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

Kuch bhi nahi badal raha: $B$ constant, $A$ constant, $\theta$ constant. Toh $\frac{d{\Phi }_B}{dt}=0$ aur $\mathcal{E}=-\frac{d{\Phi }_B}{dt}=0\text{V}.$ Humne kya kiya: teeno knobs ($B$, $A$, $\theta$) check kiye. Kyun: EMF ko change se matlab hai, size se nahi. Ek bahut bada lekin frozen flux bhi zero deta hai.

Sirf $B$ badal raha hai, toh "changing field" knob use karo: $|\mathcal{E}|=A\left|\frac{dB}{dt}\right|=A\cdot \frac{\Delta B}{\Delta t}.$ $\theta =0$ se kyun faida: $\cos 0=1$, toh ${\Phi }_B=BA$ cleanly milta hai. $\frac{\Delta B}{\Delta t}=\frac{0.60-0.20}{2.0}=0.20\text{T/s},|\mathcal{E}|=0.050\times 0.20=0.010\text{V}=10\text{mV}.$

Har turn same flux link karta hai, aur pushes series mein add hote hain: $|\mathcal{E}|=NA\left|\frac{dB}{dt}\right|=50\times 0.050\times 0.20=0.50\text{V}.$ Humne kya kiya: ek turn ke answer ko $N=50$ se multiply kiya. Kyun: $N$ turns = $N$ loops stack hue, EMFs sum hote hain.

Figure dekho: moving rod (red) $\frac{dA}{dt}=Lv$ rate se nayi area sweep karta hai. Sirf area knob ghoomta hai, toh $\mathcal{E}=B\frac{dA}{dt}=BvL=0.30\times 4.0\times 0.25=0.30\text{V}.$ $BvL$ kyun: yahi Motional EMF ka result hai — har charge pe force $qvB$, charge per work $=BvL$.

EMF ek battery ki tarah kaam karta hai jo $R$ se current drive karta hai (Ohm's law): $I=\frac{\mathcal{E}}{R}=\frac{0.30}{0.60}=0.50\text{A}.$ $P=\mathcal{E}I=0.30\times 0.50=0.15\text{W}.$ $P=\mathcal{E}I$ kyun: yeh deliver hoti electrical power hai — aur, energy conservation se, bilkul wahi mechanical power jo tumhara haath Lenz drag se ladne ke liye supply karta hai.

Fixed loop, fixed angle → sirf $B$ badalta hai, lekin $\cos \theta$ permanently flux ko scale karta hai: ${\Phi }_B=BA\cos \theta \Rightarrow |\mathcal{E}|=A\cos \theta \left|\frac{dB}{dt}\right|.$ $\cos 60^{\circ }=0.5,|\mathcal{E}|=0.040\times 0.5\times 1.5=0.030\text{V}=30\text{mV}.$ $\cos \theta$ derivative ke bahar kyun: angle change nahi ho raha, toh yeh ek constant multiplier hai, variable nahi.

Hum kya karte hain: differentiate karo, kyunki $B$ ab $t$ pe non-linearly depend karta hai — ek constant "rise per second" ab exist nahi karta. ${\Phi }_B=AB(t)=0.020\times 0.10t^2=0.0020t^2.$ $|\mathcal{E}|=\left|\frac{d{\Phi }_B}{dt}\right|=0.0020\times 2t=0.0040t.$ $t=3.0$s pe: $|\mathcal{E}|=0.0040\times 3.0=0.012\text{V}=12\text{mV}$. Calculus kyun, $\Delta B/\Delta t$ nahi: field accelerate ho rahi hai, toh slope moment to moment badalta hai. Derivative instantaneous rate deta hai.

Step 1 — kya change ho raha hai: outward flux badh raha hai. Step 2 — Lenz kehta hai oppose karo: induced current ko apna field page ke andar banani chahiye loop ke andar, taaki growth se lad sake. Step 3 — right-hand rule: loop ke andar page ke andar field matlab current clockwise circulate karti hai. Rod mein hi (loop ka right edge), clockwise matlab current rod mein neeche flow karti hai. Force se check: current neeche × $\vec{B}$ page se bahar deta hai force $I\vec{L}\times \vec{B}$ jo left point karta hai — rightward pull ko oppose karta hai. Sahi: nature resist karti hai, bilkul jaisa Lenz's Law maangti hai.

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

• $0\le t<2$: 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. Signs kyun flip hote hain: jab flux badhta hai toh EMF use oppose karta hai (ek sign); jab flux girth hai toh EMF use prop up karne ki koshish karta hai (opposite sign). Flat beech mein kuch induce nahi hota.

$\theta =\omega t$ ke saath, ${\Phi }_B=BA\cos (\omega t)$, toh $\mathcal{E}=-N\frac{d}{dt}\left[BA\cos \omega t\right]=NBA\omega \sin \omega t.$ (a) Peak (jab $\sin \omega t=1$, yaani loop edge-on to $\vec{B}$): ${\mathcal{E}}_0=NBA\omega =200\times 0.40\times 0.015\times 100=120\text{V}.$ (b) $\theta =30^{\circ }$ pe: $\sin 30^{\circ }=0.5$, $\mathcal{E}={\mathcal{E}}_0\sin 30^{\circ }=120\times 0.5=60\text{V}.$ Sine kyun (cosine nahi): flux ek cosine hai; uska rate of change ek sine hai. Flux sabse tezi se badalta hai jahan cosine steepest hoti hai — $\theta =90^{\circ }$ pe, jab loop edge-on face karta hai. Electric Generators and AC dekho.

Peak EMF se peak current: $I_0=\frac{{\mathcal{E}}_0}{R}=\frac{120}{30}=4.0\text{A}.$ Peak instantaneous power: $P_0={\mathcal{E}}_0{I}_0=120\times 4.0=480\text{W}.$ "Peak" kyun: dono EMF aur current $\sin \omega t$ ki tarah oscillate karte hain; yeh unki maximum values hain, jo simultaneously reach hoti hain (pure resistance, no phase lag).

Ab area knob changing radius ke zariye ghoomta hai. $A=\pi r^2$, toh chain rule se $\frac{dA}{dt}=2\pi r\frac{dr}{dt}.$ $|\mathcal{E}|=B\left|\frac{dA}{dt}\right|=B\cdot 2\pi r\left|\frac{dr}{dt}\right|=0.50\times 2\pi \times 0.15\times \mathrm{0.10.}$ $|\mathcal{E}|=0.50\times 0.09425=0.0471\text{V}\approx 47\text{mV}.$ Chain rule kyun: area $r$ pe depend karta hai, aur $r$ $t$ pe depend karta hai. Hum dono rates link karte hain: $\frac{dA}{dt}=\frac{dA}{dr}\frac{dr}{dt}$.

Charge current ko time ke upar integrate karna hai: $q=\int Idt=\int \frac{\mathcal{E}}{R}dt=\int \frac{1}{R}\left(-\frac{d{\Phi }_B}{dt}\right)dt.$ $dt$ cancel ho jata hai — hum flux ko directly integrate karte hain: $q=-\frac{1}{R}{\int }_{{\Phi }_1}^{{\Phi }_2}d{\Phi }_B=\frac{{\Phi }_1-{\Phi }_2}{R}\Rightarrow |q|=\frac{|{\Phi }_2-{\Phi }_1|}{R}.$ Numerically: $|q|=\frac{|0.10-0.30|}{5.0}=\frac{0.20}{5.0}=0.040\text{C}$. Speed kyun matter nahi karta: fast change → thodi der ke liye bada EMF; slow change → lamba time ke liye chhota EMF. Current curve ke neeche area (charge) dono mein same hota hai — yeh sirf net flux swing pe depend karta hai.

Length $s$ ka trailing edge sliding rod ki tarah act karta hai; overlap area $sv$ rate se shrink hota hai. (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) Field mein current-carrying edge ek drag force feel karta hai $F=BIs=0.80\times 0.32\times 0.10=0.0256$N (pull ko oppose karta hai, Lenz's Law se). Constant $v$ pe chalane ke liye tumhara haath supply karta hai $P_{\text{mech}}=Fv=0.0256\times 2.0=0.0512$W. (d) Electrical power dissipated: $P_{\text{elec}}=I^{2}R=(0.32{)}^2\times 0.50=0.0512$W. Equal. ✓ Kyun match karna zaroori hai: koi kinetic energy gain nahi hoti (constant speed), aur koi energy store nahi hoti — toh tumhare haath ka har joule $R$ mein heat banta hai. Woh equality hi energy conservation hai, isliye minus sign (Lenz) optional nahi hai.

Do knobs ek saath ghoom rahe hain, toh ${\Phi }_B=B(t)A\cos (\omega t)$ pe product rule use karo: $\frac{d{\Phi }_B}{dt}=A[\underset{\text{field term}}{\underbrace{\frac{dB}{dt}\cos \omega t}}-\underset{\text{rotation term}}{\underbrace{B\omega \sin \omega t}}].$ $t=0$ pe: $\cos 0=1$, $\sin 0=0$, $B(0)=B_0=0.20$, $\frac{dB}{dt}=B_{0}k=0.20\times 0.50=0.10$T/s. $\frac{d{\Phi }_B}{dt}{|}_0=0.030[0.10\times 1-0.20\times 10\times 0]=0.030\times 0.10=\mathrm{0.0030.}$ $|\mathcal{E}|=0.0030\text{V}=3.0\text{mV}.$ Rotation term $t=0$ pe kyun vanish hota hai: $\theta =0$ pe loop field ke square-on face karta hai, jahan tilting ke against flux momentarily flat hota hai ($\sin 0=0$). Sirf field-growth term us instant pe survive karta hai. Ek moment baad rotation term jaag uthta hai.

Hamesha teen sawaal karo order mein: (1) Kya change ho raha hai — $B$, $A$, ya $\theta$? (2) Kya change time mein linear hai — average rate, ya differentiate karna hoga? (3) Kya mujhe sign/direction chahiye — toh Lenz aur energy conservation invoke karo.