Magnetic force on current-carrying conductor: Worked examples

Before anything, let me re-earn every symbol so a newcomer can start from line one.

Definition The symbols, in plain words

$I$ — the current, measured in amperes (A). "How much charge flows per second."
$L$ — a little arrow of length equal to the wire's length $L$ (in metres), pointing the way conventional current flows. So $L$ knows both how long the wire is and which way it points.
$B$ — the magnetic field, an arrow at every point of space telling you which way a compass would point and how strong the field is, measured in tesla (T).
$θ$ — the angle between the wire arrow $L$ and the field arrow $B$ , always the interior angle you sweep to go from one arrow to the other, so $0^{\circ} \leq θ \leq 18 0^{\circ}$ . It is never negative and never more than a straight angle — if a picture looks like $20 0^{\circ}$ , you have simply drawn the outside angle; take $36 0^{\circ}$ minus it. Both $sin θ$ and $sin (18 0^{\circ} - θ)$ are equal, so an obtuse tilt gives the same force size as its acute mirror.
$\times$ — the cross product (see Cross product (vectors)): it takes two arrows and makes a third arrow that is perpendicular to both, with size $= (size of first) (size of second) sin θ$ . That $sin θ$ is exactly where the $sin$ in $F = B I L sin θ$ comes from.

Figure — Magnetic force on current-carrying conductor

Look at figure s01: the wire arrow $L$ (cyan) and field $B$ (white) meet at the interior angle $θ$ . The amber dashed arrow is the part of $L$ that is perpendicular to $B$ — its length is $L sin θ$ . Only that perpendicular part earns any force.

The scenario matrix

Every problem this topic can throw is one (or a combination) of these cells:

Cell	What varies	Degenerate / limiting version
A	Wire $⊥$ field, $θ = 9 0^{\circ}$	max force $F = B I L$
B	Wire at a general angle $0 < θ < 9 0^{\circ}$	uses $sin θ$
C	Wire $∥$ field, $θ = 0^{\circ}$ or $18 0^{\circ}$	zero force
D	Find the direction (Left-Hand Rule / cross product)	into/out of page cases
E	Full component/vector input ( $L$ and $B$ as $\overset{x}{^}, \overset{y}{^}, \overset{z}{^}$ )	any $θ$ hidden inside
F	Curved / bent wire, uniform $B$	replace by straight end-to-end vector
G	Closed loop, uniform $B$	net force $= 0$ (limiting case of F)
H	Word problem (a real device, e.g. a rail/motor bar)	connect force to motion/mass

The eight examples below are each labelled with the cell(s) they cover, so you can see the whole map gets painted.

Worked examples

Worked example Example 4 — Cell D (pure direction, into/out of page)

Current flows to the right ( $+ \overset{x}{^}$ ) through a wire. The field points into the page ( $- \overset{z}{^}$ , where $+ \overset{z}{^}$ = out of page). Which way is the force on the wire?

Forecast: cross product $\Rightarrow$ perpendicular to both — so the force is vertical (up or down). Guess which.

Set axes: $\overset{x}{^}$ = right, $\overset{y}{^}$ = up the page, $\overset{z}{^}$ = out of the page. Why this step? We need a fixed frame so "into the page" has a symbol: $B \propto - \overset{z}{^}$ .
Compute $L \times B \propto \overset{x}{^} \times (- \overset{z}{^})$ . Use $\overset{x}{^} \times \overset{z}{^} = - \overset{y}{^}$ , so $\overset{x}{^} \times (- \overset{z}{^}) = + \overset{y}{^}$ . Why this step? The cross-product table of the unit arrows is the direction rule, written out.
Force points $+ \overset{y}{^}$ = up the page. Why this step? $F = I L \times B$ shares the direction of $L \times B$ .

Verify with the hand: Fleming's Left-Hand Rule — First finger into the page, seCond finger to the right, thuMb points up. ✓ Matches the algebra. In figure s02 the cyan current arrow points right, the white crosses show $B$ diving into the page, and the amber arrow is the resulting upward force — check that all three are mutually perpendicular.

Worked example Example 5 — Cell E (full vectors, angle hidden inside)

A wire segment is $L = (0.30 \overset{x}{^} + 0.40 \overset{y}{^}) m$ carrying $I = 5.0 A$ in a field $B = 0.20 \overset{z}{^} T$ . Find the force vector and its magnitude.

Forecast: $L$ lies in the page-plane, $B$ points out of it, so $L ⊥ B$ — this is secretly a Cell A (perpendicular) problem with $θ = 9 0^{\circ}$ . Expect $F = B I L$ with $L = ∣ L ∣$ , i.e. $F = 0.20 \times 5.0 \times 0.50 = 0.50 N$ .

Length of the wire: $L = 0.3 0^{2} + 0.4 0^{2} = 0.25 = 0.50 m$ . Why this step? We need $∣ L ∣$ both to test the perpendicular claim and to predict the magnitude via the shortcut $F = B I L$ .
Cross product component-wise, with $B = 0.20 \overset{z}{^}$ . Use $\overset{x}{^} \times \overset{z}{^} = - \overset{y}{^}$ and $\overset{y}{^} \times \overset{z}{^} = + \overset{x}{^}$ : $= 5.0\big[0.080\,\hat x - 0.060\,\hat y\big].$$ *Why this step?* Doing the cross product on components delivers direction and size in one shot — no need to hunt for $\theta$ by hand.$
$F = (0.40 \overset{x}{^} - 0.30 \overset{y}{^}) N$ . Why this step? We just multiply the bracket by $5.0$ : $5.0 \times 0.080 = 0.40$ and $5.0 \times 0.060 = 0.30$ . This is the final force vector — writing it out lets us read off both components a reader might need for a follow-up (e.g. equilibrium).
Magnitude: $∣ F ∣ = 0.4 0^{2} + 0.3 0^{2} = 0.16 + 0.09 = 0.25 = 0.50 N$ . Why this step? The magnitude is what the shortcut $F = B I L$ predicts, so computing it from the vector is our cross-check that the component work is correct.

Verify: the shortcut said $F = B I L = 0.20 \times 5.0 \times 0.50 = 0.50 N$ , and the component route also gives $0.50 N$ ✓. Independent perpendicularity checks: $F$ has no $\overset{z}{^}$ part, so $F ⊥ B$ ✓; and the dot product $F \cdot L = (0.40) (0.30) + (- 0.30) (0.40) + 0 = 0.12 - 0.12 = 0$ , so $F ⊥ L$ ✓. Both perpendicularities are exactly what a magnetic force must satisfy.

Worked example Example 6 — Cell F (curved wire, uniform field)

A semicircular wire of radius $R = 0.10 m$ carries $I = 3.0 A$ in a uniform field $B = 0.40 T$ pointing out of the page. The straight diameter joins the two ends. Find the net force on the curved half.

Forecast: integrating force along a curve sounds horrible — but the shortcut says a curved wire in a uniform field feels the same force as the straight wire joining its ends. That straight wire is the diameter, length $2 R$ .

Effective length vector: $L_{eff}$ = straight arrow from start to end = the diameter, magnitude $2 R = 0.20 m$ . Why this step? Because $F = I (\int d ℓ) \times B$ and $\int d ℓ$ is just the end-to-end displacement — the loop's curviness cancels out in a uniform field.
The diameter lies in the page, $B$ is out of the page, so they are perpendicular: $F = B I (2 R) = 0.40 \times 3.0 \times 0.20 = 0.24 N$ . Why this step? $θ = 9 0^{\circ}$ between diameter and field, so $sin θ = 1$ .

Verify: units $T \cdot A \cdot m = N$ ✓. The force is directed perpendicular to the diameter (Left-Hand Rule on the equivalent straight wire). In figure s03 the cyan curve is the real wire, the amber arrow is the effective straight diameter $L_{eff}$ , the white dots are $B$ coming out of the page, and the white dashed arrow is the net force pushing the curve away from its chord.

Worked example Example 7 — Cell G (closed loop, net force zero)

Now the same semicircle is closed by adding the straight diameter wire carrying the same current $I$ back the other way. What is the net force on the whole closed loop in the uniform field?

Forecast: the loop is closed, and the shortcut said $\int d ℓ = 0$ around any closed path. Predict net force = zero.

Curved half feels $0.24 N$ one way (Example 6). Why this step? We already computed it.
The straight diameter (length $2 R$ , perpendicular to $B$ ) feels $F = B I (2 R) = 0.40 \times 3.0 \times 0.20 = 0.24 N$ — but current runs the opposite way along it, so its force is $0.24 N$ in the opposite direction. Why this step? Opposite current direction flips $L$ , hence flips the force.
Net force $= 0.24 - 0.24 = 0 N$ .

Verify: matches $\oint d ℓ = 0 \Rightarrow F_{net} = I (\oint d ℓ) \times B = 0$ . ✓ (The loop can still feel a torque — that is what spins an Electric motor and deflects a Moving-coil galvanometer; see Torque on a current loop.)

Active recall

Recall Which cell is a wire given as

$L$ and $B$ in $\overset{x}{^}, \overset{y}{^}, \overset{z}{^}$ components? Cell E — use the cross product on components directly; you never need to find $θ$ .

Recall Why does a curved wire in a uniform field reduce to a straight one?

Because $F = I (\int d ℓ) \times B$ and $\int d ℓ$ is just the start-to-end displacement — the curviness integrates away. A closed loop gives $\oint d ℓ = 0$ , hence zero net force.

Predict-then-check

For Cell C (wire parallel to

B

), force is zero because

sin 0^{\circ} = 0

Cell A vs Cell B ratio

Example 2 at

3 0^{\circ}

is exactly half of Example 1 at

9 0^{\circ}

since

sin 3 0^{\circ} / sin 9 0^{\circ} = 0.5

Connections

Parent topic (Hinglish) — the law we are drilling.
Lorentz force on a moving charge — where $F = I L \times B$ is born.
Cross product (vectors) — the machinery behind Cells D and E.
Torque on a current loop — why Cell G's zero net force still spins motors.
Electric motor / Moving-coil galvanometer — Cell G in the real world.
Force between two parallel currents — each wire is a Cell A in the other's field.
Drift velocity and current — supplies $I = n A q v_{d}$ behind the whole law.