Force on Current-Carrying Wire in Magnetic Field — AP Physics 2 Study Guide

For: AP Physics 2 candidates sitting AP Physics 2.

Covers: Derivation of net force on a current-carrying wire, the $F=BIL\sin \theta$ force formula, right-hand rule for direction, force between parallel wires, and problem-solving in uniform and non-uniform magnetic fields for AP Physics 2.

You should already know: Lorentz force on moving point charges in magnetic fields. Current as the rate of charge flow in a conductor. Right-hand rule for cross product direction.

A note on the practice questions: All worked questions in the "Practice Questions" section below are original problems written by us in the AP Physics 2 style for educational use. They are not reproductions of past College Board / Cambridge / IB papers and may differ in wording, numerical values, or context. Use them to practise the technique; cross-check with official mark schemes for grading conventions.

1. What Is Force on Current-Carrying Wire in Magnetic Field?

When a 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: Magnetism and Electromagnetic Induction, which accounts for 17–23% of the total AP Physics 2 exam score. It appears regularly in both multiple-choice (MCQ) and free-response (FRQ) sections, often paired with concepts from electric circuits or Newtonian mechanics to create multi-concept problems. Unlike force on a single point charge, force on a wire depends on macroscopic properties of the wire (length, current) rather than the velocity of individual charges, making it easy to measure experimentally and use in practical devices like electric motors and loudspeakers. AP exam questions frequently test both magnitude calculation and direction determination, as well as application to parallel current-carrying wires, a common context for conceptual and quantitative problems.

2. Magnitude and Direction of Force on a Straight Wire

The force on a current-carrying wire is derived from summing the Lorentz force on individual charge carriers. For a straight wire of length $L$ carrying current $I$ in a uniform magnetic field $B$, let $n$ be the number density of charge carriers, $A$ the cross-sectional area, and $v_d$ the drift velocity of charges. Total charge in the wire segment is $Q=nALq$, and total force is $F=Q{v}_{d}B\sin \theta$, where $\theta$ is the angle between the current direction and the magnetic field vector. Substituting the definition of current $I=nAq{v}_d$, we get the general force formula: $$F=BIL\sin \theta$$ If the wire is parallel to the magnetic field, $\theta =0^{\circ }$ so $\sin \theta =0$, and the net force is zero. If the wire is perpendicular to the field, $\theta =90^{\circ }$ so $\sin \theta =1$, and force is maximized at $F=BIL$. Direction is given by the cross product rule: $$\vec{F}=I\vec{L}\times \vec{B}$$ where $\vec{L}$ is a vector pointing in the direction of current with magnitude equal to wire length. The right-hand rule for cross products applies: point fingers along $\vec{L}$, curl toward $\vec{B}$, and your thumb points in the direction of $\vec{F}$.

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

Exam tip: Always measure $\theta$ between the current direction and the magnetic field vector, not between the force and the field. Double-check geometry before plugging in values to avoid using the wrong angle.

3. Force Between Two Parallel Current-Carrying Wires

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 straight parallel wires separated by distance $r$, carrying currents $I_1$ and $I_2$, the magnetic field produced by $I_1$ at the location of $I_2$ is $B=\frac{{\mu }_0{I}_1}{2\pi r}$, where ${\mu }_0=4\pi \times 10^{-7}$ T·m/A is the permeability of free space. This field is always perpendicular to $I_2$, so $\sin \theta =1$, and the total force on a length $L$ of $I_2$ is: $$F=\frac{{\mu }_0{I}_1{I}_{2}L}{2\pi r}$$ Force per unit length is $\frac{F}{L}=\frac{{\mu }_0{I}_1{I}_2}{2\pi r}$. For direction: currents in the same direction attract each other, and currents in opposite directions repel each other. This rule comes directly from the right-hand rule for cross products, and 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 formula for force per unit length: $\frac{F}{L}=\frac{{\mu }_0{I}_1{I}_2}{2\pi r}$.
2. Substitute values: ${\mu }_0=4\pi \times 10^{-7}$ T·m/A, $I_1=2.0$ A, $I_2=5.0$ A, $r=0.50$ m.
3. Simplify: $\frac{F}{L}=\frac{(4\pi \times 10^{-7})(2.0)(5.0)}{2\pi (0.50)}=4.0\times 10^{-6}$ N/m.
4. Direction: currents are opposite, so the interaction is repulsive.
5. Confirm with Newton's third law: force per unit length on Wire 1 is equal in magnitude to force per unit length on Wire 2, as expected.

Exam tip: Don't rely solely on the "same attract, opposite repel" memory rule. Re-derive direction with the right-hand rule on the exam to avoid mixing up direction for non-vertical wires.

4. Net Force on Curved Wires in Uniform Magnetic Fields

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 the force over all small segments of the wire: $d\vec{F}=Id\vec{l}\times \vec{B}=I\left(\int d\vec{l}\right)\times \vec{B}$, and $\int d\vec{l}$ between two points is just the displacement vector between the endpoints. For a closed loop (where the start and endpoint are the same), the net displacement is zero, so the net force on the entire loop in a uniform field is always zero. Note that this rule only applies to uniform magnetic fields; non-uniform fields require integration over each segment.

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. The straight-line distance between the endpoints of the semicircle is the diameter: $L_{\text{net}}=2r=2(0.20)=0.40$ m.
2. The magnetic field is perpendicular to the diameter (the net displacement vector), so $\theta =90^{\circ }$ and $\sin \theta =1$.
3. Substitute into the net force formula for curved wires: $F=BI{L}_{\text{net}}=(0.30\text{T})(4.0\text{A})(0.40\text{m})=0.48$ N.
4. Direction is given by the right-hand rule for the straight displacement vector, same as for a straight wire across the endpoints.

Exam tip: Do not use the total length of the curved wire for this calculation. The net force depends only on the endpoint separation for uniform fields.

5. Common Pitfalls (and how to avoid them)

• Wrong move: Using $\cos \theta$ instead of $\sin \theta$ in the force formula, where $\theta$ is the angle between current and B. Why: Students confuse this formula with magnetic flux, which uses $\cos \theta$, or misidentify the angle between vectors. Correct move: Always draw the current and B vectors, measure the smallest angle between them, and confirm that force is zero when parallel (which matches $\sin 0^{\circ }=0$) to check your work.
• Wrong move: Using the total length of a curved wire instead of the straight endpoint distance to find net force in a uniform field. Why: Students assume total length is always used, forgetting the endpoint rule for curved wires. Correct move: For any non-straight wire in a uniform field, first calculate the straight-line distance between the two endpoints, and use that value for $L$.
• Wrong move: Memorizing the parallel wire direction rule backwards, getting attraction/repulsion reversed. 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 cross product rule to get force direction.
• Wrong move: Calculating a non-zero net force for a closed current loop in a uniform magnetic field. Why: Students calculate force on individual segments but forget to add opposing force vectors from opposite sides of the loop. Correct move: Remember that net force on any closed loop in uniform B is always zero; if you get a non-zero result, you made a direction error when adding vectors.
• Wrong move: Using the wire's own magnetic field to calculate the force on the wire, especially in parallel wire problems. Why: Students confuse the source of the force, which always comes from an external field. Correct move: Always label which field is acting: for parallel wires, the force on wire 1 comes from the field produced by wire 2, not wire 1's own field.

6. Practice Questions (AP Physics 2 Style)

Question 1 (Multiple Choice)

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? (A) To the right (B) Into the plane of the page (C) Upward (toward the top of the page) (D) Downward (toward the bottom of the page)

Worked Solution: We use the cross product rule $\vec{F}=I\vec{L}\times \vec{B}$, where $\vec{L}$ points right (direction of current) and $\vec{B}$ points out of the page. Point the fingers of your right hand along $\vec{L}$, then curl them toward $\vec{B}$ (out of the page): your thumb points upward. Using the palm rule confirms this: open right hand, fingers along current, palm faces out of the page (to let B pass through), thumb points up. Correct answer: C.

Question 2 (Free Response)

A 0.10 kg rigid straight rod of length 1.0 m is hinged at its lower end, and carries a current of 5.0 A upward along the rod. The rod is held stationary in a vertical position in a uniform horizontal magnetic field of 0.20 T directed into the page. (a) Calculate the magnitude of the magnetic force on the rod. (b) What is the direction of the magnetic force? (c) Calculate the net torque about the hinge from the magnetic force and the weight of the rod, and state whether the rod rotates clockwise or counterclockwise when released.

Worked Solution: (a) The rod is vertical, so current is perpendicular to the horizontal magnetic field: $\theta =90^{\circ }$, $\sin \theta =1$. Substitute into $F=BIL$: $F=(0.20\text{T})(5.0\text{A})(1.0\text{m})=1.0\text{N}$. (b) Using the right-hand rule: $\vec{L}$ upward, $\vec{B}$ into the page. $\vec{L}\times \vec{B}$ points left, so force direction is horizontal to the left. (c) Weight acts at the center of mass, 0.5 m from the hinge. Torque from weight (clockwise): ${\tau }_g=rmg=(0.5\text{m})(0.10\text{kg}\times 9.8{\text{m/s}}^2)=0.49\text{Ncdotpm clockwise}$. Torque from magnetic force (counterclockwise): ${\tau }_B=rF=(0.5\text{m})(1.0\text{N})=0.50\text{Ncdotpm counterclockwise}$. Net torque: ${\tau }_{net}=0.50-0.49=0.01\text{Ncdotpm counterclockwise}$. The rod rotates counterclockwise when released.

Question 3 (Application / Real-World Style)

A loudspeaker uses a 10-turn circular coil of radius 1.5 cm placed in a 0.10 T uniform magnetic field perpendicular to the plane of the coil. The coil carries a current of 0.50 A. What is the net force on the entire coil, and what is the magnitude of the force on a single straight diameter segment of one turn of the coil? Interpret your result in the context of how a loudspeaker works.

Worked Solution:

1. The entire coil is a closed loop in a uniform magnetic field, so net force on the entire coil is $0\text{N}$.
2. For a single diameter segment, diameter $L=2r=2(0.015\text{m})=0.030\text{m}$.
3. Force on the segment: $F=BIL=(0.10\text{T})(0.50\text{A})(0.030\text{m})=1.5\times 10^{-3}\text{N}$.
4. Interpretation: Even though net force on the whole coil is zero, opposite segments of the coil experience forces in opposite directions, creating a net torque that causes the coil (and the attached speaker cone connected to it) to vibrate, producing sound waves.

7. Quick Reference Cheatsheet

8. 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 the behavior of common electromagnetic devices, which are frequent FRQ topics on the AP exam. This topic also connects directly to the Lorentz force on point charges, linking macroscopic electromagnetic behavior to the motion of individual charges, a core theme across AP Physics 2. Beyond magnetism, force on current-carrying wires is used in concepts like electromagnetic propulsion, which are common contexts for application questions.

• Torque on Current-Carrying Loops
• Magnetic Fields from Current-Carrying Wires
• Faraday's Law of Electromagnetic Induction
• Lorentz Force on Moving Charges