🧲ap physics 2 (2025) review

12.3 Magnetism and Current-Carrying Wires

Verified for the 2026 AP Physics 2 (2025) examCitation:

Magnetic Field Produced by a Current-Carrying Wire

Current-carrying wires generate magnetic fields around them, creating one of the most fundamental relationships in electromagnetism. This relationship demonstrates how moving electric charges produce magnetic effects. 🧲

  • Magnetic field vectors around a long, straight, current-carrying wire form concentric circles perpendicular to the wire
  • Field vectors are tangent to these circles at every point
  • No field component points toward, away from, or parallel to the wire itself

The magnetic field strength at any point near a long, straight, current-carrying wire depends on two key factors: the current magnitude and the distance from the wire.

  • Field magnitude is directly proportional to current (double the current, double the field strength)

  • Field magnitude is inversely proportional to perpendicular distance from wire's central axis (double the distance, half the field strength)

  • Calculate using the equation: B=μ02πIrB=\frac{\mu_{0}}{2 \pi} \frac{I}{r}

    Where:

    • BB = magnetic field strength (in teslas, T)
    • μ0\mu_0 = permeability of free space constant (4π×1074π × 10^{-7} T·m/A)
    • II = current in the wire (in amperes, A)
    • rr = perpendicular distance from wire (in meters, m)

To determine the direction of a current-carrying wire's magnetic field, we use the right-hand rule:

  1. Point your thumb in the direction of conventional current (positive to negative)
  2. Your fingers naturally curl in the direction of the magnetic field
  3. This creates a circular pattern around the wire

For a current loop (like a circular wire), the magnetic field at its center points along the loop's axis. Again, the right-hand rule helps determine the specific direction - curl your fingers in the direction of the current flow, and your thumb points in the field direction at the center.

When multiple current-carrying wires are present in a region, the net magnetic field at any location is found by vector addition of the individual fields from each wire. This means accounting for both magnitude and direction of each contributing field.

Force on a Current-Carrying Wire in a Magnetic Field

When a current-carrying wire is placed in an external magnetic field, it experiences a force. This force is the basis for electric motors and many other electromagnetic devices. ⚡

The force magnitude depends on four critical factors:

  1. Current in the wire (more current means more force)
  2. Length of wire within the magnetic field (longer wire means more force)
  3. Magnetic field strength (stronger field means more force)
  4. Angle between the current direction and magnetic field direction (maximum force when perpendicular)

This relationship is quantified by the equation:

FB=IBsinθF_{B}=I \ell B \sin \theta

Where:

  • FBF_B = magnetic force on the wire (in newtons, N)
  • II = current in the wire (in amperes, A)
  • \ell = length of wire in the magnetic field (in meters, m)
  • BB = external magnetic field strength (in teslas, T)
  • θ\theta = angle between current and field directions (in degrees or radians)

The force is maximum when the current and magnetic field are perpendicular (θ=90°\theta = 90°, sin90°=1\sin 90° = 1), and zero when they're parallel (θ=0°\theta = 0°, sin0°=0\sin 0° = 0).

To determine the direction of the magnetic force on the current-carrying wire, we use another version of the right-hand rule:

  1. Point your fingers in the direction of the conventional current
  2. Orient your palm facing the direction of the magnetic field
  3. Your extended thumb now points in the direction of the force

This rule helps visualize the three-dimensional relationship between current, field, and force, which are all perpendicular to each other in the maximum force scenario.

Practice Problem 1: Magnetic Field from a Wire

A long, straight wire carries a current of 5.0 A. Calculate the magnitude of the magnetic field at a distance of 10 cm from the wire. The permeability of free space is μ0=4π×107\mu_0 = 4\pi \times 10^{-7} T·m/A.

Solution

To find the magnetic field strength at a point near a long, straight current-carrying wire, we use the equation:

B=μ02πIrB=\frac{\mu_{0}}{2 \pi} \frac{I}{r}

Given:

  • Current, I=5.0I = 5.0 A
  • Distance from wire, r=10r = 10 cm = 0.10 m
  • Permeability of free space, μ0=4π×107\mu_0 = 4\pi \times 10^{-7} T·m/A

Substituting these values:

B=4π×1072π5.00.10B=\frac{4\pi \times 10^{-7}}{2 \pi} \frac{5.0}{0.10}

B=4π×1072π×50B=\frac{4\pi \times 10^{-7}}{2 \pi} \times 50

B=2×107×50B= 2 \times 10^{-7} \times 50

B=1.0×105B= 1.0 \times 10^{-5} T

The magnetic field at a distance of 10 cm from the wire is 1.0×1051.0 \times 10^{-5} T or 10 μT.

Practice Problem 2: Force on a Current-Carrying Wire

A straight wire of length 25 cm carries a current of 3.0 A. The wire is placed in a uniform magnetic field of 0.50 T, with the current direction making a 30° angle with the magnetic field. Calculate the magnitude of the magnetic force on the wire.

Solution

To find the magnetic force on a current-carrying wire in a magnetic field, we use the equation:

FB=IBsinθF_{B}=I \ell B \sin \theta

Given:

  • Current, I=3.0I = 3.0 A
  • Wire length, =25\ell = 25 cm = 0.25 m
  • Magnetic field strength, B=0.50B = 0.50 T
  • Angle between current and field, θ=30°\theta = 30°

Substituting these values:

FB=3.0×0.25×0.50×sin30°F_{B}= 3.0 \times 0.25 \times 0.50 \times \sin 30°

FB=3.0×0.25×0.50×0.5F_{B}= 3.0 \times 0.25 \times 0.50 \times 0.5

FB=0.1875F_{B}= 0.1875 N

The magnetic force on the wire is 0.1875 N or approximately 0.19 N.

Frequently Asked Questions

What is the magnetic field around a current-carrying wire and how does it work?

A steady current in a long, straight wire makes a magnetic field of concentric circles around the wire—the field lines are tangent to circles centered on the wire and have no component along the wire (CED 12.3.A.1.i). The magnitude at distance r is B = μ0/(2π) · I/r, so B grows with current and falls off as 1/r (CED 12.3.A.1.ii). Use the right-hand rule (thumb along current, fingers curl) to find direction; for a single loop the field at the center points along the loop’s axis (CED 12.3.A.1.iii–iv). When multiple wires or loops contribute, add the B vectors (superposition, CED 12.3.A.1.v). On the exam you may need to apply B = μ0 I/(2πr), use the right-hand rule, or add fields vectorially. For a quick review, check the Topic 12.3 study guide (https://library.fiveable.me/ap-physics-2-revised/unit-4/3-magnetism-and-current-carrying-wires/study-guide/TzaJHQDjTp5y8iuQ), the unit overview (https://library.fiveable.me/ap-physics-2-revised/unit-12), and practice problems (https://library.fiveable.me/practice/ap-physics-2-revised).

Why does current in a wire create a magnetic field in the first place?

Because a current is just lots of charges moving together, it produces a magnetic field the same way a moving charge does. At the AP level you don’t need the full math derivation—you use the Biot–Savart law or Ampère’s law to show that moving charges create B-fields. For a long straight wire the result is simple and testable (CED 12.3.A.1): the field lines are concentric circles around the wire, the direction follows the right-hand rule, and the magnitude follows B = μ0/(2π) · I/r (so B ∝ I and B ∝ 1/r). You can also think of it conceptually: motion of charge changes the electromagnetic field in space, and that changing field has a direction (magnetic) perpendicular to the motion—which is exactly what the Biot–Savart/Ampère descriptions predict. For review and AP-style practice see the Topic 12.3 study guide (https://library.fiveable.me/ap-physics-2-revised/unit-4/3-magnetism-and-current-carrying-wires/study-guide/TzaJHQDjTp5y8iuQ), the Unit 12 overview (https://library.fiveable.me/ap-physics-2-revised/unit-12), and lots of practice problems (https://library.fiveable.me/practice/ap-physics-2-revised).

How do I use the right-hand rule for magnetic fields - I keep getting confused about which direction to point my thumb?

Think of two different right-hand rules and pick the one that fits the question. 1) Magnetic field around a long straight wire (CED 12.3.A.1.iii): point your right-hand thumb in the direction of the current I. Your curled fingers show the direction of B (concentric circles around the wire). That thumb = current, fingers = field. 2) Field on the axis of a current loop (CED 12.3.A.1.iv): curl your fingers around the loop in the direction of the current; your right-hand thumb then points along the loop’s axis in the direction of the magnetic field at the center. 3) Force on a current-carrying wire (CED 12.3.B.1.ii): use the force rule F = Iℓ × B. Point your right-hand thumb along the current (Iℓ), align your fingers with B; the palm (or the direction your thumb would be pushed toward) gives the magnetic force direction on a positive current element. If you mix them up, ask “am I finding B from I, or F from I and B?” and pick the matching rule. For more practice and AP-style questions, see the Topic 12.3 study guide (https://library.fiveable.me/ap-physics-2-revised/unit-4/3-magnetism-and-current-carrying-wires/study-guide/TzaJHQDjTp5y8iuQ) and the full unit/practice problem sets (https://library.fiveable.me/ap-physics-2-revised/unit-12, https://library.fiveable.me/practice/ap-physics-2-revised).

What's the difference between the magnetic field around a straight wire versus a loop of wire?

Short answer: they’re shaped and scaled differently. Around a long, straight wire the magnetic field lines are concentric circles centered on the wire, tangent at every point, with no component along the wire. Its magnitude falls off like 1/r: B = μ0 I / (2πr). Use the right-hand rule (thumb = current, curled fingers = B direction). (CED 12.3.A.1.i–ii, 12.3.A.1.iii) A single current loop produces field lines that circle the wire locally but give a net field along the loop’s axis at the center—the center field points along that axis (right-hand rule for loops) and for a circular loop Bcenter = μ0 I / (2R). Far from the loop the field looks like a magnetic dipole. (CED 12.3.A.1.iv) On the exam you may need to add fields from multiple wires (vector addition/superposition) or use Ampère/Biot–Savart for more complex shapes. For a focused review, see the Topic 12.3 study guide (https://library.fiveable.me/ap-physics-2-revised/unit-4/3-magnetism-and-current-carrying-wires/study-guide/TzaJHQDjTp5y8iuQ) and Unit 12 overview (https://library.fiveable.me/ap-physics-2-revised/unit-12). For lots of practice, check the practice problems page (https://library.fiveable.me/practice/ap-physics-2-revised).

I don't understand the equation B = μ₀I/2πr - what does each variable actually represent?

B = μ0 I / (2π r) gives the magnitude of the magnetic field a distance r from a long, straight current-carrying wire. Here's what each symbol means: - B: magnetic field magnitude at the point you care about (units: tesla, T). - μ0: the permeability of free space (μ0 = 4π × 10^−7 T·m/A). It’s a constant that sets the strength of magnetic effects in vacuum. - I: the electric current in the wire (units: amperes, A). The field strength is directly proportional to |I|—double I, double B. - 2π: part of the geometric factor that comes from the circular field lines around the wire. - r: the perpendicular distance from the wire’s central axis to the point (units: meters). B falls off as 1/r, so farther away means weaker field. This formula assumes a long, straight wire (so fields form concentric circles and no component along the wire) and gives the magnitude only; the direction is found with the right-hand rule (Topic 12.3.A in the CED). For a clear AP-aligned review, see the Topic 12.3 study guide (https://library.fiveable.me/ap-physics-2-revised/unit-4/3-magnetism-and-current-carrying-wires/study-guide/TzaJHQDjTp5y8iuQ) and more practice problems at (https://library.fiveable.me/practice/ap-physics-2-revised).

Why is the magnetic field inversely proportional to distance from the wire?

Think of the wire as a source that makes magnetic field lines form concentric circles. As you move farther away, those same field lines are spread over a larger circle (circumference 2πr), so the field strength at each point gets weaker. Mathematically, symmetry + Ampère’s law gives ∮B·dl = μ0I_enclosed, and for a circular path centered on a long straight wire B is constant on that path, so B(2πr) = μ0I → B = (μ0/2π)·(I/r). That 1/r comes directly from the circumference factor. You can also derive the same result from the Biot–Savart law, but Ampère’s law is faster here because of symmetry. This is exactly the CED essential knowledge (12.3.A.1.ii)—memorize B = μ0I/(2πr) and use the right-hand rule for direction. For more review and practice on Topic 12.3, see the Fiveable study guide (https://library.fiveable.me/ap-physics-2-revised/unit-4/3-magnetism-and-current-carrying-wires/study-guide/TzaJHQDjTp5y8iuQ) and practice problems (https://library.fiveable.me/practice/ap-physics-2-revised).

How do you add magnetic fields from multiple wires using vectors?

Think of each wire’s B field as a vector at the point you care about. For long straight wires use B = μ0/(2π) · I/r for the magnitude, and get the direction with the right-hand rule (field lines are tangent to concentric circles around each wire—CED 12.3.A.1.i–iii). Because magnetic fields obey superposition (CED 12.3.A.1.v), add the vectors from each wire head-to-tail or by components. Quick recipe: - For each wire, compute B_i = μ0/(2π) · I_i / r_i. - Use the right-hand rule to get the direction of B_i at your point (into/out of page or some angle). - If directions aren’t the same, resolve each B_i into x and y components, sum ΣB_x and ΣB_y, then magnitude = sqrt((ΣB_x)^2+(ΣB_y)^2) and direction = arctan(ΣB_y/ΣB_x). - Special case: if two fields are opposite along the same line, subtract magnitudes. This is exactly what AP asks you to do on vector-addition problems in Topic 12.3—practice with examples from the topic study guide (https://library.fiveable.me/ap-physics-2-revised/unit-4/3-magnetism-and-current-carrying-wires/study-guide/TzaJHQDjTp5y8iuQ) and more problems at (https://library.fiveable.me/practice/ap-physics-2-revised).

What happens to the force on a current-carrying wire when you change the angle between the current and magnetic field?

The magnetic force on a current-carrying wire depends on the angle θ between the current direction and the magnetic field through F = I ℓ B sin θ (CED 12.3.B.1.i). So: - If θ = 90° (current perpendicular to B) you get the maximum force, Fmax = IℓB. - If θ = 0° or 180° (current parallel/anti-parallel to B) the force is zero because sin 0 = 0. - For intermediate angles the force scales with sin θ (e.g., going from 30° to 60° changes sin θ from 0.5 to 0.866, so the force increases by a factor 1.73). The force direction is given by the right-hand rule (CED 12.3.B.1.ii): point your thumb along the current, fingers along B, then your palm gives the force direction on the wire. For more review on wires, fields, and exam-style problems see the Topic 12.3 study guide (https://library.fiveable.me/ap-physics-2-revised/unit-4/3-magnetism-and-current-carrying-wires/study-guide/TzaJHQDjTp5y8iuQ) and practice problems (https://library.fiveable.me/practice/ap-physics-2-revised).

Can someone explain the force equation F = IℓB sin θ in simple terms?

F = IℓB sinθ tells you the magnetic force on a straight piece of wire of length ℓ carrying current I that sits in a magnetic field B. Think of it like this: - I is how many charges per second move through the wire; larger I → bigger force. - ℓ is the length of the wire that’s actually inside the field; a longer wire feels more force. - B is the magnetic field strength; stronger field → bigger force. - sinθ accounts for angle θ between the direction of current (along the wire) and the magnetic field. If current is perpendicular to B (θ = 90°), sinθ = 1 and the force is maximal. If current is parallel to B (θ = 0°), sinθ = 0 and there’s no force. Units: I (A)·ℓ (m)·B (T) gives newtons (N). Direction: use the right-hand rule for current-in-wire—point your thumb along I, fingers along B; your palm gives the force direction. This is exactly what Topic 12.3 (12.3.B.1.i) expects you to know for the AP exam. For a clear refresher and examples, check the Topic 12.3 study guide (https://library.fiveable.me/ap-physics-2-revised/unit-4/3-magnetism-and-current-carrying-wires/study-guide/TzaJHQDjTp5y8iuQ) and practice problems (https://library.fiveable.me/practice/ap-physics-2-revised).

I'm confused about when to use the right-hand rule for magnetic field direction versus force direction - are they different?

Short answer: yes—they’re related but different uses of right-hand rules. - To find the magnetic field direction from a long straight current (CED 12.3.A.1.iii), point your right thumb along the current I and curl your fingers; the fingers show B (and remember B = μ0 I / 2πr for magnitude). - To find the magnetic force on a current or moving charge (CED 12.3.B.1.ii), use the cross-product rule: for a positive charge, point your index finger in the velocity v (or for a wire point your thumb along I), point your middle finger in B, and your thumb (or palm) gives the force direction F (use FB = IℓB sinθ for magnitude). For electrons (negative charges) the force is opposite the RH result. So one RH rule → field around wire; another RH orientation (v × B or Iℓ × B) → force. Practice problems on the Topic 12.3 study guide (https://library.fiveable.me/ap-physics-2-revised/unit-4/3-magnetism-and-current-carrying-wires/study-guide/TzaJHQDjTp5y8iuQ) and the Unit 12 overview (https://library.fiveable.me/ap-physics-2-revised/unit-12) help cement this for the exam. For more practice, see Fiveable’s problem set collection (https://library.fiveable.me/practice/ap-physics-2-revised).

Why does the magnetic field form circles around a straight wire instead of just pointing in one direction?

Think of a long straight wire as the symmetry axis of the problem. By symmetry there’s no preferred direction along or radially toward/away from the wire, only around it—so the magnetic field must lie tangent to circles centered on the wire (CED 12.3.A.1.i). The Biot–Savart law or Ampère’s law gives the magnitude B = μ0/(2π) · I/r (CED 12.3.A.1.ii), and the right-hand rule tells you the sense of those circular field lines: thumb along current, fingers curl the B-field (CED 12.3.A.1.iii). If the field pointed in one fixed direction instead of wrapping around, it would break the wire’s cylindrical symmetry and contradict those laws. This circular geometry is what you use on the exam when adding fields from multiple wires or finding forces on current segments (see Topic 12.3 study guide for quick review and practice: https://library.fiveable.me/ap-physics-2-revised/unit-4/3-magnetism-and-current-carrying-wires/study-guide/TzaJHQDjTp5y8iuQ).

How do I determine the magnetic field at the center of a current loop?

For a single circular loop of radius R carrying current I, use the Biot–Savart result: B at the center = μ0 I / (2R) (direction along the loop’s axis). For a coil with N closely spaced turns, multiply by N: B = μ0 N I / (2R). The direction is found with the right-hand rule: curl your fingers along the current; your thumb points along B at the center (along the loop’s axis)—that matches CED 12.3.A.1.iv. Don’t use the long-straight-wire formula (μ0 I / 2πr) for a loop. For on-axis points away from the center you’d integrate Biot-Savart; for many turns in a solenoid, use B ≈ μ0 n I. If you want a quick review or practice problems tied to the CED, check the Topic 12.3 study guide (https://library.fiveable.me/ap-physics-2-revised/unit-4/3-magnetism-and-current-carrying-wires/study-guide/TzaJHQDjTp5y8iuQ) and the unit practice set (https://library.fiveable.me/practice/ap-physics-2-revised).

What's the relationship between current direction and magnetic field direction in a wire?

The magnetic field around a straight current-carrying wire forms concentric circles around the wire and its direction is set by the right-hand rule: point your thumb along the current (direction of positive charge flow) and your curled fingers show the direction of B (CED 12.3.A.1.i, 12.3.A.1.iii). Magnitude falls off as B = μ0/(2π) · I/r (CED 12.3.A.1.ii). For a loop, use the right-hand rule with your fingers following current—your thumb points along the loop’s axis toward the field at the center (CED 12.3.A.1.iv). Remember: reversing the current reverses the magnetic field direction. This matters on the exam for predicting field superposition and magnetic force on wires (F = IℓB sinθ and direction by the right-hand rule, CED 12.3.B.1.i–ii). For a focused review, see the Topic 12.3 study guide (https://library.fiveable.me/ap-physics-2-revised/unit-4/3-magnetism-and-current-carrying-wires/study-guide/TzaJHQDjTp5y8iuQ) and practice problems (https://library.fiveable.me/practice/ap-physics-2-revised).

If I have two parallel wires with current, how do I find the total magnetic field at a point between them?

Take the B from each long straight wire (Biot–Savart / Ampère result): B = (μ0/2π)·I/r, where r is the perpendicular distance from that wire to your point. Use the right-hand rule to get each B’s direction (fields circle each wire). At a point between the wires the fields are in the plane and usually point opposite or same depending on current directions, so treat them as vectors and add. Procedure: 1. For wire 1: compute B1 = μ0/(2π) · I1 / r1 and use right-hand rule to get its sign/direction. 2. For wire 2: compute B2 = μ0/(2π) · I2 / r2 and get its direction. 3. Add vectorially: Btotal = B1 + B2 (if they’re opposite, subtract magnitudes and keep the sign of the larger). Example: if currents are opposite so both fields at the point point in same direction, magnitudes add. If they produce opposite directions at the point, subtract. This is exactly AP Topic 12.3: use B ∝ I/r, right-hand rule, and superposition (CED 12.3.A.1.ii, .iii, .v). For a worked example like MCQ 10, see the Topic 12.3 study guide (https://library.fiveable.me/ap-physics-2-revised/unit-4/3-magnetism-and-current-carrying-wires/study-guide/TzaJHQDjTp5y8iuQ) and practice problems (https://library.fiveable.me/practice/ap-physics-2-revised).

Why do we need to know about magnetic fields from current-carrying wires in real life applications?

You need this stuff because current-carrying wires are how we make and control magnetic fields in nearly every real electromagnetic device. Knowing the shape (concentric circles for a long straight wire), magnitude (B = μ0/2π · I/r), and direction (right-hand rule) of those fields lets you predict how devices behave: motors and loudspeakers (force on a wire, F = IℓB sinθ), solenoids and electromagnets (fields add by superposition), transformers, MRI scanners, and power-line interactions all depend on these ideas. On the AP exam you’ll be asked to describe fields, add fields from multiple wires, and compute magnetic forces or torques (CED 12.3.A & 12.3.B). If you can use the Biot–Savart/Ampère intuition, apply vector addition, and use the right-hand rules reliably, you’ll handle both MCQs and FRQ parts that test these learning objectives. For a concise review and worked examples, see the Topic 12.3 study guide (https://library.fiveable.me/ap-physics-2-revised/unit-4/3-magnetism-and-current-carrying-wires/study-guide/TzaJHQDjTp5y8iuQ) and practice problems (https://library.fiveable.me/practice/ap-physics-2-revised).