Force on Current-Carrying Wire in Magnetic Field

AP Physics 2· AP Physics 2 CED — Magnetism and Electromagnetic Induction· 14 min read

1. Core Origin of Force on Current-Carrying Wires★★☆☆☆⏱ 3 min

When current flows through a wire, it consists of countless moving charged electrons, each of which experiences a Lorentz force when placed in an external magnetic field. The net sum of these individual forces on all charges equals the total force acting on the wire as a whole.

This topic is a core component of Unit 5, accounting for 17–23% of the total AP Physics 2 exam score, and appears regularly in both multiple-choice and free-response sections, often paired with electric circuits or Newtonian mechanics.

📘 Definition

Magnetic Force on a Current-Carrying Wire

Net macroscopic magnetic force on a wire from an external magnetic field, arising from the sum of Lorentz forces on individual moving charge carriers in the wire

Example:

This effect powers common practical devices including electric motors and loudspeakers.

2. Magnitude and Direction: Straight Wires in Uniform Fields★★☆☆☆⏱ 4 min

The force formula is derived from summing Lorentz forces on individual charge carriers. Substituting the definition of current $I = n A q v_{d}$ gives the general result:

F = BIL\sin\theta

Where $B$ is magnetic field strength, $I$ is current, $L$ is wire length, and $θ$ is the angle between the current direction and magnetic field vector. If the wire is parallel to the field, force is zero; if perpendicular, force is maximized at $F = B I L$ .

The vector form of the rule is:

\vec{F} = I \vec{L} \times \vec{B}

$L$ is a vector pointing in the direction of current. Use the right-hand rule: point fingers along $L$ , curl toward $B$ , thumb points to the direction of $F$ .

📐 Worked Example

A 2.5 m long straight wire carries a current of 3.0 A through a uniform magnetic field of magnitude 0.40 T. The angle between the wire and the magnetic field is 30°. Calculate the magnitude of the force, and confirm the force when the wire is rotated parallel to the field.

1
Identify known values:
2
$L = 2.5\ \text{m}, I = 3.0\ \text{A}, B = 0.40\ \text{T}, \theta = 30^\circ$
3
Write the force formula for a straight wire:
4
$F = BIL\sin\theta$
5
Substitute values:
6
$F = (0.40)(3.0)(2.5)(\sin 30^\circ)$
7
Simplify: $sin 3 0^{\circ} = 0.5$ , so $F = 1.5$ N
8
When parallel to $B$ , $θ = 0^{\circ}$ , $sin 0^{\circ} = 0$ , so force is $0$ N, as expected.

Exam tip:

Always measure $θ$ between the current direction and the magnetic field vector, not between the force and the field. Double-check geometry before plugging in values.

3. Force Between Two Parallel Current-Carrying Wires★★★☆☆⏱ 3 min

Two parallel current-carrying wires exert force on each other: one wire produces a magnetic field that acts on the current in the second wire, and vice versa.

For two long parallel wires separated by distance $r$ , carrying currents $I_{1}$ and $I_{2}$ , the magnetic field from $I_{1}$ at $I_{2}$ is $B = \frac{μ _{0} I _{1}}{2 π r}$ . This field is always perpendicular to $I_{2}$ , so $sin θ = 1$ , giving force per unit length:

\frac{F}{L} = \frac{\mu_0 I_1 I_2}{2\pi r}

Direction rule: currents in the same direction attract, opposite directions repel. This rule is the basis for the formal definition of the ampere.

📐 Worked Example

Two parallel wires are separated by 0.50 m. Wire 1 carries 2.0 A upward, and Wire 2 carries 5.0 A downward. What is the force per unit length on Wire 1, and is the interaction attractive or repulsive?

1
Write the force per unit length formula:
2
$\frac{F}{L} = \frac{\mu_0 I_1 I_2}{2\pi r}$
3
Substitute values: $μ_{0} = 4 π \times 1 0^{- 7}$ T·m/A, $I_{1} = 2.0$ A, $I_{2} = 5.0$ A, $r = 0.50$ m
4
Simplify:
5
$\frac{F}{L} = \frac{(4\pi \times 10^{-7})(2.0)(5.0)}{2\pi (0.50)} = 4.0 \times 10^{-6}\ \text{N/m}$
6
Currents are opposite, so the interaction is repulsive.
7
Newton's third law confirms magnitude is equal for both wires.

Exam tip:

Don't rely solely on the 'same attract, opposite repel' memory rule. Re-derive direction with the right-hand rule to avoid mistakes.

4. Net Force on Curved Wires and Closed Loops★★★☆☆⏱ 3 min

For any curved wire in a uniform magnetic field, the net force equals the force that would act on a straight wire connecting the two endpoints of the curved wire, carrying the same current. This comes from integrating force over all small segments, where the integral of $d l$ equals the net displacement between endpoints.

For a closed loop (start and endpoint are the same), net displacement is zero, so net force on the entire loop in a uniform field is always zero. This rule only applies to uniform magnetic fields.

📐 Worked Example

A semicircular wire of radius 0.20 m carries a current of 4.0 A, placed in a uniform 0.30 T magnetic field perpendicular to the plane of the semicircle. What is the net force on the semicircular wire?

1
Straight-line distance between endpoints is the diameter:
2
$L_{\text{net}} = 2r = 2(0.20) = 0.40\ \text{m}$
3
Field is perpendicular to the diameter, so $θ = 9 0^{\circ}$ , $sin θ = 1$
4
Substitute into force formula:
5
$F = BI L_{\text{net}} = (0.30)(4.0)(0.40) = 0.48\ \text{N}$
6
Direction follows the right-hand rule for the net displacement vector.

Exam tip:

Do not use the total length of the curved wire for this calculation. Net force depends only on endpoint separation for uniform fields.

5. AP-Style Concept Check★★★★☆⏱ 4 min

✓ Quick check

Test your understanding of direction rules:

A straight horizontal wire carries a constant current $I$ directed to the right. The wire is placed in a uniform magnetic field that points directly out of the plane of the page. What is the direction of the net magnetic force on the wire?
- To the right
- Into the plane of the page
- Upward (toward the top of the page)
- Downward (toward the bottom of the page)
Reveal answer
2 —
Using $F = I L \times B$ : $L$ right, $B$ out. Right-hand rule gives upward direction, so this is correct.

📐 Worked Example

A 0.10 kg rigid straight rod of length 1.0 m is hinged at its lower end, carries 5.0 A upward along the rod, and is held stationary vertically in a uniform 0.20 T horizontal magnetic field directed into the page. (a) Find magnetic force magnitude, (b) find direction, (c) find net torque about the hinge.

1
(a) Current is perpendicular to field, so $θ = 9 0^{\circ}$ , $sin θ = 1$ :
2
$F = BIL = (0.20)(5.0)(1.0) = 1.0\ \text{N}$
3
(b) Right-hand rule: $L$ upward, $B$ into page, $L \times B$ points left. Force direction is horizontal left.
4
(c) Weight acts at center of mass (0.5 m from hinge):
5
$\tau_g = (0.5)(0.10 \times 9.8) = 0.49\ \text{N·m clockwise}$
6
$\tau_B = (0.5)(1.0) = 0.50\ \text{N·m counterclockwise}$
7
Net torque: $τ_{n e t} = 0.01$ N·m counterclockwise, so the rod rotates counterclockwise when released.

6. Common Pitfalls

Wrong move:

Using $cos θ$ instead of $sin θ$ in the force formula

Why:

Students confuse this formula with magnetic flux (which uses $cos θ$ ) or misidentify the angle between vectors

Correct move:

Draw the current and $B$ vectors, measure the smallest angle between them, confirm force is zero when parallel (matches $sin 0^{\circ} = 0$ ) to check your work

Wrong move:

Using total curved length instead of straight endpoint distance for net force on curved wires

Why:

Students assume total length is always used, forgetting the endpoint rule for uniform fields

Correct move:

For any non-straight wire in a uniform field, use the straight-line distance between endpoints for $L$

Wrong move:

Memorizing parallel wire direction rule backwards, reversing attraction/repulsion

Why:

Students rely on memory instead of confirming with first principles

Correct move:

Always find $B$ from the first wire at the location of the second, then apply the right-hand rule to get direction

Wrong move:

Calculating non-zero net force for a closed loop in a uniform magnetic field

Why:

Students calculate force on individual segments but forget to add opposing force vectors

Correct move:

Remember net force on any closed loop in uniform $B$ is always zero; a non-zero result indicates a direction error

Wrong move:

Using the wire's own magnetic field to calculate force on the wire

Why:

Students confuse the source of the force, which always comes from an external field

Correct move:

For parallel wires, force on wire 1 comes from the field of wire 2, not wire 1's own field

7. Quick Reference Cheatsheet

Category	Formula	Notes
Force on straight wire (uniform B)	$F = B I L sin θ$	$θ$ = angle between current and $B$ ; $L$ = wire length
Force vector direction	$F = I L \times B$	$L$ points in current direction; use right-hand cross product rule
Force per unit length (parallel wires)	$\frac{F}{L} = \frac{μ _{0} I _{1} I _{2}}{2 π r}$	$r$ = separation; same currents attract, opposite repel
Permeability of free space	$μ_{0} = 4 π \times 1 0^{- 7} T \cdotp m/A$	Provided on AP equation sheet
Net force on curved wire (uniform B)	$F = B I L_{net} sin θ$	$L_{net}$ = straight distance between endpoints
Net force on closed loop (uniform B)	$F_{net} = 0$	Net torque may still be non-zero for motors

When this came up on past exams

AI-estimated based on syllabus patterns — cross-check with official past papers for accuracy. Use only as revision-focus signals.

2022 · MCQ
Force direction on current-carrying wire
2023 · FRQ
Torque on hinged current-carrying rod

Going deeper

Official ResourceAP Physics 2 Equation SheetContains all formulas provided on exam day

What's Next

This topic is the foundational prerequisite for understanding torque on current-carrying loops, the next core concept in Unit 5, and is essential for analyzing electric motors, generators, and galvanometers. Without mastering the direction and magnitude of force on a current-carrying wire, you cannot correctly calculate torque on loops or analyze common electromagnetic devices, which are frequent FRQ topics on the AP exam. This topic connects directly to the Lorentz force on point charges, linking macroscopic electromagnetic behavior to individual charge motion, a core theme across AP Physics 2.