Magnetic force on charge — F = qv × B

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WHAT is the law?

The magnitude: $|\vec{F}| = |q|\,v\,B\,\sin\theta$ where $\theta$ is the angle between $\vec{v}$ and $\vec{B}$.

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WHY a cross product? (Derivation from first principles)

We don't postulate the formula — we read it off from experiments, then encode the pattern.

Observation 1 — Need motion. A stationary charge in a magnet's field feels no magnetic force. So $\vec{F}\propto v$ (it vanishes when $v=0$).

Observation 2 — Need a sideways component. Fire a charge along the field lines: no force. Fire it across: maximum force. So the effective speed is the part of $\vec v$ perpendicular to $\vec B$, i.e. $v\sin\theta$. Hence $F\propto v\sin\theta$.

Observation 3 — Force is perpendicular to both. Measure the deflection: it's always $\perp\vec v$ and $\perp\vec B$. The unique vector operation that (a) gives a result perpendicular to two inputs and (b) has magnitude $\propto \sin\theta$ is the cross product.

Observation 4 — Proportional to charge, sign matters. Double $q$ → double force; flip the sign of $q$ → flip the direction.

Bundle all four into one vector statement: $\boxed{\vec{F} = q\,\vec{v}\times\vec{B}}$ The constant of proportionality is exactly $1$ in SI — because the tesla is defined to make it so. So the formula is half experiment, half a clean choice of units.

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HOW to get the direction — right-hand rule

Component form (the foolproof method):

\begin{vmatrix} \hat i & \hat j & \hat k\\ v_x & v_y & v_z\\ B_x & B_y & B_z\end{vmatrix} =(v_yB_z-v_zB_y)\,\hat i+(v_zB_x-v_xB_z)\,\hat j+(v_xB_y-v_yB_x)\,\hat k$$ --- ## Circular motion: the most important consequence If $\vec v\perp\vec B$ (so $\sin\theta=1$), the force has constant magnitude $qvB$ and is always perpendicular to motion → uniform **circular motion**. **Derive the radius.** Centripetal requirement: $\dfrac{mv^2}{r}=qvB$. Cancel one $v$: $$\boxed{r=\frac{mv}{qB}}$$ **Period** (time for one loop): distance $2\pi r$ at speed $v$: $$T=\frac{2\pi r}{v}=\frac{2\pi m}{qB},\qquad \omega=\frac{qB}{m}$$ > [!intuition] Surprise of cyclotron motion > $T$ has **no $v$ in it!** A fast particle traces a big circle, a slow one a small circle, but both finish a lap in the **same time**. This is why cyclotrons work. The frequency $\omega=qB/m$ is called the ==cyclotron frequency==. If $\vec v$ has a component along $\vec B$ too, that component feels **no force**, so the particle drifts forward while circling → a **helix**. --- ## Worked Examples > [!example] 1 — Plug into the magnitude formula > A proton ($q=1.6\times10^{-19}\,$C) moves at $v=2\times10^6$ m/s at $30^\circ$ to a field $B=0.5$ T. Find $|\vec F|$. > > $|\vec F|=qvB\sin\theta=(1.6\times10^{-19})(2\times10^6)(0.5)\sin30^\circ$ > *Why this step?* Magnitude only needs $\sin\theta$, no vector bookkeeping. > $=(1.6\times10^{-19})(2\times10^6)(0.5)(0.5)=8\times10^{-14}\,$N. > [!example] 2 — Full vector cross product > $q=+2$ C, $\vec v=(3,0,0)$ m/s, $\vec B=(0,4,0)$ T. Find $\vec F$. > > $\vec v\times\vec B=(v_yB_z-v_zB_y,\ v_zB_x-v_xB_z,\ v_xB_y-v_yB_x)=(0-0,\ 0-0,\ 3\cdot4-0)=(0,0,12)$. > *Why this step?* Only $v_xB_y$ survives, giving a $+\hat k$ result — force points out of the page. > $\vec F=q(\vec v\times\vec B)=2(0,0,12)=(0,0,24)\,$N. > [!example] 3 — Electron, watch the sign > An electron ($q=-1.6\times10^{-19}$ C) moves in $+\hat x$ through $\vec B$ in $+\hat z$. Direction of force? > > $\hat x\times\hat z=-\hat y$. So $\vec v\times\vec B$ points in $-\hat y$. > *Why this step?* $\hat x\times\hat z=-\hat y$ (cyclic order is $x\to y\to z$; going $x\to z$ is backwards → minus). > Multiply by negative charge: $\vec F\propto q(-\hat y)=(-)(-\hat y)=+\hat y$. The electron deflects in $+\hat y$. > [!example] 4 — Radius of an electron's circle > Electron, $v=1\times10^7$ m/s, $B=0.01$ T, $m=9.1\times10^{-31}$ kg. > $r=\dfrac{mv}{qB}=\dfrac{(9.1\times10^{-31})(10^7)}{(1.6\times10^{-19})(0.01)}\approx 5.7\times10^{-3}\,$m $\approx 5.7$ mm. > *Why this step?* Used $r=mv/qB$ with $|q|$ (radius is a magnitude). --- ## Common Mistakes (Steel-manned) > [!mistake] "Magnetic force does work and speeds the particle up." > **Why it feels right:** it's a force, and forces usually change speed (Newton training). > **The fix:** $P=\vec F\cdot\vec v=0$ because $\vec F\perp\vec v$ *always*. It changes **direction**, not **speed**. Only an electric field does work on a charge. > [!mistake] "Forgetting the $\sin\theta$ / using the full $v$." > **Why it feels right:** in the simplest problems $\vec v\perp\vec B$, so $\sin\theta=1$ and you never see it. > **The fix:** force depends only on the perpendicular part $v\sin\theta$. If $\vec v\parallel\vec B$, force $=0$. > [!mistake] "Electron curves the same way as a proton." > **Why it feels right:** same right-hand rule, same $\vec v$ and $\vec B$. > **The fix:** the right-hand rule gives $\vec v\times\vec B$; for **negative** charge you flip the result. Electron and proton with the same velocity curve in **opposite** directions. > [!mistake] "Period depends on speed." > **Why it feels right:** faster things usually finish laps quicker. > **The fix:** $T=2\pi m/qB$ — speed cancels. Faster ⇒ bigger circle, same lap time. --- > [!recall]- Feynman: explain to a 12-year-old > Imagine you're sliding on ice and an invisible hand always shoves you **sideways**, never forward or backward. You can't go faster or slower, but you keep getting turned — so you go in a **circle**. A magnet does exactly this to a moving charged ball. If the ball isn't moving, the hand doesn't bother with it at all. And the harder you throw the ball, the bigger the circle it carves — but funnily, it always takes the **same amount of time** to go around once. > [!mnemonic] Remember it > **"Sideways Steering, No Speeding."** Magnetic force = a steering wheel (turns you), never a gas pedal (no work). And **FvB** order in $\vec F=q\vec v\times\vec B$ — *"Force from velocity-cross-field."* --- ## Active Recall #flashcards/physics What is the full vector formula for magnetic force on a charge? ::: $\vec F = q\,\vec v\times\vec B$ What is the magnitude of the magnetic force? ::: $|F| = |q|\,v\,B\sin\theta$ Why can a magnetic force never change a particle's speed? ::: Because $\vec F\perp\vec v$, so power $\vec F\cdot\vec v=0$; kinetic energy stays constant. When is the magnetic force on a moving charge zero? ::: When $v=0$, or when $\vec v\parallel\vec B$ ($\sin\theta=0$). Radius of circular motion for $\vec v\perp\vec B$? ::: $r=\dfrac{mv}{qB}$ Period of cyclotron motion, and what's surprising about it? ::: $T=\dfrac{2\pi m}{qB}$; it is independent of speed $v$. Cyclotron (angular) frequency? ::: $\omega=\dfrac{qB}{m}$ How does an electron's curving direction compare to a proton's (same $\vec v,\vec B$)? ::: Opposite, because of the negative charge flipping $\vec v\times\vec B$. What shape does motion take if $\vec v$ has components both along and across $\vec B$? ::: A helix (circle + drift along $\vec B$). SI unit of B and its definition? ::: Tesla; $1\,\text{T}=1\,\text{N}/(\text{A·m})$. --- ## Connections - [[Lorentz force law]] — full $\vec F=q(\vec E+\vec v\times\vec B)$, adds the electric part. - [[Cross product]] — the math engine behind the direction. - [[Centripetal force and circular motion]] — supplies $mv^2/r$ used to get $r$. - [[Cyclotron]] and [[Mass spectrometer]] — devices built on $r=mv/qB$. - [[Magnetic force on a current-carrying wire]] — $\vec F=I\vec L\times\vec B$, the same law summed over moving charges. - [[Velocity selector]] — balancing $qE$ against $qvB$. ## 🖼️ Concept Map ```mermaid flowchart TD LAW["F = q v x B"] MOTION["Charge must move"] CROSS["Cross product v x B"] MAG["Magnitude q v B sin theta"] PERP["Force perp to v and B"] RHR["Right-hand rule"] SIGN["Sign of q flips direction"] WORK["No work done"] SPEED["Speed constant, path bends"] TESLA["Tesla defined so constant = 1"] MOTION -->|F vanishes if v=0| LAW LAW -->|direction from| CROSS CROSS -->|gives| MAG CROSS -->|result| PERP CROSS -->|direction via| RHR LAW -->|proportional to q| SIGN PERP -->|F dot v = 0| WORK WORK -->|KE constant| SPEED LAW -->|SI units| TESLA ``` ## 🔊 Hinglish (regional understanding) > [!intuition]- Hinglish mein samjho > Dekho, magnetic force ka ek hi funda yaad rakho: yeh sirf **moving** charge pe lagti hai, aur hamesha **sideways** dhakka deti hai. Agar charge khada hai (v=0) ya field ke parallel ja raha hai, to force zero — kuch nahi hota. Formula hai $\vec F = q\,\vec v\times\vec B$, aur magnitude $|F|=qvB\sin\theta$. Yeh $\sin\theta$ wala part important hai: sirf velocity ka woh hissa jo field ke perpendicular hai, wahi force banata hai. > > Sabse mast baat — magnetic force kabhi **work nahi karti**. Kyun? Kyunki force hamesha velocity ke perpendicular hoti hai, to power $\vec F\cdot\vec v=0$. Iska matlab speed kabhi nahi badalti, sirf **direction** ghoomti hai. Isiliye charge ek **circle** mein ghoomne lagta hai jab $\vec v\perp\vec B$ ho. Yeh ek steering wheel jaisa hai, gas pedal nahi. > > Circle ka radius nikalna easy hai: centripetal force $mv^2/r$ ko magnetic force $qvB$ ke barabar rakho, ek $v$ cancel, mil gaya $r=mv/qB$. Aur period $T=2\pi m/qB$ — isme $v$ hai hi nahi! Matlab fast particle bada circle banayega, slow chhota, par dono ek round same time mein complete karenge. Yeh cyclotron ka secret hai. > > Do galtiyaan bilkul mat karna: (1) electron aur proton same direction mein nahi mudte — negative charge right-hand rule ka answer ulta kar deta hai. (2) $\sin\theta$ kabhi bhoolna mat, warna parallel case galat aayega. Right-hand rule lagao: fingers $\vec v$ pe, curl towards $\vec B$, thumb dega $\vec v\times\vec B$ — positive charge ke liye wahi force, negative ke liye opposite. ![[audio/1.8.20-Magnetic-force-on-charge-—-F-=-qv-×-B.mp3]]