Lenz's law — opposing induced current: Worked examples

1.8.27 · D3 · Physics › Electromagnetism › Lenz's law — opposing induced current

Kuch bhi shuru karne se pehle, ek symbol contract taaki hum kabhi notation chupa ke na le jayein:

Definition Woh symbols jo hum use karenge (har ek yahan ek baar clearly define kiya gaya hai)

$Φ_{B}$ = magnetic flux — ise loop se guzarne wali magnetic field lines ki ginti samjho. Bada loop ya strong field ya lines seedha loop se takra rahi hain → zyada lines andar → bada $Φ_{B}$ . Poora build Magnetic flux mein hai. Uniform field ke liye formula: $Φ_{B} = B A cos θ$ .
$B$ = external magnetic field ki strength (unit: tesla, T).
$A$ = loop ka area jisme se field guzarti hai (unit: $m^{2}$ ).
$θ$ = field $B$ aur loop ki normal (ek imaginary arrow jo loop ke face se seedha bahar nikalta hai) ke beech ka angle.
$N$ = turns ki ginti (wire ke loops jo ek coil mein stack hote hain). $N$ identical loops har ek same flux pakadta hai, isliye total $N$ guna bada ho jaata hai — isliye coil ke liye EMF mein $N$ multiply hota hai.
$\frac{d Φ _{B}}{d t}$ = flux kitni tezi se change ho raha hai per second. $\frac{d}{d t}$ ka matlab hai "time ke saath change ki rate". Positive = flux badh raha hai; negative = flux ghut raha hai; zero = flux frozen hai.
$ε$ = induced EMF (electrical "push", volts mein) jo $ε = - \frac{d Φ _{B}}{d t}$ se milta hai (Faraday's law of induction).
$R$ = loop ki resistance (ohms, $Ω$ ); induced current hai $I = \frac{∣ ε ∣}{R}$ .

Scenario matrix

Lenz's law jo bhi karta hai woh yeh decide karta hai ki loop kis taraf ladta hai. Sirf teen cheezein flux $Φ_{B} = B A cos θ$ ko change kar sakti hain: field $B$ , area $A$ , ya angle $θ$ . Ise change ke sign aur degenerate cases ke saath combine karo, aur yeh hai poora map:

#	Cell (case class)	Kya change hota hai	Sign of $d Φ_{B} / d t$	Example
C1	Field badhta hai	$B ↑$ , $A, θ$ fixed	$+$	Ex 1
C2	Field ghatta hai	$B ↓$	$-$	Ex 2
C3	Area badhta hai (moving conductor)	$A ↑$	$+$	Ex 3
C4	Rotation (angle sweep karta hai)	$θ$ changes	oscillating $\pm$	Ex 4
C5a	Degenerate: field constant, koi motion nahi	kuch nahi	$0$ (koi current nahi!)	Ex 5a
C5b	Degenerate: motion field ke parallel	$θ = 9 0^{\circ}$ hamesha	$0$	Ex 5b
C6	Limit: bahut fast vs bahut slow motion	$v \to \infty$ , $v \to 0$	$v$ ke saath scale karta hai	Ex 6
C7	Real-world word problem	falling magnet / brake	$-$ phir damping	Ex 7
C8	Exam twist: loop shrinks / sign trap	$A ↓$ ya reversed $B$	flip ho jaata hai	Ex 8

Ab hum har cell fill karte hain.

Cell C1 — Field badhta hai

Worked example Example 1 — Electromagnet ko ramp up karna (cell C1)

Radius $r = 0.10 m$ ka ek flat circular loop flat, face-up pada hai, ek uniform field ke andar jo seedha upar se guzar raha hai ( $θ = 0$ ). Field ko $B_{0} = 0.20 T$ se $B_{1} = 0.80 T$ tak $t = 0.30 s$ mein steadily badha diya jaata hai. Loop ki resistance $R = 2.0 Ω$ hai. $∣ ε ∣$ , current, aur current ki direction (upar se dekha jaaye) nikalo.

Forecast: field upar hai aur badh raha hai. Guess karo: kya loop ka induced current us upar wale field ke saath push karta hai ya against? Padhne se pehle apna guess likho.

Loop ka area. $A = π r^{2} = π (0.10)^{2} = 0.0314 m^{2}$ . Yeh step kyun? Flux ke liye area chahiye; field seedha guzar rahi hai isliye $cos θ = cos 0 = 1$ aur $Φ_{B} = B A$ .
Flux change ki rate. Sirf $B$ change hota hai, isliye $\frac{d Φ _{B}}{d t} = A \frac{d B}{d t} = A \frac{B _{1} - B _{0}}{t} = 0.0314 \times \frac{0.80 - 0.20}{0.30}$ . Yeh step kyun? Faraday ko rate chahiye; $A$ constant hai isliye woh sirf $B$ ke slope ko multiply karta hai. $\frac{d B}{d t} = \frac{0.60}{0.30} = 2.0 T/s$ , isliye $\frac{d Φ _{B}}{d t} = 0.0628 Wb/s$ .
EMF magnitude. $∣ ε ∣ = 0.0628 V = 62.8 mV$ .
Current. $I = \frac{∣ ε ∣}{R} = \frac{0.0628}{2.0} = 0.0314 A = 31.4 mA$ .
Direction (Lenz). Upar wala flux increasing hai → loop increase ko oppose karta hai → uska induced field loop ke andar neeche point karna chahiye → Right-hand rule se (fingers curl karo taaki thumb neeche point kare), current clockwise viewed from above chalta hai. Neeche figure dekho.

Verify: units — $T \cdot m^{2} / s = Wb/s = V$ ✓. Agar tumhara forecast yeh kehta tha ki "loop field ke saath push karta hai," toh woh increase mein madad karta → free runaway energy → forbidden. Sahi answer ise oppose karta hai. ✓

Figure — Lenz's law — opposing induced current

Figure 1 (cell C1). Blue up-arrows growing external field $B$ hain. Loop increase ko apna induced field (red, neeche pointing) loop ke andar bana ke fight karta hai. Right hand ko is tarah curl karo ki thumb neeche point kare toh yellow clockwise current milta hai, upar se dekha jaaye.

Cell C2 — Field ghatta hai

Worked example Example 2 — Electromagnet ko off karna (cell C2)

Same loop jaise Ex 1 mein ( $A = 0.0314 m^{2}$ , $R = 2.0 Ω$ ). Ab field $0.80 T$ se $0$ tak $0.20 s$ mein drop ho jaati hai. Current aur direction nikalo.

Forecast: field upar thi, ab khatam ho rahi hai. Ab loop kis taraf fight karega — Ex 1 ke ulta?

Rate. $\frac{d Φ _{B}}{d t} = A \frac{0 - 0.80}{0.20} = 0.0314 \times (- 4.0) = - 0.1256 Wb/s$ . Yeh step kyun? Negative isliye kyunki flux gir raha hai. Sign direction ke liye matter karta hai.
EMF. $∣ ε ∣ = 0.1256 V$ .
Current. $I = \frac{0.1256}{2.0} = 0.0628 A = 62.8 mA$ .
Direction (Lenz). Upar wala flux decreasing hai → loop use prop up karna chahta hai → induced field upar point karta hai (dying field ke saath same) → RHR se counterclockwise viewed from above milta hai — bilkul Ex 1 ka ulta.

Verify: $d Φ_{B} / d t$ ka sign $+$ (Ex 1) se $-$ (Ex 2) ho gaya, isliye current direction flip ho gayi. Yeh parent note ka key point hai: change ko oppose karo, field ko nahi. ✓ Yahan "change ko oppose karna" matlab hai induced field ko $B$ ke saath point karna. ✓

Cell C3 — Area badhta hai (moving conductor)

Worked example Example 3 — Sliding rod, full numbers (cell C3)

$L = 0.50 m$ lambi ek rod rails par $v = 3.0 m/s$ se slide karti hai ek field $B = 0.40 T$ mein jo page ke andar ja raha hai. Loop resistance $R = 1.5 Ω$ hai. $∣ ε ∣$ , $I$ , drag force, power, aur current direction nikalo. (Deep quantitative build Motional EMF and sliding rod mein hai.)

Forecast: rod slide karne se enclosed area badh raha hai. Current direction (clockwise / counterclockwise) compute karne se pehle guess karo.

Flux. Rod position $x$ par ho toh area $= Lx$ , isliye $Φ_{B} = B Lx$ . Yeh step kyun? Field uniform hai aur loop ke perpendicular hai ( $θ = 0$ ), isliye $Φ_{B} = B A = B Lx$ .
Rate. $\frac{d Φ _{B}}{d t} = B L \frac{d x}{d t} = B Lv$ . Sirf $x$ change hota hai. $∣ ε ∣ = B Lv = 0.40 \times 0.50 \times 3.0 = 0.60 V$ .
Current. $I = \frac{∣ ε ∣}{R} = \frac{0.60}{1.5} = 0.40 A$ .
Direction (Lenz). Flux page ke andar ja raha hai aur badh raha hai → induced field loop ke andar page se bahar point karna chahiye → RHR ⇒ current counterclockwise. Neeche figure dekho.
Drag force. Current-carrying rod field mein $F = B I L = 0.40 \times 0.40 \times 0.50 = 0.080 N$ feel karta hai, motion ke against directed. Yeh step kyun? Lenz guarantee karta hai ki force woh oppose kare jo tum kar rahe ho — tumhe push karna padega.
Power balance. Mechanical power in $= F v = 0.080 \times 3.0 = 0.24 W$ . Electrical power dissipated $= I^{2} R = (0.40)^{2} \times 1.5 = 0.24 W$ .

Verify: $F v = I^{2} R = 0.24 W$ — exactly match karta hai. Yeh Conservation of energy hai: jo bhi joule tum push karo woh heat ban jaata hai. ✓ Units: $T \cdot m \cdot m/s = V$ ✓.

Figure 2 (cell C3). Blue "×" marks field $B$ ko page ke andar jaate hue dikhate hain. Jab yellow rod speed $v$ se right slide karti hai (green), enclosed area — aur isliye andar jaata flux — badhta hai. Lenz red counterclockwise current force karta hai (loop ke andar page se bahar field banata hai), aur rod ko red drag force milta hai jo $v$ ke against wapas point karta hai.

Cell C4 — Rotation (angle sweep karta hai)

Worked example Example 4 — Rotating loop (AC generator) (cell C4)

Area $A = 0.020 m^{2}$ aur $N = 50$ turns ki ek coil field $B = 0.30 T$ mein angular speed $ω = 100 rad/s$ se spin karti hai. Uski normal field ke saath angle $θ = ω t$ banati hai. Peak EMF nikalo aur oscillating sign explain karo.

Forecast: ab kuch enter ya exit nahi ho raha; coil bas ghoom rahi hai. Spin mein kahan EMF sabse bada hai, aur kahan zero hai? Padhne se pehle guess karo.

Flux vs angle. $Φ_{B} = N B A cos θ = N B A cos (ω t)$ . Yeh step kyun? Rotation field aur normal ke beech ke angle ko change karta hai, isliye yahan $cos θ$ factor kaam karta hai — $B$ ya $A$ nahi. Factor $N$ isliye aata hai kyunki saare $N$ turns same flux pakadते hain.
Rate. $\frac{d Φ _{B}}{d t} = - N B A ω sin (ω t)$ , isliye $ε = N B A ω sin (ω t)$ . Yeh step kyun? $cos$ ka derivative $- sin$ hota hai; $- d Φ/ d t$ se minus aur derivative se minus milake clean $+ sin$ milta hai.
Peak EMF. $sin$ ka maximum $1$ hai: $ε_{m a x} = N B A ω = 50 \times 0.30 \times 0.020 \times 100 = 30 V$ .
Zero kahan, max kahan? Jab coil ka face field ke perpendicular ho ( $θ = 0$ , flux maximum) toh rate $sin θ = 0$ hai, isliye $ε = 0$ . Jab face field ke parallel ho ( $θ = 9 0^{\circ}$ , flux zero) toh rate sabse fast hai, isliye $ε$ peak par hai. Sign har half turn mein flip hota hai — isliye yeh alternating current hai.

Verify: peak EMF $= 30 V$ . Sanity check: units $T \cdot m^{2} \cdot s^{- 1} = V$ ✓. Yeh famous counterintuitive fact — EMF zero hai jab flux maximum hai — seedha " $ε$ rate track karta hai, value nahi" se nikalta hai. ✓

Cell C5a aur C5b — Degenerate cases (bilkul koi current nahi)

Cell C6 — Limiting behaviour (fast vs slow)

Worked example Example 6 — Drag speed ke saath kaise scale karta hai (cell C6, limits)

Ex 3 ka sliding rod lo ( $B = 0.40 T$ , $L = 0.50 m$ , $R = 1.5 Ω$ ). Drag force ko $v$ ke function ke roop mein likho, phir $v = 0$ , $v = 3.0 m/s$ , aur $v = 30 m/s$ par evaluate karo.

Forecast: kya speed double karne se drag double hota hai, ya zyada? Aur $v = 0$ par?

General drag. Ex 3 se: $I = \frac{B Lv}{R}$ aur $F = B I L = \frac{B ^{2} L ^{2} v}{R}$ . Yeh step kyun? Dono results ko chain karo — force $v$ mein linear hai.
Coefficient. $\frac{B ^{2} L ^{2}}{R} = \frac{( 0.40 ) ^{2} ( 0.50 ) ^{2}}{1.5} = \frac{0.16 \times 0.25}{1.5} = 0.02667 N\cdots/m$ .
Evaluate.
- $v = 0$ : $F = 0$ — rest par rod ko koi magnetic drag nahi lagta. (Degenerate limit, C5a se match karta hai.)
- $v = 3.0$ : $F = 0.02667 \times 3.0 = 0.080 N$ — Ex 3 se match karta hai ✓.
- $v = 30$ : $F = 0.02667 \times 30 = 0.80 N$ — das guna speed, das guna drag (linear).

Verify: parent ne warn kiya tha "faster ⇒ instant stop" ke against. Kyunki $F \propto v$ hai (na $v^{2}$ ya zyada), drag smoothly badhta hai aur sirf energy remove karta hai — yeh kabhi motion reverse nahi karta ya energy create nahi karta. Jab $v \to \infty$ toh finite $v$ ke liye force bada lekin finite hai; jab $v \to 0$ toh yeh kuch nahi ho jaata. ✓

Cell C7 — Real-world word problem

Worked example Example 7 — Copper pipe mein girta magnet (cell C7)

Ek chota magnet ek vertical copper pipe mein drop kiya jaata hai. $g$ par accelerate karne ki bajay, yeh dheere dheere neeche drift karta hai almost constant "terminal" speed par. Lenz's law se explain karo aur terminal condition estimate karo.

Forecast: ek magnet copper mein same size ki plastic pipe se slower kyun hota hai? Mechanism guess karo.

Falling = changing flux. Jab magnet girta hai, pipe ke har ring-shaped slice ka flux neeche usse badh ta hai (magnet approach kar raha hai) aur har slice upar ka ghatta hai (magnet ja raha hai). Dono changes copper walls mein eddy currents drive karte hain. Yeh step kyun? Copper conductor hai, isliye induced current ke loops circulate ho sakte hain — yahi Lenz ko chahiye.
Lenz direction. Magnet ke neeche current ek aisa face banata hai jo use repel karta hai (approach fight karta hai); upar, ek aisa face jo use attract karta hai (departure fight karta hai). Dono upar act karte hain — fall ko oppose karne wala retarding force.
Terminal condition. Magnet tab tak speed up karta hai jab tak magnetic drag $F_{drag} (v)$ (jo $v$ ke saath badhta hai, bilkul Ex 6 jaisa) gravity ko balance na kar de: $F_{drag} (v_{term}) = m g$ . Uske baad, net force $= 0$ , isliye speed constant hai.

Verify: Plastic pipe mein koi current flow nahi kar sakta ( $R = \infty$ , isliye $I = 0$ ), koi drag nahi — magnet $g$ par girta hai. Copper mein drag real hai, aur saari lost gravitational energy pipe mein heat ban jaati hai (Conservation of energy). Yeh wahi physics hai jo Eddy currents and magnetic braking mein train brakes aur drop-tower rides mein use hoti hai. ✓

Cell C8 — Exam twist (sign trap)

Worked example Example 8 — Shrinking loop reversed-field trap ke saath (cell C8)

Ek square loop, side $s$ , ek field $B = 0.50 T$ mein pada hai jo page se bahar point kar raha hai. Koi loop ko squeeze karta hai taaki uski side $s_{0} = 0.20 m$ se $s_{1} = 0.10 m$ tak $t = 0.40 s$ mein shrink ho jaye. $R = 0.80 Ω$ . Current magnitude aur direction nikalo. Phir twist ka jawab do: agar field page ke andar point kare, kya current direction change hogi?

Forecast: loop shrink ho rahi hai — kya yeh flux ka increase hai ya decrease? Aur current kis taraf jaata hai? Guess karo.

Flux. $Φ_{B} = B s^{2}$ (field seedha guzar rahi hai, $cos 0 = 1$ ).
Rate. $\frac{d Φ _{B}}{d t} = B \frac{d ( s ^{2} )}{d t}$ . Average slope use karte hue: $\frac{s _{1}^{2} - s _{0}^{2}}{t} = \frac{0.01 - 0.04}{0.40} = \frac{- 0.03}{0.40} = - 0.075 m^{2} / s$ . Isliye $\frac{d Φ _{B}}{d t} = 0.50 \times (- 0.075) = - 0.0375 Wb/s$ — flux decreasing hai (chota loop kam lines pakadta hai). Yeh step kyun? Area $s^{2}$ ke saath girta hai; shrinking decrease hai chahe tum "kuch kar" rahe ho.
EMF aur current. $∣ ε ∣ = 0.0375 V$ , $I = \frac{0.0375}{0.80} = 0.0469 A \approx 46.9 mA$ .
Direction (Lenz). Page se bahar flux decreasing hai → loop use prop up karta hai → induced field page se bahar point karta hai → RHR ⇒ current counterclockwise.
The twist. Agar field page ke andar point kare, toh flux (page ke andar) bhi decrease ho raha hai → induced field page ke andar point karna chahiye use prop up karne ke liye → RHR ⇒ current clockwise — ulti direction.

Verify: current magnitude dono field orientations mein identical hai ( $∣ ε ∣$ sirf $∣ d Φ_{B} / d t ∣$ par depend karta hai), lekin direction reverse hoti hai jab field reverse hoti hai. Trap yeh hai ki sochna "shrinking loop = kuch badhta hai." Nahi: fixed field ke saath area shrinking = decreasing flux. ✓

Recall Self-test: solve karne se pehle cell ka naam batao

A coil sits in a field that is constant in size but the coil is slowly tilted from face-on to edge-on. Increase ya decrease of flux? Kaun sa cell? ::: Decrease ( $cos θ$ $1$ se $0$ tak girta hai); yeh C4-type (angle change) hai jo C5b degenerate zero-flux limit ki taraf slide kar raha hai. Sliding rod achanak rok di jaati hai. Rok ne ke ek instant baad induced current kya hai? ::: Zero — $v = 0$ isliye $∣ ε ∣ = B Lv = 0$ (C6 ka $v \to 0$ limit). Current direction kaun sa quantity flip karta hai: $B$ ka sign, ya $d Φ_{B} / d t$ ka sign, ya dono? ::: Dono — sirf ek ko akele reverse karne se current flip hoti hai; dono ko reverse karne se woh unchanged rehti hai (Ex 8).

Connections

Faraday's law of induction — har example mein use hone wala $∣ ε ∣ = ∣ d Φ_{B} / d t ∣$ provide karta hai.
Magnetic flux — woh $Φ_{B} = B A cos θ$ jo humne poore time differentiate kiya.
Right-hand rule — har "change ko oppose karo" field ko current direction mein convert kiya.
Motional EMF and sliding rod — Ex 3 aur Ex 6 ke peeche ki poori theory.
Eddy currents and magnetic braking — Ex 7 ka mechanism.
Conservation of energy — Ex 3 mein verified power balance $F v = I^{2} R$ .
Self-inductance and back-EMF — Lenz apni khud ki changing current wali coil par apply hota hai.