Magnetic flux

Hold a wire loop near a bar magnet and tilt it slowly. Nothing about the magnet changes, yet the "amount of field" threading the loop clearly grows and shrinks as you turn it — maximum when the loop faces the magnet square-on, zero when you turn it edge-on so the field lines skim past without ever crossing the opening. That threading amount is magnetic flux, $\phi_B$ , a single number built from both the field's strength and the surface's orientation. It sounds like a bookkeeping detail, but it is the single quantity whose change drives every electric generator, transformer and induction stove — the next chapter, Electromagnetic Induction, is really just the study of $d\phi_B/dt$ . Get flux solid here — the definition, the correct angle, the sign, the closed-surface rule — and Faraday's and Lenz's laws will read as consequences rather than new mysteries.

The core insight

Magnetic flux $\phi_B = \vec B\cdot\vec A = BA\cos\theta$ is the field-times-effective-area threading a surface, maximum when $\vec B$ is along the surface normal and zero when $\vec B$ lies in the surface's plane.

The core idea

Flux is defined for a flat surface described by an area vector $\vec A$ — a vector whose magnitude equals the surface's area and whose direction is the outward normal (perpendicular to the surface). Placed in a uniform field $\vec B$ , the magnetic flux through it is the dot product of the two vectors:

\phi_B = \vec B\cdot\vec A = BA\cos\theta

where: $\phi_B$ is the magnetic flux (weber, Wb), $B$ is the magnitude of the field (tesla, T), $A$ is the area of the surface (m²), and $\theta$ is the angle between $\vec B$ and the normal to the surface (not the plane of the surface — this mix-up is the single most common error in this topic). The unit works out as $1\ \text{Wb} = 1\ \text{T}\,\cdot\,\text{m}^2$ . Three orientation facts follow directly from the cosine:

Face-on ( $\theta = 0$ ): the normal is parallel to $\vec B$ , so $\cos\theta = 1$ and flux is maximum, $\phi_B = BA$ .
Edge-on ( $\theta = 90°$ ): $\vec B$ lies in the plane of the loop, $\cos\theta = 0$ , and flux is zero — no field lines cross the opening at all.
In between: flux varies smoothly as $BA\cos\theta$ , so rotating a loop from face-on to edge-on sweeps the flux continuously from its maximum down to zero. This continuous sweep is precisely what a rotating generator coil exploits to produce a changing flux, hence an induced EMF, in the next chapter.

Flux is also a signed quantity — the sign depends on which way you choose the normal to point, a convention you fix once at the start of a problem and never flip mid-calculation. For a coil of $N$ tightly wound turns, each turn links essentially the same flux, so the total flux linkage is

N\phi_B = NBA\cos\theta

This $N\phi_B$ combination is what actually appears in the self- and mutual-inductance formulas you will meet in the induction unit, so it is worth treating flux linkage as a distinct, nameable quantity now rather than re-deriving it later. For a curved surface or a non-uniform field, the rigorous definition is the surface integral $\phi_B = \int \vec B\cdot d\vec A$ , summing the dot product over small patches $d\vec A$ of the surface. This syllabus, however, restricts calculation to uniform $\vec B$ through flat loops, where the integral collapses to the single product $BA\cos\theta$ above.

Figure

A loop rotating from face on to edge on showing flux falling from maximum to zero as the angle to the field grows

Derivation AWhy flux is BA cos theta, not just BA

Show that for a flat loop of area $A$ tilted at angle $\theta$ to a uniform field $\vec B$ , the flux threading it is $\phi_B = BA\cos\theta$ rather than the full $BA$ , by finding the effective area the field actually crosses once the loop is turned away from face-on.

Step 1. Start from the face-on case, where every field line that reaches the loop's outline passes straight through it, so the flux is simply field strength times area, $\phi_B = BA$ . This is the $\theta = 0$ case and matches the definition with $\cos 0 = 1$ .

Step 2. Now tilt the loop through angle $\theta$ about an axis in its own plane, keeping the field direction fixed. The loop's true area $A$ has not changed, but the silhouette it presents to the oncoming field lines has shrunk, because the loop is now edge-on by an amount that depends on $\theta$ .

Step 3. Project the loop onto a plane perpendicular to $\vec B$ . Basic geometry of a tilted flat shape gives the projected (effective) area as $A_{\text{eff}} = A\cos\theta$ : at $\theta = 0$ the projection equals the full area, and at $\theta = 90°$ the projection collapses to a line of zero area, matching the edge-on picture where the loop presents no silhouette to the field at all.

Step 4. Since only the field lines that fall within this effective silhouette actually cross the loop, the flux is the field strength times the effective area, $\phi_B = B\times A_{\text{eff}} = BA\cos\theta$ . Recognising that $A\cos\theta$ is exactly the dot product $\vec B\cdot\vec A/B$ shows this geometric projection argument and the vector definition $\phi_B = \vec B\cdot\vec A$ are the same statement, which is why the cosine belongs with the angle to the normal, not the angle to the plane.

Worked example

A square coil of side $0.2\ \text{m}$ with $N = 50$ turns sits in a uniform field $B = 0.5\ \text{T}$ , with its normal making an angle of $60°$ with the field. Find the flux through one turn and the total flux linkage.

Step 1. Find the coil's area. $A = (0.2)^2 = 0.04\ \text{m}^2$ .

Step 2. Identify the angle and apply $\phi_B = BA\cos\theta$ . With $\theta = 60°$ , $\cos 60° = 0.5$ , so $\phi_B = 0.5\times0.04\times0.5$ .

Step 3. Compute the flux per turn. $0.5\times0.04 = 0.02$ , then $0.02\times0.5 = 0.01$ , giving $\phi_B = 0.01\ \text{Wb}$ .

Step 4. Scale up by the number of turns for flux linkage. $N\phi_B = 50\times0.01\ \text{Wb}$ , so the flux per turn is $0.01\ \text{Wb}$ and the total flux linkage is $\mathbf{0.5\ \text{Wb}}$ .

Competitive stretch

Suppose the same coil is now rotated so its normal sweeps from $\theta = 60°$ to $\theta = 90°$ . Since $\cos 90° = 0$ , the flux linkage falls all the way from $0.5\ \text{Wb}$ to exactly zero — even though the field, the coil's area and its turns have not changed at all. This is the essential picture behind a generator: turning a coil steadily changes $\theta$ , which changes $\phi_B$ continuously, and it is that continuous change (not the flux value itself) that will induce an EMF once you reach the Faraday's-law chapter.

Common traps

Using the angle between $\vec B$ and the plane of the loop instead of the angle between $\vec B$ and the normal — these two angles are complementary, so using the wrong one silently swaps $\cos\theta$ for $\sin\theta$ .
Writing the unit of flux as tesla — tesla (T) is the unit of the field $B$ ; the unit of flux $\phi_B$ is the weber (Wb), with $1\ \text{Wb} = 1\ \text{T}\,\cdot\,\text{m}^2$ .
Treating flux as always positive. It is a signed quantity fixed by the chosen direction of the normal, and that sign becomes essential once Lenz's law asks which way an induced current opposes a change in flux.
Forgetting the factor of $N$ when a question asks for total flux linkage through a multi-turn coil rather than the flux through a single turn.

Ask the tutor: Want to see why rotating a coil turns this constant flux formula into a time-varying one that induces an EMF — ask the AI tutor to walk Faraday's law forward from $\phi_B = BA\cos\theta$ with $\theta$ changing steadily in time.

Why is flux defined using the normal to the surface and not the surface itself

Because a flat surface by itself only fixes an area, not an orientation in three dimensions — the same patch of area can be tilted many ways relative to a field. The outward normal is the one direction that uniquely captures "which way the surface is facing," so pairing it with the area into a vector $\vec A$ lets the dot product $\vec B\cdot\vec A$ correctly weight the field by how squarely it strikes the surface.

If flux can be negative, does that mean the field is somehow negative too

No — the field strength $B$ stays positive; it is the cosine that turns negative once $\theta$ exceeds $90°$ , meaning the chosen normal now points roughly against the field rather than with it. A negative flux just means the field is threading the surface through the back face relative to your chosen normal direction, not that the physical field has changed sign.

Why is the total flux through any closed surface always zero

Because magnetic field lines never begin or end — there are no magnetic monopoles for them to start or stop at, unlike electric field lines which begin and end on charges. Every field line that dips into a closed surface (say, an imaginary balloon) must eventually come back out somewhere else on that same surface, so whatever flux enters is exactly cancelled by the flux leaving, making the net always zero. This qualitative fact is sometimes called Gauss's law for magnetism.

Magnetic flux

The core idea

Worked example

Competitive stretch

Common traps

Before this

Where this leads