Exercises — Faraday's law — EMF = −dΦ - dt

Nothing changes: $B$ constant, $A$ constant, $\theta$ constant. So $\frac{d{\Phi }_B}{dt}=0$ and $\mathcal{E}=-\frac{d{\Phi }_B}{dt}=0\text{V}.$ What we did: checked all three knobs ($B$, $A$, $\theta$). Why: EMF cares about change, not size. A giant but frozen flux still gives zero.

Only $B$ changes, so keep the "changing field" knob: $|\mathcal{E}|=A\left|\frac{dB}{dt}\right|=A\cdot \frac{\Delta B}{\Delta t}.$ Why $\theta =0$ helps: $\cos 0=1$, so ${\Phi }_B=BA$ cleanly. $\frac{\Delta B}{\Delta t}=\frac{0.60-0.20}{2.0}=0.20\text{T/s},|\mathcal{E}|=0.050\times 0.20=0.010\text{V}=10\text{mV}.$

Each turn links the same flux, and the pushes add in series: $|\mathcal{E}|=NA\left|\frac{dB}{dt}\right|=50\times 0.050\times 0.20=0.50\text{V}.$ What we did: multiplied the one-turn answer by $N=50$. Why: $N$ turns = $N$ loops stacked, EMFs sum.

Look at the figure: the moving rod (red) sweeps out new area at rate $\frac{dA}{dt}=Lv$. Only the area knob turns, so $\mathcal{E}=B\frac{dA}{dt}=BvL=0.30\times 4.0\times 0.25=0.30\text{V}.$ Why $BvL$: this is exactly the Motional EMF result — force $qvB$ on each charge, work per charge $=BvL$.

The EMF acts like a battery driving current through $R$ (Ohm's law): $I=\frac{\mathcal{E}}{R}=\frac{0.30}{0.60}=0.50\text{A}.$ $P=\mathcal{E}I=0.30\times 0.50=0.15\text{W}.$ Why $P=\mathcal{E}I$: this is the electrical power delivered — and, by energy conservation, exactly the mechanical power your hand supplies to fight the Lenz drag.

Fixed loop, fixed angle → only $B$ changes, but $\cos \theta$ scales the flux permanently: ${\Phi }_B=BA\cos \theta \Rightarrow |\mathcal{E}|=A\cos \theta \left|\frac{dB}{dt}\right|.$ $\cos 60^{\circ }=0.5,|\mathcal{E}|=0.040\times 0.5\times 1.5=0.030\text{V}=30\text{mV}.$ Why $\cos \theta$ stays outside the derivative: the angle isn't changing, so it's a constant multiplier, not a variable.

What we do: differentiate, because $B$ now depends on $t$ non-linearly — a constant "rise per second" no longer exists. ${\Phi }_B=AB(t)=0.020\times 0.10t^2=0.0020t^2.$ $|\mathcal{E}|=\left|\frac{d{\Phi }_B}{dt}\right|=0.0020\times 2t=0.0040t.$ At $t=3.0$s: $|\mathcal{E}|=0.0040\times 3.0=0.012\text{V}=12\text{mV}$. Why calculus and not $\Delta B/\Delta t$: the field accelerates, so the slope changes moment to moment. The derivative gives the instantaneous rate.

Step 1 — what's changing: outward flux is growing. Step 2 — Lenz says oppose: the induced current must create its own field pointing into the page inside the loop, to fight the growth. Step 3 — right-hand rule: a field into the page inside the loop means the current circulates clockwise. In the rod itself (right edge of the loop), clockwise means current flows downward through the rod. Check with force: current down × $\vec{B}$ out of page gives a force $I\vec{L}\times \vec{B}$ pointing left — opposing the rightward pull. Good: nature resists, exactly as Lenz's Law demands.

$\mathcal{E}=-\frac{d{\Phi }_B}{dt}=-(\text{slope})$ on each straight piece.

• $0\le t<2$: slope $=+0.10$Wb/s → $\mathcal{E}=-0.10$V.
• $2\le t<5$: slope $=0$ (flat) → $\mathcal{E}=0$V.
• $5\le t<9$: slope $=-0.05$Wb/s → $\mathcal{E}=+0.05$V. Why signs flip: when flux rises the EMF opposes it (one sign); when flux falls the EMF tries to prop it up (opposite sign). The flat middle induces nothing.

With $\theta =\omega t$, ${\Phi }_B=BA\cos (\omega t)$, so $\mathcal{E}=-N\frac{d}{dt}\left[BA\cos \omega t\right]=NBA\omega \sin \omega t.$ (a) Peak (when $\sin \omega t=1$, i.e. loop edge-on to $\vec{B}$): ${\mathcal{E}}_0=NBA\omega =200\times 0.40\times 0.015\times 100=120\text{V}.$ (b) At $\theta =30^{\circ }$: $\sin 30^{\circ }=0.5$, $\mathcal{E}={\mathcal{E}}_0\sin 30^{\circ }=120\times 0.5=60\text{V}.$ Why the sine (not cosine): the flux is a cosine; its rate of change is a sine. Flux changes fastest where the cosine is steepest — at $\theta =90^{\circ }$, when the loop faces edge-on. See Electric Generators and AC.

Peak current from peak EMF: $I_0=\frac{{\mathcal{E}}_0}{R}=\frac{120}{30}=4.0\text{A}.$ Peak instantaneous power: $P_0={\mathcal{E}}_0{I}_0=120\times 4.0=480\text{W}.$ Why "peak": both EMF and current oscillate as $\sin \omega t$; these are their maximum values, reached simultaneously (pure resistance, no phase lag).

Now the area knob turns via a changing radius. $A=\pi r^2$, so by the chain rule $\frac{dA}{dt}=2\pi r\frac{dr}{dt}.$ $|\mathcal{E}|=B\left|\frac{dA}{dt}\right|=B\cdot 2\pi r\left|\frac{dr}{dt}\right|=0.50\times 2\pi \times 0.15\times \mathrm{0.10.}$ $|\mathcal{E}|=0.50\times 0.09425=0.0471\text{V}\approx 47\text{mV}.$ Why the chain rule: area depends on $r$, and $r$ depends on $t$. We link the two rates: $\frac{dA}{dt}=\frac{dA}{dr}\frac{dr}{dt}$.

Charge is current integrated over time: $q=\int Idt=\int \frac{\mathcal{E}}{R}dt=\int \frac{1}{R}\left(-\frac{d{\Phi }_B}{dt}\right)dt.$ The $dt$ cancels — we integrate the flux directly: $q=-\frac{1}{R}{\int }_{{\Phi }_1}^{{\Phi }_2}d{\Phi }_B=\frac{{\Phi }_1-{\Phi }_2}{R}\Rightarrow |q|=\frac{|{\Phi }_2-{\Phi }_1|}{R}.$ Numerically: $|q|=\frac{|0.10-0.30|}{5.0}=\frac{0.20}{5.0}=0.040\text{C}$. Why speed doesn't matter: a fast change → big EMF for a short time; a slow change → small EMF for a long time. The area under the current curve (the charge) is the same either way — it depends only on the net flux swing.

The trailing edge of length $s$ acts as the sliding rod; overlap area shrinks at rate $sv$. (a) $\mathcal{E}=Bvs=0.80\times 2.0\times 0.10=0.16\text{V}$. (b) $I=\frac{\mathcal{E}}{R}=\frac{0.16}{0.50}=0.32\text{A}$. (c) The current-carrying edge in the field feels a drag force $F=BIs=0.80\times 0.32\times 0.10=0.0256$N (opposing the pull, by Lenz's Law). To move at constant $v$ your hand supplies $P_{\text{mech}}=Fv=0.0256\times 2.0=0.0512$W. (d) Electrical power dissipated: $P_{\text{elec}}=I^{2}R=(0.32{)}^2\times 0.50=0.0512$W. Equal. ✓ Why they must match: no kinetic energy is gained (constant speed), and no energy is stored — so every joule your hand does becomes heat in $R$. That equality is energy conservation, which is why the minus sign (Lenz) is not optional.

Two knobs turn at once, so use the product rule on ${\Phi }_B=B(t)A\cos (\omega t)$: $\frac{d{\Phi }_B}{dt}=A[\underset{\text{field term}}{\underbrace{\frac{dB}{dt}\cos \omega t}}-\underset{\text{rotation term}}{\underbrace{B\omega \sin \omega t}}].$ At $t=0$: $\cos 0=1$, $\sin 0=0$, $B(0)=B_0=0.20$, $\frac{dB}{dt}=B_{0}k=0.20\times 0.50=0.10$T/s. $\frac{d{\Phi }_B}{dt}{|}_0=0.030[0.10\times 1-0.20\times 10\times 0]=0.030\times 0.10=\mathrm{0.0030.}$ $|\mathcal{E}|=0.0030\text{V}=3.0\text{mV}.$ Why the rotation term vanishes at $t=0$: at $\theta =0$ the loop faces the field square-on, where flux is momentarily flat against tilting ($\sin 0=0$). Only the field-growth term survives at that instant. A moment later the rotation term wakes up.

Always ask three questions in order: (1) What is changing — $B$, $A$, or $\theta$? (2) Is the change linear in time — average rate, or must I differentiate? (3) Do I need the sign/direction — then invoke Lenz and energy conservation.