^2 B^2 \omega^2}{R} \frac{\pi}{4\omega} = \mathbf{\frac{\pi N^2 A^2 B^2 \omega}{4R}}$.
**No.** Equipotential surfaces can only exist if the field is conservative (i.e., the work done along a closed path is exactly zero, $\oint \vec{E} \cdot d\vec{l} = 0$). Since a time-varying magnetic field produces a non-conservative rotational electric field, a unique scalar potential $V$ cannot be defined globally, making equipotential surfaces physically impossible.
As the magnet falls, it induces circulating eddy currents in the pipe walls. By Lenz's law, these currents create an upward magnetic force $\vec{F}_m$ opposing gravity. Because $\vec{F}_m \propto v$ (since $E \propto v$ and $I \propto E$), as velocity increases, $\vec{F}_m$ increases. When $\mathbf{F_m = mg}$, net acceleration is zero, and it falls with a constant terminal velocity.
For $I(t) = I_0 \sin(\omega t)$, current is increasing from $t=0$ to $T/4$. Induced current in small loop is counter-clockwise (opposing). Two concentric loops with opposite currents repel each other radially. However, due to perfect symmetry, the net translational force on the entire small loop is **exactly zero**. It only experiences a symmetric inward radial compression (pinch).
Graph of $a$ vs $t$: Starts at $a=g$. As it approaches the ring, $a < g$ (repulsion, upward force). At the exact center of the ring, flux change is momentarily zero, so induced current vanishes, and $a = g$ exactly. As it falls below, it is attracted upward (opposing departure), so $a < g$ again. Finally, far away, $a \to g$. It's a "W" shaped dip below the $a=g$ line.
$E = Blv \implies I = \frac{Blv}{R}$.
Opposing magnetic force $F_m = I l B = \frac{B^2 l^2 v}{R}$. External power $P_{ext} = F_{ext} v = F_m v = \mathbf{\frac{B^2 l^2 v^2}{R}}$.
Joulean heat rate $P_{heat} = I^2 R = \left(\frac{Blv}{R}\right)^2 R = \mathbf{\frac{B^2 l^2 v^2}{R}}$.
Thus, $P_{ext} = P_{heat}$, strictly proving macroscopic energy conservation.
AC current creates an alternating flux. The induced EMF in the ring lags the flux by $90^\circ$ ($d/dt$ of sine is cosine). Due to the ring's self-inductance, the induced current lags the EMF further. This total phase difference ($>90^\circ$) ensures the magnetic force (Lorentz force between core field and ring current) has a highly dominant repulsive time-average, physically levitating the ring.
The complete loop supports induced current, producing magnetic braking. Its magnet falls slowly. The cut loop has $R \to \infty$, so induced current is zero, providing no magnetic braking. Its magnet falls at pure $g$ (faster).
Thermodynamically, the complete loop system converts $mg\Delta h$ into kinetic energy + Joule heat. The cut loop converts $mg\Delta h$ entirely into kinetic energy.
Entering the $B$ field increases local flux, inducing eddy currents that repel the field source. Exiting decreases flux, inducing attracting eddy currents. Both interactions oppose the velocity vector $\vec{v}$. Since EMF $\propto \frac{d\Phi}{dt} \propto v$, and Current $I \propto EMF \propto v$, the resulting Lorentz force $F = \int I(d\vec{l} \times \vec{B})$ scales as $I \times B$, making $F_{drag} \propto v$.
The outward flux is decreasing. By Lenz's law, the induced current must produce an outward magnetic field to supplement the collapsing field. By the Right-Hand grip rule, an outward field requires a counter-clockwise current. A counter-clockwise current loop has its magnetic dipole moment vector pointing **out of the page** ($+\hat{k}$).
Removing EMF attempts a $dI/dt \to -\infty$. The inductor induces a massive forward EMF ($E = -L \frac{dI}{dt}$) trying to keep $I$ flowing, conserving its magnetic field energy $\frac{1}{2}LI^2$. It prevents infinite current decay by ionizing the switch gap (sparking), maintaining a conductive plasma path until the energy is dissipated as heat/light in the arc.
Approaching North pole increases flux. Lenz's law dictates the coil face must act as a North pole to repel it, meaning current attempts to flow counter-clockwise. This pushes positive charge to the capacitor plate connected to the terminal that conventional current would exit from. Thus, the plate connected to the **anti-clockwise exiting terminal** becomes positively charged.
Motional EMF $E = BvL$. Charge on capacitor $q = CE = CBvL$.
Current $I = \frac{dq}{dt} = CBL \frac{dv}{dt} = CBL a$.
Equation of motion: $mg - F_m = ma \implies mg - I L B = ma$.
$mg - (CBL a) L B = ma \implies mg = ma + CB^2 L^2 a \implies a = \mathbf{\frac{mg}{m + CB^2 L^2}}$. (Acceleration is constant, less than $g$).
EMF $E = BvL$. For inductor, $E = L_{ind} \frac{dI}{dt} \implies BvL = L_{ind} \frac{dI}{dt}$.
Differentiating EOM ($ma = mg - ILB$): $m \frac{d^2v}{dt^2} = -L B \frac{dI}{dt} = -L B \left(\frac{BvL}{L_{ind}}\right) = -\frac{B^2 L^2}{L_{ind}} v$.
This is SHM $\frac{d^2v}{dt^2} + \omega^2 v = 0$ with angular frequency $\mathbf{\omega = \frac{BL}{\sqrt{m L_{ind}}}}$.
The rod is split into two segments: length $x$ and length $(L-x)$.
EMF of part 1: $E_1 = \int_0^x B \omega r dr = \frac{1}{2} B \omega x^2$.
EMF of part 2: $E_2 = \int_0^{L-x} B \omega r dr = \frac{1}{2} B \omega (L-x)^2$.
Since they oppose each other relative to the ends, net EMF $E_{net} = E_2 - E_1 = \frac{1}{2} B \omega [ (L-x)^2 - x^2 ] = \mathbf{\frac{1}{2} B \omega L (L - 2x)}$.
The effective length of the conductor cutting the field is the straight-line distance between its ends, which is the diameter $2R$.
Area swept formula: $\Phi(t) = B_0 \cdot \frac{1}{2}\pi R^2 \cos(\omega t)$.
$E = -\frac{d\Phi}{dt} = \mathbf{\frac{1}{2} \pi R^2 B_0 \omega \sin(\omega t)}$.
*Alternatively, using effective length $L=2R$ rotating: $E = \frac{1}{2} B_0 \omega (2R)^2 = 2 B_0 \omega R^2$ (peak).*
$F_m = -IlB = -\left(\frac{Blv}{R}\right)lB = -\frac{B^2 l^2 v}{R}$.
$m \frac{dv}{dt} = -\frac{B^2 l^2}{R} v \implies m \frac{dv}{dx} \frac{dx}{dt} = m v \frac{dv}{dx} = -\frac{B^2 l^2}{R} v \implies m \frac{dv}{dx} = -\frac{B^2 l^2}{R}$.
Integrate $\int_{v_0}^0 dv = -\frac{B^2 l^2}{mR} \int_0^x dx \implies -v_0 = -\frac{B^2 l^2}{mR} x \implies x = \mathbf{\frac{m R v_0}{B^2 l^2}}$.
Consider an infinitesimal radial element $dr$ at distance $r$. Velocity $v = \omega r$.
Motional EMF across $dr$ is $dE = B_0 v dr = B_0 (\omega r) dr$.
Total EMF from center to rim is $E = \int_0^R B_0 \omega r dr = \mathbf{\frac{1}{2} B_0 \omega R^2}$. (The disc acts as infinite parallel spokes).
Let side $a$ be horizontal. EMF $E = Bav$. Current $I = Bav/R$.
Upward magnetic force $F_m = I a B = \frac{B^2 a^2 v}{R}$.
At terminal velocity $v_t$, net force is zero: $mg - \frac{B^2 a^2 v_t}{R} = 0$.
$v_t = \mathbf{\frac{m g R}{B^2 a^2}}$.
At distance $r$, electron velocity $\vec{v} = \omega r \hat{\theta}$. Lorentz force $\vec{F}_m = -e(\vec{v} \times \vec{B})$. If $\vec{B} = B\hat{k}$, $\vec{F}_m = -e(\omega r \hat{\theta} \times B\hat{k}) = -e\omega r B \hat{r}$.
In steady state, internal electrostatic force exactly balances it: $-e\vec{E}_{in} + \vec{F}_m = 0 \implies -e\vec{E}_{in} - e\omega r B \hat{r} = 0$.
$\vec{E}_{in} = \mathbf{-\omega B r \hat{r}}$. The field increases linearly from the center.
Wire is along y-axis. Element $dy$ at $y$. Velocity is $v_0\hat{i}$. $\vec{B} = B_0 y \hat{k}$.
$dE = (\vec{v} \times \vec{B}) \cdot d\vec{l} = (v_0\hat{i} \times B_0 y \hat{k}) \cdot (dy\hat{j}) = (-v_0 B_0 y \hat{j}) \cdot (dy\hat{j}) = -v_0 B_0 y dy$.
Magnitude $E = \int_0^L v_0 B_0 y dy = v_0 B_0 \left[\frac{y^2}{2}\right]_0^L = \mathbf{\frac{1}{2} B_0 v_0 L^2}$.
Component of $B$ perpendicular to rail plane is $B_{\perp} = B \cos\theta$.
EMF $E = B_{\perp} l v = B \cos\theta \cdot l \cdot v$. Current $I = \frac{B \cos\theta \cdot l \cdot v}{R}$.
Magnetic force along incline (upward) $F_m = I l B_{\perp} = \frac{B^2 l^2 v \cos^2\theta}{R}$.
Balance gravity component: $mg \sin\theta = F_m \implies v_t = \mathbf{\frac{mg R \sin\theta}{B^2 l^2 \cos^2\theta}}$.
Consider a cylindrical shell of radius $r$, thickness $dr$. EMF $e = \pi r^2 \frac{dB}{dt} = \pi r^2 B_0 \omega \cos\omega t$.
Resistance $dR = \rho \frac{2\pi r}{L dr}$. Power $dP = \frac{e^2}{dR} = \frac{(\pi r^2 B_0 \omega \cos\omega t)^2 L dr}{2\pi \rho r} = \frac{\pi L B_0^2 \omega^2 \cos^2\omega t}{2\rho} r^3 dr$.
Average $\cos^2\omega t = 1/2$. $\langle dP \rangle = \frac{\pi L B_0^2 \omega^2}{4\rho} r^3 dr$. Total $\langle P \rangle = \int_0^R \langle dP \rangle = \frac{\pi L B_0^2 \omega^2 R^4}{16\rho}$.
Volume $V = \pi R^2 L$. Power per unit volume $\frac{\langle P \rangle}{V} = \mathbf{\frac{B_0^2 \omega^2 R^2}{16\rho}}$.
An alternating current creates an alternating internal magnetic field. By Faraday's/Lenz's law, this changing field induces eddy currents within the bulk conductor itself. These internal eddy currents circulate such that they logically oppose the main current at the center of the wire and reinforce it at the skin (surface). Thus, net current density is forced exponentially outward.
Motional EMF in the disc is $E \propto v$. Eddy current $I = E/R \propto v/R$.
Lorentz force on this current is $F_d \propto I B \propto (v/R) B \implies F_d = -c v$ (where $c$ is a positive constant).
EOM: $ma = -cv - kx \implies m\ddot{x} + c\dot{x} + kx = 0$. This standard differential equation describes a damped harmonic oscillator, yielding solutions enveloped by $e^{-(c/2m)t}$.
Empirical Steinmetz equation: $P_e = K_e f^2 B_{max}^2 V t^2$ (where $t$ is lamination thickness).
Ferrites are ceramic compounds of transition metals with oxygen. Crucially, they are highly magnetic but simultaneously **electrical insulators** (immense resistivity $\rho$). High $\rho$ prevents eddy currents entirely, making them vastly superior for high-frequency (RF) cores where $f^2$ losses would otherwise vaporize silicon steel.
Power loss in a sheet of thickness $t$ is $P \propto t^2$ (from $e \propto Area \propto t$, $R \propto 1/t \implies P=e^2/R \propto t^3$ per sheet. Volume is $\propto t$). Total power for solid core of thickness $T$ is $P_0 \propto T^2$.
If split into $n$ laminations, thickness $t = T/n$. Power per lamination $p \propto (T/n)^2$.
Total power $P_{new} = n \times p \propto n(T/n)^2 = \frac{T^2}{n}$. Thus $P_{new} = \mathbf{P_0 / n^2}$.
Superconducting magnets on the train move rapidly over conducting guideway coils. The changing flux induces immense eddy currents in the guideway. By Lenz's law, these currents create a magnetic pole directly opposing the train's magnet. The vertical component of this repulsion provides **levitation**, while the horizontal component produces **magnetic drag** opposing forward motion.
The standard "skin depth" or penetration depth formula derived from Maxwell's equations in conductors is: $\mathbf{\delta = \sqrt{\frac{2}{\omega \mu \sigma}}}$.
Higher frequencies ($\omega$), higher permeability ($\mu$), and higher conductivity ($\sigma$) drastically reduce penetration depth, making induction heating intensely localized at the surface.
**Increase.** The induced EMF $e \propto \frac{d\Phi}{dt} \propto \omega$. Eddy current $I_e \propto \omega$. The repulsive magnetic force $F \propto I_{coil} I_{coin}$. However, due to self-inductance of the coin, higher frequency creates a near $180^\circ$ phase shift, aligning peak opposing currents simultaneously, exponentially maximizing the repulsive Lorentz force and jump height.
**No.** The braking force $F_d$ is strictly proportional to velocity $v$ ($F_d = -cv$). As velocity approaches zero, the braking force also approaches zero asymptotically. The equation of motion $dv/dt = -(c/m)v$ yields $v(t) = v_0 e^{-(c/m)t}$. Mathematically, $v$ only reaches exactly zero at $t = \infty$.
A voltage coil and a current coil create two AC magnetic fluxes $\Phi_1, \Phi_2$ that are spatially separated and phase-shifted. Each induces eddy currents $I_1, I_2$ in the disk. The flux $\Phi_1$ interacts with $I_2$, and $\Phi_2$ interacts with $I_1$. The cross-interaction creates a net, non-zero unidirectional time-averaged Lorentz torque, spinning the meter disk proportional to total power.
Field of wire at distance $r$ is $B = \frac{\mu_0 I}{2\pi r}$.
Flux through a strip of width $dr$, length $b$ is $d\Phi = B \cdot (b dr) = \frac{\mu_0 I b}{2\pi r} dr$.
Total flux $\Phi = \int_{x_0}^{x_0+a} \frac{\mu_0 I b}{2\pi r} dr = \frac{\mu_0 I b}{2\pi} \ln\left(\frac{x_0+a}{x_0}\right)$.
Since $\Phi = MI$, Mutual Inductance $M = \mathbf{\frac{\mu_0 b}{2\pi} \ln\left(1 + \frac{a}{x_0}\right)}$.
Assume current $I$ flows in the large loop. Field at its center is roughly uniform over the small loop area ($r \ll R$). $B = \frac{\mu_0 I}{2R}$.
Flux through small loop $\Phi_{small} = B \times (\pi r^2) = \left( \frac{\mu_0 I}{2R} \right) \pi r^2$.
Since $\Phi_{small} = M I$, $M = \mathbf{\frac{\mu_0 \pi r^2}{2R}}$.
Using energy method: Magnetic field exists only between $a$ and $b$, $B = \frac{\mu_0 I}{2\pi r}$.
Energy density $u = \frac{B^2}{2\mu_0} = \frac{\mu_0 I^2}{8\pi^2 r^2}$.
Total Energy $U = \int u dV = \int_a^b \left( \frac{\mu_0 I^2}{8\pi^2 r^2} \right) (2\pi r l dr) = \frac{\mu_0 I^2 l}{4\pi} \int_a^b \frac{dr}{r} = \frac{\mu_0 I^2 l}{4\pi} \ln(b/a)$.
Equating to $\frac{1}{2}LI^2$, $L = \frac{\mu_0 l}{2\pi} \ln(b/a)$. Thus $L/l = \mathbf{\frac{\mu_0}{2\pi} \ln(b/a)}$.
Field inside toroid at distance $r$ from axis: $B = \frac{\mu_0 N I}{2\pi r}$.
Flux through single turn cross-section: $\phi = \int_a^b B (h dr) = \int_a^b \frac{\mu_0 N I h}{2\pi r} dr = \frac{\mu_0 N I h}{2\pi} \ln(b/a)$.
Total flux $\Phi_{total} = N\phi = \frac{\mu_0 N^2 I h}{2\pi} \ln(b/a)$.
$L = \frac{\Phi_{total}}{I} = \mathbf{\frac{\mu_0 N^2 h}{2\pi} \ln(b/a)}$.
Flux $\Phi_2 = \oint_2 \vec{A}_1 \cdot d\vec{l}_2$. Vector potential $\vec{A}_1 = \frac{\mu_0 I_1}{4\pi} \oint_1 \frac{d\vec{l}_1}{r_{12}}$.
$\Phi_2 = \frac{\mu_0 I_1}{4\pi} \oint_2 \oint_1 \frac{d\vec{l}_1 \cdot d\vec{l}_2}{r_{12}}$.
Since $M_{12} = \Phi_2 / I_1$, $M_{12} = \frac{\mu_0}{4\pi} \oint_1 \oint_2 \frac{d\vec{l}_1 \cdot d\vec{l}_2}{r_{12}}$.
The dot product is commutative ($d\vec{l}_1 \cdot d\vec{l}_2 = d\vec{l}_2 \cdot d\vec{l}_1$), proving $M_{12} = M_{21}$ strictly by geometric symmetry.
Let flux per turn be $\phi$. $L_1 = N_1 (N_1 \phi / I_1) \implies \phi/I_1 = L_1 / N_1^2$.
Perfect coupling means all flux from 1 links 2. $\Phi_{21} = N_2 (N_1 \phi)$.
$M = \Phi_{21} / I_1 = N_1 N_2 (\phi / I_1) = N_1 N_2 (L_1 / N_1^2) = \frac{N_2}{N_1} L_1$.
Similarly $M = \frac{N_1}{N_2} L_2$. Multiplying both equations: $M^2 = L_1 L_2 \implies \mathbf{M = \sqrt{L_1 L_2}}$.
Voltage $V = L_1 \frac{dI_1}{dt} + M \frac{dI_2}{dt}$ and $V = L_2 \frac{dI_2}{dt} + M \frac{dI_1}{dt}$. Total $I = I_1 + I_2$.
Solving for $dI_1/dt$ and $dI_2/dt$, adding them gives $\frac{dI}{dt} = V \frac{L_1 + L_2 - 2M}{L_1 L_2 - M^2}$.
Since $V = L_{eq} \frac{dI}{dt}$, inversion yields $\mathbf{L_{eq} = \frac{L_1 L_2 - M^2}{L_1 + L_2 - 2M}}$.
Current $I(t) = I_0(1 - e^{-Rt/L})$. Energy $U(t) = \frac{1}{2} L I(t)^2 = U_{max} (1 - e^{-Rt/L})^2$.
Set $U(t) = 0.25 U_{max} \implies (1 - e^{-Rt/L})^2 = 0.25 \implies 1 - e^{-Rt/L} = 0.5 \implies e^{-Rt/L} = 0.5$.
$-Rt/L = \ln(1/2) = -\ln 2 \implies \mathbf{t = \frac{L}{R} \ln 2}$.
Consider a slice $dx$. Turn density $n(x) = n_0 + \alpha x$. Inductance of this slice $dL = \mu_0 [n(x)]^2 A dx$.
Total $L = \int_0^l \mu_0 A (n_0 + \alpha x)^2 dx = \mu_0 A \int_0^l (n_0^2 + 2n_0\alpha x + \alpha^2 x^2) dx$.
$L = \mathbf{\mu_0 A \left( n_0^2 l + n_0 \alpha l^2 + \frac{1}{3} \alpha^2 l^3 \right)}$.
Field inside $B = \mu_0 n I$. Field outside $B \approx 0$.
Energy density $u_B = \frac{B^2}{2\mu_0} = \frac{(\mu_0 n I)^2}{2\mu_0} = \frac{\mu_0 n^2 I^2}{2}$.
Total Energy $U = u_B \times Volume = \left( \frac{\mu_0 n^2 I^2}{2} \right) (A l)$.
We know $U = \frac{1}{2} L I^2$. Equating terms: $\frac{1}{2} L I^2 = \frac{1}{2} \mu_0 n^2 I^2 A l \implies \mathbf{L = \mu_0 n^2 A l}$. Matches perfectly.
Flux $\Phi(t) = NAB \cos(\theta) = NAB \cos(\frac{1}{2}\alpha t^2)$.
$E = -\frac{d\Phi}{dt} = NAB \sin(\frac{1}{2}\alpha t^2) \cdot \frac{d}{dt}(\frac{1}{2}\alpha t^2) = \mathbf{NAB\alpha t \sin(\frac{1}{2}\alpha t^2)}$.
Both the **Amplitude ($NAB\alpha t$)** and the **Frequency ($\omega = \alpha t$)** vary linearly and increase continuously with time (a "chirp" signal).
$E = NAB\omega\sin\omega t$. Current $I = \frac{E}{R} = \frac{NAB\omega}{R}\sin\omega t$.
Magnetic torque on coil $\tau_m = M \times B = (NIA) B \sin\omega t = \frac{N^2 A^2 B^2 \omega}{R} \sin^2\omega t$.
To maintain constant $\omega$, external prime mover must exactly balance this resisting torque: $\tau_{mech}(t) = \mathbf{\frac{N^2 A^2 B^2 \omega}{R} \sin^2\omega t}$.
Full cycle: $\langle e \rangle = \frac{1}{T} \int_0^T E_0 \sin(\omega t) dt = \frac{E_0}{T\omega} [-\cos(\omega t)]_0^{2\pi/\omega} = -\frac{E_0}{2\pi} (\cos 2\pi - \cos 0) = 0$.
Half cycle: $\langle e \rangle_{1/2} = \frac{1}{T/2} \int_0^{T/2} E_0 \sin(\omega t) dt = \frac{2\omega}{2\pi} E_0 [-\frac{\cos(\omega t)}{\omega}]_0^{\pi/\omega} = -\frac{E_0}{\pi}(\cos\pi - \cos 0) = -\frac{E_0}{\pi}(-1 - 1) = \mathbf{\frac{2}{\pi} E_0}$.
It uses three separate identical armature coils rigidly mounted $120^\circ$ ($2\pi/3$ rad) apart on the same rotor. Rotating in the same $B$ field, they produce three AC voltages of equal magnitude but mutually phase-shifted by $120^\circ$.
$e_1(t) = E_0 \sin(\omega t)$
$e_2(t) = E_0 \sin(\omega t - 120^\circ)$
$e_3(t) = E_0 \sin(\omega t - 240^\circ)$ (or $+120^\circ$).
Generator EMF $e(t) = E_0 \sin\omega t$. Circuit equation: $L \frac{di}{dt} + R_a i = E_0 \sin\omega t$.
This is a standard RL driven circuit. The steady-state particular solution is $i(t) = I_0 \sin(\omega t - \phi)$.
Impedance $Z = \sqrt{R_a^2 + (\omega L)^2}$. Steady-state amplitude $I_0 = \mathbf{\frac{E_0}{\sqrt{R_a^2 + \omega^2 L^2}}}$.
Form Factor = $V_{rms} / V_{avg}$. Peak Factor = $V_{peak} / V_{rms}$.
For pure sine wave: $V_{rms} = E_0/\sqrt{2} \approx 0.707 E_0$. $V_{avg} = \frac{2}{\pi}E_0 \approx 0.637 E_0$.
Form Factor = $\frac{E_0/\sqrt{2}}{2E_0/\pi} = \frac{\pi}{2\sqrt{2}} \approx \mathbf{1.11}$.
Peak Factor = $\frac{E_0}{E_0/\sqrt{2}} = \mathbf{\sqrt{2} \approx 1.414}$.
$\Phi(t) = N A B(t) \cos(\omega t) = N A (B_0 \cos\omega_0 t) \cos\omega t$.
Using trig identity: $\Phi(t) = \frac{1}{2}NAB_0 [\cos((\omega_0+\omega)t) + \cos((\omega_0-\omega)t)]$.
$E = -\frac{d\Phi}{dt} = \mathbf{\frac{1}{2}NAB_0 [(\omega_0+\omega)\sin((\omega_0+\omega)t) + (\omega_0-\omega)\sin((\omega_0-\omega)t)]}$.
The beat frequencies are unequivocally **$(\omega_0 + \omega)$** and **$|\omega_0 - \omega|$**.
When supplying heavy current to a load, the armature coil acts as an electromagnet. The alternating magnetic field generated by the armature itself interacts with the primary stationary field (field poles). This "armature reaction" distorts the main flux pattern (cross-magnetization) and can weaken it (demagnetization), causing the terminal EMF to drop under load.
From Q52, $\tau_{mech} = \frac{E_0^2}{R\omega} \sin^2\omega t$. Mech Power $P_{mech} = \tau_{mech}\omega = \frac{E_0^2}{R} \sin^2\omega t$.
$\langle P_{mech} \rangle = \frac{E_0^2}{R} \langle\sin^2\omega t\rangle = \frac{E_0^2}{2R}$.
Elec Power $P_{elec} = i^2 R = (\frac{E_0}{R}\sin\omega t)^2 R = \frac{E_0^2}{R}\sin^2\omega t$.
$\langle P_{elec} \rangle = \frac{E_0^2}{2R}$. Thus, $\mathbf{\langle P_{mech} \rangle = \langle P_{elec} \rangle = \frac{E_0^2}{2R}}$, verifying energy conservation.
Peak EMF $E_0 = NAB\omega$. The mathematical derivation of Faraday's flux linkage relies strictly on the total enclosed area vector ($A$), totally independent of the specific geometric perimeter shape (square, rectangular, or circular). Therefore, the peak EMF generated remains **exactly invariant (unchanged)**.
