Magnetic flux Φ = ∫B·dA: Worked examples — Physics deep dive

This page is a drill. We take the flux idea — $Φ = B A cos θ$ and its integral big brother $Φ = \int_{S} B \cdot d A$ — and hit it with every kind of question it can produce. Before anything, one reminder in plain words:

Everything below is just those two knobs, turned to every possible setting.

The scenario matrix

Here is the complete list of case-classes flux problems come in. Every cell gets at least one fully worked example further down.

#	Case class	The distinguishing feature	Example
C1	Face-on ( $θ = 0$ )	Field $∥$ normal, flux is maximal $= B A$	Ex 1
C2	Tilted ( $0 < θ < 9 0^{\circ}$ )	Partial projection, $cos θ$ between 0 and 1	Ex 2
C3	Edge-on ( $θ = 9 0^{\circ}$ ) — degenerate	Field lies in the surface, flux $= 0$	Ex 3
C4	Obtuse angle ( $9 0^{\circ} < θ \leq 18 0^{\circ}$ ) — negative flux	Field pierces the back of the surface, $cos θ < 0$	Ex 4
C5	Non-uniform field	$B$ varies over the surface → must integrate	Ex 5
C6	Closed surface — degenerate	Everything that enters must leave, $\oint = 0$	Ex 6
C7	Multi-turn coil	Flux linkage $N Φ$ , easy to forget	Ex 7
C8	Real-world word problem	Translate words → $B, A, θ$	Ex 8
C9	Exam twist: angle-with-surface trap	They give angle to the plane, not the normal	Ex 9
C10	Limiting behaviour	What happens as $θ$ sweeps $0 \to 18 0^{\circ}$	Ex 10

The sign of $cos θ$ is the spine of this whole table, so let us make that visible first.

Worked examples

C1 — Face-on loop (maximum flux)

C2 — Tilted loop (partial projection)

C3 — Edge-on loop (degenerate, zero flux)

C4 — Obtuse angle (NEGATIVE flux)

Worked example Ex 4

Same loop and field, now the normal points $θ = 12 0^{\circ}$ from $B$ — i.e. the field enters through the back face of the loop (relative to the chosen normal direction).

Forecast: the flux number will do something the parent note never showed. Guess its sign.

$cos 12 0^{\circ} = - 0.5$ . Why this step? Beyond $9 0^{\circ}$ the field pierces the surface from the side opposite to the normal, so its component along the normal is negative.
$Φ = B A cos 12 0^{\circ} = 0.80 \times 0.0314 \times (- 0.5) = - 0.0126 Wb$ . Why this step? The dot product carries the sign automatically.

Verify: The sign is just a bookkeeping label for "which way the lines cross the chosen normal." Its magnitude $0.0126 Wb$ equals the flux you'd get at $θ = 6 0^{\circ}$ (since $∣ cos 12 0^{\circ} ∣ = ∣ cos 6 0^{\circ} ∣ = 0.5$ ) — a nice symmetry check on the $s 01$ curve. ✓

Why it matters: Faraday's law and Lenz's Law care about the change in $Φ$ , and flux flipping from positive to negative is a real change that drives current.

C5 — Non-uniform field (must integrate)

Worked example Ex 5

A field points along $\hat{z}$ but grows with position: $B = (0.20 + 0.10 x) \hat{z} T$ ( $x$ in metres). Find the flux through the square $0 \leq x \leq 3$ , $0 \leq y \leq 2$ lying in the $x y$ -plane, normal $\hat{z}$ .

Forecast: you can't use one $B A$ — why not? Guess whether the answer is bigger or smaller than using the field at $x = 0$ .

$B \cdot d A = (0.20 + 0.10 x) d A$ because field $∥$ normal. Why this step? When both vectors are along $\hat{z}$ the dot product is just the product of magnitudes — see Surface and Line Integrals.
Set up the double integral over the square: $Φ = \int_{0}^{2} \int_{0}^{3} (0.20 + 0.10 x) d x d y .$ Why this step? $B$ changes with $x$ , so we slice into strips where $B$ is locally constant, then sum (integrate).
Inner integral: $\int_{0}^{3} (0.20 + 0.10 x) d x = [0.20 x + 0.05 x^{2}]_{0}^{3} = 0.60 + 0.45 = 1.05$ . Why this step? Antiderivative of a constant is linear; of $x$ is $x^{2} /2$ , and $0.10/2 = 0.05$ .
Outer integral: $\int_{0}^{2} 1.05 d y = 1.05 \times 2 = 2.10 Wb$ . Why this step? Nothing depends on $y$ , so integrating over $y$ just multiplies by the width $2$ .

Verify: If the field were constant at its average value over $0 \leq x \leq 3$ , that average is $0.20 + 0.10 (1.5) = 0.35 T$ , giving $0.35 \times (3 \times 2) = 2.10 Wb$ — exact match, because the field is linear so the average trick is exact. ✓

C6 — Closed surface (degenerate, always zero)

Worked example Ex 6

A uniform field $B = 1.5 T$ points along $\hat{x}$ through a closed cube of side $0.20 m$ . Find the total flux out of the cube.

Forecast: two faces are perpendicular to the field. Guess the total before summing.

Only the two faces $⊥ \hat{x}$ have flux; the other four are edge-on ( $θ = 9 0^{\circ}$ , zero each). Why this step? Edge-on faces contribute nothing ( $cos 9 0^{\circ} = 0$ ).
Face area $A = (0.20)^{2} = 0.040 m^{2}$ . On the entry face the outward normal points $- \hat{x}$ , so $θ = 18 0^{\circ}$ : $Φ_{in} = B A cos 18 0^{\circ} = 1.5 \times 0.040 \times (- 1) = - 0.060 Wb$ . Why this step? Outward normal opposes the field where lines enter → negative flux.
On the exit face the outward normal is $+ \hat{x}$ , $θ = 0$ : $Φ_{out} = 1.5 \times 0.040 \times (+ 1) = + 0.060 Wb$ . Why this step? Lines leave along the outward normal → positive flux.
Total: $\oint B \cdot d A = - 0.060 + 0.060 + 0 = 0 Wb$ . Why this step? Sum every face.

Verify: This is exactly Gauss's Law for Magnetism: no monopoles, every line entering also leaves, so a closed surface always gives $0$ — contrast with Electric Flux where enclosed charge makes it nonzero. ✓

C7 — Multi-turn coil (flux linkage)

C8 — Real-world word problem

C9 — Exam twist: angle-to-the-surface trap

Worked example Ex 9

A loop of area $A = 0.020 m^{2}$ in $B = 0.60 T$ is tilted so the field makes $2 5^{\circ}$ with the plane of the loop. Find $Φ$ . (Note the wording!)

Forecast: they gave the angle to the plane, not the normal. Guess whether you plug $cos 2 5^{\circ}$ or something else.

Convert: angle to normal $θ = 9 0^{\circ} - 2 5^{\circ} = 6 5^{\circ}$ . Why this step? The normal is $9 0^{\circ}$ from the plane, so an angle $2 5^{\circ}$ from the plane is $6 5^{\circ}$ from the normal. The formula's $θ$ is always from the normal.
$Φ = B A cos 6 5^{\circ} = 0.60 \times 0.020 \times 0.4226 = 0.00507 Wb$ . Why this step? Now $θ$ is correctly measured from the normal.

Verify: Using the wrong angle $cos 2 5^{\circ} = 0.906$ would give $0.0109 Wb$ — over double the truth. The equivalent right way is $B A sin (angle-to-plane) = 0.60 \times 0.020 \times sin 2 5^{\circ} = 0.00507 Wb$ , which matches, confirming $cos θ = sin (plane angle)$ . ✓

C10 — Limiting behaviour (sweep the angle)

Recall Self-test: which cell is this?

A field lies at $7 0^{\circ}$ to the plane of a loop. Which matrix cell, and what's $θ$ ? ::: C9 (plane-angle trap); $θ = 9 0^{\circ} - 7 0^{\circ} = 2 0^{\circ}$ from the normal. Flux through a sealed sphere in any field? ::: C6; always $0$ by Gauss's law for magnetism. A loop at $θ = 13 5^{\circ}$ from the normal — sign of flux? ::: C4; negative, since $cos 13 5^{\circ} < 0$ . Why can't you use $B A$ when $B = 0.1 x \overset{z}{^}$ ? ::: C5; $B$ varies with position, so integrate strip by strip.

Worked examples — Magnetic flux Φ = ∫B·dA