A long straight wire carries 10 A. Find the field 0.20 m away (_0 = 4π10^-7). [2 points]

United StatesPhysics C: Electricity and MagnetismSyllabus dot point

How does the Biot-Savart law give the magnetic field of a current, and how is it applied to wires and loops?

Topic 12.3 Magnetic Fields of Current-Carrying Wires and the Biot-Savart Law: use the Biot-Savart law to find the field of current elements, straight wires and loops.

A calculus-based answer to AP Physics C E&M Topic 12.3, covering the Biot-Savart law, the field of a current element, integration for a straight wire and a circular loop on its axis, and the force between parallel wires.

Generated by Claude Opus 4.812 min answerUpdated 2026-06-04

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Quick answer

The Biot-Savart law gives the magnetic field of a current element: $d\vec{B} = \dfrac{\mu_0}{4\pi}\dfrac{I\,d\vec{\ell}\times\hat{r}}{r^2}$ , where $\mu_0 = 4\pi\times10^{-7}$ T m/A, $d\vec{\ell}$ points along the current, and $\hat{r}$ points from the element to the field point. You integrate $d\vec{B}$ over the whole current to get $\vec{B}$ , exactly as you integrate $d\vec{E}$ for a charge distribution. Two standard results follow: a long straight wire gives $B = \dfrac{\mu_0 I}{2\pi r}$ (circling the wire, $\propto 1/r$ ), and a circular loop gives $B = \dfrac{\mu_0 I}{2R}$ at its center (and $B = \dfrac{\mu_0 I R^2}{2(x^2+R^2)^{3/2}}$ on its axis). Use the right-hand rule (thumb along the current, fingers curl in the field direction) for direction. Parallel currents attract; antiparallel currents repel.

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What this topic is asking
The Biot-Savart law
Field of a straight wire
Field of a circular loop
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What this topic is asking

The College Board (Topic 12.3) wants you to use the Biot-Savart law to find the magnetic field produced by a current: the field of a single current element, and by integration the field of a straight wire and a circular loop. This is the magnetic analogue of integrating Coulomb's law over a charge distribution.

The Biot-Savart law

This is the magnetic counterpart of the point-charge field. The cross product means each element's field circles around the current, and the $1/r^2$ falloff is the same form as Coulomb's law. To find the total field, integrate over the whole current path.

Field of a straight wire

Integrating the Biot-Savart law along a long straight wire gives a field that circles the wire:

B = \frac{\mu_0 I}{2\pi r}

at perpendicular distance $r$ . The field lines are concentric circles; the right-hand rule (thumb along the current, fingers curling) gives their direction. Note the $1/r$ dependence, weaker than a point charge's $1/r^2$ .

Field of a circular loop

For a circular loop of radius $R$ carrying current $I$ , the field on the axis a distance $x$ from the center is

B = \frac{\mu_0 I R^2}{2(x^2 + R^2)^{3/2}}

and at the center ( $x = 0$ ) this reduces to

B = \frac{\mu_0 I}{2R}

The derivation parallels the ring's electric field: every element is the same distance from the center, and the components add along the axis.

Field at the center of a circular loop

A circular loop of radius $0.050$ m carries a current of $3.0$ A. Find the magnetic field at its center. Use $\mu_0 = 4\pi\times10^{-7}$ T m/A.

step 1 Select the central-field formula

At the center, $B = \dfrac{\mu_0 I}{2R}$ (the on-axis formula with $x = 0$ ).

step 2 Substitute the values

$B = \dfrac{(4\pi\times10^{-7})(3.0)}{2(0.050)} = \dfrac{(1.257\times10^{-6})(3.0)}{0.10}$ .

step 3 Evaluate

$B = \dfrac{3.77\times10^{-6}}{0.10} = 3.77\times10^{-5}$ T.

step 4 Find the direction

By the right-hand rule, curling the fingers along the current, the field at the center points perpendicular to the loop's plane, out of the side the fingers curl toward.

Try this

Q1. A long straight wire carries $10$ A. Find the field $0.20$ m away ( $\mu_0 = 4\pi\times10^{-7}$ ). [2 points]

Cue. $B = \dfrac{\mu_0 I}{2\pi r} = \dfrac{(4\pi\times10^{-7})(10)}{2\pi(0.20)} = 1.0\times10^{-5}$ T.

Q2. State how the field at the center of a loop changes if the current is doubled. [1 point]

Cue. It doubles: $B = \dfrac{\mu_0 I}{2R} \propto I$ .

Exam-style practice questions

Practice questions written in the style of College Board exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

AP 2023 (style)1 marksSection I (multiple choice). The magnetic field a distance

r

from a long straight wire carrying current

I

B = \dfrac{\mu_0 I}{2\pi r}

. If the distance is tripled, the field becomes (A)

9

times larger (B)

3

times larger (C) one third (D) one ninth. Justify your reasoning.

Show worked answer →

A 1-point MCQ on the straight-wire field. The answer is (C).

$B = \dfrac{\mu_0 I}{2\pi r}$ varies as $1/r$ . Tripling $r$ divides the field by three. The trap is (D): treating the dependence as inverse-square ( $1/r^2$ ), which applies to a point charge's electric field, not a straight wire's magnetic field.

AP 2022 (style)6 marksSection II (FRQ, derivation). A circular loop of radius

R

carries current

I

. (a) Write the Biot-Savart law for a current element. (b) Use it to derive the magnetic field at the center of the loop. (c) State how the field at the center would change if the radius were doubled at the same current.

Show worked answer →

A 6-point FRQ deriving the loop's central field from Biot-Savart.

(a) Biot-Savart (1 point): $d\vec{B} = \dfrac{\mu_0}{4\pi}\dfrac{I\,d\vec{\ell}\times\hat{r}}{r^2}$ .
(b) Derivation (4 points): at the center, every element $d\ell$ is a distance $R$ away with $d\vec{\ell}\perp\hat{r}$ , so $dB = \dfrac{\mu_0}{4\pi}\dfrac{I\,d\ell}{R^2}$ , all in the same direction (out of the loop plane). Integrating around the loop, $\int d\ell = 2\pi R$ , gives $B = \dfrac{\mu_0 I}{4\pi R^2}(2\pi R) = \dfrac{\mu_0 I}{2R}$ .
(c) Doubling $R$ (1 point): $B = \dfrac{\mu_0 I}{2R} \propto \dfrac{1}{R}$ , so doubling the radius halves the central field.

Markers reward the Biot-Savart statement, the integration with constant $r = R$ , and the $1/R$ dependence.

Related dot points

Sources & how we know this

AP Physics C: Electricity and Magnetism Course and Exam Description — College Board (2024)