Topic 12.3: Magnetic Fields of Current-Carrying Wires and the Biot-Savart Law

What Topic 12.3 requires

Topic 12.3 is the first of the two calculus topics in Unit 12, and it has no counterpart by this title in the algebra-based course. Two learning objectives, four essential-knowledge statements, three equations, and one boundary statement.

12.3.A, describe the magnetic field produced by a current-carrying wire.

• 12.3.A.1 the Biot-Savart law defines the magnitude and direction of a magnetic field created by an electrical current. The relevant equation is

• 12.3.A.2 the magnetic field vectors around a small segment of a current-carrying wire are tangent to concentric circles centered on that wire. The field has no component toward, away from, or parallel to the segment of the current-carrying wire.
• 12.3.A.3 the Biot-Savart law can be used to derive the magnitudes and directions of magnetic fields around segments of current-carrying wires, for example at the center of a circular loop of wire. The derived equation is

12.3.B, describe the force exerted on current-carrying wires by a magnetic field.

• 12.3.B.1 a magnetic field will exert a force on a current-carrying wire. The relevant equation is

The suggested skills are 1.C, create qualitative sketches of graphs that represent features of a model or the behavior of a physical system; 2.A, derive a symbolic expression from known quantities by selecting and following a logical mathematical pathway; 2.D, predict new values or factors of change of physical quantities using functional dependence between variables; and 3.C, justify or support a claim using evidence from experimental data, physical representations, or physical principles or laws.

Two of the three equations are printed on the AP Physics C: E&M equation sheet: the Biot-Savart law itself and the force integral. The third, the field at the centre of a loop, carries the Derived Equation label and is not on the sheet. The CED's Required Equations page defines that label: not all equations in the framework appear on the sheet, and many "are provided for reference and guidance, or to demonstrate the final results of derivations expected of students on the exam. These equations are denoted as 'Derived Equations.'"

So $\mu_0 I / 2R$ is not a formula you are given. It is a result you are expected to be able to produce.

The boundary statement, quoted whole

One boundary statement sits under 12.3.B, and it governs the whole topic. Here it is in full, exception clause included:

> "AP Physics C: Electricity & Magnetism only expects students to perform quantitative analysis of certain cases of current-carrying conductors using the Biot-Savart law, such as at a location along the perpendicular bisector of a straight conductor, at a location along the central axis of a circular loop, or at the center of a segment of a circular loop."

Read it twice, because two phrases do most of the work.

"only expects... quantitative analysis of certain cases" is the fence. You are not going to be asked to integrate Biot-Savart for an arbitrarily shaped conductor at an arbitrary point.

"such as" is the gap in the fence. The three named geometries illustrate the class rather than closing it. A question about the field at the centre of a square loop, or along the axis of a short stack of coaxial loops, is built from the same pieces and is not excluded by the wording.

The three named cases, and what each is worth to you:

What the statement rules out in practice is the field of a loop at a point off its axis, and the field of a finite straight wire at a point not on its perpendicular bisector. Both are real integrals; neither is on this exam.

One more limit that is not in the boundary statement but is set by the booklet. The Table of Information's CALCULUS box prints exactly five integral rules: the power rule with its $n \neq -1$ exception, $\int e^{ax}\,dx$, $\int dx/(x+a)$, $\int \cos(ax)\,dx$ and $\int \sin(ax)\,dx$. The antiderivative you need for a straight-wire Biot-Savart integral is not among them. So on a free-response question, producing the correct integrand, variable and limits is itself the work that skill 2.A names, and "derive a symbolic expression by selecting and following a logical mathematical pathway" is satisfied by a correctly set-up integral. The CED's own sample free-response set contains an instruction to "Derive, but do not solve" a differential equation on another topic, which tells you the register.

Biot-Savart, piece by piece

Every symbol earns its place, and knowing what each does is what lets you set an integral up rather than pattern-match one.

• $d\vec{B}$ is the contribution of one infinitesimal piece of the conductor. The whole field is the vector sum of all of them, which is what makes this a superposition law rather than a formula.
• $I\,d\vec{\ell}$ is the current element. Its magnitude is the current times the length of the piece, and its direction is the direction of conventional current through that piece. The exam conventions box on the sheet fixes that as the direction in which positive charges would drift.
• $\hat{r}$ is a unit vector pointing from the current element to the field point. Getting this direction backwards flips the answer, and it is the most common setup error.
• $r$ is the distance from that element to the field point, and it changes as you move along the conductor. It is almost never constant, which is why you cannot pull it out of the integral in general.
• $1/r^2$ is the inverse-square falloff, the same shape as Coulomb's law and the reason the two laws look like cousins.
• $\mu_0/4\pi$ is exactly $1.0 \times 10^{-7}$, since the constants table prints $\mu_0 = 4\pi \times 10^{-7}$ exactly rather than as a decimal. Use the combination, not the two factors.
• The [cross product](/glossary/cross-product) does the geometry. Its magnitude is $d\ell \sin\theta$ with $\theta$ the angle between the element and $\hat{r}$, so an element contributes nothing at a point that lies along its own line, and most at a point off to the side. That is exactly statement 12.2.A.1.iii from the previous topic, now with an equation attached.

Statement 12.3.A.2 is the geometric summary you should be able to recite: the field vectors around a small segment are tangent to concentric circles centred on that wire, and the field has no component toward, away from, or parallel to the segment. Three forbidden directions and one allowed one. That single sentence rules out any answer that has the field pointing at the wire, away from it, or along it.

The corresponding right-hand rule: point the right thumb along the conventional current and the fingers curl in the direction of $\vec{B}$ on the circles around it. Say it the same way every time.

The procedure for any Biot-Savart problem, and it is worth having as a fixed routine:

1. Draw the conductor and the field point, and choose a variable to run along the conductor.
2. Write $d\vec{\ell}$ in terms of that variable, with its direction.
3. Write $\hat{r}$ and $r$ in terms of the same variable, from the element to the field point.
4. Take the cross product and identify which components survive. Symmetry usually kills all but one; say which symmetry, because that sentence carries credit.
5. Set the limits from the physical extent of the conductor.
6. Integrate, and check the limiting cases.

Step 4 is where the marks are. Statement 12.3.A.3 says the law "can be used to derive the magnitudes and directions" of fields around segments, so a direction argument is part of the required answer, not a decoration on it.

Case one: the perpendicular bisector of a straight conductor

This is the first geometry the boundary statement names, and it is the one that connects Topic 12.3 to Topic 12.4, because taking its long-wire limit reproduces the derived equation that Ampere's law produces in one line.

Put the wire along the $x$-axis running from $x = -a$ to $x = +a$, with current $I$ in the $+x$ direction, and put the field point $P$ on the $y$-axis a perpendicular distance $d$ from the middle of the wire.

For an element at position $x$:

• $d\vec{\ell} = dx\,\hat{i}$,
• the vector from the element to $P$ is $-x\,\hat{i} + d\,\hat{j}$, of length $r = \sqrt{x^2 + d^2}$,
• so $d\vec{\ell} \times \hat{r} = \dfrac{dx}{r}\left(\hat{i} \times (-x\,\hat{i} + d\,\hat{j})\right) = \dfrac{d\,dx}{r}\,\hat{k}$.

Every element contributes in the same direction, $+\hat{k}$, out of the page. That is the symmetry argument: there is nothing to cancel, so this is a magnitude problem. Then

Three readings of that result, and each is a 2.D functional-dependence question waiting to be asked.

The long-wire limit. Let the wire grow, $a \gg d$. Then $a/\sqrt{a^2+d^2} \to 1$ and

which is exactly the derived equation at 12.4.A.1.i. So the two topics agree, as they must, and you now have two independent routes to the same standard result. Ampere's law gets there in one line if the wire is infinite; Biot-Savart gets there by integration and also handles the finite case that Ampere cannot.

The geometric reading. The factor $a/\sqrt{a^2+d^2}$ is the sine of the angle subtended at $P$ by half the wire. A finite wire always gives less field than an infinite one at the same distance, and the shortfall is that sine.

The falloff. Close to a long wire the field goes as $1/d$, not as $1/d^2$, even though the law that built it has an inverse square in it. The inverse square belongs to a single element; the $1/d$ emerges after adding up a line of them. That distinction is worth stating in a justification, because it is the kind of thing skill 3.C rewards.

One honest caveat about the physical setup: a genuinely finite current-carrying segment with nothing attached at its ends cannot exist, since charge would pile up. The boundary statement names the case anyway, treating the segment as part of a larger circuit whose other parts are far away or contribute nothing at $P$. Say that out loud in a derivation if there is room; it is the sort of modelling assumption a 3.C justification can be built on.

Case two: the axis of a circular loop

The second named geometry, and the one that produces the derived equation at 12.3.A.3.

Take a circular loop of radius $R$ carrying current $I$, lying in the $xy$-plane and centred at the origin, and find the field at a point on the $z$-axis a distance $z$ from the centre.

Two facts about the geometry make it work.

1. Every element is the same distance from the field point, $r = \sqrt{R^2 + z^2}$. So $1/r^2$ is a constant and comes out of the integral.
2. Every element is perpendicular to its own $\hat{r}$. The element points around the loop; $\hat{r}$ lies in the plane containing the axis and that element. So $\left| d\vec{\ell} \times \hat{r} \right| = d\ell$ with no sine factor.

Each element therefore contributes a field of magnitude $\dfrac{\mu_0 I}{4\pi}\dfrac{d\ell}{R^2+z^2}$, tilted away from the axis. The components perpendicular to the axis cancel in pairs between elements on opposite sides of the loop, which is the symmetry argument here, and only the axial components survive. The fraction of each that is axial is $R/\sqrt{R^2+z^2}$, so

Notice that the integral itself was trivial: $\oint d\ell$ is just the circumference. All the work was in steps 1, 2 and the symmetry argument, which is a good illustration of where the marks actually sit in a Biot-Savart derivation.

At the centre, $z = 0$:

which is the derived equation 12.3.A.3 prints, produced rather than recalled.

Far away, $z \gg R$: the $R^2$ in the denominator becomes negligible and $B \to \mu_0 I R^2 / 2z^3$, falling as the inverse cube. This is worth connecting back to Topic 12.1. Statement 12.1.B.1.iv says only that the field of a magnetic dipole decreases with increasing distance and prints no exponent, and a current loop is the simplest magnetic dipole there is. Here you have derived an exponent for one specific geometry along one specific line. The two are consistent: the CED declines to print a general exponent, and this is a particular case you built yourself from a printed law. Do not run the argument in reverse and start quoting an inverse cube for arbitrary dipoles at arbitrary angles.

Sketching it, which is skill 1.C. The graph of $B$ against $z$ along the axis starts at $\mu_0 I / 2R$, is flat at $z = 0$ because the function is even in $z$ and so has zero slope there, then falls, steepening and eventually approaching an inverse cube. Getting the zero slope at the origin right is the mark that separates a correct sketch from a plausible one.

Arcs. For a segment of a circular loop subtending angle $\theta$ at the centre, the same integral with $\oint d\ell$ replaced by the arc length $R\theta$ gives $B = \mu_0 I \theta / 4\pi R$ at the centre, with $\theta$ in radians. That is the boundary statement's third named case, and setting $\theta = 2\pi$ recovers the loop result, which is a free check on any arc answer.

The force on a wire is an integral, and when that matters

This is 12.3.B.1, and it is on the sheet. It is the Topic 12.2 force law with the moving charges reorganised into a current: instead of one charge with a velocity, you have a line of charge in motion, and $I\,d\vec{\ell}$ plays the part $q\vec{v}$ played there.

When the field is uniform, the integral collapses. Pull the constants out:

and $\int d\vec{\ell}$ over a path is just the straight vector from the start of the path to its end, however the wire wanders in between. So:

• A straight wire of length $\ell$ gives the familiar $\vec{F} = I\vec{\ell} \times \vec{B}$, of magnitude $I\ell B \sin\theta$.
• A bent or curved wire in a uniform field feels the same force as a straight wire joining its two endpoints. A semicircular arc of radius $R$ behaves like a straight wire of length $2R$, not of length $\pi R$.
• A closed loop in a uniform field has $\oint d\vec{\ell} = \vec{0}$, so the net force on it is zero. There can still be a net torque, which is what turns a motor, but the net force is zero.

That last result is one of the highest-value single facts in the unit, and it is a genuine calculus result rather than a memorised rule. It also matters in Unit 13: statement 13.3.A.2 says that when current is induced in a conducting loop, magnetic forces are only exerted on the segments of the loop that are within the external magnetic field. A loop half in and half out of a field region is not in a uniform field over its whole length, the cancellation fails, and a net force appears. That is exactly why induction problems put the loop at the edge of a field region.

When the field is not uniform, the integral is real. Then $\vec{B}$ depends on position along the wire and stays inside. The standard case in this course is a wire sitting in the field of another wire, where $B$ varies as $1/r$ across it, and that is the next section.

A note on the sheet, because it prevents a specific mistake. The C: E&M sheet prints $\vec{F}_B = \int I(d\vec{\ell} \times \vec{B})$ and does not print $F = I\ell B \sin\theta$. That algebra-based form is on the AP Physics 2 sheet, at its statement 12.3.B.1.i. In this course it is a special case you produce by noting the field is uniform and the wire is straight. If either condition fails, saying so is part of the answer.

Two parallel wires: a derivation, not a printed rule

Every textbook has a boxed formula for the force per unit length between two parallel current-carrying wires. The AP Physics C: Electricity and Magnetism course description does not. The word "parallel" appears exactly once in the whole of Unit 12's required content, in statement 12.3.A.2, describing the directions the field of a wire does not point. There is no essential-knowledge statement about the force between two wires, and no such equation is printed on the equation sheet. The same is true in the AP Physics 2 course description.

That does not make the situation unexaminable. It makes it a derivation, and the CED settles the question itself in two places.

The unit's essential questions include "Why does the deflection of a pair of parallel conducting wires depend on the direction of current in the wires?" So the phenomenon is squarely in the course.

The CED's own sample free-response Question 3 is built on it. Two vertical wires carry currents in opposite directions, one is attached to a force sensor, and the separation and current can be varied. It is worth 10 points, is the Experimental Design and Analysis question, and aligns to learning objectives 11.3.B, 12.3.B and 12.4.A. Its scoring guideline composes the relationship in exactly two steps, using nothing that is not already in the framework:

The first factor is 12.3.B.1 for a straight wire in the field of the other, and the second is the derived equation at 12.4.A.1.i. Note the alignment tags on that question: 12.3.B and 12.4.A together, which is the CED confirming that the parallel-wire force is the two topics used in sequence rather than a rule of its own.

So the derivation to be able to run, in four lines:

1. Wire 1 carries $I_1$ and produces a field at wire 2's location of magnitude $B_1 = \mu_0 I_1 / 2\pi d$, tangent to a circle around wire 1, so perpendicular to wire 2.
2. Wire 2 carries $I_2$ through that field over a length $L$. Since $\vec{B}$ has the same magnitude and direction all along wire 2, the field is effectively uniform over it and the integral collapses: $F = I_2 L B_1$.
3. Substituting, $F = \dfrac{\mu_0 I_1 I_2 L}{2\pi d}$, and the force per unit length is $\dfrac{\mu_0 I_1 I_2}{2\pi d}$.
4. Direction, which is the part the essential question is asking about. Run the right-hand rule twice: once for the field wire 1 makes at wire 2, once for the force on wire 2 in that field. Currents in the same direction attract; currents in opposite directions repel. That is the opposite sense to two like electric charges, which repel, and the contrast is a favourite comparison.

The CED's worked solution to that question is also a useful calibration on lab data. From a best-fit line through force against current squared, with $L = 1.4$ m and $d = 0.005$ m, its example response extracts $\mu_0 = 1.48 \times 10^{-6}\ \mathrm{N/A^2}$, and its scoring note says linear regression on the given data gives $1.47 \times 10^{-6}\ \mathrm{N/A^2}$. Both are around 18 percent above the accepted $4\pi \times 10^{-7}$, and the question awards full credit anyway. Extracting a constant from a slope is the skill; matching the textbook value is not.

Where AP Physics 2 stops, and how 12.3 is tested

AP Physics 2 has a Topic 12.3 called Magnetism and Current-Carrying Wires, and it covers the same physical situations with the calculus removed. The contrast is sharp enough to be worth a table.

The row that captures the whole difference is the second. The algebra-based course hands you the field of a wire; the calculus-based course makes you produce it. Physics 2 prints $B = \mu_0 I / 2\pi r$ as a relevant equation and puts it on its sheet. Physics C prints the same expression as a derived equation and keeps it off the sheet, so every time you need it you either integrate Biot-Savart along the wire or apply Ampere's law to a circular loop.

The [AP Physics 2 Topic 12.3 page](/ap-physics-2/unit-12-magnetism-and-electromagnetism/12-3-magnetism-and-current-carrying-wires) is for AP Physics 2 students, and this page is for AP Physics C: Electricity and Magnetism students. If you are in the algebra-based course, the formula is yours to use and the integral is not on your exam. If you are in Physics C, that page is a fast way to fix the geometry and the right-hand rules before you come back and do the integrals.

How the topic is examined. One of the fifteen sample multiple-choice questions in the CED aligns to 12.3.B, testing essential knowledge 12.3.B.1 with skill 2.C, and one of the four sample free-response questions is the parallel-wires experiment described above.

The four suggested skills tell you the shapes.

1. Skill 2.A, derive a symbolic expression. Set up and evaluate a Biot-Savart integral for one of the named geometries. This is the highest-value skill on the exam, carrying 25 to 30 percent of the multiple-choice weighting, and it is listed for three of Unit 12's four topics.
2. Skill 2.D, predict new values or factors of change using functional dependence. Double the current, halve the radius, double the distance. Every result on this page is a product of powers, so factor-of-change questions are cheap to set and quick to answer if you have the symbolic form.
3. Skill 1.C, qualitative sketches. The axial field of a loop against $z$, flat at the origin. The field of a long wire against distance, an inverse-first-power curve rather than an inverse square. This is the only Unit 12 topic besides 12.4 where a graph sketch is on the skill list.
4. Skill 3.C, justify with evidence. Use the symmetry argument, or 12.3.A.2's three forbidden directions, to justify a claim about the field's direction.

Three traps:

• Reaching for Ampere's law on a finite wire or a single loop. No Amperian loop has the symmetry, so it is Biot-Savart or nothing. The reverse error is integrating Biot-Savart along an infinite wire when Ampere's law gives it in a line.
• Using the arc length in a uniform-field force problem. $\int d\vec{\ell}$ is the straight vector between the endpoints, so a curved wire in a uniform field feels less force than its length suggests.
• Pointing $\hat{r}$ the wrong way. It runs from the current element to the field point, never the reverse.