Magnetostatics: Ampere & Biot-Savart (IPhO E&M)

IPhO E&M lesson on magnetostatics: when to use Ampere's law vs Biot-Savart, plus high-yield field and force results.

International Physics Olympiad preparation

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Magnetostatics is the “steady current” limit: currents do not change with time, so magnetic fields are time-independent. IPhO magnetostatics problems are usually won by choosing the right loop (Ampere) or the right geometry split (Biot-Savart), plus a clean right-hand rule for direction.

If symmetry can make | vector B| constant on a loop and tangential everywhere, use Ampere: ∮ vector B · d vector ℓ = μ₀ I_enc.
If not, use Biot-Savart and superposition: d vector B = (μ₀/4π)(I d vector ℓ × r hat)/r².
Always do direction first with a right-hand rule, then compute the magnitude.
Check scaling with distance and limit cases (far away, finite loops behave like dipoles).

1. Definitions (Must Know)

Lorentz force:
vector F = q vector v × vector B.
Force on a current element:
d vector F = I d vector ℓ × vector B.
Biot-Savart law (steady currents):
d vector B = (μ₀/4π)(I d vector ℓ × r hat)/r².
Ampere’s law (magnetostatics form):
∮ vector B · d vector ℓ = μ₀ I_enc.
∇ · vector B = 0 (no magnetic monopoles).
Superposition: magnetic fields add as vectors.

2. Key Ideas (What Earns Marks)

Ampere works when symmetry makes the integral trivial. Common symmetries: infinite straight wire, infinite sheet of current, long solenoid, toroid, uniformly current-filled cylinder.
Biot-Savart always works but can be slow. Use known building blocks (finite wire segment, circular arc, full loop) and add them.
Choose I_enc carefully. For distributed current density vector J:
I_enc = ∬ vector J · d vector A.
Direction is a mark. Write one clean direction sentence: “By the right-hand rule, vector B is azimuthal around the wire,” etc.
Force between currents is high yield. Parallel wires and loops often appear as “find the direction of motion” questions.

3. Detailed Explanations

A. Infinite straight wire (and the “azimuthal” field).
For a long straight wire carrying current I, symmetry forces vector B to be tangent to circles centered on the wire and depend only on r. Ampere loop is a circle of radius r:

B(2π r) = μ₀ I ⇒ B(r) = (μ₀ I)/(2π r).

B. Inside a uniformly current-filled cylinder.
A solid wire of radius R carries total current I uniformly (constant current density). For radius r inside the wire, enclosed current is proportional to area:

I_enc(r) = I(π r²)/(π R²) = Ir²/R².

B(r) = (μ₀ I_enc)/(2π r) = ((μ₀ I)/(2π R²))r (inside).

Outside, it returns to μ₀ I/(2π r).

C. Long solenoid and toroid (Ampere classics).
For a long solenoid with n turns per unit length and current I, Ampere with a rectangular loop gives

B ≈ μ₀ n I inside, B ≈ 0 outside (ideal long solenoid).

For an ideal toroid with N turns, inside the core (at radius r between inner and outer radii),

B(r) = (μ₀ N I)/(2π r),

and outside the core the field is small (ideal model gives zero).

D. Biot-Savart building blocks.
Two results you can reuse constantly:

Finite straight wire: field magnitude at perpendicular distance a from the wire is
B = ((μ₀ I)/(4π a))(sin θ₁ + sin θ₂),
where θ₁ and θ₂ are the angles subtended by the ends as seen from the point.
Circular arc of radius R subtending angle φ at the center:
B_center = ((μ₀ I)/(4π R))φ,
with direction given by the right-hand rule.

E. Force between two long parallel currents.
Wire 1 creates B = μ₀ I₁/(2π d) at the location of wire 2, so force per unit length is

F/L = I₂ B = (μ₀ I₁ I₂)/(2π d).

Same direction currents attract, opposite direction currents repel.

4. Common Mistakes

Using Ampere in a geometry with no symmetry (where vector B is not constant along the chosen loop).
Forgetting that I_enc can be piecewise (coaxial cables, thick conductors).
Getting the direction wrong by mixing right-hand rules (field around a wire vs force on a wire).
Dropping vector nature too early: vector B adds as a vector, not as a scalar.
Confusing the “inside solenoid is uniform” idealization with short solenoids (edge fields matter).

5. Exam Tips

Write the Ampere loop on your diagram and label where vector B∥ d vector ℓ and where it is perpendicular (so the dot product vanishes).
In Biot-Savart, look for symmetry that cancels components (only one component survives).
Dimensional check:
[B] = T = N A⁻¹ m⁻¹.
If the problem asks for a force direction, compute vector B direction first, then use vector F = I vector ℓ × vector B.

6. Worked Examples

Example 1 (coaxial cable: piecewise B(r)).
An inner conductor of radius a carries current + I uniformly. A concentric outer cylindrical shell from radius b to c carries current -I uniformly. Find B(r) for all r.

Solution sketch

Use a circular Ampere loop of radius r.

For r up to a (inside inner conductor):

I_enc = Ir²/a² ⇒ B = ((μ₀ I)/(2π a²))r.

For r between a and b (between conductors): I_enc = I, so

B = (μ₀ I)/(2π r).

For r between b and c (inside the outer shell): enclosed current is the inner I plus a fraction of the outer -I proportional to area:

I_enc = I-I(r²-b²)/(c²-b²).

B = (μ₀/(2π r))[I-I(r²-b²)/(c²-b²)].

For r beyond c: net enclosed current is zero, so B = 0.

Example 2 (Biot-Savart: field on the axis of a circular loop).
A circular loop of radius R carries current I. Find the magnetic field on its axis a distance z from the center.

Solution sketch

By symmetry, transverse components cancel and only the axial component survives. The standard Biot-Savart integration gives

B(z) = (μ₀ I R²)/(2(R² + z²)^(3/2)),

directed along the axis by the right-hand rule.

Two checks:

At z = 0: B(0) = μ₀ I/(2R).
For large z: B ∝ 1/z³ (magnetic dipole scaling).

Example 3 (force per length between parallel wires).
Two long parallel wires are separated by distance d and carry currents I and 2I in the same direction. Find the force per unit length and the direction of the force.

Solution sketch

Field from wire 1 at wire 2:

B = (μ₀ I)/(2π d).

Force per unit length on wire 2:

F/L = (2I)B = (μ₀ (2I)I)/(2π d) = (μ₀ I²)/(π d).

Same-direction currents attract, so the force is toward the other wire.

7. Mind Stretchers

Mind stretcher: a semicircular wire with two semi-infinite arms

A wire consists of a semicircle of radius R connected to two straight semi-infinite arms tangent to the semicircle at its ends (all in one plane). The current is I. Find the magnetic field magnitude at the center of the semicircle.

Solution idea: add three contributions: semicircle plus two semi-infinite tangents.

Semicircle contribution (half of a full loop):

Bₛₑₘᵢ = (μ₀ I)/4R.

Each semi-infinite straight arm at distance R contributes half of the infinite-wire field:

Bₐᵣₘ = (μ₀ I)/(4π R).

The two arms add in the same direction (right-hand rule gives the same out-of-plane sense).

Total:

B = Bₛₑₘᵢ + 2Bₐᵣₘ = ((μ₀ I)/4R)(1 + 2/π).

8. Practice

Re-derive B(r) for a long wire, inside a solid wire, inside a solenoid, and inside a toroid from Ampere.
Memorize one Biot-Savart building block (finite wire or circular arc) and practice using superposition.
Do one “piecewise enclosed current” coaxial problem and one force-between-wires problem.

Syllabus and review details

No official syllabus alignment is listed for this lesson.