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:

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

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

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

5. 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:

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

4. Enclosed current is $I_{\text{enc}} = I$.
5. Apply Ampere’s law:

$B(2\pi r) = \mu_0 I$

6. 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 $r<R$.

Key idea: Enclosed current grows with area because current density is uniform.

1. Uniform current density magnitude:

$J = \frac{I}{\pi R^2}$

2. For an Amperian circle of radius $r<R$, enclosed current is

$I_{\text{enc}} = J\pi r^2 = I\frac{r^2}{R^2}$

3. Ampere’s law with circular loop gives

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

4. Solve:

$B = \frac{\mu_0 I}{2\pi R^2}r$

So inside a uniform current-carrying solid wire, $B$ increases linearly with $r$.

Important limitation (context you should know)

For time-varying electric fields, the magnetostatic form can fail unless you use the Maxwell correction (displacement current). The generalized form is

$\oint \vec{B}\cdot d\vec{\ell} = \mu_0\left(I_{\text{enc}} + \epsilon_0\frac{d\Phi_E}{dt}\right)$

In many Unit 4 problems, currents are steady, so the simpler form is what you apply.

Exam Focus

• Typical question patterns:
  • “Use Ampere’s law to find $B(r)$ for an infinite wire / cylindrical conductor / solenoid / toroid.”
  • “Find the enclosed current from a given current density $J$ or surface current density.”
  • “Multiple wires: determine $I_{\text{enc}}$ with signs using the loop orientation.”
• Common mistakes:
  • Picking an Amperian loop where $B$ is not constant (then treating it as constant anyway).
  • Forgetting that only the tangential component contributes to $\vec{B}\cdot d\vec{\ell}$.
  • Using $I$ instead of $I_{\text{enc}}$ when the loop does not enclose the full current (especially inside conductors).

Magnetic Fields of Common Configurations

Why “common configurations” matter

AP Physics C: E&M problems often involve a small set of geometries where the magnetic field is either (1) a standard result from Biot–Savart, or (2) quickly derived using Ampere’s law and symmetry. Learning these isn’t about memorizing random formulas; it’s about recognizing which law is the smartest tool and what symmetry tells you about direction and spatial dependence.

A helpful habit: before writing equations, sketch field lines and ask:

• What symmetry does the current distribution have (cylindrical, planar, toroidal)?
• Can I choose an Amperian loop where $B$ is constant and tangent?
• If not, can I reduce Biot–Savart using cancellations and geometry?

Notation and “source” quantities

In addition to total current $I$, you may see currents described by densities:

Infinite straight wire (Ampere)

For an infinitely long straight wire:

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

• Direction: tangent to circles centered on the wire (right-hand rule).
• Dependence: decreases as $1/r$.

What students often miss: “infinite” really means “long enough that edge effects are negligible” at the distances you care about.

Finite straight wire (Biot–Savart result)

For a straight segment, the field at a point a perpendicular distance $r$ away can be written in terms of angles to the endpoints:

$B = \frac{\mu_0 I}{4\pi r}(\sin\theta_1 + \sin\theta_2)$

Here $\theta_1$ and $\theta_2$ are measured from the perpendicular dropped to the wire to the lines connecting the field point to each end.

• This is a classic case where Ampere’s law is not convenient because symmetry is not enough.
• In the infinite-wire limit, $\theta_1\to 90^\circ$ and $\theta_2\to 90^\circ$, giving back $\mu_0 I/(2\pi r)$.

Circular loop (Biot–Savart standard results)

At the center of a single loop of radius $R$:

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

On the axis at distance $x$ from the center:

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

For $N$ turns, multiply by $N$.

Direction: along the axis, given by curling fingers in current direction; thumb gives $\vec{B}$ direction through the loop.

Common misconception: thinking the field points “around the loop” at the center because the current is circular. At the center, the net field is perpendicular to the plane of the loop.

Long solenoid (Ampere with idealization)

A solenoid is a long coil of wire. For an ideal “long” solenoid (length much greater than radius, tightly wound), the field inside is approximately uniform and parallel to the axis, and the field outside is small.

If $n$ is turns per unit length and the current is $I$, then inside the solenoid:

$B = \mu_0 n I$

Why this happens (conceptual): the fields from many turns add constructively inside, while outside they largely cancel due to symmetry and the return paths.

How Ampere’s law derives it (outline): Choose a rectangular Amperian loop with one side of length $\ell$ inside the solenoid (parallel to the axis) and one side outside. Inside, $\vec{B}$ is roughly constant and parallel to the inside segment; outside, $B$ is near zero for an ideal long solenoid. Enclosed current equals the number of turns crossing the surface, $n\ell$, times $I$.

That gives

$B\ell = \mu_0 (n\ell I)$

so

$B = \mu_0 n I$

Where students slip: using this for short solenoids without acknowledging that the outside field and nonuniformity can be significant.

Toroid (Ampere and circular symmetry)

A toroid is like a solenoid bent into a donut. With $N$ turns carrying current $I$, Ampere’s law with a circular Amperian loop of radius $r$ (centered on the toroid) gives a field inside the core:

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

This holds for $r$ values within the toroid’s windings (inside the “donut material”). For an ideal toroid, the field outside is approximately zero because the magnetic field lines largely stay confined within the core.

Key difference from a straight solenoid: the toroid’s field is not uniform in $r$; it decreases as $1/r$.

Infinite current sheet (surface current density)

An idealized infinite sheet carrying uniform surface current density $K$ produces a uniform magnetic field on each side of the sheet:

$B = \frac{\mu_0 K}{2}$

The direction is opposite on the two sides, determined by a right-hand rule using the current direction.

This configuration is a favorite for testing whether you can combine:

• symmetry (field must be parallel to the sheet and uniform in magnitude),
• Ampere’s law with a rectangular loop straddling the sheet,
• and careful direction reasoning.

Superposition in multi-wire or composite setups

Many exam problems combine these building blocks. The governing idea is always superposition:

$\vec{B}_{\text{net}} = \sum \vec{B}_i$

For instance, for two long parallel wires, you compute the field from each wire at the point and add as vectors. Often the math is simple but the direction is the whole point.

Worked example: two long parallel wires (superposition + right-hand rule)

Two infinite wires are separated by distance $d$ and carry currents $I$ and $I$ in the same direction. Find the magnetic field magnitude midway between them.

1. Midpoint is at distance $r = d/2$ from each wire.
2. Each wire alone gives magnitude

$B_1 = \frac{\mu_0 I}{2\pi (d/2)} = \frac{\mu_0 I}{\pi d}$

Same for $B_2$.

3. Direction: at the midpoint, the right-hand rule shows the fields from the two wires point in opposite directions (because you are on opposite sides relative to each wire).
4. Net field magnitude:

$B_{\text{net}} = 0$

If the currents were opposite directions, the fields at the midpoint would add instead.

This is a classic “direction over algebra” situation.

Worked example: field inside a toroid at a given radius

A toroid has $N$ turns and carries current $I$. Find $B$ at radius $r$ inside the windings.

1. Choose a circular Amperian loop of radius $r$ centered on the toroid.
2. By symmetry, $\vec{B}$ is tangent and constant on the loop.
3. Ampere’s law:

$\oint \vec{B}\cdot d\vec{\ell} = B(2\pi r) = \mu_0 I_{\text{enc}}$

4. Enclosed current is $I_{\text{enc}} = NI$ because the surface spans all $N$ turns.
5. Solve:

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

Exam Focus

• Typical question patterns:
  • “Recognize the geometry and write down or derive $B(r)$ for wire/solenoid/toroid/sheet.”
  • “Combine fields from multiple sources (superposition) with careful direction analysis.”
  • “Inside vs outside regions: give piecewise expressions (for example, inside a conductor vs outside).”
• Common mistakes:
  • Using the solenoid formula $B = \mu_0 n I$ outside its validity (short solenoid or far outside region).
  • Forgetting region dependence (for example, inside a solid wire $B\propto r$, outside $B\propto 1/r$).
  • Adding magnitudes when vectors actually cancel due to opposite directions (especially in multi-wire setups).