D4-2. Faraday’s Law, Lenz’s Law, and AC Generators

Example 1: Full Transient Solutions for Electromagnetic Braking Systems

Problem: A square wire loop of mass $m$, total circuit resistance $R$, and side length $L$ is dropped from rest under gravity. It falls vertically down the $-y$ direction, passing into a region containing a uniform horizontal magnetic field $\vec{B}$ perpendicular to the loop's surface plane. The field zone has a sharp horizontal top boundary. Derive the expression for the loop's velocity $v(t)$ while it is partially inside the field.

Solution:

Step 1: Determine flux and current as a function of velocity
Let $y(t)$ define the vertical entry depth of the loop into the field zone. The active flux tracking inside the loop is $\Phi(t) = B \cdot A(t) = B L y(t)$. By Faraday's Law, the induced e.m.f. matches: $$\varepsilon(t) = \frac{d\Phi}{dt} = BL\frac{dy}{dt} = BLv(t)$$ Using Ohm's Law, the induced current loop value is: $$I(t) = \frac{\varepsilon(t)}{R} = \frac{BLv(t)}{R}$$

Step 2: Construct the differential equation via Newton's Second Law
The lower horizontal wire segment carrying current $I(t)$ experiences an upward magnetic Lorentz force $F_B = ILB$. Substituting our current expression: $$F_B(t) = \left(\frac{BLv(t)}{R}\right)LB = \frac{B^2L^2}{R}v(t)$$ Defining downwards as our positive direction for kinematic tracking: $$\Sigma F_y = mg - F_B(t) = m\frac{dv}{dt} \implies mg - \frac{B^2L^2}{R}v = m\frac{dv}{dt}$$ Isolating the derivative yields a first-order linear differential equation: $$\frac{dv}{dt} = g - \left(\frac{B^2L^2}{mR}\right)v$$

Step 3: Separate variables and integrate
Let the damping constant be $\gamma = \frac{B^2L^2}{mR}$. Separating variables for integration from rest ($v=0$ at $t=0$): $$\frac{dv}{g - \gamma v} = dt \implies \int_{0}^{v} \frac{1}{g - \gamma v}\,dv = \int_{0}^{t}dt$$ $$-\frac{1}{\gamma}\ln\left(\frac{g-\gamma v}{g}\right) = t \implies 1 - \frac{\gamma v}{g} = e^{-\gamma t}$$ $$v(t) = \frac{g}{\gamma}\left(1 - e^{-\gamma t}\right)$$ $$\mathbf{v(t) = \frac{mgR}{B^2L^2}\left(1 - e^{-\frac{B^2L^2}{mR}t}\right)}$$

Example 2: Generator Coupled to a Series Inductive-Reactive Load

Problem: An AC armature coil with $N$ turns and area $A$ rotates at a constant angular frequency $\omega$ within a uniform field $B$. The generator terminals are hooked up to a series load possessing an internal resistance $R$ and a self-inductance $L$. Find the steady-state current equation $I(t)$ and evaluate the peak mechanical counter-torque $\tau_{\text{peak}}$ required to sustain uniform rotation.

Solution:

Step 1: Set up Kirchhoff's Loop Rule for the inductive circuit
The source e.m.f. is $\varepsilon(t) = \varepsilon_0\sin(\omega t) = BAN\omega\sin(\omega t)$. The reactive loop differential equation is: $$\varepsilon(t) - L\frac{dI}{dt} = IR \implies L\frac{dI}{dt} + IR = BAN\omega\sin(\omega t)$$

Step 2: Solve via complex impedance methods
The inductive reactance is $X_L = \omega L$. The total vector circuit impedance magnitude $Z$ is: $$Z = \sqrt{R^2 + (\omega L)^2}$$ Because the inductor naturally opposes current variations, the current waveform lags behind the driving source voltage by a phase angle $\phi$: $$I(t) = I_0\sin(\omega t - \phi)$$ $$I_0 = \frac{\varepsilon_0}{Z} = \frac{BAN\omega}{\sqrt{R^2 + (\omega L)^2}} \qquad \text{and} \qquad \phi = \tan^{-1}\left(\frac{\omega L}{R}\right)$$

Step 3: Calculate the instantaneous and peak torque components
The counter-torque exerted by the magnetic field on the spinning loop is $\tau_B(t) = NI(t)AB\sin(\omega t)$. Substituting our phase-lagged current: $$\tau_B(t) = NAB\left[\frac{BAN\omega}{\sqrt{R^2+(\omega L)^2}}\sin(\omega t - \phi)\right]\sin(\omega t) = \frac{B^2A^2N^2\omega}{\sqrt{R^2+(\omega L)^2}}\sin(\omega t)\sin(\omega t - \phi)$$ To evaluate the peak torque value required from the driving engine, we track the maximum amplitude of this product over a cycle. Using trigonometric identities, the peak amplitude simplifies to: $$\mathbf{\tau_{\text{peak}} = \frac{B^2A^2N^2\omega}{2\sqrt{R^2+(\omega L)^2}}\left(1 + \frac{R}{\sqrt{R^2+(\omega L)^2}}\right)}$$

Example 3: Induction via Coupled Multivariable Calculus Changes

Problem: A flexible single-turn circular loop of wire has an initial radius $r_0$ at $t=0$. It is placed inside a uniform magnetic field directed perpendicularly through its plane. An external mechanical mechanism causes the radius to shrink at a constant linear rate $\alpha$ (so $\frac{dr}{dt} = -\alpha$). Concurrently, the magnetic field decaying over time is defined by $B(t) = B_0 e^{-\lambda t}$. Determine the total induced e.m.f. ($\varepsilon(t)$) in the loop as a function of time.

Solution:

Since both the field strength $B(t)$ and the loop area $A(t)$ are moving variables with time, we must track the total derivative of the flux expression to solve Faraday's Law.

Step 1: Frame the time-dependent area and flux expressions
The instantaneous radius of the contracting circle at time $t$ is: $$r(t) = r_0 - \alpha t$$ The surface area expression is given by: $$A(t) = \pi [r(t)]^2 = \pi (r_0 - \alpha t)^2$$ The combined variable magnetic flux linking through the loop structure ($N=1$) is: $$\Phi(t) = B(t)A(t) = \left(B_0 e^{-\lambda t}\right) \cdot \left(\pi (r_0 - \alpha t)^2\right) = \pi B_0 e^{-\lambda t}(r_0 - \alpha t)^2$$

Step 2: Apply the multivariable chain rule differentiation
Using the product rule to differentiate the flux with respect to time: $$\varepsilon(t) = -\frac{d\Phi}{dt} = -\left[ \frac{dB}{dt}A(t) + B(t)\frac{dA}{dt} \right]$$ Evaluate the individual component derivatives: $$\frac{dB}{dt} = -\lambda B_0 e^{-\lambda t}$$ $$\frac{dA}{dt} = 2\pi(r_0 - \alpha t) \cdot \frac{d}{dt}(r_0 - \alpha t) = -2\pi\alpha(r_0 - \alpha t)$$

Step 3: Combine and simplify the final equation
Substitute these derivatives back into the expanded Faraday equation: $$\varepsilon(t) = -\left[ \left(-\lambda B_0 e^{-\lambda t}\right)\pi(r_0 - \alpha t)^2 + \left(B_0 e^{-\lambda t}\right)\big(-2\pi\alpha(r_0 - \alpha t)\big) \right]$$ Factoring out common terms ($\pi B_0 e^{-\lambda t}(r_0 - \alpha t)$) yields: $$\mathbf{\varepsilon(t) = \pi B_0 e^{-\lambda t}(r_0 - \alpha t)\Big[ \lambda(r_0 - \alpha t) + 2\alpha \Big]}$$