Magnetic force on charge — F = qv × B: Visual walkthrough

1.8.20 · D2 · Physics › Electromagnetism › Magnetic force on charge — F = qv × B

Kisi bhi symbol se pehle, yeh poora cast hai, simple words mein:

Step 1 — Still charge kuch feel nahi karta

KYA. Ek charged ball ko, jo move nahi kar rahi, magnetic field ke andar rakh do. Dekho. Woh hilti nahi.

YEH KYUN MATTER KARTA HAI. Yeh force kis cheez pe depend karta hai, iska yeh hamara pehla clue hai. Agar force $F = (something) \times q \times B$ hoti — bina motion ke mention ke — toh still charge jump karta — lekin karta nahi. Toh speed ek factor zaroor hai, aur woh factor force ko off kar deta hai jab ball still hoti hai.

PICTURE. Figure mein field lines left-to-right ja rahi hain. Ball unpe baithi hai, bilkul still, aur force meter zero read karta hai.

Figure — Magnetic force on charge — F = qv × B

"Off when still" encode karne ka sabse chhota tarika yeh hai ki force ko speed ke proportional banao: $F \propto v$

$F$ = woh push jo hum dhundh rahe hain.
$v$ = ball ki speed. Jab $v = 0$ , poori right side $0$ hai, toh $F = 0$ . Exactly wahi jo humne dekha. ✓

Step 2 — Firing ki direction sab kuch change kar deti hai

KYA. Ab hum ball ko throw karte hain. Pehle hum ise field lines ke along throw karte hain (parallel). Phir bhi kuch nahi! Phir hum ise field lines ke across throw karte hain (perpendicular). Ab yeh deflect karta hai — aur yeh sabse strong deflection hai jo hum pa sakte hain.

KYUN. Toh force velocity ke sab hisse ki parwah nahi karta — sirf us hisse ki jo field ke across hai. Humein ek mathematical dial chahiye jo $0$ read kare jab arrow $B$ ke parallel ho aur maximum read kare jab perpendicular ho.

PICTURE. Teen throws draw ki gayi hain: parallel (no force), $4 5^{\circ}$ (thodi force), perpendicular (full force). Velocity arrow aur field ke beech ke angle ko $θ$ kehte hain (Greek letter "theta", bas us angle ka naam hai).

Jo dial exactly yeh kaam karta hai woh angle ki sine hai: $F \propto v sin θ$

$θ$ = $v$ aur $B$ ke beech ka angle.
$sin θ$ = "velocity ka kitna hissa field ke across point karta hai." Figure dekho: $v sin θ$ velocity arrow ki shadow ki length hai jo field lines ke perpendicular hai.

Step 3 — Yeh push karta hai kahan? Dono ke perpendicular

KYA. Deflection ki actual direction measure karo. Yeh kabhi forward nahi, kabhi backward nahi, kabhi field ke along nahi. Yeh hamesha velocity ke sideways AUR field ke sideways ek saath hoti hai — seedha us plane se bahar nikalta hai jo dono arrows share karte hain.

KYUN yeh maths pin down karta hai. Ab humein ek aisi single operation chahiye jo do arrows ( $v$ aur $B$ ) khaaye aur ek teesra arrow perpendicular dono ke bahar nikale, jiska length $sin θ$ ki tarah bade. Poore vector maths mein exactly ek aisa tool hai, aur use Cross product kehte hain, $\times$ se likha jaata hai.

PICTURE. Velocity arrow aur field arrow ek table pe flat hain; force arrow seedha upar table se bahar khada hai, dono ke right angles par.

$v \times B ⟶ an arrow ⊥ v and ⊥ B, length = v B sin θ$

$v \times B$ = " $v$ aur $B$ ka cross product" — ise padhte hain "in dono ka perpendicular partner."
Iska length Steps 1–2 ka $v B sin θ$ pehle se carry karta hai. Isliye yeh ek symbol do alag observations ko replace kar leta hai.

Step 4 — Apne haath se direction padhna (right-hand rule)

KYA. Yeh pata karne ke liye ki $v \times B$ kis taraf point karta hai, apna right hand use karo: ungliyan $v$ ke along point karo, unhe $B$ ki taraf curl karo, aur tumhara thumb $v \times B$ ke along point karega.

HAATH KYUN? "Dono ke perpendicular" do opposite directions allow karta hai (table se upar ya neeche). Right-hand rule woh agreed-upon tie-breaker hai jo ek choose karta hai, consistently, poori physics mein.

PICTURE. Ek right hand: ungliyan magenta $v$ arrow se violet $B$ arrow ki taraf sweep karti hain; orange thumb $v \times B$ ke roop mein upar aata hai.

Bookkeeping-proof version (haath kabhi backwards nahi jaata) component determinant hai:

(v_yB_z-v_zB_y)\,\hat i+(v_zB_x-v_xB_z)\,\hat j+(v_xB_y-v_yB_x)\,\hat k$$ - $v_x,v_y,v_z$ = velocity ke teen pieces $x,y,z$ directions ke along. - $\hat i,\hat j,\hat k$ = unit arrows (length 1) $x,y,z$ ke along — "signposts." - Har bracket answer arrow ka ek coordinate hai. Numbers daalo, vector milega, haath ki zaroorat nahi. --- ## Step 5 — Charge scale karta hai aur arrow flip kar sakta hai **KYA.** Charge $q$ double karo → force double ho jaati hai. Positive charge ki jagah **negative** charge rakho (same $\vec v$, same $\vec B$) → force bilkul opposite direction mein flip ho jaati hai. **KYUN.** Toh charge $q$ final multiplier hai: iska **size** force ko scale karta hai, aur iska **sign** decide karta hai ki hum right-hand-rule arrow rakhein ya use reverse karein. **PICTURE.** Do balls identically fire ki gayi hain: positive (magenta) ek taraf curve karta hai, negative (violet) mirror-opposite taraf curve karta hai. ![[deepdives/dd-physics-1.8.20-d2-s05.png]] Step 3 ke perpendicular arrow ko signed charge se multiply karo: $$\boxed{\ \vec F = q\,(\vec v\times\vec B)\ }$$ - $q$ = signed charge. Agar $q>0$, $\vec F$ $\vec v\times\vec B$ ke saath align hota hai. Agar $q<0$, minus use flip kar deta hai. - $\vec v\times\vec B$ = Steps 3–4 se length $vB\sin\theta$ wala perpendicular arrow. Har observation ab ek statement ke andar hai. Proportionality constant exactly $1$ hai **kyunki** tesla isi tarah define kiya gaya hai — aadha experiment, aadha units ka tidy choice. --- ## Step 6 — Edge cases: formula sab mein survive karna chahiye **KYA.** Woh corners check karo jahan cheezein break ho sakti hain. **KYUN.** Jis law pe aap trust karte ho use *sahi kuch nahi* dena chahiye jab kuch nahi hona chahiye, aur har angle pe sensibly behave karna chahiye — sirf easy $90^\circ$ case mein nahi. **PICTURE.** Char mini-scenes: (a) $v=0$ still ball → $F=0$; (b) $\vec v\parallel\vec B$ → $F=0$; (c) $\vec v\perp\vec B$ → full force $qvB$; (d) $q=0$ neutral ball → $F=0$. ![[deepdives/dd-physics-1.8.20-d2-s06.png]] $$|\vec F| = |q|\,v\,B\,\sin\theta$$ | Case | Kya hota hai | Kyun (kaunsa factor $0$ hai) | |---|---|---| | $v=0$ (still) | $F=0$ | $v=0$ use khatam kar deta hai (Step 1) | | $\vec v\parallel\vec B$ ($\theta=0^\circ$) | $F=0$ | $\sin0^\circ=0$ (Step 2) | | $\vec v\perp\vec B$ ($\theta=90^\circ$) | $F=|q|vB$ | $\sin90^\circ=1$ (max) | | $q=0$ (neutral) | $F=0$ | push karne ke liye koi charge nahi (Step 5) | | $q<0$ | force reverse ho jaati hai | sign arrow flip karta hai (Step 5) | Har ek scenario cover hai — reader kabhi aisi situation nahi milega jo humne skip kiya ho. --- ## Step 7 — Formula *kya karta hai*: circle mein bend karna **KYA.** Ek charge ko $\vec B$ ke perpendicular fire karo. Force $qvB$ size mein constant hai aur hamesha motion ke sideways point karta hai, toh ball ko hamesha turn karta rehta hai → ek **circle**. (Ise tirse fire karo aur along-field wala hissa untouched chalta rehta hai → ek **helix**.) **KYUN.** Ek push jo hamesha velocity ke perpendicular ho [[Centripetal force and circular motion|centripetal force]] ki definition hai. Magnetic force = centripetal need set karne se hum circle ka *size predict* kar sakte hain. **PICTURE.** Ek ball radius $r$ ke circle mein loop kari hai; force arrow hamesha centre ki taraf point karta hai, velocity hamesha loop ke along. Ek dashed helix slanted case dikhata hai. ![[deepdives/dd-physics-1.8.20-d2-s07.png]] $$\underbrace{\frac{mv^2}{r}}_{\text{centripetal need}}=\underbrace{qvB}_{\text{magnetic supply}}\;\;\Longrightarrow\;\;\boxed{r=\frac{mv}{qB}},\qquad T=\frac{2\pi m}{qB}$$ - $m$ = ball ki mass. $r$ = uske circle ka radius. - Dono sides se ek $v$ cancel karo $r$ paane ke liye. Notice karo $T$ (ek loop ka time) mein **koi $v$ nahi** — fast ho ya slow, same lap time. Yeh [[Cyclotron]] ka heart hai, aur radius formula [[Mass spectrometer]] ka engine hai. > [!example] $r$ ka quick numeric check > Electron: $v=1\times10^7$ m/s, $B=0.01$ T, $m=9.1\times10^{-31}$ kg, $|q|=1.6\times10^{-19}$ C. > $$r=\frac{(9.1\times10^{-31})(10^7)}{(1.6\times10^{-19})(0.01)}\approx 5.7\times10^{-3}\ \text{m}=5.7\ \text{mm}.$$ --- ## Ek-picture summary ![[deepdives/dd-physics-1.8.20-d2-s08.png]] Ek frame mein poori kahani hai: velocity andar, field andar, right-hand rule unhe ek perpendicular force mein turn karta hai, charge ka sign direction choose karta hai, aur result path ko radius $mv/qB$ ke circle mein curl kar deta hai. > [!recall]- Feynman retelling — poora walkthrough plain words mein > Ek magnet ek ball ko ignore karta hai jo bas wahan baithi ho — toh *movement* ka matter karna zaroori hai (Step 1). Agar tum ball ko magnet ke grain ke *along* roll karo, phir bhi kuch nahi; ise *across* roll karo aur woh sabse zyada swerve karti hai — toh sirf motion ka sideways hissa count karta hai, jo angle ki $\sin$ hai (Step 2). Swerve kabhi forward ya back nahi hoti — woh seedha bahar nikti hai, motion aur grain *dono* ke perpendicular, aur exactly ek maths gadget hai jo yeh karta hai: cross product (Step 3). Yeh padhne ke liye ki "seedha bahar" ke do directions mein se kaunsa hai, tum apna right hand use karte ho — ungliyan velocity se field ki taraf, thumb answer hai (Step 4). Bada charge, bada dhakka; *negative* charge opposite taraf dhakka deta hai (Step 5). Corners test karo: no motion, motion-along-field, ya no charge sab zero dete hain, aur perpendicular maximum deta hai — formula har ek pass karta hai (Step 6). Aakhir mein, ek always-sideways dhakka ball ko ek perfect circle mein ghoomata rehta hai jiska size $mv/qB$ hai, aur — yeh delightful surprise hai — har ball ghoomne mein same time leta hai (Step 7). Yeh hai $\vec F = q\,\vec v\times\vec B$, bilkul wahi se built jo humne dekha. > [!mnemonic] Ise bahar le jao > **See → Sideways → Sign → Circle.** *See* karo ki use move karna hai (v), yeh *Sideways* push karta hai (cross product), charge ka *Sign* ise flip karta hai, aur always-sideways matlab woh *Circle* mein jaata hai. --- ## Active Recall Magnetic force speed $v$ pe depend kyun karna chahiye? ::: Kyunki stationary charge koi force feel nahi karta; factor $v=0$ par vanish hona chahiye. $|F|=|q|vB\sin\theta$ mein $\cos\theta$ nahi, $\sin\theta$ kyun? ::: Sirf $\vec v$ ka field ke across wala hissa deflect karta hai; $\sin\theta$ woh perpendicular portion hai (parallel throw $\sin0^\circ=0$ deta hai). Kaunsa operation do inputs ke perpendicular arrow deta hai jiska magnitude $\propto\sin\theta$ ho? ::: Cross product $\vec v\times\vec B$. $q$ ka sign $\vec F$ ki direction ke saath kya karta hai? ::: Positive charge $\vec v\times\vec B$ ke saath follow karta hai; negative charge opposite point karta hai (minus use flip karta hai). Har woh case list karo jahan magnetic force zero hoti hai. ::: $v=0$, ya $\vec v\parallel\vec B$ ($\theta=0$), ya $q=0$. Perpendicular firing circle kyun banata hai? ::: Constant-size force jo hamesha $\vec v$ ke perpendicular ho centripetal hoti hai, toh woh path ko continuously turn karti hai. $mv^2/r=qvB$ se radius kya hai? ::: $r=mv/qB$. Lap time $T=2\pi m/qB$ surprising kyun hai? ::: Isme koi $v$ nahi — har charge same time mein orbit karta hai speed se independent. --- ## Connections - Parent: [[Magnetic force on charge — F = qv × B (index 1.8.20)|1.8.20 · F = qv × B]] - Tool used: [[Cross product]], packaged in the [[Lorentz force law]] - Circle result feeds: [[Centripetal force and circular motion]], [[Cyclotron]], [[Mass spectrometer]] - Sibling ideas: [[Magnetic force on a current-carrying wire]], [[Velocity selector]]