Biot-Savart Law

AP Physics C: Electricity and Magnetism· AP Physics C: E&M CED — Magnetic Fields· 14 min read

1. Core Definition and Vector Form★★☆☆☆⏱ 4 min

The Biot-Savart Law is the fundamental empirical law describing the magnetic field generated by a steady (time-invariant) current distribution. It is the magnetic analog of Coulomb's Law for static electric fields, and forms the foundation of all classical magnetostatics. This topic appears regularly on both AP Physics C: E&M multiple-choice and free-response exam sections.

📘 Definition

Biot-Savart Law

$d B$

Relates the infinitesimal magnetic field $d B$ at a field point to an infinitesimal source current element $I d l$ , where $d l$ points in the direction of current flow.

Example:

Total magnetic field is found by integrating $d B$ over all current elements in the source distribution.

d\vec{B} = \frac{\mu_0 I}{4\pi} \frac{d\vec{l} \times \hat{r}}{r^2}

Where:

$μ_{0} = 4 π \times 1 0^{- 7} T \cdotp m/A$ is the permeability of free space, a defined constant
$I$ = magnitude of current in the source
$\overset{r}{^}$ = unit vector pointing from the current element to the field point
$r$ = straight-line distance between the current element and field point

The magnitude of $d B$ is $d B = \frac{μ _{0} I d l s i n θ}{4 π r ^{2}}$ , where $θ$ is the angle between $d l$ and $\overset{r}{^}$ . Direction is given by the right-hand rule for cross products, perpendicular to both $d l$ and $\overset{r}{^}$ .

📐 Worked Example

A current element $I d l = 2.0 A \cdotp mm \hat{i}$ is located at the origin. What are the direction and magnitude of $d B$ at the point $(3.0 cm, 4.0 cm, 0)$ ?

1
Find $r$ and $\overset{r}{^}$ from the current element to the field point:
$\vec{r} = 0.03\hat{i} + 0.04\hat{j}, \quad r = 0.05 \, \text{m}, \quad \hat{r} = 0.6\hat{i} + 0.8\hat{j}$
2
Calculate the cross product $d l \times \overset{r}{^}$ (convert $d l$ to meters):
$d\vec{l} \times \hat{r} = (2.0 \times 10^{-3} \hat{i}) \times (0.6\hat{i} + 0.8\hat{j}) = 1.6 \times 10^{-3} \, \text{A·m} \, \hat{k}$
3
Substitute into Biot-Savart, using $\frac{\mu_0}{4\pi} = 1 \times 10^{-7} , \text{T·m/A}:
$|d\vec{B}| = 1 \times 10^{-7} \frac{1.6 \times 10^{-3}}{(0.05)^2} = 6.4 \times 10^{-8} \, \text{T}$
4
The direction of $d B$ is $+ \hat{k}$ , which points out of the $x y$ -plane.

Exam tip:

Always label the direction of $d l$ and $\overset{r}{^}$ on your diagram before calculating the cross product; reversing $\overset{r}{^}$ flips the sign of $d B$ , an easy mistake that costs points on FRQs.

2. Magnetic Field from Straight Wires★★★☆☆⏱ 4 min

One of the most common AP Physics C applications of Biot-Savart is deriving the magnetic field from a straight current-carrying wire. For a straight wire of total length $2 a$ , with a field point at perpendicular distance $R$ on the wire's perpendicular bisector, symmetry ensures all $d B$ point in the same direction, allowing us to integrate magnitudes directly.

B = \frac{\mu_0 I}{4\pi R} \frac{2a}{\sqrt{a^2 + R^2}}

For an infinite wire, $a \to \infty$ , so the expression simplifies to the widely-used result:

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

This result is only valid for infinite thin wires, and cannot be used for wires of explicitly given finite length.

📐 Worked Example

A straight power line of total length 20 m carries 100 A of current. What is the magnitude of $B$ at a point 5 m from the wire, along the perpendicular bisector?

1
Identify variables: total length $2 a = 20 m$ , so $a = 10 m$ , $R = 5 m$ , $I = 100 A$ .
2
Substitute into the finite wire formula:
$B = \frac{\mu_0 I}{4\pi R} \frac{2a}{\sqrt{a^2 + R^2}}$
3
Plug in values ( $\frac{μ _{0}}{4 π} = 1 \times 1 0^{- 7}$ ):
$B = 1 \times 10^{-7} \frac{100}{5} \frac{20}{\sqrt{10^2 + 5^2}} = (2 \times 10^{-6}) \frac{20}{11.18} ≈ 3.6 \times 10^{-6} \, \text{T}$
4
This result is slightly smaller than the infinite wire approximation of $4.0 \times 1 0^{- 6} T$ , which makes physical sense because the finite ends contribute less field.

Exam tip:

If the field point is not on the perpendicular bisector, adjust your integral limits to match the start and end of the wire; don’t just use the perpendicular bisector formula by default.

3. Magnetic Field from Circular Current Loops★★★☆☆⏱ 4 min

Another standard AP exam question asks for the magnetic field along the central axis of a circular current loop. For a loop of radius $R$ carrying current $I$ , symmetry tells us that the perpendicular (x/y) components of $d B$ from opposite current elements cancel, leaving only the axial component. Integrating around the full loop gives the result:

B_z = \frac{\mu_0 I R^2}{2(R^2 + z^2)^{3/2}}

At the center of the loop, $z = 0$ , so this simplifies to a common special case that is tested frequently:

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

The direction of $B$ along the axis is given by the right-hand rule: curl your fingers along the direction of current, your thumb points in the direction of $B$ along the axis.

📐 Worked Example

A circular wire loop of radius 15 cm carries 2.0 A of current. What is $B$ at the center of the loop, and at a point 15 cm along the axis from the center?

1
For the center, $z = 0$ , so use the simplified center formula:
$B_{center} = \frac{\mu_0 I}{2R} = \frac{4\pi \times 10^{-7} \cdot 2.0}{2 \cdot 0.15} ≈ 8.4 \times 10^{-6} \, \text{T}$
2
For $z = 0.15 m$ , use the full axial formula:
$B_z = \frac{\mu_0 I R^2}{2(R^2 + z^2)^{3/2}}$
3
Substitute $R = z = 0.15 m$ :
$B_z = \frac{4\pi \times 10^{-7} \cdot 2.0 \cdot (0.15)^2}{2 \cdot (2(0.15)^2)^{3/2}} ≈ 3.0 \times 10^{-6} \, \text{T}$
4
As expected, the field decreases as we move away from the center along the axis.

Exam tip:

Always use symmetry to cancel non-axial components before integrating; this eliminates half of your work automatically and avoids integrating terms that sum to zero.

4. AP-Style Additional Worked Example★★★★☆⏱ 2 min

✓ Quick check

Test your understanding of Biot-Savart direction:

A current element $I d l$ points along the negative $y$ -axis at the origin. What is the direction of $d B$ at the point $(0, 0, + z)$ on the positive $z$ -axis?
- $+ x$
- $- x$
- $+ y$
- $- y$
Reveal answer
1 —
The cross product $(- d l \hat{j}) \times \hat{k} = - d l \hat{i}$ , so direction is negative $x$ -axis.

📐 Worked Example

A wire is bent into a square of side length $L$ , carrying counterclockwise current $I$ . (a) Derive an expression for the magnitude of $B$ at the center of the square. (b) State the direction of $B$ at the center. (c) Compare the magnitude to that of a circular loop with the same total wire length and same current. Which is larger?

1
By symmetry, all four sides contribute equal magnitude $B$ in the same direction, so calculate for one side and multiply by 4. For one side, $R = L /2$ , $a = L /2$ :
$B_{\text{per side}} = \frac{\mu_0 I}{4\pi R} \frac{2a}{\sqrt{a^2 + R^2}} = \frac{\mu_0 I}{2\sqrt{2} \pi L}$
2
Multiply by 4 sides for total $B$ :
$B_{\text{total}} = \frac{2\sqrt{2} \mu_0 I}{\pi L} ≈ 0.900 \frac{\mu_0 I}{L}$
3
By right-hand rule for counterclockwise current, $B$ points out of the plane of the square.
4
For the circular loop, total length $4 L = 2 π R ⟹ R = 2 L / π$ , so $B$ at center is:
$B_{\text{circle}} = \frac{\mu_0 I}{2R} = \frac{\pi \mu_0 I}{4L} ≈ 0.785 \frac{\mu_0 I}{L}$
5
The magnetic field at the center of the square is larger than that of the circular loop.

5. Common Pitfalls

Wrong move:

Pointing $\overset{r}{^}$ from the field point to the current element, instead of from the element to the field point.

Why:

Students confuse source-test point order across different E&M laws, leading to reversed direction.

Correct move:

Always label that $\overset{r}{^}$ points from the current element (source) to the field point on your diagram before starting any calculation.

Wrong move:

Using the infinite wire formula $B = \frac{μ _{0} I}{2 π R}$ for a wire of explicitly given finite length.

Why:

The infinite wire result is highly memorable, so students default to it even when the problem specifies a finite wire.

Correct move:

Always check if the problem describes the wire as infinite; if length is given, use the finite wire Biot-Savart result.

Wrong move:

Adding magnitudes of $d B$ when they point in different directions, instead of integrating vector components separately.

Why:

Students get used to symmetric problems where all $d B$ point the same direction, and forget to check direction for asymmetric distributions.

Correct move:

For any non-symmetric distribution, split $d B$ into x, y, z components before integrating, then combine components at the end.

Wrong move:

Omitting the $sin θ$ term from the magnitude of $d B$ , assuming it is always 1.

Why:

Most symmetric problems have $θ = 9 0^{\circ}$ so $sin θ = 1$ , leading students to forget the angle dependence.

Correct move:

Always write $d B = \frac{μ _{0} I d l s i n θ}{4 π r ^{2}}$ explicitly, even if $sin θ$ simplifies to 1.

Wrong move:

Using the center of loop formula $B = \frac{μ _{0} I}{2 R}$ for off-center points along the loop axis.

Why:

The center formula is simple to memorize, so students overapply it.

Correct move:

Use the full axial formula $B = \frac{μ _{0} I R ^{2}}{2 ( R ^{2} + z ^{2} ) ^{3/2}}$ for any axial point at distance $z$ from the center, only substituting $z = 0$ for the center.

6. Quick Reference Cheatsheet

Category	Formula	Notes
General Biot-Savart Law	$d B = \frac{μ _{0} I}{4 π} \frac{d l \times r ^}{r ^{2}}$	$\overset{r}{^}$ from current element to field point; integrate over all sources
Infinitesimal B Magnitude	$d B = \frac{μ _{0} I d l s i n θ}{4 π r ^{2}}$	$θ$ = angle between $d l$ and $\overset{r}{^}$
Finite Straight Wire (perpendicular bisector)	$B = \frac{μ _{0} I}{4 π R} \frac{2 a}{a ^{2} + R ^{2}}$	$2 a$ = total wire length, $R$ = perpendicular distance
Infinite Straight Wire	$B = \frac{μ _{0} I}{2 π R}$	Only valid for infinite thin wires
B on Axis of Circular Loop	$B = \frac{μ _{0} I R ^{2}}{2 ( R ^{2} + z ^{2} ) ^{3/2}}$	$R$ = loop radius, $z$ = distance from center
B at Center of Circular Loop	$B = \frac{μ _{0} I}{2 R}$	Special case: $z = 0$
Permeability of Free Space	$μ_{0} = 4 π \times 1 0^{- 7} T \cdotp m/A$	Exact defined constant for all calculations

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.

2023 · MCQ
Direction of B from current element
2022 · FRQ
B at center of square current loop
2021 · MCQ
B on axis of circular loop

What's Next

Next you will learn Ampère’s Law, a simpler symmetry-based shortcut for calculating magnetic fields from highly symmetric current distributions. Ampère’s Law is derived directly from the Biot-Savart Law for steady currents, so mastering the integration and symmetry arguments in this sub-topic is required to correctly identify when Ampère’s Law can be applied, and to validate its results. This topic also forms the foundation for understanding magnetic dipoles, magnetic force on current-carrying wires, and eventually Faraday’s Law of induction, where you will calculate magnetic flux through loops for induced emf problems. Without mastering how B fields are generated from current distributions, you will struggle with flux calculations and the application of Faraday’s Law in Unit 5.