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

Before we start, one reminder of the engines we lean on (all built in the parent note):

Formula The workhorses

Magnitude: $∣ F ∣ = ∣ q ∣ v B sin θ$ . Here $θ$ is the ==angle between the velocity $v$ and the field $B$ ==, measured as the opening between the two arrows when their tails are placed together. It ranges over $0^{\circ} \leq θ \leq 18 0^{\circ}$ . Vector: $F = q v \times B$ , with components $v \times B = (v_{y} B_{z} - v_{z} B_{y}, v_{z} B_{x} - v_{x} B_{z}, v_{x} B_{y} - v_{y} B_{x})$ Electric force (for the crossed-field problems in L4): a charge in an electric field $E$ also feels $F_{E} = q E$ . Here $E$ is the electric field, whose magnitude $E$ is measured in volts per metre (V/m); this force points along $E$ for a positive charge and does not need any motion. The full Lorentz force law simply adds the two: $F = q E + q v \times B$ . Circle: $r = \frac{m v}{∣ q ∣ B}, T = \frac{2 π m}{∣ q ∣ B}, ω = \frac{∣ q ∣ B}{m}$ , where $m$ is the mass of the moving particle (in kg). The $∣ q ∣$ (charge magnitude) keeps $r$ , $T$ , $ω$ positive even for a negative charge like the electron. A tool refresher lives in Cross product and Lorentz force law.

Intuition WHY the magnetic force makes a

circle (where $r, T, ω$ come from) Take $v ⊥ B$ , so $sin θ = 1$ and the force size is a constant $∣ q ∣ v B$ . The key fact (from the parent note): this force is always perpendicular to $v$ . A force that is constant in size and always perpendicular to the motion is exactly the recipe for uniform circular motion — it never speeds the particle up or down (it does no work), it only swings the velocity arrow around. This is the same role gravity plays for a satellite: a perpendicular "centre-seeking" (centripetal) pull. See Centripetal force and circular motion. To find the radius, set the magnetic force equal to the centripetal requirement $\frac{m v ^{2}}{r}$ (the inward force any circle of radius $r$ needs at speed $v$ ): $\frac{m v ^{2}}{r} = ∣ q ∣ v B \Rightarrow r = \frac{m v}{∣ q ∣ B} .$ One lap is a distance $2 π r$ covered at speed $v$ , so $T = \frac{2 π r}{v} = \frac{2 π m}{∣ q ∣ B}$ , and the turning rate is $ω = \frac{2 π}{T} = \frac{∣ q ∣ B}{m}$ .

Since $ω = \frac{∣ q ∣ B}{m}$ is the turning rate (radians per second) and one full loop is $2 π$ radians, the period is simply $T = \frac{2 π}{ω} = \frac{2 π m}{∣ q ∣ B}$ . We will lean on this $ω$ – $T$ bridge in L3 and L5.

Constants used throughout:

Proton / elementary charge $e = 1.6 \times 1 0^{- 19} C$ .
Proton mass $m_{p} = 1.67 \times 1 0^{- 27} kg$ ; electron mass $m_{e} = 9.1 \times 1 0^{- 31} kg$ .

Level 1 — Recognition

Recall Solution L1·Q1

WHAT we use: $∣ F ∣ = ∣ q ∣ v B sin θ$ . The force dies whenever any of $q$ , $v$ , $B$ , or $sin θ$ is zero.

(a) $v = 0$ : force $= 0$ . A still charge is invisible to a magnet — no motion, no push.
(b) $v ∥ B$ : angle $θ = 0^{\circ}$ , so $sin 0^{\circ} = 0$ → force $= 0$ . Riding along the field lines, nothing pushes you sideways.
(c) $v$ anti-parallel: $θ = 18 0^{\circ}$ , so $sin 18 0^{\circ} = 0$ → force $= 0$ too. Moving straight against the field lines is just as force-free as moving along them.
(d) $v ⊥ B$ : $θ = 9 0^{\circ}$ , $sin 9 0^{\circ} = 1$ → force is maximum, $= ∣ q ∣ v B$ . Not zero. Answer: zero in cases (a), (b) and (c).

Recall Solution L1·Q2

WHAT we use: $∣ F ∣ = ∣ q ∣ v B sin θ$ . Notice it uses $∣ q ∣$ — the size of the charge, not its sign. A proton has $q = + e$ ; an electron has $q = - e$ . Both have $∣ q ∣ = e$ . Same $v$ , same $B$ , same $θ = 9 0^{\circ}$ . Answer: the force magnitudes are equal. (Their directions are opposite, and their circles have different radii because masses differ — but the force size is identical.)

Level 2 — Application

Recall Solution L2·Q1

WHAT/WHY: $v ⊥ B$ means $sin θ = 1$ , so we just multiply. $∣ F ∣ = ∣ q ∣ v B sin θ = (3.0 \times 1 0^{- 6}) (5.0 \times 1 0^{4}) (0.20) (1)$ $= 3.0 \times 1 0^{- 6} \times 1.0 \times 1 0^{4} = 3.0 \times 1 0^{- 2} N = 0.030 N .$ Answer: $∣ F ∣ = 0.030 N$ .

Recall Solution L2·Q2

WHY the $sin$ : only the part of $v$ across the field matters; that part is $v sin θ$ . $∣ F ∣ = (3.2 \times 1 0^{- 19}) (1.0 \times 1 0^{6}) (0.40) sin 3 0^{\circ}$ With $sin 3 0^{\circ} = 0.5$ : $= (3.2 \times 1 0^{- 19}) (1.0 \times 1 0^{6}) (0.40) (0.5) = 6.4 \times 1 0^{- 14} N .$ Answer: $∣ F ∣ = 6.4 \times 1 0^{- 14} N$ .

Recall Solution L2·Q3

WHAT we use: the component formula $v \times B = (v_{y} B_{z} - v_{z} B_{y}, v_{z} B_{x} - v_{x} B_{z}, v_{x} B_{y} - v_{y} B_{x})$ . Plug $v_{x} = 0, v_{y} = 2, v_{z} = 0$ and $B_{x} = 0, B_{y} = 0, B_{z} = 3$ :

$x$ : $v_{y} B_{z} - v_{z} B_{y} = 2 \cdot 3 - 0 = 6$
$y$ : $v_{z} B_{x} - v_{x} B_{z} = 0 - 0 = 0$
$z$ : $v_{x} B_{y} - v_{y} B_{x} = 0 - 0 = 0$ So $v \times B = (6, 0, 0)$ , and $F = q (v \times B) = 1 \cdot (6, 0, 0) = (6, 0, 0)$ N. Answer: $F = (6, 0, 0) N$ — a force of $6$ N along $+ \overset{x}{^}$ .

Level 3 — Analysis

Recall Solution L3·Q1

Step 1 — get $v \times B$ first (ignore sign of charge for now). Using the right-hand rule above, $\overset{x}{^} \times \overset{y}{^} = \overset{z}{^}$ : fingers along $v$ (right), curl toward $B$ (up), thumb comes out of the board. So $v \times B$ points in $+ \overset{z}{^}$ (out of the board). Step 2 — apply the charge. $F = q (v \times B)$ with $q = - e < 0$ flips the direction: $+ \overset{z}{^} \to - \overset{z}{^}$ . Answer: the force points in $- \overset{z}{^}$ (into the board). A proton here would go the opposite way, $+ \overset{z}{^}$ (out of the board).

Recall Solution L3·Q2

WHY $∣ q ∣$ : radius is a length (positive), so we use $∣ q ∣ = e$ for both. Here $m$ is each particle's own mass. Proton: $r_{p} = \frac{m _{p} v}{e B} = \frac{( 1.67 \times 1 0 ^{- 27} ) ( 2.0 \times 1 0 ^{6} )}{( 1.6 \times 1 0 ^{- 19} ) ( 0.50 )}$ . Numerator $= 3.34 \times 1 0^{- 21}$ ; denominator $= 8.0 \times 1 0^{- 20}$ ; $r_{p} = 4.175 \times 1 0^{- 2} m \approx 4.2 cm$ . Electron: $r_{e} = \frac{m _{e} v}{e B} = \frac{( 9.1 \times 1 0 ^{- 31} ) ( 2.0 \times 1 0 ^{6} )}{8.0 \times 1 0 ^{- 20}} = 2.275 \times 1 0^{- 5} m \approx 23 μ m$ . Ratio: $\frac{r _{p}}{r _{e}} = \frac{m _{p}}{m _{e}} = \frac{1.67 \times 1 0 ^{- 27}}{9.1 \times 1 0 ^{- 31}} \approx 1835$ . Answer: $r_{p} \approx 4.2$ cm, $r_{e} \approx 23 μ m$ ; the proton's circle is $\sim 1835 \times$ bigger (heavier ⇒ harder to bend). Compare with Centripetal force and circular motion.

Recall Solution L3·Q3

WHAT we use: the turning rate $ω = \frac{∣ q ∣ B}{m}$ (radians per second) depends only on $∣ q ∣$ , $B$ , and the particle mass $m$ — not on speed. One loop is $2 π$ radians, so $T = \frac{2 π}{ω} = \frac{2 π m}{∣ q ∣ B}$ , and speed drops out. $T = \frac{2 π ( 1.67 \times 1 0 ^{- 27} )}{( 1.6 \times 1 0 ^{- 19} ) ( 0.30 )} = \frac{1.049 \times 1 0 ^{- 26}}{4.8 \times 1 0 ^{- 20}} \approx 2.19 \times 1 0^{- 7} s .$ This is the same for both. The fast one carves a bigger circle, the slow one a smaller circle, and they arrive back together. Answer: neither — both take $T \approx 2.2 \times 1 0^{- 7}$ s. This is the heart of the Cyclotron.

Level 4 — Synthesis

Recall Solution L4·Q1

WHAT balances what: the electric force is $F_{E} = q E$ (in $+ \overset{y}{^}$ , from the formula box). The magnetic force is $F_{B} = q v \times B$ . $v \times B = (v, 0, 0) \times (0, 0, B)$ . Component $y$ : $v_{z} B_{x} - v_{x} B_{z} = 0 - v B = - v B$ . So $v \times B = (0, - v B, 0)$ , giving $F_{B}$ in $- \overset{y}{^}$ for positive $q$ . Straight line ⇒ net force zero ⇒ the two cancel in $\overset{y}{^}$ : $q E = q v B \Rightarrow v = \frac{E}{B} .$ The charge and its sign cancel — every charge that survives has the same speed. $v = \frac{3.0 \times 1 0 ^{4}}{0.10} = 3.0 \times 1 0^{5} m/s .$ Answer: $v = 3.0 \times 1 0^{5}$ m/s. This is exactly the Velocity selector used before a Mass spectrometer.

Recall Solution L4·Q2

WHAT we use: rearrange $r = \frac{m v}{∣ q ∣ B ^{'}}$ (here $m$ is the ion's mass, what we want) to solve for it: $m = \frac{∣ q ∣ B ^{'} r}{v} = \frac{( 1.6 \times 1 0 ^{- 19} ) ( 0.20 ) ( 0.080 )}{3.0 \times 1 0 ^{5}} .$ Numerator $= (1.6 \times 1 0^{- 19}) (0.016) = 2.56 \times 1 0^{- 21}$ . Divide by $3.0 \times 1 0^{5}$ : $m = 8.53 \times 1 0^{- 27} kg .$ Answer: $m \approx 8.5 \times 1 0^{- 27}$ kg (roughly $5$ atomic mass units). This is how a Mass spectrometer weighs atoms — measure $r$ , know $v, B^{'}, q$ , read off $m$ .

Level 5 — Mastery

Recall Solution L5·Q1

Idea — split $v$ into two pieces.

$v_{⊥} = v sin θ$ (across $B$ ) → causes the circling.
$v_{∥} = v cos θ$ (along $B$ ) → feels no force, just drifts forward. Numbers: $v_{⊥} = 2.0 \times 1 0^{6} sin 6 0^{\circ} = 2.0 \times 1 0^{6} (0.8660) = 1.732 \times 1 0^{6}$ m/s. $v_{∥} = 2.0 \times 1 0^{6} cos 6 0^{\circ} = 2.0 \times 1 0^{6} (0.5) = 1.0 \times 1 0^{6}$ m/s. (a) Radius uses only the circling part $v_{⊥}$ (here $m = m_{p}$ is the proton's mass): $r = \frac{m _{p} v _{⊥}}{e B} = \frac{( 1.67 \times 1 0 ^{- 27} ) ( 1.732 \times 1 0 ^{6} )}{( 1.6 \times 1 0 ^{- 19} ) ( 0.50 )} = \frac{2.892 \times 1 0 ^{- 21}}{8.0 \times 1 0 ^{- 20}} = 3.62 \times 1 0^{- 2} m \approx 3.6 cm .$ (b) Pitch = (time for one turn) $\times v_{∥}$ . Use the $ω$ – $T$ bridge: the turning rate is $ω = \frac{e B}{m _{p}}$ , so one turn takes $T = \frac{2 π}{ω} = \frac{2 π m _{p}}{e B}$ — speed-independent, exactly as in L3·Q3. $T = \frac{2 π ( 1.67 \times 1 0 ^{- 27} )}{( 1.6 \times 1 0 ^{- 19} ) ( 0.50 )} = \frac{1.049 \times 1 0 ^{- 26}}{8.0 \times 1 0 ^{- 20}} = 1.311 \times 1 0^{- 7} s .$ $pitch = v_{∥} T = (1.0 \times 1 0^{6}) (1.311 \times 1 0^{- 7}) = 0.1311 m \approx 13 cm .$ Answer: $r \approx 3.6$ cm, pitch $\approx 13$ cm. The particle spirals — circle across $B$ , steady march along $B$ .

Recall Solution L5·Q2

Step 1 — cross product $(v_{y} B_{z} - v_{z} B_{y}, v_{z} B_{x} - v_{x} B_{z}, v_{x} B_{y} - v_{y} B_{x})$ :

$x$ : $(2) (2) - (3) (- 1) = 4 + 3 = 7$
$y$ : $(3) (0) - (1) (2) = 0 - 2 = - 2$
$z$ : $(1) (- 1) - (2) (0) = - 1 - 0 = - 1$ So $v \times B = (7, - 2, - 1)$ . Step 2 — multiply by charge $q = - 2$ (flips and scales): $F = - 2 (7, - 2, - 1) = (- 14, 4, 2)$ N. Step 3 — magnitude: $∣ F ∣ = (- 14)^{2} + 4^{2} + 2^{2} = 196 + 16 + 4 = 216 \approx 14.7$ N. Sanity check (from Cross product): $F$ must be $⊥ v$ . Dot: $(- 14) (1) + (4) (2) + (2) (3) = - 14 + 8 + 6 = 0$ . ✓ Answer: $F = (- 14, 4, 2)$ N, $∣ F ∣ = 216 \approx 14.7$ N.

Active Recall

Recall Quick self-check

Force zero when? ::: $v = 0$ , or $θ = 0^{\circ}$ ( $v ∥ B$ ), or $θ = 18 0^{\circ}$ ( $v$ anti-parallel to $B$ ) — all give $sin θ = 0$ . Same-speed proton vs electron in same $B$ : equal force magnitude? ::: Yes — magnitude uses $∣ q ∣$ , and $∣ q ∣$ is equal. Velocity-selector pass-through speed? ::: $v = E / B$ (charge cancels). Mass from a spectrometer? ::: $m = ∣ q ∣ B^{'} r / v$ , where $m$ is the ion's mass. Helix radius uses which speed? ::: Only $v_{⊥} = v sin θ$ . Is the cyclotron period speed-dependent? ::: No — $ω = ∣ q ∣ B / m$ has no $v$ , so $T = 2 π / ω = 2 π m / (∣ q ∣ B)$ is speed-independent. Why does the magnetic force make a circle at all? ::: It is constant in size and always $⊥ v$ — the exact recipe for uniform circular motion (a centripetal force).

Connections

Parent: 1.8.20 Magnetic force on charge — F = qv × B (Hinglish)
Tools: Cross product, Lorentz force law
Applications: Velocity selector, Mass spectrometer, Cyclotron
Related motion: Centripetal force and circular motion, Magnetic force on a current-carrying wire