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

This is your training ground for Magnetic Flux. Work each problem before opening its solution. The levels climb like stairs: first you recognise the formula, then apply it, then analyse trickier setups, then synthesise several ideas, and finally reach mastery problems that combine flux with Faraday's Law of Induction, Lenz's Law and Gauss's Law for Magnetism.

Everything rests on one formula and one caution:

Let us first pin down the vocabulary you will read in every problem, so no symbol is a stranger.

In the picture, the blue arrows are the field $B$ , the green disc is the surface, the yellow arrow is its normal $\hat{n}$ , and the red angle between them is $θ$ . Keep this image in your head for every problem.

Level 1 — Recognition

Can you pick out $B$ , $A$ , $θ$ and plug in?

L1.1 — Straight-on square

A square loop of side $0.30 m$ lies with its normal along a uniform field $B = 0.80 T$ . Find $Φ$ .

Recall Solution L1.1

WHAT: identify each quantity. WHY: normal along field means $θ = 0^{\circ}$ , the head-on case where every line passes.

Area: $A = (0.30)^{2} = 0.090 m^{2}$ (square area = side²).
$cos 0^{\circ} = 1$ .
$Φ = B A cos θ = 0.80 \times 0.090 \times 1 = 0.072 Wb$ .

L1.2 — Edge-on ring

A circular ring of radius $0.10 m$ is held edge-on so the field lies flat in the plane of the ring, $B = 1.2 T$ . Find $Φ$ .

Recall Solution L1.2

WHAT: edge-on means the field is in the surface, so the normal is $⊥$ to $B$ , giving $θ = 9 0^{\circ}$ .

$cos 9 0^{\circ} = 0$ .
$Φ = B A cos 9 0^{\circ} = 0 Wb$ . WHAT IT LOOKS LIKE: lines skim along the ring like rain sliding along a page held sideways — none pierce.

Level 2 — Application

Now the angle is not $0$ or $9 0^{\circ}$ — you must use $cos θ$ honestly, and sometimes read $θ$ off a picture.

L2.1 — Tilted loop

A rectangular loop $0.40 m \times 0.25 m$ sits in a uniform field $B = 0.60 T$ . Its normal makes $3 0^{\circ}$ with $B$ . Find $Φ$ .

Recall Solution L2.1

$A = 0.40 \times 0.25 = 0.10 m^{2}$ .
$θ = 3 0^{\circ}$ (already given from the normal — good).
$cos 3 0^{\circ} = \frac{3}{2} \approx 0.8660$ .
$Φ = 0.60 \times 0.10 \times 0.8660 = 0.0520 Wb$ .

L2.2 — Angle given from the surface, not the normal

A flat coil of area $0.050 m^{2}$ makes an angle of $4 0^{\circ}$ between the field and the plane of the coil. Field $B = 0.90 T$ . Find $Φ$ .

Recall Solution L2.2

WHAT: the problem gives the angle to the plane, call it $α = 4 0^{\circ}$ . WHY convert: the formula needs the angle from the normal. The normal is $9 0^{\circ}$ from the plane, so $θ = 9 0^{\circ} - α = 9 0^{\circ} - 4 0^{\circ} = 5 0^{\circ} .$ WHAT IT LOOKS LIKE: in the figure, $α$ (green) hugs the surface, $θ$ (red) reaches up to the yellow normal; together they fill the right angle.

$cos 5 0^{\circ} \approx 0.6428$ .
$Φ = 0.90 \times 0.050 \times 0.6428 = 0.0289 Wb$ .

Level 3 — Analysis

Field or surface varies — the integral earns its keep. Also: sign and direction of the normal.

L3.1 — Field that grows across the surface

A field points along $\hat{z}$ with magnitude $B = (0.50 x) \hat{z} T$ ( $x$ in metres). It passes through a rectangle in the $x y$ -plane spanning $0 \leq x \leq 3$ , $0 \leq y \leq 2$ . The normal is $\hat{z}$ . Find $Φ$ .

Recall Solution L3.1

WHY integrate: $B$ depends on $x$ , so no single $B A$ works — slice into thin strips where $B$ is nearly constant, sum, take the limit.

Since $B ∥ \hat{n}$ (both along $\hat{z}$ ), $B \cdot d A = 0.50 x d A$ . $Φ = \int_{0}^{2} \int_{0}^{3} 0.50 x d x d y = 0.50 [\frac{x ^{2}}{2}]_{0}^{3} [y]_{0}^{2} = 0.50 \times \frac{9}{2} \times 2.$
$Φ = 0.50 \times 4.5 \times 2 = 4.5 Wb$ .

L3.2 — Field growing in the perpendicular direction

Same rectangle in the $x y$ -plane ( $0 \leq x \leq 3$ , $0 \leq y \leq 2$ ), but now $B = (0.50 z) \hat{z} T$ . Find $Φ$ .

Recall Solution L3.2

WHAT's the catch: the whole rectangle lies at $z = 0$ . So on the surface $z = 0$ everywhere, meaning $B = 0.50 \times 0 = 0$ at every patch. $Φ = \int_{S} 0 d A = 0 Wb .$ WHY it matters: the field varies along $z$ , but the surface never samples any $z \neq = 0$ . A varying field only raises flux if the surface actually sits where the field is nonzero.

L3.3 — Reversing the normal

For the setup of L3.1, suppose we choose the normal to point along $- \hat{z}$ instead. What is $Φ$ now?

Recall Solution L3.3

WHY: flipping $\hat{n}$ flips the sign of $B \cdot d A$ everywhere, since now field and normal are anti-parallel ( $θ = 18 0^{\circ}$ , $cos 18 0^{\circ} = - 1$ ). $Φ = - 4.5 Wb .$ The sign of flux depends on which way you call "out" — a bookkeeping choice. Magnitude is fixed; sign follows your chosen normal. This choice is exactly what fixes the sign in Faraday's Law of Induction.

Level 4 — Synthesis

Combine flux with turns, with closed surfaces, and with rates of change.

L4.1 — Coil with many turns and a rate of change

A coil of $N = 200$ turns and area $A = 0.020 m^{2}$ sits with normal parallel to a field that rises uniformly from $0$ to $0.60 T$ in $0.30 s$ . Find (a) the final flux linkage $N Φ$ , and (b) the magnitude of the induced EMF $∣ E ∣ = N \frac{d Φ}{d t}$ .

Recall Solution L4.1

(a) WHAT: flux through one turn, then multiply by $N$ because each turn links the same flux.

$Φ = B A cos 0 = 0.60 \times 0.020 \times 1 = 0.012 Wb$ .
Flux linkage $= N Φ = 200 \times 0.012 = 2.4 Wb$ .

(b) WHY a derivative: EMF answers "how fast is flux changing?" — that is the rate of change, a derivation from Faraday's Law of Induction.

$\frac{d Φ}{d t} = A cos 0 \cdot \frac{d B}{d t} = 0.020 \times \frac{0.60}{0.30} = 0.020 \times 2.0 = 0.040 Wb/s$ .
$∣ E ∣ = N \frac{d Φ}{d t} = 200 \times 0.040 = 8.0 V$ .

L4.2 — Closed surface (Gauss for magnetism)

A cube of side $0.10 m$ sits in a uniform field $B = 2.0 T$ pointing along $+ \hat{x}$ . Find the total flux through the whole closed surface of the cube.

Recall Solution L4.2

WHY zero without computing six faces: Gauss's Law for Magnetism says $\oint B \cdot d A = 0$ for any closed surface — field lines have no start or end, so every line entering the cube leaves it.

Face at $x = 0$ : outward normal is $- \hat{x}$ , flux $= - B \cdot (0.10)^{2} = - 0.020 Wb$ .
Face at $x = 0.10$ : outward normal is $+ \hat{x}$ , flux $= + 0.020 Wb$ .
Other four faces: normal $⊥ \hat{x}$ , so $θ = 9 0^{\circ}$ , each contributes $0$ .
Total $= - 0.020 + 0.020 + 0 = 0 Wb$ . ✓

Level 5 — Mastery

Multi-step reasoning: rotating loops, and reading direction from Lenz's law.

L5.1 — Rotating loop, flux as a function of time

A loop of area $A = 0.040 m^{2}$ in a uniform field $B = 0.50 T$ rotates so that the angle between its normal and the field is $θ (t) = ω t$ with $ω = 100 rad/s$ . (a) Write $Φ (t)$ . (b) Find the peak EMF for one turn, $∣ E ∣_{m a x}$ .

Recall Solution L5.1

(a) WHAT: substitute $θ (t) = ω t$ into $Φ = B A cos θ$ . $Φ (t) = B A cos (ω t) = 0.50 \times 0.040 \times cos (100 t) = 0.020 cos (100 t) Wb .$

(b) WHY a derivative and why it peaks: EMF $= - \frac{d Φ}{d t}$ . Differentiating $cos$ gives $- sin$ : $E = - \frac{d Φ}{d t} = B A ω sin (ω t) .$

The $sin$ swings between $- 1$ and $+ 1$ , so the peak magnitude is $B A ω$ : $∣ E ∣_{m a x} = B A ω = 0.50 \times 0.040 \times 100 = 2.0 V .$ WHAT IT LOOKS LIKE: flux is maximal when the loop faces the field ( $cos = 1$ ), but the EMF is maximal a quarter-turn later, when the loop is edge-on and flux is changing fastest ( $sin = 1$ ).

L5.2 — Direction of induced current (Lenz)

In L5.1, at the instant the loop's flux is decreasing (moving from face-on toward edge-on), which way does the induced current flow — to reinforce or oppose the falling flux?

Recall Solution L5.2

WHY oppose: Lenz's Law says the induced current always fights the change in flux (energy conservation — nature resists free lunches).

Flux is falling, so the induced current flows in the direction whose own magnetic field points the same way as $B$ through the loop — trying to prop the flux up, i.e. to maintain it.
It does not reinforce the change; it opposes the change (here, opposes the decrease by adding flux).

Wrap-up

Related building blocks: Surface and Line Integrals, Magnetic Field B, Electric Flux, Inductance.