1.8.21 · Physics › Electromagnetism
Current-carrying wire basically ek nadī hai bhagte hue charges ki . Hum jaante hain ki ek akela moving
charge ek force feel karta hai F = q v × B magnetic field mein. Ek current-carrying wire
mein trillions aise charges saath milke chal rahe hote hain — toh wire as a whole ek force feel karta hai.
WHY it matters: yahi principle hai har electric motor, loudspeaker, aur
moving-coil galvanometer ka. Electricity ko → motion mein convert karo.
Intuition HUM KYA KAR RAHE HAIN
Hum F = B I L yaad nahi karenge. Hum ise build karenge ek charge par force se, phir wire ki
length L mein saare charges add karke.
Setup. Ek seedhi wire lo jiska cross-sectional area A hai, length L hai, current I carry kar rahi hai. Maano
n = free charge carriers per unit volume, har ek ka charge q hai, drift velocity
v d se bhaag rahe hain.
Step 1 — Ek carrier par force.
F 1 = q v d × B
Yeh step kyun? Yeh Lorentz force hai, ek matra magnetic force law jo humein chahiye.
Step 2 — Carriers count karo. Segment mein carriers ki sankhya:
N = n × ( volume ) = n A L
Yeh step kyun? Total force = (force per carrier) × (kitne carriers hain).
Step 3 — Total force.
F = N F 1 = n A L ( q v d × B )
Step 4 — Current ko pehchano. Drift-current relation yaad karo:
I = n A q v d ⇒ n A q v d = I L
jahan L ek vector hai jiska magnitude L hai aur jo conventional current direction ki taraf point karta hai.
Yeh step kyun? Isse hum microscopic cheezein ( n , A , q , v d ) ko macroscopic,
measurable current I se replace kar sakte hain.
Ek bent/curved wire ke liye hum infinitesimal pieces ko add karte hain:
d F = I d ℓ × B ⇒ F = I ∫ d ℓ × B
Uniform B mein closed/curved wire ke liye neat shortcut
F = I ( ∫ d ℓ ) × B . Integral ∫ d ℓ simply
start se end tak straight vector hai. Toh ek curved wire wohi force feel karta hai jo uske endpoints ko join karne wali straight wire feel karti. Ek closed loop (∫ d ℓ = 0 ) uniform field mein zero net force feel karta hai (phir bhi torque feel kar sakta hai!).
Intuition Yahan LEFT hand kyun?
Force hai F = I L × B — ek cross product, toh F dono wire aur field ke perpendicular
hoti hai. Fleming's Left-Hand Rule bas isi cheez ka haath-gesture encoding hai.
Mnemonic Fleming's Left-Hand Rule (wire par force ke liye / motor)
LEFT haath par thumb, First finger, seCond finger ko mutually perpendicular stretch karo:
F irst finger → F ield B
seC ond finger → C urrent I
thuM b → M otion / Force F
("FBI ": Field–By first finger, current–seCond... force–thumb.)
Worked example Example 1 — straight wire, perpendicular field
Ek wire jis ki length 0.50 m hai, I = 4.0 A current carry karti hai, field
B = 0.30 T ke right angles par. Force find karo.
Solve. θ = 9 0 ∘ , toh F = B I L = 0.30 × 4.0 × 0.50 = 0.60 N .
Yeh step kyun? Perpendicular ⇒ sin 9 0 ∘ = 1 , maximum-force case.
Worked example Example 2 — wire at an angle
Same wire lekin field se θ = 3 0 ∘ par.
Solve. F = B I L sin θ = 0.30 × 4.0 × 0.50 × sin 3 0 ∘ = 0.60 × 0.5 = 0.30 N .
Yeh step kyun? Sirf wire ka B ke perpendicular component "field feel" karta hai; parallel
component kuch contribute nahi karta.
Worked example Example 3 — left hand se direction
Current east ki taraf flow kar rahi hai, field north ki taraf point karti hai (dono horizontal). Force kis taraf hai?
Solve. First finger north, second finger east → thumb upar (zameen se bahar) point karta hai.
Yeh step kyun? E ^ × ... verify karte hain: maano x ^ = east, y ^ = north,
z ^ = up. L × B ∝ x ^ × y ^ = z ^ = up. ✓ Haath se match karta hai.
Worked example Example 4 — semicircular wire (shortcut use karta hai)
Ek semicircular wire jis ka radius R hai, current I carry karti hai uniform field B mein jo page se bahar hai.
Diameter (straight chord) dono ends ko join karti hai.
Solve. Curved part par net force I L eff × B ke barabar hai jahan
L eff = diameter ke across straight vector, magnitude 2 R .
Toh F = I ( 2 R ) B , diameter ke perpendicular direction mein.
Yeh step kyun? Uniform B mein, F = I ( ∫ d ℓ ) × B aur ∫ d ℓ bas
end-to-end displacement = 2 R hai. Ek tedha integral bachata hai.
Common mistake "Force field ke along hoti hai, jaise
B se push milti hai."
Kyun sahi lagta hai: gravity mein, force field g ke along hoti hai. Toh hum expect karte hain ki B khud ke along push kare. Fix: magnetic force ek cross product hai — yeh dono I aur B ke perpendicular hoti hai. B ke parallel wire ko zero force milti hai, maximum nahi.
Common mistake "Ek closed current loop uniform field mein sideways dhakela jata hai."
Kyun sahi lagta hai: loop ki har side clearly ek force feel karti hai. Fix: opposite sides par equal-and-opposite forces lagte hain, toh net force zero hoti hai (∮ d ℓ = 0 ). Jo bachta hai woh hai ek torque — yahi motor ko spin karta hai.
Common mistake Wire par force ke liye right hand use karna.
Kyun sahi lagta hai: right-hand rule har jagah aata hai (jaise positive charge ke liye v × B ,
wire ka field). Fix: Fleming's Left -Hand Rule motor rule hai. (Right-hand = generator
rule / v × B ki direction directly.) Dono consistent hain — bas gestures mix mat karo.
sin θ bhool jaana.
Kyun sahi lagta hai: F = B I L famous formula hai. Fix: woh sirf θ = 9 0 ∘ ke liye hai.
General law hai F = B I L sin θ .
Recall Forecast-then-verify: aage padhne se pehle predict karo ki
B ke parallel wire par force kya hogi.
Prediction: zero. Verify: sin 0 ∘ = 0 ⇒ F = 0 . ✓ Carriers ki drift velocity
B ke along hai, aur v ∥ B ⇒ v × B = 0 .
Recall Feynman — ek 12-saal ke bache ko explain karo
Socho ek bheed hai bache ki jo ek tezi se chalti hawa mein bhaag rahe hain. Hawa sirf bache ko sideways dhakeli hai jab woh hawa ke across bhaagta hai — agar tum seedha hawa mein bhaago, toh yeh khaas "magnetic wind" tumhe nahi dhakeli. Current wali wire mein bhagte hue electric charges ki bheed hoti hai. Charges ko magnetic field ke across line up karo aur poori wire sideways dhakeli jaati hai. Wire ko field ke along ghuma do aur koi nahi dhakela jaata. Yahi sideways dhakka motors ko spin karta hai.
Straight current wire par force (vector form)? B se θ angle par wire par force ka magnitude?F = B I L sin θ
B ke parallel wire ko force kyun nahi lagti?Wire par force ki direction kis rule se milti hai? Fleming's Left-Hand Rule (F=Field=first finger, C=Current=second, M=Motion=thumb).
Uniform field mein closed current loop par net force?Zero (
∮ d ℓ = 0 ); lekin net torque ho sakta hai.
q v × B se I L × B par kaise jaate hain?n A q v d = I L use karo aur force per carrier ko
N = n A L carriers se multiply karo.
Uniform B mein do endpoints ke beech curved wire par force? Same jaise ends ko join karne wali straight wire par:
F = I L eff × B .
0.5 m wire, 4 A, 0.3 T, perpendicular par force?F = B I L = 0.60 N.
Lorentz force on a moving charge — parent law jisse humne yeh derive kiya.
Drift velocity and current — I = n A q v d provide karta hai.
Torque on a current loop — loop ki net force zero hoti hai lekin torque motors chalata hai.
Electric motor / Moving-coil galvanometer — direct applications.
Force between two parallel currents — har wire doosri ke field mein baithi hai.
Cross product (vectors) — perpendicularity ke peeche ki geometry.
generalise to curved wire
direction via cross product
Zero net force, may have torque
Motors, loudspeakers, galvanometers