AP Physics C: E&M Unit 4 Notes — How Currents Create Magnetic Fields

Biot–Savart Law

What it is

The Biot–Savart law is the “from-the-source” way to calculate magnetic fields: it tells you how a small piece of current-carrying wire contributes a small magnetic field at a point in space. You then add up (integrate) the contributions from all pieces of the current distribution to get the total magnetic field.

Conceptually, it plays a role similar to Coulomb’s law in electrostatics:

  • Coulomb’s law: a point charge makes an electric field, and you add contributions from all charges.
  • Biot–Savart law: a current element makes a magnetic field, and you add contributions from all current elements.

This matters because many realistic current shapes (loops, arcs, finite wires) don’t have enough symmetry for Ampere’s law to work cleanly. Biot–Savart is more general for steady currents, even if the math can get heavier.

How it works (the physics behind the formula)

A few qualitative facts are built into Biot–Savart:

  1. More current produces more magnetic field. Doubling the current doubles the field.
  2. Closer current produces a stronger field. The contribution falls off with distance.
  3. Direction is perpendicular. The magnetic field contribution from a current element is perpendicular to both the direction of current and the line from the element to the field point.

That last point is the big conceptual difference from electric fields: electric fields point radially away from charges; magnetic fields “wrap around” currents.

The Biot–Savart law (mathematical statement)

For a steady current I in a wire, the magnetic field contribution d\vec{B} at a field point from a small wire element d\vec{\ell} is

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

An equivalent and often convenient form is

d\vec{B} = \frac{\mu_0}{4\pi}\frac{I\,d\vec{\ell} \times \vec{r}}{r^3}

Here’s what each symbol means:

  • \mu_0 is the permeability of free space.
  • I is the current.
  • d\vec{\ell} is a vector pointing along the direction of the current with magnitude equal to a tiny length of wire.
  • \vec{r} points from the current element to the field point; r is its magnitude.
  • \hat{r} is the unit vector in the \vec{r} direction.
  • The cross product encodes direction and the sine of the angle between d\vec{\ell} and \vec{r}.

A very common simplification is to take magnitudes when symmetry makes direction easy:

dB = \frac{\mu_0}{4\pi}\frac{I\,d\ell\sin\theta}{r^2}

where \theta is the angle between d\vec{\ell} and \vec{r}.

Direction: the right-hand rule and the cross product

Because d\vec{B} is proportional to d\vec{\ell} \times \hat{r}, you find the direction using the right-hand rule for a cross product:

  • Point your fingers along d\vec{\ell} (direction of current).
  • Curl them toward \hat{r} (from element to the field point).
  • Your thumb points in the direction of d\vec{B}.

A common confusion is mixing up \hat{r} (from source to field point) with the radial direction from the origin of coordinates. In Biot–Savart, \vec{r} is always drawn from the current element to the observation point.

Superposition: adding up contributions

Magnetic fields add by vector superposition:

\vec{B}_{\text{total}} = \int d\vec{B}

You integrate along the current path (or through a volume/surface if the current is distributed). The hardest part is usually setting up geometry correctly: expressing r, \theta, and the direction of d\vec{B} in terms of an integration variable.

Worked example 1: field at the center of a circular loop

Goal: Find the magnetic field magnitude at the center of a circular loop of radius R carrying current I.

Reasoning before math: Every current element is the same distance R from the center. Also, by symmetry, horizontal components of d\vec{B} cancel, and all contributions point along the loop’s axis (given by the right-hand rule).

  1. Distance from any element to center: r = R.
  2. For each element, d\vec{\ell} is tangent to the loop and \vec{r} points radially inward, so the angle is \theta = 90^\circ and \sin\theta = 1.
  3. Magnitude contribution:

dB = \frac{\mu_0}{4\pi}\frac{I\,d\ell}{R^2}

  1. Integrate around the loop, where \int d\ell = 2\pi R:

B = \frac{\mu_0}{4\pi}\frac{I}{R^2}(2\pi R)

  1. Simplify:

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

If the loop has N turns (a tightly wound coil), the field scales linearly:

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

Worked example 2: field on the axis of a circular loop (useful extension)

At a point on the axis a distance x from the center of a loop of radius R, Biot–Savart gives

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

Direction is along the axis, set by the right-hand rule. This is a common “set up the integral or use the known result” scenario.

Exam Focus
  • Typical question patterns:
    • “A loop/arc/finite wire carries current I. Find \vec{B} at a special point (center, axis, or perpendicular bisector).”
    • “Use symmetry to argue which components cancel before integrating.”
    • “Multiple currents present: compute each contribution and superpose.”
  • Common mistakes:
    • Treating r as constant when it changes with the integration variable.
    • Getting the cross-product direction backward by drawing \vec{r} in the wrong direction.
    • Forgetting that you must add fields as vectors (sign and direction matter), not just magnitudes.

Ampere’s Law

What it is

Ampere’s law connects the magnetic field around a closed loop to the amount of current passing through that loop. In its simplest AP Physics C: E&M form (steady currents), it says the circulation of \vec{B} around a closed path equals \mu_0 times the enclosed current.

Ampere’s law is powerful because, with the right symmetry, it lets you find \vec{B} without doing a Biot–Savart integral. The tradeoff is that it only becomes “plug-and-chug” when you can argue that \vec{B} has constant magnitude and a simple direction along your chosen loop.

The law (integral form) and what it really means

For steady currents,

\oint \vec{B}\cdot d\vec{\ell} = \mu_0 I_{\text{enc}}

Interpretation:

  • The left side is a closed-loop line integral. You choose an Amperian loop (any closed path you like).
  • The dot product \vec{B}\cdot d\vec{\ell} means only the component of \vec{B} tangent to the loop contributes.
  • I_{\text{enc}} is the net current through the surface bounded by the loop (with sign based on an orientation convention).

A subtle but important point: the equation is true for any loop, but it only helps when symmetry makes the left side easy to evaluate.

Choosing an Amperian loop: the symmetry strategy

You typically pick a loop so that one or more of these happen:

  1. \vec{B} is parallel to d\vec{\ell} along the loop, so \vec{B}\cdot d\vec{\ell} = B\,d\ell.
  2. B is constant on portions (or all) of the loop.
  3. \vec{B} is perpendicular to d\vec{\ell} on some segments, so those segments contribute zero to the integral.

This is why Ampere’s law shines for highly symmetric current distributions: infinite straight wires, ideal solenoids, and toroids.

Orientation and the sign of enclosed current

Ampere’s law uses a right-hand convention:

  • If you traverse the loop in a chosen direction, your right thumb points along the positive normal of the surface.
  • Currents piercing the surface in the thumb direction count positive; opposite direction counts negative.

A common exam trap is including magnitudes of currents without signs. If two currents pass through the surface in opposite directions, they partially cancel in I_{\text{enc}}.

Relationship to Biot–Savart

Both laws describe the same physics for magnetostatics (steady currents):

  • Biot–Savart is more direct and generally applicable but often requires integration.
  • Ampere’s law is often simpler but depends on symmetry.

You can think of Ampere’s law as a “global” constraint on \vec{B}, while Biot–Savart builds \vec{B} from local contributions.

Worked example 1: magnetic field of an infinite straight wire

Setup: A long straight wire carries current I. Find B(r) at distance r from the wire.

Physics first: Field lines form circles around the wire (right-hand rule). By cylindrical symmetry, B depends only on r and is constant on a circle centered on the wire.

  1. Choose a circular Amperian loop of radius r centered on the wire.
  2. Along the loop, \vec{B} is tangential and parallel to d\vec{\ell}, so \vec{B}\cdot d\vec{\ell} = B\,d\ell.
  3. Evaluate the integral:

\oint \vec{B}\cdot d\vec{\ell} = \oint B\,d\ell = B(2\pi r)

  1. Enclosed current is I_{\text{enc}} = I.
  2. Apply Ampere’s law:

B(2\pi r) = \mu_0 I

  1. Solve:

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

Direction is given by the right-hand rule: curl fingers with current; fingers show \vec{B} direction.

Worked example 2: inside a long cylindrical conductor with uniform current density

Setup: A solid wire of radius R carries total current I uniformly across its cross-section. Find B(r) for r