Magnetic Flux: Study Guide (Physics C: E&M 13.1)

Course note. This guide is for the calculus-based Physics C: Electricity and Magnetism course, Topic 13.1. The algebra-based Physics 2 course meets magnetic flux for flat surfaces in a uniform field; here you also need the surface integral ∫B·dA.

Constant used throughout: μ₀ = 4π × 10⁻⁷ T·m/A.

What magnetic flux measures

Picture a wire loop held in a magnetic field. Some field lines pass through the loop and some miss it. Magnetic flux, Φ_B, measures how much field passes through the surface that the loop encloses.

Three things control it:

• the strength of the field, B;
• the area of the surface, A;
• the orientation of the surface. Only the part of B perpendicular to the surface passes through it.

Flux is a scalar. It has a size and a sign, but no direction. Its SI unit is the weber (Wb), where 1 Wb = 1 T·m².

You already know this idea from electric flux in Topic 8.5. Magnetic flux uses the same mathematics with B in place of E. The big difference comes later: in Topic 13.2 you will see that a changing magnetic flux induces an emf.

The area vector and the sign of flux

Every flat surface has an area vector A:

• its size is the area of the surface;
• its direction is perpendicular (normal) to the surface.

A flat open surface, such as the inside of a loop, has two possible normals. You choose one and state it. The choice changes the sign of the flux, not its size.

On a closed surface (a box, a sphere, a can with both ends on) there is no choice: the area vector always points outward.

The sign of the flux comes from the dot product B·A:

Angle θ between B and A	B·A	Flux
0° ≤ θ < 90°	positive	positive: B passes through in the direction of A
θ = 90°	zero	zero: B runs along the surface
90° < θ ≤ 180°	negative	negative: B passes through against A

For a loop of wire, a useful convention links the normal to a direction around the loop: curl the fingers of your right hand around the loop in the chosen direction, and your thumb gives A. You will use this in Topic 13.2 to connect the sign of the flux to the direction of an induced current.

Flat surfaces in a uniform field

If B is the same everywhere on a flat surface, the flux is a single dot product:

Φ_B = B·A = BA cos θ

Here θ is the angle between B and the area vector, not between B and the surface. If a problem gives the angle between the field and the plane of the loop, subtract it from 90° first.

In component form, if B = B_x î + B_y ĵ + B_z k̂ and the surface lies in the xy-plane with A = A k̂, then Φ_B = B_z A. Only the component along the normal counts.

Worked example 1: turning a coil in a uniform field

Question. A flat rectangular coil, 0.40 m by 0.25 m, can turn on an axle that is perpendicular to a uniform horizontal field of 0.060 T. Choose A along the field when the coil faces the field squarely. Find the flux through the coil when the angle between B and A is (a) 0°, (b) 60°, (c) 90° and (d) 180°. (e) By how much does the flux change between (a) and (d)?

1. Area: A = 0.40 m × 0.25 m = 0.10 m². Maximum flux: BA = (0.060 T)(0.10 m²) = 6.0 × 10⁻³ Wb.
2. (a) θ = 0°: Φ_B = 6.0 × 10⁻³ Wb × cos 0° = 6.0 × 10⁻³ Wb.
3. (b) θ = 60° (the coil’s plane is now 30° from the field, as in Figure 1): Φ_B = 6.0 × 10⁻³ × 0.5 = 3.0 × 10⁻³ Wb.
4. (c) θ = 90° (the plane of the coil lies along the field): Φ_B = 0.
5. (d) θ = 180° (the coil has turned half a revolution): Φ_B = −6.0 × 10⁻³ Wb.
6. (e) ΔΦ_B = (−6.0 × 10⁻³) − (6.0 × 10⁻³) = −1.2 × 10⁻² Wb.

Check. The size of the flux is never more than BA. A half turn reverses the sign but keeps the size, because the field now passes through the coil from the other face. The change in (e) is twice the maximum flux, not zero: forgetting the sign is a common error.

Interpretation. Turning a coil changes the flux even though B and A stay the same size. That is how a generator works, which you will meet in Topic 13.2.

Flux as a surface integral

When B changes from place to place on the surface, or the surface is curved, a single product is not enough. Split the surface into small patches dA. On each patch B is nearly constant, so the patch contributes B·dA. Add them all:

Φ_B = ∫ B·dA

In practice you choose patches on which B is constant:

• strips of width dx when B depends only on x (dA = ℓ dx for a strip of length ℓ);
• rings of radius r and width dr when B depends only on distance from a centre (dA = 2πr dr).

If B is uniform and the surface is flat, the integral reduces to B·A, so the uniform-field formula is just a special case.

Worked example 2: a rectangle beside a long wire

Question. A long straight wire carries a steady current I = 25 A. A rectangular loop lies in the same plane, with two sides parallel to the wire (Figure 2). The near side is d = 0.020 m from the wire, the loop is b = 0.060 m wide and ℓ = 0.15 m long. (a) Find an expression for the flux through the loop. (b) Evaluate it. (c) Compare with the estimate “field at the middle × area”.

(a) Set up the integral. At distance x from the wire, B = μ₀I/(2πx), into the page. B is the same along a strip parallel to the wire, so use strips of width dx and area dA = ℓ dx. Take A into the page, so B·dA = B dA (positive).

Φ_B = ∫ from d to d+b of [μ₀I/(2πx)] ℓ dx = (μ₀Iℓ/2π) ∫ from d to d+b of dx/x = (μ₀Iℓ/2π) ln[(d + b)/d]

(b) Numbers. μ₀Iℓ/(2π) = (2 × 10⁻⁷ T·m/A)(25 A)(0.15 m) = 7.5 × 10⁻⁷ T·m². The ratio (d + b)/d = 0.080/0.020 = 4, and ln 4 = 1.386. So Φ_B = 7.5 × 10⁻⁷ × 1.386 = 1.04 × 10⁻⁶ Wb.

(c) Comparison. At the middle of the loop, x = 0.050 m, B = (2 × 10⁻⁷)(25) ÷ 0.050 = 1.0 × 10⁻⁴ T. Multiplying by the area (0.060 m × 0.15 m = 9.0 × 10⁻³ m²) gives 9.0 × 10⁻⁷ Wb, about 13% too small. The field near the wire (2.5 × 10⁻⁴ T at the near side) is much stronger than at the far side (6.25 × 10⁻⁵ T), and the 1/x curve is not symmetric about the middle, so you must integrate.

Check. If the loop is very narrow (b ≪ d), then ln(1 + b/d) ≈ b/d, and the result becomes μ₀Iℓb/(2πd): the field at the loop times its area, as expected for an almost uniform field. Units: T·m/A × A × m = T·m², which is a weber.

Flux through a solenoid

In Topic 12.4 you used Ampère’s law to show that the field inside a long ideal solenoid is uniform, B = μ₀nI, along the axis, and nearly zero outside. So the flux through one turn of radius R is:

Φ_B = μ₀nI × πR²

For example, n = 2000 turns/m, I = 1.5 A and R = 0.020 m give B = 3.77 × 10⁻³ T and Φ_B = 4.74 × 10⁻⁶ Wb per turn.

Two consequences are often tested:

• A coaxial loop inside the solenoid, with radius r < R, has flux B × πr². Only its own area counts.
• A loop around the outside of the solenoid, of any larger radius, has the same flux as one turn, B × πR². The field outside is nearly zero, so the extra area adds nothing.

Closed surfaces: zero net magnetic flux

Magnetic field lines always form closed loops. There are no magnetic monopoles where lines could start or stop. So every field line that enters a closed surface must also leave it, and:

∮ B·dA = 0 for every closed surface

This is Gauss’s law for magnetism (Topic 12.1). Compare Gauss’s law for electric fields, ∮E·dA = q_enc/ε₀: electric field lines can start and end on charges, but magnetic field lines cannot.

A useful result follows. Take any two open surfaces bounded by the same loop, for example a flat disc and a bowl. Together they form a closed surface, so the flux through the bowl equals the flux through the disc (measured in the same direction). You can always pick the easiest surface to calculate the flux through a loop.

Common misconceptions

• Using the angle to the surface. In Φ_B = BA cos θ, θ is measured from the normal. If the field makes 30° with the plane, use θ = 60°.
• “Flux is a vector.” It is a scalar; it has a sign because of the dot product, not a direction.
• Ignoring the sign when a coil turns over. Turning a loop half a revolution changes the flux from +BA to −BA. The change is 2BA, not zero.
• Using the loop’s area outside a solenoid. Only the area where B is non-zero counts. A wide loop around a solenoid has the same flux as one turn.
• Multiplying B at the middle by the area when B varies. Integrate B·dA. The middle-value estimate is only safe if B changes linearly across the surface.
• “Zero net flux means no field.” The net flux through a closed surface is always zero, even inside a strong magnet. It tells you that lines enter and leave in equal amounts, not that B = 0.
• Confusing flux with field. A strong field through a tiny loop can give less flux than a weak field through a large loop.

Where this leads

Magnetic flux is the starting point for the whole of Unit 13. In Topic 13.2, Electromagnetic Induction, Faraday’s law says that the induced emf equals the rate of change of this flux, and Lenz’s law gives its direction. Practise now with the practice questions, then use the revision notes and the topic checklist.