Magnetic force on current-carrying conductor: Worked examples

1.8.21 · D3 · Physics › Electromagnetism › Magnetic force on current-carrying conductor

Shuru karne se pehle, main har symbol ko dobara explain karta hoon taaki koi bhi newcomer pehli line se samajh sake.

Definition Symbols, seedhi boli mein

$I$ — current, amperes (A) mein measure hoti hai. "Ek second mein kitna charge flow karta hai."
$L$ — ek chhota sa arrow jiska length wire ki lambai $L$ (metres mein) ke barabar hai, aur jo conventional current ke direction mein point karta hai. Toh $L$ ko pata hai kitni lambi wire hai aur kis taraf point karti hai.
$B$ — magnetic field, space ke har point par ek arrow jo batata hai ki compass kis taraf point karega aur field kitni strong hai, tesla (T) mein measure hoti hai.
$θ$ — wire arrow $L$ aur field arrow $B$ ke beech ka angle, hamesha woh interior angle jo tum ek arrow se doosre par jaane ke liye sweep karte ho, toh $0^{\circ} \leq θ \leq 18 0^{\circ}$ . Yeh kabhi bhi negative nahi hota aur kabhi bhi straight angle se zyada nahi — agar picture mein $20 0^{\circ}$ dikhta hai, toh tumne sirf bahar wala angle draw kiya hai; $36 0^{\circ}$ mein se wo minus kar do. $sin θ$ aur $sin (18 0^{\circ} - θ)$ dono barabar hain, toh ek obtuse tilt bhi wahi force size deta hai jitna uska acute mirror.
$\times$ — cross product (dekho Cross product (vectors)): yeh do arrows leta hai aur ek teesra arrow banata hai jo dono ke perpendicular hota hai, jiska size $= (size of first) (size of second) sin θ$ hota hai. Wahi $sin θ$ hai jahan se $F = B I L sin θ$ mein $sin$ aata hai.

Figure — Magnetic force on current-carrying conductor

Figure s01 dekho: wire arrow $L$ (cyan) aur field $B$ (white) interior angle $θ$ par milte hain. Amber dashed arrow woh hissa hai $L$ ka jo $B$ ke perpendicular hai — uski length $L sin θ$ hai. Sirf wahi perpendicular part koi force kamaata hai.

The scenario matrix

Is topic ki har problem inheen cells mein se ek (ya combination) hoti hai:

Cell	Kya vary karta hai	Degenerate / limiting version
A	Wire $⊥$ field, $θ = 9 0^{\circ}$	max force $F = B I L$
B	Wire ek general angle par $0 < θ < 9 0^{\circ}$	$sin θ$ use karta hai
C	Wire $∥$ field, $θ = 0^{\circ}$ ya $18 0^{\circ}$	zero force
D	Direction nikalna (Left-Hand Rule / cross product)	into/out of page cases
E	Full component/vector input ( $L$ aur $B$ as $\overset{x}{^}, \overset{y}{^}, \overset{z}{^}$ )	koi bhi $θ$ andar chhupa hua
F	Curved / bent wire, uniform $B$	straight end-to-end vector se replace karo
G	Closed loop, uniform $B$	net force $= 0$ (F ka limiting case)
H	Word problem (ek real device, jaise rail/motor bar)	force ko motion/mass se connect karo

Neeche ke aath examples mein se har ek us cell ke saath labeled hai jo wo cover karta hai, toh tum dekh sako ki poora map paint ho raha hai.

Worked examples

Worked example Example 4 — Cell D (pure direction, into/out of page)

Current right ki taraf ( $+ \overset{x}{^}$ ) flow kar rahi hai ek wire mein. Field into the page point karti hai ( $- \overset{z}{^}$ , jahaan $+ \overset{z}{^}$ = out of page). Wire par force kis taraf hai?

Forecast: cross product $\Rightarrow$ dono ke perpendicular — toh force vertical hogi (upar ya neeche). Guess karo kaun sa.

Axes set karo: $\overset{x}{^}$ = right, $\overset{y}{^}$ = page par upar, $\overset{z}{^}$ = page se bahar. Yeh step kyun? Humein ek fixed frame chahiye taaki "into the page" ka ek symbol ho: $B \propto - \overset{z}{^}$ .
Compute karo $L \times B \propto \overset{x}{^} \times (- \overset{z}{^})$ . Use karo $\overset{x}{^} \times \overset{z}{^} = - \overset{y}{^}$ , toh $\overset{x}{^} \times (- \overset{z}{^}) = + \overset{y}{^}$ . Yeh step kyun? Unit arrows ki cross-product table hi direction rule hai, likha hua.
Force $+ \overset{y}{^}$ = page par upar point karti hai. Yeh step kyun? $F = I L \times B$ ka direction $L \times B$ jaisa hi hota hai.

Verify hand se: Fleming's Left-Hand Rule — Pehli ungli page ke andar, doosri ungli right ki taraf, angootha upar point karta hai. ✓ Algebra se match karta hai. Figure s02 mein cyan current arrow right point kar raha hai, white crosses $B$ ko page ke andar jaate dikhate hain, aur amber arrow resulting upward force hai — check karo ki teeno mutually perpendicular hain.

Worked example Example 5 — Cell E (full vectors, angle andar chhupa hua)

Ek wire segment hai $L = (0.30 \overset{x}{^} + 0.40 \overset{y}{^}) m$ jo $I = 5.0 A$ carry karta hai field $B = 0.20 \overset{z}{^} T$ mein. Force vector aur uski magnitude nikalo.

Forecast: $L$ page-plane mein hai, $B$ us se bahar point karta hai, toh $L ⊥ B$ — yeh secretly ek Cell A (perpendicular) problem hai jisme $θ = 9 0^{\circ}$ hai. Expect karo $F = B I L$ jahan $L = ∣ L ∣$ , yaani $F = 0.20 \times 5.0 \times 0.50 = 0.50 N$ .

Wire ki length: $L = 0.3 0^{2} + 0.4 0^{2} = 0.25 = 0.50 m$ . Yeh step kyun? Humein $∣ L ∣$ chahiye perpendicular claim test karne ke liye bhi aur shortcut $F = B I L$ se magnitude predict karne ke liye bhi.
Cross product component-wise, jahan $B = 0.20 \overset{z}{^}$ . Use karo $\overset{x}{^} \times \overset{z}{^} = - \overset{y}{^}$ aur $\overset{y}{^} \times \overset{z}{^} = + \overset{x}{^}$ : $= 5.0\big[0.080\,\hat x - 0.060\,\hat y\big].$$ *Yeh step kyun?* Components par cross product karne se direction aur size ek saath milte hain — $\theta$ manually dhundhne ki zaroorat nahi.$
$F = (0.40 \overset{x}{^} - 0.30 \overset{y}{^}) N$ . Yeh step kyun? Hum bracket ko sirf $5.0$ se multiply karte hain: $5.0 \times 0.080 = 0.40$ aur $5.0 \times 0.060 = 0.30$ . Yeh final force vector hai — ise likhne se hum dono components padh sakte hain jo kisi follow-up ke liye chahiye ho sakte hain (jaise equilibrium).
Magnitude: $∣ F ∣ = 0.4 0^{2} + 0.3 0^{2} = 0.16 + 0.09 = 0.25 = 0.50 N$ . Yeh step kyun? Magnitude wahi hai jo shortcut $F = B I L$ predict karta hai, toh ise vector se compute karna humara cross-check hai ki component work sahi hai.

Verify: shortcut ne kaha tha $F = B I L = 0.20 \times 5.0 \times 0.50 = 0.50 N$ , aur component route bhi $0.50 N$ deta hai ✓. Independent perpendicularity checks: $F$ mein koi $\overset{z}{^}$ part nahi, toh $F ⊥ B$ ✓; aur dot product $F \cdot L = (0.40) (0.30) + (- 0.30) (0.40) + 0 = 0.12 - 0.12 = 0$ , toh $F ⊥ L$ ✓. Dono perpendicularity exactly wahi hain jo ek magnetic force zaroori satisfy kare.

Worked example Example 6 — Cell F (curved wire, uniform field)

Radius $R = 0.10 m$ ki ek semicircular wire $I = 3.0 A$ carry karti hai uniform field $B = 0.40 T$ mein jo page se bahar point karta hai. Straight diameter dono ends ko jodata hai. Curved half par net force nikalo.

Forecast: ek curve ke along force integrate karna bahut mushkil lagta hai — lekin shortcut kehta hai ki uniform field mein ek curved wire ko wahi force lagti hai jo uske ends ko jodne wali straight wire ko lagti hai. Woh straight wire diameter hai, length $2 R$ .

Effective length vector: $L_{eff}$ = start se end tak straight arrow = diameter, magnitude $2 R = 0.20 m$ . Yeh step kyun? Kyunki $F = I (\int d ℓ) \times B$ aur $\int d ℓ$ sirf end-to-end displacement hai — loop ki curviness uniform field mein cancel ho jaati hai.
Diameter page mein hai, $B$ page se bahar hai, toh yeh perpendicular hain: $F = B I (2 R) = 0.40 \times 3.0 \times 0.20 = 0.24 N$ . Yeh step kyun? Diameter aur field ke beech $θ = 9 0^{\circ}$ hai, toh $sin θ = 1$ .

Verify: units $T \cdot A \cdot m = N$ ✓. Force diameter ke perpendicular direction mein hai (equivalent straight wire par Left-Hand Rule). Figure s03 mein cyan curve real wire hai, amber arrow effective straight diameter $L_{eff}$ hai, white dots $B$ ko page se bahar aate dikhate hain, aur white dashed arrow net force hai jo curve ko uske chord se door push kar rahi hai.

Worked example Example 7 — Cell G (closed loop, net force zero)

Ab wahi semicircle band kar diya straight diameter wire add karke jo same current $I$ ko doosri taraf carry karta hai. Uniform field mein poore closed loop par net force kya hai?

Forecast: loop band hai, aur shortcut ne kaha tha $\int d ℓ = 0$ kisi bhi closed path ke around. Net force = zero predict karo.

Curved half ko $0.24 N$ ek taraf lagti hai (Example 6). Yeh step kyun? Yeh hum pehle hi compute kar chuke hain.
Straight diameter (length $2 R$ , $B$ ke perpendicular) ko $F = B I (2 R) = 0.40 \times 3.0 \times 0.20 = 0.24 N$ lagti hai — lekin current us mein opposite taraf run karta hai, toh uski force $0.24 N$ opposite direction mein hai. Yeh step kyun? Opposite current direction $L$ ko flip karta hai, isliye force flip hoti hai.
Net force $= 0.24 - 0.24 = 0 N$ .

Verify: $\oint d ℓ = 0 \Rightarrow F_{net} = I (\oint d ℓ) \times B = 0$ se match karta hai. ✓ (Loop phir bhi ek torque feel kar sakta hai — wahi cheez hai jo ek Electric motor ko spin karti hai aur ek Moving-coil galvanometer ko deflect karti hai; dekho Torque on a current loop.)

Worked example Example 8 — Cell H (word problem: ek sliding rail)

Ek horizontal conducting bar jiska mass $m = 0.020 kg$ aur length $L = 0.25 m$ hai, vertical field $B = 0.80 T$ (upar point karta hua) mein do frictionless rails par rakha hai. Kaun sa current $I$ bar par magnetic force ko bar ke weight ke barabar kar deta hai, yaani use lift karta hai? $g = 9.8 m/ s^{2}$ lo.

Forecast: hum chahte hain $B I L = m g$ , toh $I$ weight ke saath scale karta hai aur $B$ aur $L$ ke inversely.

Balance karne wala weight: $W = m g = 0.020 \times 9.8 = 0.196 N$ . Yeh step kyun? Magnetic force ko bar lift karne ke liye yahi match karna hoga.
Bar par force (bar $⊥$ field): $F = B I L$ . Set karo $B I L = m g$ . Yeh step kyun? Perpendicular geometry, toh $sin 9 0^{\circ} = 1$ ke saath poori force.
Solve karo: $I = \frac{m g}{B L} = \frac{0.196}{0.80 \times 0.25} = \frac{0.196}{0.20} = 0.98 A$ .

Verify: plug back karo — $F = B I L = 0.80 \times 0.98 \times 0.25 = 0.196 N = m g$ ✓. Units $\frac{kg \cdot m/ s ^{2}}{T \cdot m} = A$ ✓.

Active recall

Recall Kaun sa cell hai jab wire

$L$ aur $B$ $\overset{x}{^}, \overset{y}{^}, \overset{z}{^}$ components mein diye gaye hoon? Cell E — seedha components par cross product use karo; tumhe kabhi $θ$ dhundhna nahi padega.

Recall Uniform field mein ek curved wire straight wire mein kyun reduce ho jaati hai?

Kyunki $F = I (\int d ℓ) \times B$ aur $\int d ℓ$ sirf start-to-end displacement hai — curviness integrate ho ke khatam ho jaati hai. Ek closed loop $\oint d ℓ = 0$ deta hai, isliye net force zero hoti hai.

Predict-then-check

Cell C (wire parallel to

B

) ke liye, force zero hai kyunki

sin 0^{\circ} = 0

Cell A vs Cell B ratio

Example 2 at

3 0^{\circ}

bilkul Example 1 at

9 0^{\circ}

ka aadha hai kyunki

sin 3 0^{\circ} / sin 9 0^{\circ} = 0.5

Connections

Parent topic (Hinglish) — woh law jise hum drill kar rahe hain.
Lorentz force on a moving charge — jahaan se $F = I L \times B$ paida hota hai.
Cross product (vectors) — Cells D aur E ke peechhe ki machinery.
Torque on a current loop — kyun Cell G ka zero net force phir bhi motors spin karta hai.
Electric motor / Moving-coil galvanometer — Cell G real world mein.
Force between two parallel currents — har wire doosre ke field mein ek Cell A hai.
Drift velocity and current — $I = n A q v_{d}$ poore law ke peechhe supply karta hai.