Lenz's law — opposing induced current: Exercises

1.8.27 · D4 · Physics › Electromagnetism › Lenz's law — opposing induced current

Do symbols jinhe hum baar baar use karte hain:

$Φ_{B}$ = magnetic flux, loop ke andar se guzarne wale field ka "amount." Uniform field ke liye yeh hai $Φ_{B} = B A cos θ$ , jahan $B$ field strength hai (tesla, T mein), $A$ loop ka area hai (in $m^{2}$ ), aur $θ$ $B$ aur loop ke normal (loop ke face se seedha bahar nikalti arrow) ke beech ka angle hai. Dekho Magnetic flux.
$ε = - \frac{d Φ _{B}}{d t}$ = Faraday's law with the Lenz sign. $\frac{d}{d t}$ rate of change hai — "flux har second kitni tezi se badal raha hai." Dekho Faraday's law of induction.

Level 1 — Recognition

L1·Q1

Ek bar magnet ka N-pole ek coil ke bilkul bahar bilkul still rakha gaya hai. Induced current bolo.

Recall Solution

KYA change ho raha hai? Kuch bhi nahi — magnet still hai, toh $Φ_{B}$ constant hai, toh $\frac{d Φ _{B}}{d t} = 0$ . Isliye $ε = - \frac{d Φ _{B}}{d t} = 0$ aur induced current zero hai. Trap yeh sochna hai ki "paas mein magnet hai, toh current zaroor hogi." Lenz change pe respond karta hai, presence pe nahi. Answer: koi current nahi.

L1·Q2

Loop ke through flux decrease ho raha hai. Induced field loop ke andar external field $B$ ke saath point karta hai ya against?

Recall Solution

Flux decrease hona matlab loop jo uske paas tha woh kho raha hai. Loss ko oppose karne ke liye, woh flux ko maintain karne ki koshish karta hai, toh induced field $B$ ke saath same direction mein point karta hai (isse support karta hai). Answer: $B$ ke saath.

L1·Q3

Ek sentence mein batao, Lenz's law aapko kaunsi quantity deta hai jo akele magnitude $\frac{d Φ _{B}}{d t}$ nahi deta.

Recall Solution

Induced current ki direction (sense) — yaani $ε = - \frac{d Φ _{B}}{d t}$ mein minus sign ka physical meaning.

Level 2 — Application

L2·Q1

Figure dekho. Ek uniform field $B$ ek fixed square loop ke through page ke andar point kar raha hai, aur iska strength time ke saath increase ho raha hai. Induced current kis taraf flow karta hai — clockwise (CW) ya counterclockwise (CCW)?

Figure — Lenz's law — opposing induced current

Recall Solution

Step 1 (external field): page ke andar ( $\times$ symbols). $\times$ kyun? " $\times$ " ek arrow ki tail hai jo aapse dur ja rahi hai — page ke andar. Step 2 (trend): $B$ increase ho raha hai ⇒ into-page flux increase ho raha hai. Step 3 (oppose): zyada into-page flux ko oppose karne ke liye, loop ke andar induced field page ke bahar point karna chahiye. Step 4 (Right-hand rule): apna right thumb page ke bahar point karo; curled fingers counterclockwise circle karti hain. Toh induced current CCW hai. Answer: counterclockwise.

L2·Q2

Same loop aur same into-the-page field, lekin ab field strength decrease ho raha hai. Current ki direction?

Recall Solution

Into-page flux ab gir raha hai. Girne ko oppose karne ke liye, loop into-page flux ko maintain karne ki koshish karta hai ⇒ induced field page ke andar point karta hai. Right thumb page ke andar ⇒ fingers clockwise circle karti hain. Answer: clockwise — bilkul L2·Q1 ka ulta, kyunki trend reverse hua jabki field ki direction nahi. Yahi parent note ki pehli common mistake ka core hai.

L2·Q3

Length $L = 0.50 m$ ka ek rod rails par $v = 4.0 m/s$ se field $B = 0.20 T$ (page ke andar) mein slide karta hai. Circuit resistance $R = 2.0 Ω$ hai. $∣ ε ∣$ , current $I$ , aur drag force $F$ nikalo.

Recall Solution

Motional EMF (dekho Motional EMF and sliding rod): jab rod move karta hai, enclosed area badhta hai, toh $Φ_{B} = B Lx$ aur $∣ ε ∣ = \frac{d Φ _{B}}{d t} = B L \frac{d x}{d t} = B Lv = (0.20) (0.50) (4.0) = 0.40 V .$ Current: $I = \frac{∣ ε ∣}{R} = \frac{0.40}{2.0} = 0.20 A .$ Current-carrying rod par force: $F = B I L = (0.20) (0.20) (0.50) = 0.020 N .$ Lenz ke according yeh force motion ko oppose karti hai (drag). Answers: $0.40 V$ , $0.20 A$ , $0.020 N$ .

Level 3 — Analysis

L3·Q1

L2·Q3 ke rod ke liye energy conservation verify karo: dikhao ki jo mechanical power tum supply karte ho woh electrical power dissipated ke barabar hai.

Recall Solution

Mechanical power (tum drag $F$ ke against speed $v$ se push karte ho): $P_{mech} = F v = \frac{B ^{2} L ^{2} v ^{2}}{R} .$ Plug in: $P_{mech} = \frac{( 0.20 ) ^{2} ( 0.50 ) ^{2} ( 4.0 ) ^{2}}{2.0} = \frac{0.04 \times 0.25 \times 16}{2.0} = \frac{0.16}{2.0} = 0.080 W .$ Electrical power (resistor mein heat): $P_{elec} = I^{2} R = (0.20)^{2} (2.0) = 0.080 W .$ Dono match karte hain. Har joule jo tum push in karte ho woh resistor heat ke roop mein wapas aata hai — yeh hi Conservation of energy hai jo minus sign pahne hua hai. Answer: dono $0.080 W$ , barabar.

L3·Q2

Mass $m = 0.010 kg$ ka ek magnet ek vertical copper tube se gir ke ek terminal (constant) speed $v_{t}$ tak pahunchta hai. $v_{t}$ par magnetic drag exactly gravity ko balance karta hai. Agar drag $F_{drag} = k v$ follow karta hai jahan $k = 0.20 N\cdots/m$ hai, $v_{t}$ nikalo. ( $g = 9.8 m/s^{2}$ lo.)

Recall Solution

Drag kyun hota hai? Girta hua magnet tube ke rings ke through flux change karta hai, eddy currents induce karta hai. Lenz ke according yeh fall ko oppose karte hain — ek retarding force. Terminal condition: acceleration zero hai, toh drag = weight: $k v_{t} = m g \Rightarrow v_{t} = \frac{m g}{k} = \frac{( 0.010 ) ( 9.8 )}{0.20} = 0.49 m/s .$ Answer: $v_{t} = 0.49 m/s$ . Isse zyada tez ⇒ drag > weight ⇒ slow ho jaata hai; isse kam ⇒ weight > drag ⇒ speed up ho jaata hai. Yeh $v_{t}$ par settle ho jaata hai — drag self-correcting hai lekin fall kabhi reverse nahi karta (parent note ki doosri mistake).

L3·Q3

Ek loop ka normal ek uniform field $B = 0.50 T$ ke saath angle $θ$ banata hai; area $A = 0.020 m^{2}$ hai. Loop ko rotate kiya jaata hai taaki $θ$ $0^{\circ}$ (face-on) se $9 0^{\circ}$ (edge-on) tak $0.10 s$ mein constant rate se jaye. Average EMF magnitude nikalo.

Recall Solution

Start par flux ( $θ = 0^{\circ}$ , $cos 0 = 1$ ): $Φ_{i} = B A cos 0^{\circ} = (0.50) (0.020) (1) = 0.010 Wb .$ End par flux ( $θ = 9 0^{\circ}$ , $cos 9 0^{\circ} = 0$ ): $Φ_{f} = 0$ . Average EMF: $∣ ε_{avg} ∣ = \frac{Φ _{f} - Φ _{i}}{Δ t} = \frac{0.010}{0.10} = 0.10 V .$ Lenz ke according current flux ke loss ko oppose karne ke liye flow karta hai, yaani woh field ko loop se thread karte rehne ki koshish karta hai. Answer: $0.10 V$ .

Level 4 — Synthesis

L4·Q1

Coil se magnet bahar kheenche jaane ka poora picture banao. Ek N-pole coil se door kheeencha jaata hai. (a) Kya near-face flux increase ya decrease hota hai? (b) Coil ka near face kaun sa magnetic pole ban jaata hai? (c) Magnet ki taraf se dekhe current ki direction? (d) Kya coil magnet ko pull ya push karti hai?

Recall Solution

(a) Jab magnet jaata hai, distance badhta hai, coil par $B$ girta hai, toh flux (jo coil ke andar point kar raha tha, kyunki field lines N-pole se nikalti hain) decrease hota hai. (b) Loss ko oppose karne ke liye, coil inward flux ko maintain karne ki koshish karti hai ⇒ iska induced field coil ke andar magnet ki taraf point karta hai ⇒ near face south pole ban jaata hai (field lines S mein enter karti hain). (c) Induced field coil ke andar (magnet wali side ke observer se door) point karne ke liye, Right-hand rule magnet se dekhe clockwise current deta hai. (d) Ek near-face S-pole jaate N-pole ke saamne attract karta hai — coil flux maintain karne ke liye magnet ko wapas kheenchne ki koshish karti hai. Answers: decrease; south; clockwise; pull karta hai (attracts).

L4·Q2

Side $a = 0.10 m$ ( $A = a^{2}$ ) ka ek single square loop ek aise field mein apne plane ko perpendicular rakh ke baitha hai jo linearly ramp karta hai: $B (t) = B_{0} + α t$ jahan $B_{0} = 0.30 T$ aur $α = 2.0 T/s$ hai. Resistance $R = 0.50 Ω$ hai. Induced current magnitude nikalo, aur batao kya yeh time par depend karta hai.

Recall Solution

Flux (face-on, $cos θ = 1$ ): $Φ_{B} = B (t) A = (B_{0} + α t) a^{2} .$ Rate of change: $\frac{d Φ _{B}}{d t} = α a^{2}$ — constant $B_{0}$ gayab ho jaata hai (iska derivative zero hai), sirf changing part $α t$ bachta hai. $∣ ε ∣ = α a^{2} = (2.0) (0.10)^{2} = 0.020 V, I = \frac{∣ ε ∣}{R} = \frac{0.020}{0.50} = 0.040 A .$ Kyunki $α$ constant hai, $I$ time mein constant hai ( $t$ se independent). Offset $B_{0}$ aur elapsed time current mein kabhi enter nahi karte. Answer: $I = 0.040 A$ , time-independent.

Level 5 — Mastery

L5·Q1 (degenerate case)

Ek loop uniform, unchanging field ke through constant velocity se move karta hai, poori tarah field region ke andar rehta hai, apne plane ko $B$ ke perpendicular rakh ke. Kya koi induced current hai? Flux se justify karo.

Recall Solution

Flux $= B A cos 0^{\circ} = B A$ . Jab loop translate karta hai, $B$ har jagah same hai, $A$ fixed hai, aur orientation fixed hai, toh $Φ_{B}$ kabhi change nahi hota: $\frac{d Φ _{B}}{d t} = 0$ . Isliye $ε = 0$ aur koi induced current nahi. Sirf motion kaafi nahi — tumhe flux change hona chahiye. Yeh classic degenerate case hai jo real understanding ko pattern-matching se alag karta hai. Answer: zero. (Contrast: L2·Q3 ka sliding rod current isliye rakhta hai kyunki enclosed area badhta hai — wahan loop ki boundary expand ho rahi hai, toh flux change hota hai.)

L5·Q2 (edge-crossing, sign flip)

Same loop ab is tarah move karta hai ki iska leading edge field region se bahar nikal raha hai field-free space mein, jabki trailing edge abhi andar hai. Describe karo ki induced current ki direction entry ke moment par exit ke moment se kaise compare karti hai, aur kyun.

Recall Solution

Field mein enter karte waqt: loop ka woh area jo field ke andar hai woh badhta hai, toh loop ke through flux increase hota hai ⇒ induced current increase ko oppose karta hai (iska field incoming $B$ se ladhta hai) ⇒ ek definite sense, maan lo CCW ( $B$ ke liye page ke bahar). Field se bahar nikalte waqt: ab inside-field area shrink hota hai, flux decrease hota hai ⇒ induced current loss ko oppose karne ke liye reverse hota hai (iska field $B$ ko prop up karta hai) ⇒ opposite sense (CW). Dono moments mein loop par induced force motion ko oppose karti hai (entry par aur exit par drag), toh tumhe dono baar kaam karna padta hai — phir se Conservation of energy. Answer: current direction exit par entry ke ulta hoti hai; drag dono mein motion ko oppose karta hai.

L5·Q3 (limiting behaviour / self-inductance)

Ek coil battery se maintain steady current le raha hai. Battery suddenly switch off ho jaati hai. Lenz ki idea use karke, switch-off ke baad ke instant mein coil ke self-induced current ki direction explain karo, aur kyun switch par ek spark jump kar sakta hai.

Recall Solution

Coil apne through khud jo flux banata tha woh steady tha; switch off karna isse fast zero par drop karne ki koshish karta hai, toh $\frac{d Φ _{B}}{d t}$ large aur negative hai. Lenz ke according coil apne khud ke flux ke is loss ko oppose karta hai, toh yeh ek self-induced current original current ke same direction mein drive karta hai, flux ko zinda rakhne ki koshish karte hua — yeh back-EMF hai. Kyunki change almost instantaneous hai, $\frac{d Φ _{B}}{d t}$ bahut bada hai, ek large EMF produce karta hai jo opening switch ke tiny air gap ke across current push kar sakta hai — ek spark. Answer: current original direction mein continue karta hai; large back-EMF gap ko ionize karta hai aur spark deta hai.

Connections

Faraday's law of induction — $∣ ε ∣$ supply karta hai; Lenz yahan har solution mein use hone wala sign supply karta hai.
Magnetic flux — $Φ_{B} = B A cos θ$ jo humne poore mein differentiate kiya.
Right-hand rule — har "induced-field direction" ko ek current sense mein convert kiya.
Motional EMF and sliding rod — L2·Q3 / L3·Q1.
Eddy currents and magnetic braking — L3·Q2 terminal-velocity magnet.
Conservation of energy — L3·Q1 mein power balance aur drag arguments.
Self-inductance and back-EMF — L5·Q3 switch-off spark.
Lenz's law — opposing induced current (index 1.8.27) — parent topic.