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

Before anything, a one-line reminder of the two tools we lean on:

Here $\hat{i},\hat{j},\hat{k}$ are just the three unit arrows pointing along $x$, $y$, $z$ (each one metre long, at right angles to the other two). "Out of the page" toward you $=+\hat{k}$; "into the page" $=-\hat{k}$.

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The scenario matrix

Every problem this topic can throw is one of these cells. Each worked example below is tagged with the cell(s) it covers.

Cell	Situation	What's tricky	Example
A	$\vec{v}\perp \vec{B}$, positive charge	baseline magnitude + direction	Ex 1
B	General angle $0<\theta <90^{\circ }$	must use $\sin \theta$, not full $v$	Ex 2
C	Negative charge (electron)	flip the cross-product result	Ex 3
D	Degenerate: $\vec{v}\parallel \vec{B}$, $v=0$, and $B=0$	force is zero	Ex 4
E	Full 3-D vectors, all components nonzero	determinant bookkeeping	Ex 5
F	Circular motion — find $r$, $T$	$\vec{v}\perp \vec{B}$ limit, speed cancels	Ex 6
G	Helix: $\vec{v}$ has along + across parts	split velocity, pitch	Ex 7
H	Real machine: Velocity selector	balance electric vs magnetic (vectors!)	Ex 8
I	Real machine: Mass spectrometer	chain $r=mv/qB$ into a mass	Ex 9
J	Exam twist: angle past $90^{\circ }$, sign & limit	direction reverses across $\theta =180^{\circ }$	Ex 10

The matrix has 10 cells across 10 examples. Cover them all and you've covered the topic.

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

Steps.

1. Magnitude: $|\vec{F}|=|q|vB\sin 90^{\circ }=(1.6\times 10^{-19})(3\times 10^6)(0.2)(1)$. Why this step? $\vec{v}\perp \vec{B}$ means $\sin \theta =1$, the maximum — this is the simplest cell, no wasted force. $=9.6\times 10^{-14}$ N.

2. Direction via cross product: $\vec{v}=(3\times 10^6,0,0)$, $\vec{B}=(0,0,0.2)$. $\vec{v}\times \vec{B}=(v_y{B}_z-v_z{B}_y,v_z{B}_x-v_x{B}_z,v_x{B}_y-v_y{B}_x)=(0,-(3\times 10^6)(0.2),0)=(0,-6\times 10^5,0)$. Why this step? Only the middle term survives; it's negative, so the arrow points along $-\hat{y}$ (downward).

3. Multiply by charge (positive, so no flip): $\vec{F}=q(\vec{v}\times \vec{B})$ points in $-\hat{y}$.

Read the figure: the blue $\vec{v}$ arrow points right ($+\hat{x}$), the green $\odot$ marks $\vec{B}$ coming out of the page at you, and the red $\vec{F}$ arrow drops straight down ($-\hat{y}$). Notice all three are mutually at right angles — that perpendicularity is the visual signature of a cross product, and it's why this force can only steer, never speed up.

Recall Verify

Units: C · (m/s) · T $=$ C·(m/s)·(kg/(C·s)) $=$ kg·m/s² $=$ N. ✓ The magnitude $9.6\times 10^{-14}$ N is checked in VERIFY. Right-hand rule sanity: fingers along $+\hat{x}$, curl toward $+\hat{z}$... thumb points $-\hat{y}$. ✓

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

Steps.

1. $|\vec{F}|=|q|vB\sin \theta =(1.6\times 10^{-19})(2\times 10^6)(0.5)\sin 30^{\circ }$. Why this step? Only the part of $\vec{v}$ across the field matters. That perpendicular part is $v\sin \theta$; the part along $\vec{B}$ contributes nothing.

2. $\sin 30^{\circ }=0.5$, so $|\vec{F}|=(1.6\times 10^{-19})(2\times 10^6)(0.5)(0.5)=8\times 10^{-14}$ N. Why this step? We just multiply — the geometry is fully captured by that single $\sin 30^{\circ }$.

Read the figure: the blue $\vec{v}$ leans $30^{\circ }$ above the green $\vec{B}$. The orange dashed arrow is the perpendicular slice $v\sin \theta$ — that shorter arrow is the only part the field "feels". Seeing it drawn shorter than $\vec{v}$ itself is the whole reason the force is half of the perpendicular case.

Recall Verify

Compared to a $90^{\circ }$ hit at the same $v,B$, the force is $\sin 30^{\circ }=0.5$ times as strong — exactly half. If you'd wrongly used the full $v$ you'd get $1.6\times 10^{-13}$ N, double the truth. Value $8\times 10^{-14}$ N checked in VERIFY.

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Example 3 — Cell C: electron, watch the sign flip

Steps.

1. Compute $\vec{v}\times \vec{B}$ ignoring the sign of charge: $\hat{x}\times \hat{z}=-\hat{y}$. Why this step? The cross product only cares about the directions $\vec{v}$ and $\vec{B}$; charge sign comes in afterward. Cyclic order is $x\to y\to z$; going $x\to z$ is backwards, hence the minus.

2. Multiply by the negative charge: $\vec{F}\propto q(-\hat{y})=(-)(-\hat{y})=+\hat{y}$. Why this step? A negative $q$ reverses whatever the right-hand rule gives. The electron goes $+\hat{y}$; a proton (positive) would go $-\hat{y}$.

Read the figure: same blue $\vec{v}$ and same green $\vec{B}$-out-of-page for both particles, yet the red electron arrow points up ($+\hat{y}$) while the orange proton arrow points down ($-\hat{y}$). Two opposite arrows from identical inputs — that split is entirely the work of the minus sign on the charge, and it's why a mixed beam fans apart.

Recall Verify

Two opposite arrows for the same $\vec{v},\vec{B}$ — exactly the "electron curves opposite to proton" rule. If both curved the same way, a beam with mixed charges couldn't be sorted; they can, so opposite is right. ✓

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Example 4 — Cell D: the three degenerate "zero force" cases

Steps.

1. (a) $\theta =0^{\circ }\Rightarrow \sin 0^{\circ }=0\Rightarrow |\vec{F}|=|q|vB(0)=0$. Why this step? When $\vec{v}$ points along $\vec{B}$ there is no "across" component — nothing for the field to grab. 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) $v=0\Rightarrow |\vec{F}|=|q|(0)B\sin \theta =0$. Why this step? The magnetic force is born from motion. No motion, no force — a magnet ignores a still charge completely.

3. (c) $B=0\Rightarrow |\vec{F}|=|q|v(0)\sin \theta =0$; the cross product $\vec{v}\times \vec{0}=\vec{0}$. Why this step? With no field there is nothing to cross the velocity into. This is the logical twin of the $v=0$ case — either factor in $\vec{v}\times \vec{B}$ being zero kills the whole force.

Recall Verify

All three are the "sideways steering, no speeding" personality: a field can only bend a crossing motion. Parallel motion, rest, and no field at all are the three ways to have zero crossing, so each gives $0$ N. ✓

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Example 5 — Cell E: full 3-D determinant

Steps.

1. Plug into the component formula: $\vec{v}\times \vec{B}=(v_y{B}_z-v_z{B}_y,v_z{B}_x-v_x{B}_z,v_x{B}_y-v_y{B}_x)$ Why this step? With every component nonzero, the safe method is the determinant expansion — no geometry needed, no chance of a right-hand-rule slip.

2. Compute each slot:

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

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

3. Multiply by $q=+2$: $\vec{F}=(-28,2,8)$ N. Why this step? Charge is a scalar multiplier applied to every component equally.

Recall Verify

Perpendicularity self-check: $\vec{F}\cdot \vec{v}$ should be $0$ (force $\perp$ velocity always). $(-28)(1)+(2)(2)+(8)(3)=-28+4+24=0$. ✓ Also $\vec{F}\cdot \vec{B}=(-28)(0)+(2)(4)+(8)(-1)=8-8=0$. ✓ Both dot products vanish, confirming $\vec{F}$ is $\perp$ to both inputs — the cross product's signature.

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

Steps.

1. The sideways force of constant size $|q|vB$ acts as the centripetal force: $\frac{m{v}^2}{r}=|q|vB$. Why this step? $\vec{v}\perp \vec{B}$ gives a constant-magnitude force always $\perp$ to motion — the exact recipe for uniform circular motion. Here we use $|q|$ (the magnitude of the charge, $1.6\times 10^{-19}$ C) because $r$ and $T$ are sizes — the negative sign only sets which way the electron circles, not how big the circle is.

2. Cancel one $v$ and solve: $r=\frac{mv}{|q|B}=\frac{(9.1\times 10^{-31})(10^7)}{(1.6\times 10^{-19})(0.01)}$. Why this step? One $v$ cancels because centripetal need grows with $v^2$ but the force only grows with $v$. $r\approx 5.7\times 10^{-3}$ 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. Why this step? The $v$ cancels — a fast electron sweeps a bigger circle but at higher speed, so the lap time is unchanged.

Read the figure: the blue loop is the orbit; at the marked point the green $\vec{v}$ arrow points along the circle (tangent) while the red $\vec{F}$ arrow points straight to the centre. That centre-pointing force is what curls the path into the circle — and because $\vec{F}$ is always $\perp$ to $\vec{v}$ (see the right angle), it never speeds the electron up. The gray dashed $r$ is what shrinks or grows with speed while $T$ stays fixed.

Recall Verify

Answer to the forecast: doubling $v$ doubles $r$ but leaves $T$ untouched — the surprise behind the Cyclotron. Values $r\approx 5.7$ mm and $T\approx 3.57$ ns checked in VERIFY.

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

Steps.

1. Split $\vec{v}$ into a part along $\vec{B}$ and a part across it.

   • Along $\hat{z}$: $v_{\parallel }=4\times 10^5$ m/s — feels no force.
   • Across (in the $xy$-plane): $v_{\perp }=3\times 10^5$ m/s — this is what circles. Why this step? Only the crossing part is steered; the along part coasts straight. Together: a helix.

2. Radius of the circling part: $r=\frac{m{v}_{\perp }}{|q|B}=\frac{(1.67\times 10^{-27})(3\times 10^5)}{(1.6\times 10^{-19})(0.1)}\approx 3.13\times 10^{-2}$ m $=3.13$ cm. Why this step? The radius uses only $v_{\perp }$, since only that component orbits. We use $|q|$ because a radius is a size.

3. Period (same for helix as for circle — depends on $B$, not $v$): $T=\frac{2\pi m}{|q|B}=\frac{2\pi (1.67\times 10^{-27})}{(1.6\times 10^{-19})(0.1)}\approx 6.56\times 10^{-7}$ s.

4. Pitch $=$ forward drift in one period $=v_{\parallel }T=(4\times 10^5)(6.56\times 10^{-7})\approx 0.262$ m. Why this step? While the proton completes one loop (time $T$), the un-steered part slides forward at constant $v_{\parallel }$.

Read the figure: the green vertical line is $\vec{B}$ along $z$; the blue curve is the proton spiralling around it. See how it goes round in the $xy$-plane (that's the orange "circle: $v_{\perp }$" part) while steadily climbing in $z$ (the red "drift: $v_{\parallel }$" part). The vertical rise per full turn is exactly the pitch we computed — the picture makes clear it's a circle and a straight line happening at once.

Recall Verify

Forecast answer: a spiral (helix) — circle in $xy$, steady drift along $z$. Pitch $\approx 0.262$ m checked in VERIFY. If $v_{\parallel }$ were $0$ it'd be a flat circle (Cell F); if $v_{\perp }$ were $0$ it'd be a straight line (Cell D).

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

Steps.

1. Write both forces as vectors for a positive charge.

   • Electric: ${\vec{F}}_E=q\vec{E}=qE\hat{y}$ — points $+\hat{y}$.
   • Magnetic: ${\vec{F}}_B=q\vec{v}\times \vec{B}=q(v\hat{x})\times (B\hat{z})=qvB(\hat{x}\times \hat{z})=qvB(-\hat{y})$ — points $-\hat{y}$. Why this step? The geometry ($\vec{E}\perp \vec{B}\perp \vec{v}$) is exactly what makes the two forces lie on the same line ($\pm \hat{y}$) so they can cancel. If they weren't anti-parallel, no single speed could null the deflection.

2. Undeflected means the net $\hat{y}$-force is zero: $qE\hat{y}+qvB(-\hat{y})=\vec{0}\Rightarrow qE=qvB$. Why this step? "Goes straight" = the two opposing $\hat{y}$ pushes exactly balance.

3. The charge $q$ cancels from both sides (and its sign flips both forces together, so the balance survives): $E=vB\Rightarrow v=\frac{E}{B}=\frac{2\times 10^4}{0.1}=2\times 10^5$ m/s. Why this step? Both forces scale with $q$, so $q$ drops out — the selector picks by speed alone, regardless of charge magnitude or sign.

Recall Verify

Forecast answer: no — neither magnitude nor sign of $q$ matters; $v=E/B$ is charge-independent. Units: $(\text{V/m})/\text{T}=(\text{V/m})/({\text{V⋅s/m}}^2)=\text{m/s}$. ✓ Value $2\times 10^5$ m/s checked in VERIFY. (For a negative charge both ${\vec{F}}_E$ and ${\vec{F}}_B$ flip to the opposite side together, so they still cancel at the same $v$.)

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Example 9 — Cell I: mass spectrometer (chain into a mass)

Steps.

1. Rearrange $r=\frac{mv}{|q|B}$ for mass: $m=\frac{|q|Br}{v}$. Why this step? We measured $r$; everything else ($q,B,v$) is known, so the circle's size reveals the mass. We use $|q|$ because $r$ is a size — the sign of the ion only fixes which way it curls.

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. Why this step? Direct substitution — this is the whole principle of the machine: bigger mass ⇒ bigger circle.

Recall Verify

Forecast answer: $4.16\times 10^{-27}/1.67\times 10^{-27}\approx 2.5$ proton-masses — so this behaves like a light ion (about mass-2.5 in atomic units, e.g. a deuteron-ish object). Value $4.16\times 10^{-27}$ kg checked in VERIFY.

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Example 10 — Cell J: exam twist, angle past $90^{\circ }$, sign and limit together

Steps.

1. (a) Magnitude. At $150^{\circ }$: $\sin 150^{\circ }=0.5$. At $30^{\circ }$: $\sin 30^{\circ }=0.5$. Equal magnitudes. Why this step? $\sin$ is symmetric about $90^{\circ }$: $\sin (180^{\circ }-\theta )=\sin \theta$. So $150^{\circ }$ and $30^{\circ }$ give the same $|\vec{F}|$ even though the velocities point quite differently.

2. (b) Set up a 2-D model to watch the direction. Put $\vec{B}=(1,0,0)$ and $\vec{v}=(\cos \theta ,\sin \theta ,0)$. Then $\vec{v}\times \vec{B}=(0,0,-\sin \theta )$. Why this step? Writing it out isolates the single nonzero component, whose value is $-\sin \theta$ — the direction now lives entirely in the sign of $\sin \theta$.

3. Sweep $\theta$ upward and read the sign.

   • $0^{\circ }<\theta <90^{\circ }$: $\sin \theta >0$ and rising to its max at $90^{\circ }$ — force grows, points $-\hat{k}$ (into the page).
   • $\theta =90^{\circ }$: $\sin \theta =1$, force is maximum, still $-\hat{k}$.
   • $90^{\circ }<\theta <180^{\circ }$: $\sin \theta >0$ but now shrinking — force weakens, still $-\hat{k}$. (This is exactly the $150^{\circ }$ case: same size as $30^{\circ }$, same side.)
   • $\theta =180^{\circ }$: $\sin 180^{\circ }=0$ — force is zero ($\vec{v}$ now points directly opposite $\vec{B}$, a degenerate parallel case just like $\theta =0^{\circ }$).
   • $180^{\circ }<\theta <360^{\circ }$: $\sin \theta <0$ — the $\hat{k}$-component becomes $+$, so the force flips to $+\hat{k}$ (out of the page). Why this step? The steering side reverses only after the velocity swings past directly-opposite the field. And note the force passes smoothly through zero at $\theta =180^{\circ }$ — answering the "does it ever vanish?" forecast: yes, exactly when $\vec{v}\parallel \vec{B}$ again.

Read the figure: the blue curve is $\sin \theta$ across a full turn. The two orange dots at $30^{\circ }$ and $150^{\circ }$ sit at the same height ($0.5$) — that's why their force magnitudes match. The curve touches zero at $0^{\circ }$, $180^{\circ }$, $360^{\circ }$ (the parallel/anti-parallel degeneracies), and the red dashed line at $180^{\circ }$ marks where the curve dives below the axis — the moment the force direction flips sign.

Recall Verify

Forecast answers: (a) equal magnitudes at $30^{\circ }$ and $150^{\circ }$ (both $\sin =0.5$); (b) yes, the force passes through zero at $\theta =180^{\circ }$ where $\vec{v}$ is anti-parallel to $\vec{B}$, then reverses side. Checked in VERIFY: $\sin 30^{\circ }=\sin 150^{\circ }$, $\sin 180^{\circ }=0$, and the $\hat{k}$-sign flip from $\theta <180^{\circ }$ to $\theta >180^{\circ }$.

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Recall

Recall Which cell asks you to flip the cross-product result?

Cell C ::: the negative-charge (electron) case — compute $\vec{v}\times \vec{B}$ by right-hand rule, then multiply by the negative sign.

Recall Three ways to get exactly zero magnetic force

$v=0$ (at rest), $\vec{v}\parallel \vec{B}$ ($\sin \theta =0$, including anti-parallel $\theta =180^{\circ }$), or $B=0$ (no field) ::: all mean no "crossing" for the field to steer.

Recall Why does a velocity selector ignore the charge (and its sign)?

Both $qE$ and $qvB$ scale with $q$ ::: so $q$ cancels, leaving $v=E/B$; flipping the sign reverses both forces together, so they still balance.

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Connections

• 1.8.20 Magnetic force on charge — F = qv × B (Hinglish) — the parent topic
• Lorentz force law — adds the electric term $q\vec{E}$ used in Example 8
• Cross product — the machinery behind direction (Examples 1, 3, 5, 10)
• Centripetal force and circular motion — the balance in Example 6
• Cyclotron — the "period independent of speed" surprise (Example 6)
• Mass spectrometer — Example 9
• Velocity selector — Example 8
• Magnetic force on a current-carrying wire — the many-charges version