Worked examples — Magnetic force on charge — F = qv × B

1.8.20 · D3 · Physics › Electromagnetism › Magnetic force on charge — F = qv × B

Shuru karne se pehle, un do tools ka ek-line reminder jo hum use karte hain:

Yahan $\hat{i},\hat{j},\hat{k}$ sirf teen unit arrows hain jo $x$, $y$, $z$ ki taraf point karte hain (har ek ek metre lamba, dono se right angles par). "Out of the page" tumhari taraf $=+\hat{k}$; "into the page" $=-\hat{k}$.

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Scenario matrix

Is topic ke har problem ka type in cells mein se ek hai. Neeche har worked example us cell ke saath tagged hai jo woh cover karta hai.

Cell	Situation	Tricky kya hai	Example
A	$\vec{v}\perp \vec{B}$, positive charge	baseline magnitude + direction	Ex 1
B	General angle $0<\theta <90^{\circ }$	$\sin \theta$ use karna zaroori hai, poora $v$ nahi	Ex 2
C	Negative charge (electron)	cross-product result ko flip karo	Ex 3
D	Degenerate: $\vec{v}\parallel \vec{B}$, $v=0$, aur $B=0$	force zero hai	Ex 4
E	Full 3-D vectors, sabhi components nonzero	determinant bookkeeping	Ex 5
F	Circular motion — $r$, $T$ nikalo	$\vec{v}\perp \vec{B}$ limit, speed cancel ho jaati hai	Ex 6
G	Helix: $\vec{v}$ mein along + across dono parts hain	velocity split karo, pitch nikalo	Ex 7
H	Real machine: Velocity selector	electric vs magnetic balance karo (vectors!)	Ex 8
I	Real machine: Mass spectrometer	$r=mv/qB$ ko chain karke mass nikalo	Ex 9
J	Exam twist: angle past $90^{\circ }$, sign & limit	direction reverses across $\theta =180^{\circ }$	Ex 10

Matrix mein 10 cells hain 10 examples ke across. Sab cover karo aur topic cover ho gaya.

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Example 1 — Cell A: perpendicular, positive charge (baseline)

Steps.

1. Magnitude: $|\vec{F}|=|q|vB\sin 90^{\circ }=(1.6\times 10^{-19})(3\times 10^6)(0.2)(1)$. Yeh step kyun? $\vec{v}\perp \vec{B}$ matlab $\sin \theta =1$, maximum — yeh sabse simple cell hai, koi wasted force nahi. $=9.6\times 10^{-14}$ N.

2. Cross product se direction: $\vec{v}=(3\times 10^6,0,0)$, $\vec{B}=(0,0,0.2)$. $\vec{v}\times \vec{B}=(v_y{B}_z-v_z{B}_y,v_z{B}_x-v_x{B}_z,v_x{B}_y-v_y{B}_x)=(0,-(3\times 10^6)(0.2),0)=(0,-6\times 10^5,0)$. Yeh step kyun? Sirf middle term bachta hai; woh negative hai, toh arrow $-\hat{y}$ (neeche) ki taraf point karta hai.

3. Charge se multiply karo (positive, toh flip nahi): $\vec{F}=q(\vec{v}\times \vec{B})$ $-\hat{y}$ mein point karta hai.

Figure padho: blue $\vec{v}$ arrow right ($+\hat{x}$) ki taraf point karta hai, green $\odot$ $\vec{B}$ ko out of the page tumhari taraf mark karta hai, aur red $\vec{F}$ arrow seedha neeche ($-\hat{y}$) girta hai. Notice karo teeno mutually right angles par hain — woh perpendicularity cross product ka visual signature hai, aur isi liye yeh force sirf steer kar sakti hai, speed nahi badha sakti.

Recall Verify

Units: C · (m/s) · T $=$ C·(m/s)·(kg/(C·s)) $=$ kg·m/s² $=$ N. ✓ Magnitude $9.6\times 10^{-14}$ N VERIFY mein check hai. Right-hand rule sanity: fingers $+\hat{x}$ ke along, $+\hat{z}$ ki taraf curl karo... thumb $-\hat{y}$ ki taraf point karta hai. ✓

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Example 2 — Cell B: general angle, $\sin \theta$ use karna zaroori

Steps.

1. $|\vec{F}|=|q|vB\sin \theta =(1.6\times 10^{-19})(2\times 10^6)(0.5)\sin 30^{\circ }$. Yeh step kyun? $\vec{v}$ ka sirf across field wala part matter karta hai. Woh perpendicular part $v\sin \theta$ hai; $\vec{B}$ ke along wala part kuch contribute nahi karta.

2. $\sin 30^{\circ }=0.5$, toh $|\vec{F}|=(1.6\times 10^{-19})(2\times 10^6)(0.5)(0.5)=8\times 10^{-14}$ N. Yeh step kyun? Hum sirf multiply karte hain — geometry us single $\sin 30^{\circ }$ mein poori capture ho gayi hai.

Figure padho: blue $\vec{v}$ green $\vec{B}$ se $30^{\circ }$ upar lean karta hai. Orange dashed arrow perpendicular slice $v\sin \theta$ hai — woh chhota arrow hi akela part hai jise field "feel" karta hai. Use $\vec{v}$ se chhota draw dekh kar samajh aata hai kyun force perpendicular case se aadhi hai.

Recall Verify

Same $v,B$ par $90^{\circ }$ hit se compare karo, force $\sin 30^{\circ }=0.5$ times strong hai — exactly aadhi. Agar galti se poora $v$ use karte toh $1.6\times 10^{-13}$ N milta, jo sach se double hai. Value $8\times 10^{-14}$ N VERIFY mein check hai.

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Example 3 — Cell C: electron, sign flip dhyan se dekho

Steps.

1. Charge ka sign ignore karke $\vec{v}\times \vec{B}$ compute karo: $\hat{x}\times \hat{z}=-\hat{y}$. Yeh step kyun? Cross product ko sirf $\vec{v}$ aur $\vec{B}$ ki directions se matlab hai; charge sign baad mein aata hai. Cyclic order $x\to y\to z$ hai; $x\to z$ jaana ulta hai, isliye minus.

2. Negative charge se multiply karo: $\vec{F}\propto q(-\hat{y})=(-)(-\hat{y})=+\hat{y}$. Yeh step kyun? Negative $q$ right-hand rule jo deta hai usse reverse kar deta hai. Electron $+\hat{y}$ jaata hai; proton (positive) $-\hat{y}$ jaata.

Figure padho: dono particles ke liye same blue $\vec{v}$ aur same green $\vec{B}$-out-of-page hai, phir bhi red electron arrow upar ($+\hat{y}$) point karta hai jabki orange proton arrow neeche ($-\hat{y}$). Identical inputs se do opposite arrows — yeh split poori tarah charge par minus sign ka kaam hai, aur isi liye mixed beam fan apart hoti hai.

Recall Verify

Same $\vec{v},\vec{B}$ ke liye do opposite arrows — exactly "electron proton se opposite curve karta hai" wala rule. Agar dono same taraf curve karte, mixed charges wali beam sort nahi hoti; hoti hai, toh opposite sahi hai. ✓

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Example 4 — Cell D: teen degenerate "zero force" cases

Steps.

1. (a) $\theta =0^{\circ }\Rightarrow \sin 0^{\circ }=0\Rightarrow |\vec{F}|=|q|vB(0)=0$. Yeh step kyun? Jab $\vec{v}$, $\vec{B}$ ke along point karta hai toh koi "across" component nahi hota — field ke paas pakdne ko kuch nahi. Component check: $\vec{v}\times \vec{B}=(0\cdot 0-0\cdot 0,0\cdot 0.3-5\times 10^6\cdot 0,5\times 10^6\cdot 0-0\cdot 0.3)=(0,0,0)$. ✓

2. (b) $v=0\Rightarrow |\vec{F}|=|q|(0)B\sin \theta =0$. Yeh step kyun? Magnetic force motion se paida hoti hai. Motion nahi, force nahi — magnet ruke hue charge ko completely ignore karta hai.

3. (c) $B=0\Rightarrow |\vec{F}|=|q|v(0)\sin \theta =0$; cross product $\vec{v}\times \vec{0}=\vec{0}$. Yeh step kyun? Koi field nahi toh velocity cross karne ko kuch hai hi nahi. Yeh $v=0$ case ka logical twin hai — $\vec{v}\times \vec{B}$ mein koi bhi factor zero ho jaaye toh poori force zero ho jaati hai.

Recall Verify

Teeno "sideways steering, no speeding" personality wale hain: field sirf crossing motion ko bend kar sakta hai. Parallel motion, rest, aur koi field nahi — yeh teen tarike hain zero crossing ke, isliye har ek $0$ N deta hai. ✓

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Example 5 — Cell E: full 3-D determinant

Steps.

1. Component formula mein plug karo: $\vec{v}\times \vec{B}=(v_y{B}_z-v_z{B}_y,v_z{B}_x-v_x{B}_z,v_x{B}_y-v_y{B}_x)$ Yeh step kyun? Jab har component nonzero ho, safe method determinant expansion hai — geometry ki zaroorat nahi, right-hand-rule slip ka chance nahi.

2. Har slot compute karo:

   • $x$: $v_y{B}_z-v_z{B}_y=(2)(-1)-(3)(4)=-2-12=-14$
   • $y$: $v_z{B}_x-v_x{B}_z=(3)(0)-(1)(-1)=0+1=1$
   • $z$: $v_x{B}_y-v_y{B}_x=(1)(4)-(2)(0)=4-0=4$

   Toh $\vec{v}\times \vec{B}=(-14,1,4)$.

3. $q=+2$ se multiply karo: $\vec{F}=(-28,2,8)$ N. Yeh step kyun? Charge ek scalar multiplier hai jo har component par equally apply hota hai.

Recall Verify

Perpendicularity self-check: $\vec{F}\cdot \vec{v}$ zero hona chahiye (force $\perp$ velocity hamesha). $(-28)(1)+(2)(2)+(8)(3)=-28+4+24=0$. ✓ Aur $\vec{F}\cdot \vec{B}=(-28)(0)+(2)(4)+(8)(-1)=8-8=0$. ✓ Dono dot products vanish karte hain, confirm karta hai $\vec{F}$ dono inputs ke $\perp$ hai — cross product ka signature.

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Example 6 — Cell F: circular motion, radius aur period

Steps.

1. Constant size $|q|vB$ ki sideways force centripetal force ka kaam karti hai: $\frac{m{v}^2}{r}=|q|vB$. Yeh step kyun? $\vec{v}\perp \vec{B}$ ek constant-magnitude force deta hai jo hamesha motion ke $\perp$ hoti hai — yeh exactly uniform circular motion ki recipe hai. Yahan hum $|q|$ (charge ki magnitude, $1.6\times 10^{-19}$ C) use karte hain kyunki $r$ aur $T$ sizes hain — negative sign sirf yeh set karta hai ki electron kis taraf circle karta hai, circle kitna bada hoga nahi.

2. Ek $v$ cancel karo aur solve karo: $r=\frac{mv}{|q|B}=\frac{(9.1\times 10^{-31})(10^7)}{(1.6\times 10^{-19})(0.01)}$. Yeh step kyun? Ek $v$ cancel hota hai kyunki centripetal need $v^2$ ke saath badhti hai lekin force sirf $v$ ke saath. $r\approx 5.7\times 10^{-3}$ m $=5.7$ mm.

3. Period: $T=\frac{2\pi r}{v}=\frac{2\pi m}{|q|B}=\frac{2\pi (9.1\times 10^{-31})}{(1.6\times 10^{-19})(0.01)}\approx 3.57\times 10^{-9}$ s. Yeh step kyun? $v$ cancel ho jaata hai — fast electron bada circle sweep karta hai lekin higher speed par, toh lap time unchanged rehta hai.

Figure padho: blue loop orbit hai; marked point par green $\vec{v}$ arrow circle ke along (tangent) point karta hai jabki red $\vec{F}$ arrow seedha centre ki taraf point karta hai. Woh centre-pointing force hi path ko circle mein curl karti hai — aur kyunki $\vec{F}$ hamesha $\vec{v}$ ke $\perp$ hai (right angle dekho), woh electron ko kabhi speed up nahi karti. Gray dashed $r$ woh hai jo speed ke saath shrink ya grow karta hai jabki $T$ fixed rehta hai.

Recall Verify

Forecast ka jawab: $v$ double karne par $r$ double hota hai lekin $T$ unchanged rehta hai — Cyclotron ke peeche ka surprise. Values $r\approx 5.7$ mm aur $T\approx 3.57$ ns VERIFY mein check hain.

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Example 7 — Cell G: helix ($\vec{B}$ ke along + across velocity)

Steps.

1. $\vec{v}$ ko $\vec{B}$ ke along aur across parts mein split karo.

   • $\hat{z}$ ke along: $v_{\parallel }=4\times 10^5$ m/s — koi force nahi.
   • Across ($xy$-plane mein): $v_{\perp }=3\times 10^5$ m/s — yahi circle karta hai. Yeh step kyun? Sirf crossing part steer hota hai; along part seedha coast karta hai. Dono milke: ek helix.

2. Circling part ka radius: $r=\frac{m{v}_{\perp }}{|q|B}=\frac{(1.67\times 10^{-27})(3\times 10^5)}{(1.6\times 10^{-19})(0.1)}\approx 3.13\times 10^{-2}$ m $=3.13$ cm. Yeh step kyun? Radius sirf $v_{\perp }$ use karta hai, kyunki sirf woh component orbit karta hai. Hum $|q|$ use karte hain kyunki radius ek size hai.

3. Period (helix ke liye circle jaisa — $B$ par depend karta hai, $v$ par nahi): $T=\frac{2\pi m}{|q|B}=\frac{2\pi (1.67\times 10^{-27})}{(1.6\times 10^{-19})(0.1)}\approx 6.56\times 10^{-7}$ s.

4. Pitch $=$ ek period mein forward drift $=v_{\parallel }T=(4\times 10^5)(6.56\times 10^{-7})\approx 0.262$ m. Yeh step kyun? Jab proton ek loop complete karta hai (time $T$), un-steered part constant $v_{\parallel }$ par aage slide karta hai.

Figure padho: green vertical line $z$ ke along $\vec{B}$ hai; blue curve proton ka spiral hai. Dekho kaise woh $xy$-plane mein round jaata hai (woh orange "circle: $v_{\perp }$" part hai) jabki steadily $z$ mein climb karta hai (red "drift: $v_{\parallel }$" part). Har poore turn mein vertical rise exactly woh pitch hai jo humne compute ki — picture clear karti hai ki ek saath circle aur straight line ho rahi hai.

Recall Verify

Forecast ka jawab: ek spiral (helix) — $xy$ mein circle, $z$ ke along steady drift. Pitch $\approx 0.262$ m VERIFY mein check hai. Agar $v_{\parallel }$ zero hota toh flat circle hoti (Cell F); agar $v_{\perp }$ zero hoti toh straight line hoti (Cell D).

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Example 8 — Cell H: velocity selector (electric vs magnetic balance)

Steps.

1. Positive charge ke liye dono forces vectors ki tarah likho.

   • Electric: ${\vec{F}}_E=q\vec{E}=qE\hat{y}$ — $+\hat{y}$ ki taraf point karta hai.
   • Magnetic: ${\vec{F}}_B=q\vec{v}\times \vec{B}=q(v\hat{x})\times (B\hat{z})=qvB(\hat{x}\times \hat{z})=qvB(-\hat{y})$ — $-\hat{y}$ ki taraf point karta hai. Yeh step kyun? Geometry ($\vec{E}\perp \vec{B}\perp \vec{v}$) exactly wahi hai jo dono forces ko same line ($\pm \hat{y}$) par rakhti hai taaki woh cancel ho sakein. Agar anti-parallel na hote, toh koi single speed deflection null nahi kar sakti.

2. Undeflected matlab net $\hat{y}$-force zero hai: $qE\hat{y}+qvB(-\hat{y})=\vec{0}\Rightarrow qE=qvB$. Yeh step kyun? "Seedha jaana" = dono opposing $\hat{y}$ pushes exactly balance karte hain.

3. Charge $q$ dono sides se cancel ho jaata hai (aur uska sign dono forces ko saath flip karta hai, toh balance survive karta hai): $E=vB\Rightarrow v=\frac{E}{B}=\frac{2\times 10^4}{0.1}=2\times 10^5$ m/s. Yeh step kyun? Dono forces $q$ ke saath scale karte hain, toh $q$ drop ho jaata hai — selector speed se hi pick karta hai, charge magnitude ya sign se nahi.

Recall Verify

Forecast ka jawab: nahi — $q$ ki magnitude ya sign se koi fark nahi padta; $v=E/B$ charge-independent hai. Units: $(\text{V/m})/\text{T}=(\text{V/m})/({\text{V⋅s/m}}^2)=\text{m/s}$. ✓ Value $2\times 10^5$ m/s VERIFY mein check hai. (Negative charge ke liye ${\vec{F}}_E$ aur ${\vec{F}}_B$ dono saath opposite side flip ho jaate hain, toh same $v$ par phir bhi cancel karte hain.)

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Example 9 — Cell I: mass spectrometer (mass mein chain karo)

Steps.

1. $r=\frac{mv}{|q|B}$ ko mass ke liye rearrange karo: $m=\frac{|q|Br}{v}$. Yeh step kyun? Humne $r$ measure kiya; baaki sab ($q,B,v$) known hai, toh circle ki size mass reveal karti hai. Hum $|q|$ use karte hain kyunki $r$ ek size hai — ion ka sign sirf yeh fix karta hai ki woh kis taraf curl karta hai.

2. $m=\frac{(1.6\times 10^{-19})(0.5)(1.04\times 10^{-2})}{2\times 10^5}=4.16\times 10^{-27}$ kg. Yeh step kyun? Direct substitution — yahi machine ka poora principle hai: badi mass $\Rightarrow$ bada circle.

Recall Verify

Forecast ka jawab: $4.16\times 10^{-27}/1.67\times 10^{-27}\approx 2.5$ proton-masses — toh yeh ek light ion jaisa behave karta hai (atomic units mein roughly mass-2.5, jaise deuteron-ish object). Value $4.16\times 10^{-27}$ kg VERIFY mein check hai.

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Example 10 — Cell J: exam twist, $90^{\circ }$ se zyada angle, sign aur limit saath mein

Steps.

1. (a) Magnitude. $150^{\circ }$ par: $\sin 150^{\circ }=0.5$. $30^{\circ }$ par: $\sin 30^{\circ }=0.5$. Equal magnitudes. Yeh step kyun? $\sin$ $90^{\circ }$ ke baare mein symmetric hai: $\sin (180^{\circ }-\theta )=\sin \theta$. Toh $150^{\circ }$ aur $30^{\circ }$ same $|\vec{F}|$ dete hain chahe velocities quite differently point karein.

2. (b) Direction dekhne ke liye 2-D model set karo. $\vec{B}=(1,0,0)$ aur $\vec{v}=(\cos \theta ,\sin \theta ,0)$ rakho. Toh $\vec{v}\times \vec{B}=(0,0,-\sin \theta )$. Yeh step kyun? Ise likhne se single nonzero component isolate hota hai, jiska value $-\sin \theta$ hai — direction ab poori tarah $\sin \theta$ ke sign mein rehti hai.

3. $\theta$ upar sweep karo aur sign padho.

   • $0^{\circ }<\theta <90^{\circ }$: $\sin \theta >0$ aur badhta hai $90^{\circ }$ par max tak — force badhti hai, $-\hat{k}$ (into the page) point karti hai.
   • $\theta =90^{\circ }$: $\sin \theta =1$, force maximum hai, still $-\hat{k}$.
   • $90^{\circ }<\theta <180^{\circ }$: $\sin \theta >0$ lekin ab ghatta hai — force kamzor hoti hai, still $-\hat{k}$. (Exactly yahi $150^{\circ }$ case hai: $30^{\circ }$ se same size, same side.)
   • $\theta =180^{\circ }$: $\sin 180^{\circ }=0$ — force zero hai ($\vec{v}$ ab directly $\vec{B}$ ke opposite point kar raha hai, $\theta =0^{\circ }$ jaisa ek degenerate parallel case).
   • $180^{\circ }<\theta <360^{\circ }$: $\sin \theta <0$ — $\hat{k}$-component $+$ ho jaata hai, toh force $+\hat{k}$ ki taraf flip ho jaati hai (out of the page). Yeh step kyun? Steering side sirf tab reverse hoti hai jab velocity directly-opposite field se past swing karti hai. Aur force $\theta =180^{\circ }$ par smoothly zero se pass hoti hai — "kya kabhi vanish hoti hai?" forecast ka jawab deti hai: haan, exactly jab $\vec{v}\parallel \vec{B}$ phir se hota hai.

Figure padho: blue curve ek poore turn mein $\sin \theta$ hai. $30^{\circ }$ aur $150^{\circ }$ par do orange dots same height ($0.5$) par hain — isi liye unki force magnitudes match karti hain. Curve $0^{\circ }$, $180^{\circ }$, $360^{\circ }$ par zero touch karti hai (parallel/anti-parallel degeneracies), aur $180^{\circ }$ par red dashed line mark karta hai jahan curve axis ke neeche dive karti hai — woh moment jab force direction sign flip karti hai.

Recall Verify

Forecast ke jawab: (a) $30^{\circ }$ aur $150^{\circ }$ par equal magnitudes (dono $\sin =0.5$); (b) haan, force $\theta =180^{\circ }$ par zero se pass karti hai jahan $\vec{v}$, $\vec{B}$ ke anti-parallel hota hai, phir side reverse karta hai. VERIFY mein check hai: $\sin 30^{\circ }=\sin 150^{\circ }$, $\sin 180^{\circ }=0$, aur $\theta <180^{\circ }$ se $\theta >180^{\circ }$ tak $\hat{k}$-sign flip.

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Recall

Recall Kaun sa cell cross-product result flip karne ko kehta hai?

Cell C ::: negative-charge (electron) wala case — $\vec{v}\times \vec{B}$ right-hand rule se compute karo, phir negative sign se multiply karo.

Recall Magnetic force exactly zero hone ke teen tarike

$v=0$ (at rest), $\vec{v}\parallel \vec{B}$ ($\sin \theta =0$, including anti-parallel $\theta =180^{\circ }$), ya $B=0$ (koi field nahi) ::: teeno ka matlab hai koi "crossing" nahi field ke steer karne ke liye.

Recall Velocity selector charge (aur uska sign) kyun ignore karta hai?

Dono $qE$ aur $qvB$, $q$ ke saath scale karte hain ::: toh $q$ cancel ho jaata hai, $v=E/B$ bachta hai; sign flip karne se dono forces saath reverse hoti hain, toh phir bhi balance karte hain.

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Connections

• 1.8.20 Magnetic force on charge — F = qv × B (Hinglish) — parent topic
• Lorentz force law — electric term $q\vec{E}$ add karta hai jo Example 8 mein use hua
• Cross product — direction ke peeche ki machinery (Examples 1, 3, 5, 10)
• Centripetal force and circular motion — Example 6 mein balance
• Cyclotron — "period speed se independent" wala surprise (Example 6)
• Mass spectrometer — Example 9
• Velocity selector — Example 8
• Magnetic force on a current-carrying wire — many-charges wala version