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

WHAT we use: $|\vec{F}|=|q|vB\sin \theta$. Force tab khatam hoti hai jab $q$, $v$, $B$, ya $\sin \theta$ mein se koi ek zero ho.

• (a) $v=0$: force $=0$. Ek ruka hua charge magnet ko invisible hota hai — koi motion nahi, koi push nahi.
• (b) $\vec{v}\parallel \vec{B}$: angle $\theta =0^{\circ }$, toh $\sin 0^{\circ }=0$ → force $=0$. Field lines ke saath chalte hue, kuch bhi tumhe sideways push nahi karta.
• (c) $\vec{v}$ anti-parallel: $\theta =180^{\circ }$, toh $\sin 180^{\circ }=0$ → force $=0$ bhi. Field lines ke seedha against chalte hue utna hi force-free hai jitna unke saath chalna.
• (d) $\vec{v}\perp \vec{B}$: $\theta =90^{\circ }$, $\sin 90^{\circ }=1$ → force maximum hai, $=|q|vB$. Zero nahi. Answer: zero cases (a), (b) aur (c) mein.

WHAT we use: $|\vec{F}|=|q|vB\sin \theta$. Dhyan do ki ye $|q|$ use karta hai — charge ka size, uska sign nahi. A proton has $q=+e$; an electron has $q=-e$. Both have $|q|=e$. Same $v$, same $B$, same $\theta =90^{\circ }$. Answer: the force magnitudes are equal. (Unki directions opposite hain, aur unke circles ke radii alag hain kyunki masses alag hain — lekin force ki size identical hai.)

WHAT/WHY: $\vec{v}\perp \vec{B}$ matlab $\sin \theta =1$, toh hum sirf multiply karte hain. $|\vec{F}|=|q|vB\sin \theta =(3.0\times 10^{-6})(5.0\times 10^4)(0.20)(1)$ $=3.0\times 10^{-6}\times 1.0\times 10^4=3.0\times 10^{-2}\text{N}=0.030\text{N}.$ Answer: $|\vec{F}|=0.030\text{N}$.

WHY the $\sin$: sirf $\vec{v}$ ka woh part jo field ke across hai, matter karta hai; woh part $v\sin \theta$ hai. $|\vec{F}|=(3.2\times 10^{-19})(1.0\times 10^6)(0.40)\sin 30^{\circ }$ $\sin 30^{\circ }=0.5$ ke saath: $=(3.2\times 10^{-19})(1.0\times 10^6)(0.40)(0.5)=6.4\times 10^{-14}\text{N}.$ Answer: $|\vec{F}|=6.4\times 10^{-14}\text{N}$.

WHAT we use: component formula $\vec{v}\times \vec{B}=(v_y{B}_z-v_z{B}_y,v_z{B}_x-v_x{B}_z,v_x{B}_y-v_y{B}_x)$. Plug $v_x=0,v_y=2,v_z=0$ aur $B_x=0,B_y=0,B_z=3$:

• $x$: $v_y{B}_z-v_z{B}_y=2\cdot 3-0=6$
• $y$: $v_z{B}_x-v_x{B}_z=0-0=0$
• $z$: $v_x{B}_y-v_y{B}_x=0-0=0$ Toh $\vec{v}\times \vec{B}=(6,0,0)$, aur $\vec{F}=q(\vec{v}\times \vec{B})=1\cdot (6,0,0)=(6,0,0)$ N. Answer: $\vec{F}=(6,0,0)\text{N}$ — $+\hat{x}$ direction mein $6$ N ki force.

Step 1 — pehle $\vec{v}\times \vec{B}$ lo (abhi charge ka sign ignore karo). Upar wale right-hand rule se, $\hat{x}\times \hat{y}=\hat{z}$: ungliyan $\vec{v}$ ke saath (right), $\vec{B}$ ki taraf curl karo (up), thumb board se bahar aata hai. Toh $\vec{v}\times \vec{B}$ $+\hat{z}$ mein point karta hai (board se bahar). Step 2 — charge apply karo. $\vec{F}=q(\vec{v}\times \vec{B})$ mein $q=-e<0$ direction flip kar deta hai: $+\hat{z}\to -\hat{z}$. Answer: force $-\hat{z}$ mein point karti hai (board ke andar). Yahan ek proton opposite direction mein jaata, $+\hat{z}$ (board se bahar).

WHY $|q|$: radius ek length hai (positive), isliye dono ke liye $|q|=e$ use karte hain. Yahan $m$ har particle ki apni mass hai. Proton: $r_p=\frac{m_{p}v}{eB}=\frac{(1.67\times 10^{-27})(2.0\times 10^6)}{(1.6\times 10^{-19})(0.50)}$. Numerator $=3.34\times 10^{-21}$; denominator $=8.0\times 10^{-20}$; $r_p=4.175\times 10^{-2}\text{m}\approx 4.2\text{cm}$. Electron: $r_e=\frac{m_{e}v}{eB}=\frac{(9.1\times 10^{-31})(2.0\times 10^6)}{8.0\times 10^{-20}}=2.275\times 10^{-5}\text{m}\approx 23\mu \text{m}$. Ratio: $\frac{r_p}{r_e}=\frac{m_p}{m_e}=\frac{1.67\times 10^{-27}}{9.1\times 10^{-31}}\approx 1835$. Answer: $r_p\approx 4.2$ cm, $r_e\approx 23\mu \text{m}$; proton ka circle $\sim 1835\times$ bada hai (heavier ⇒ bend karna mushkil). Compare karo Centripetal force and circular motion se.

WHAT we use: turning rate $\omega =\frac{|q|B}{m}$ (radians per second) sirf $|q|$, $B$, aur particle mass $m$ par depend karta hai — speed par nahi. Ek loop $2\pi$ radians ka hai, isliye $T=\frac{2\pi }{\omega }=\frac{2\pi m}{|q|B}$, aur speed drop out ho jaati hai. $T=\frac{2\pi (1.67\times 10^{-27})}{(1.6\times 10^{-19})(0.30)}=\frac{1.049\times 10^{-26}}{4.8\times 10^{-20}}\approx 2.19\times 10^{-7}\text{s}.$ Ye dono ke liye same hai. Fast wala bada circle banata hai, slow wala chota circle, aur dono saath wapas aate hain. Answer: koi nahi — dono $T\approx 2.2\times 10^{-7}$ s lete hain. Ye Cyclotron ka core idea hai.

WHAT kya balance karta hai: electric force ${\vec{F}}_E=q\vec{E}$ hai (formula box se $+\hat{y}$ mein). Magnetic force ${\vec{F}}_B=q\vec{v}\times \vec{B}$ hai. $\vec{v}\times \vec{B}=(v,0,0)\times (0,0,B)$. Component $y$: $v_z{B}_x-v_x{B}_z=0-vB=-vB$. Toh $\vec{v}\times \vec{B}=(0,-vB,0)$, positive $q$ ke liye ${\vec{F}}_B$ $-\hat{y}$ mein deta hai. Seedhi line ⇒ net force zero ⇒ dono $\hat{y}$ mein cancel hoti hain: $qE=qvB\Rightarrow v=\frac{E}{B}.$ Charge aur uska sign cancel ho jaata hai — har woh charge jo survive karta hai uski same speed hoti hai. $v=\frac{3.0\times 10^4}{0.10}=3.0\times 10^5\text{m/s}.$ Answer: $v=3.0\times 10^5$ m/s. Ye exactly woh Velocity selector hai jo Mass spectrometer se pehle use hota hai.

WHAT we use: $r=\frac{mv}{|q|B^{\prime }}$ (yahan $m$ ion ka mass hai, jo hum chahte hain) ko rearrange karo uske liye solve karne ke liye: $m=\frac{|q|B^{\prime }r}{v}=\frac{(1.6\times 10^{-19})(0.20)(0.080)}{3.0\times 10^5}.$ Numerator $=(1.6\times 10^{-19})(0.016)=2.56\times 10^{-21}$. $3.0\times 10^5$ se divide karo: $m=8.53\times 10^{-27}\text{kg}.$ Answer: $m\approx 8.5\times 10^{-27}$ kg (roughly $5$ atomic mass units). Aise hi Mass spectrometer atoms ko weighs karta hai — $r$ measure karo, $v,B^{\prime },q$ jaano, $m$ pad lo.

Idea — $\vec{v}$ ko do pieces mein split karo.

• $v_{\perp }=v\sin \theta$ ($\vec{B}$ ke across) → circling ka cause.
• $v_{\parallel }=v\cos \theta$ ($\vec{B}$ ke along) → koi force feel nahi hoti, sirf aage drift karta hai. Numbers: $v_{\perp }=2.0\times 10^6\sin 60^{\circ }=2.0\times 10^6(0.8660)=1.732\times 10^6$ m/s. $v_{\parallel }=2.0\times 10^6\cos 60^{\circ }=2.0\times 10^6(0.5)=1.0\times 10^6$ m/s. (a) Radius sirf circling part $v_{\perp }$ use karta hai (yahan $m=m_p$ proton ki mass hai): $r=\frac{m_p{v}_{\perp }}{eB}=\frac{(1.67\times 10^{-27})(1.732\times 10^6)}{(1.6\times 10^{-19})(0.50)}=\frac{2.892\times 10^{-21}}{8.0\times 10^{-20}}=3.62\times 10^{-2}\text{m}\approx 3.6\text{cm}.$ (b) Pitch = (ek turn ka time) $\times v_{\parallel }$. $\omega$–$T$ bridge use karo: turning rate $\omega =\frac{eB}{m_p}$ hai, toh ek turn mein $T=\frac{2\pi }{\omega }=\frac{2\pi m_p}{eB}$ lagta hai — speed-independent, exactly jaise L3·Q3 mein. $T=\frac{2\pi (1.67\times 10^{-27})}{(1.6\times 10^{-19})(0.50)}=\frac{1.049\times 10^{-26}}{8.0\times 10^{-20}}=1.311\times 10^{-7}\text{s}.$ $\text{pitch}=v_{\parallel }T=(1.0\times 10^6)(1.311\times 10^{-7})=0.1311\text{m}\approx 13\text{cm}.$ Answer: $r\approx 3.6$ cm, pitch $\approx 13$ cm. Particle spiral karta hai — $\vec{B}$ ke across circle, $\vec{B}$ ke along steady march.

Step 1 — cross product $(v_y{B}_z-v_z{B}_y,v_z{B}_x-v_x{B}_z,v_x{B}_y-v_y{B}_x)$:

• $x$: $(2)(2)-(3)(-1)=4+3=7$
• $y$: $(3)(0)-(1)(2)=0-2=-2$
• $z$: $(1)(-1)-(2)(0)=-1-0=-1$ Toh $\vec{v}\times \vec{B}=(7,-2,-1)$. Step 2 — charge se multiply karo $q=-2$ (flip bhi karta hai aur scale bhi): $\vec{F}=-2(7,-2,-1)=(-14,4,2)$ N. Step 3 — magnitude: $|\vec{F}|=\sqrt{(-14{)}^2+4^2+2^2}=\sqrt{196+16+4}=\sqrt{216}\approx 14.7$ N. Sanity check (Cross product se): $\vec{F}$ $\vec{v}$ ke $\perp$ hona chahiye. Dot: $(-14)(1)+(4)(2)+(2)(3)=-14+8+6=0$. ✓ Answer: $\vec{F}=(-14,4,2)$ N, $|\vec{F}|=\sqrt{216}\approx 14.7$ N.

Force zero kab? ::: $v=0$, ya $\theta =0^{\circ }$ ($\vec{v}\parallel \vec{B}$), ya $\theta =180^{\circ }$ ($\vec{v}$ anti-parallel to $\vec{B}$) — sab $\sin \theta =0$ dete hain. Same-speed proton vs electron same $B$ mein: equal force magnitude? ::: Haan — magnitude $|q|$ use karta hai, aur $|q|$ equal hai. Velocity-selector pass-through speed? ::: $v=E/B$ (charge cancel ho jaata hai). Mass spectrometer se mass? ::: $m=|q|B^{\prime }r/v$, jahan $m$ ion ki mass hai. Helix radius kaunsi speed use karta hai? ::: Sirf $v_{\perp }=v\sin \theta$. Kya cyclotron period speed-dependent hai? ::: Nahi — $\omega =|q|B/m$ mein koi $v$ nahi, isliye $T=2\pi /\omega =2\pi m/(|q|B)$ speed-independent hai. Magnetic force circle kyun banata hai? ::: Ye size mein constant hai aur hamesha $\perp \vec{v}$ hai — uniform circular motion ki exact recipe (ek centripetal force).