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

Before any symbol, here is the whole cast, in plain words:

Step 1 — A still charge feels nothing

WHAT. Place a charged ball, not moving, inside a magnetic field. Watch it. It does not budge.

WHY it matters. This is our first clue about what the force depends on. If the force were $F = (something) \times q \times B$ with no mention of motion, a still charge would jump — but it doesn't. So speed must be a factor, and that factor must switch the force off when the ball is still.

PICTURE. In the figure the field lines run left-to-right. The ball sits on them, dead still, and the force meter reads zero.

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

The tiniest way to encode "off when still" is to make the force proportional to speed: $F \propto v$

$F$ = the push we are hunting for.
$v$ = the speed of the ball. When $v = 0$ , the whole right side is $0$ , so $F = 0$ . Exactly what we saw. ✓

Step 2 — Direction of firing changes everything

WHAT. Now we throw the ball. First we throw it along the field lines (parallel). Still nothing! Then we throw it across the field lines (perpendicular). Now it deflects — and this is the strongest deflection we can get.

WHY. So the force does not care about all of the velocity — only the part that lies across the field. We need a mathematical dial that reads $0$ when the arrow is parallel to $B$ and reads maximum when it is perpendicular.

PICTURE. Three throws are drawn: parallel (no force), $4 5^{\circ}$ (some force), perpendicular (full force). The angle between the velocity arrow and the field is called $θ$ (the Greek letter "theta", just a name for that angle).

The dial that does exactly this job is the sine of the angle: $F \propto v sin θ$

$θ$ = angle between $v$ and $B$ .
$sin θ$ = "how much of the velocity points across the field." Look at the figure: $v sin θ$ is the length of the velocity arrow's shadow perpendicular to the field lines.

Step 3 — Which way does it push? Perpendicular to both

WHAT. Measure the actual direction of the deflection. It is never forward, never backward, never along the field. It is always sideways to the velocity AND sideways to the field at the same time — sticking straight out of the plane the two arrows share.

WHY this pins down the maths. We now need a single operation that eats two arrows ( $v$ and $B$ ) and spits out a third arrow perpendicular to both, whose length grows like $sin θ$ . There is exactly one such tool in all of vector maths, and it is called the Cross product, written with a $\times$ .

PICTURE. The velocity arrow and field arrow lie flat on a table; the force arrow stands straight up out of the table, at right angles to both.

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

$v \times B$ = "cross product of $v$ and $B$ " — read it as "the perpendicular partner of these two."
Its length already carries the $v B sin θ$ from Steps 1–2 for free. That is why this one symbol replaces two separate observations.

Step 4 — Reading the direction with your hand (right-hand rule)

WHAT. To find which way $v \times B$ points, use your right hand: point fingers along $v$ , curl them toward $B$ , and your thumb points along $v \times B$ .

WHY a hand at all? "Perpendicular to both" allows two opposite directions (up or down out of the table). The right-hand rule is the agreed-upon tie-breaker that picks one, consistently, everywhere in physics.

PICTURE. A right hand: fingers sweep from the magenta $v$ arrow toward the violet $B$ arrow; the orange thumb pops up as $v \times B$ .

The bookkeeping-proof version (never gets the hand backwards) is the component determinant:

(v_yB_z-v_zB_y)\,\hat i+(v_zB_x-v_xB_z)\,\hat j+(v_xB_y-v_yB_x)\,\hat k$$ - $v_x,v_y,v_z$ = the three pieces of the velocity along the $x,y,z$ directions. - $\hat i,\hat j,\hat k$ = unit arrows (length 1) along $x,y,z$ — the "signposts." - Each bracket is one coordinate of the answer arrow. Plug numbers, get a vector, no hand needed. --- ## Step 5 — The charge scales and can flip the arrow **WHAT.** Double the charge $q$ → the force doubles. Swap a positive charge for a **negative** one (same $\vec v$, same $\vec B$) → the force flips to the exact opposite direction. **WHY.** So the charge $q$ is the final multiplier: its **size** scales the force, and its **sign** decides whether we keep the right-hand-rule arrow or reverse it. **PICTURE.** Two balls fired identically: the positive (magenta) curves one way, the negative (violet) curves the mirror-opposite way. ![[deepdives/dd-physics-1.8.20-d2-s05.png]] Multiply the perpendicular arrow from Step 3 by the signed charge: $$\boxed{\ \vec F = q\,(\vec v\times\vec B)\ }$$ - $q$ = signed charge. If $q>0$, $\vec F$ lines up with $\vec v\times\vec B$. If $q<0$, the minus flips it around. - $\vec v\times\vec B$ = the perpendicular arrow of length $vB\sin\theta$ from Steps 3–4. Every observation is now inside one statement. The constant of proportionality is exactly $1$ **because** the tesla is defined to make it so — half experiment, half a tidy choice of units. --- ## Step 6 — Edge cases: the formula must survive all of them **WHAT.** Check the corners where things might break. **WHY.** A law you trust must give the *right nothing* when nothing should happen, and behave sensibly at every angle — not just the easy $90^\circ$ case. **PICTURE.** Four mini-scenes: (a) $v=0$ still ball → $F=0$; (b) $\vec v\parallel\vec B$ → $F=0$; (c) $\vec v\perp\vec B$ → full force $qvB$; (d) $q=0$ neutral ball → $F=0$. ![[deepdives/dd-physics-1.8.20-d2-s06.png]] $$|\vec F| = |q|\,v\,B\,\sin\theta$$ | Case | What happens | Why (which factor is $0$) | |---|---|---| | $v=0$ (still) | $F=0$ | $v=0$ kills it (Step 1) | | $\vec v\parallel\vec B$ ($\theta=0^\circ$) | $F=0$ | $\sin0^\circ=0$ (Step 2) | | $\vec v\perp\vec B$ ($\theta=90^\circ$) | $F=|q|vB$ | $\sin90^\circ=1$ (max) | | $q=0$ (neutral) | $F=0$ | no charge to push (Step 5) | | $q<0$ | force reverses | sign flips arrow (Step 5) | Every single scenario is covered — the reader can never meet a situation we skipped. --- ## Step 7 — What the formula *does*: bending into a circle **WHAT.** Fire a charge perpendicular to $\vec B$. The force $qvB$ is constant in size and always points sideways to the motion, so it turns the ball forever → a **circle**. (Fire it at a slant and the along-field part sails on untouched → a **helix**.) **WHY.** A push that is always perpendicular to velocity is the definition of [[Centripetal force and circular motion|centripetal force]]. Setting magnetic force = centripetal need lets us *predict the size of the circle*. **PICTURE.** A ball loops in a circle of radius $r$; the force arrow always points to the centre, the velocity always along the loop. A dashed helix shows the slanted case. ![[deepdives/dd-physics-1.8.20-d2-s07.png]] $$\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}$$ - $m$ = mass of the ball. $r$ = radius of its circle. - Cancel one $v$ from both sides to get $r$. Notice $T$ (time for one loop) has **no $v$** in it — fast or slow, same lap time. This is the heart of the [[Cyclotron]], and the radius formula is the engine of the [[Mass spectrometer]]. > [!example] Quick numeric check of $r$ > Electron: $v=1\times10^7$ m/s, $B=0.01$ T, $m=9.1\times10^{-31}$ kg, $|q|=1.6\times10^{-19}$ C. > $$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}.$$ --- ## The one-picture summary ![[deepdives/dd-physics-1.8.20-d2-s08.png]] One frame holds the whole story: velocity in, field in, right-hand rule turns them into a perpendicular force, the charge's sign chooses the direction, and the result curls the path into a circle of radius $mv/qB$. > [!recall]- Feynman retelling — the whole walkthrough in plain words > A magnet ignores a ball that just sits there — so *movement* has to matter (Step 1). If you roll the ball *along* the magnet's grain, still nothing; roll it *across* and it swerves hardest — so only the sideways part of the motion counts, which is the $\sin$ of the angle (Step 2). The swerve is never forward or back — it juts straight out, perpendicular to *both* the motion and the grain, and there's exactly one maths gadget that does that: the cross product (Step 3). To read which of the two "straight out" directions it is, you use your right hand — fingers from velocity to field, thumb is the answer (Step 4). Bigger charge, bigger shove; a *negative* charge shoves the opposite way (Step 5). Test the corners: no motion, motion-along-field, or no charge all give zero, and perpendicular gives the maximum — the formula passes every one (Step 6). Finally, an always-sideways shove keeps turning the ball into a perfect circle whose size is $mv/qB$, and — the delightful surprise — every ball takes the same time to go around (Step 7). That's $\vec F = q\,\vec v\times\vec B$, built entirely from what we saw. > [!mnemonic] Carry it out the door > **See → Sideways → Sign → Circle.** *See* it must move (v), it pushes *Sideways* (cross product), the *Sign* of the charge flips it, and always-sideways means it goes in a *Circle*. --- ## Active Recall Why must the magnetic force depend on speed $v$? ::: Because a stationary charge feels no force; the factor must vanish when $v=0$. Why $\sin\theta$ and not $\cos\theta$ in $|F|=|q|vB\sin\theta$? ::: Only the part of $\vec v$ across the field deflects; $\sin\theta$ is that perpendicular portion (parallel throw gives $\sin0^\circ=0$). Which operation gives an arrow perpendicular to two inputs with magnitude $\propto\sin\theta$? ::: The cross product $\vec v\times\vec B$. What does the sign of $q$ do to the direction of $\vec F$? ::: A positive charge follows $\vec v\times\vec B$; a negative charge points opposite (the minus flips it). List every case where the magnetic force is zero. ::: $v=0$, or $\vec v\parallel\vec B$ ($\theta=0$), or $q=0$. Why does a perpendicular firing make a circle? ::: A constant-size force always perpendicular to $\vec v$ is centripetal, so it turns the path continuously. From $mv^2/r=qvB$, what is the radius? ::: $r=mv/qB$. Why is the lap time $T=2\pi m/qB$ surprising? ::: It contains no $v$ — every charge orbits in the same time regardless of speed. --- ## 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]] - The circle result feeds: [[Centripetal force and circular motion]], [[Cyclotron]], [[Mass spectrometer]] - Sibling ideas: [[Magnetic force on a current-carrying wire]], [[Velocity selector]]