Faraday's law — EMF = −dΦ - dt: Worked examples

Definition The four symbols we will keep using

$B$ — the magnetic field strength, measured in tesla (T). Picture it as the density of invisible strands passing through space, like the number of spaghetti strands per square metre.
$A$ — the area of the flat loop, in square metres ( $m^{2}$ ). The size of the hoop the strands pass through.
$θ$ — the angle between the field $B$ and the loop's normal $\overset{n}{^}$ (an arrow sticking straight out of the loop's face, perpendicular to it). When the loop faces the field head-on, $θ = 0$ .
$N$ — the number of turns of wire, i.e. how many times the same wire loops around before the strands must be counted again. A coil wound $50$ times has $N = 50$ , and the same flux is "felt" $50$ times over. A single loop has $N = 1$ .

Put together, the flux is $Φ_{B} = B A cos θ$ — "how many strands actually thread the hoop." Its unit is the weber, whose symbol is $Wb$ ; one weber equals $1 T \cdot m^{2}$ .

The scenario matrix

Faraday's law is $E = - \frac{d Φ _{B}}{d t}$ , and $Φ_{B} = B A cos θ$ has three knobs ( $B$ , $A$ , $θ$ ). Every problem is one or more knobs turning, plus a sign question. Here is the full grid. Ten cells; Example 4 covers two cells at once (both degenerate zeros), so nine examples fill all ten:

Cell	Case class	What changes	Special feature	Example
C1	Changing field	$B (t)$	linear ramp, area fixed	Ex 1
C2	Changing area	$A (t)$	sliding rod (motional)	Ex 2
C3	Rotating loop	$θ (t)$	sine output, sign per quadrant	Ex 3
C4	Degenerate: edge-on	$θ = 90°$	flux is zero	Ex 4
C5	Degenerate: constant	nothing changes	EMF is zero	Ex 4
C6	Sign / direction	Lenz	which way current flows	Ex 5
C7	Two knobs at once	$B (t)$ and $A (t)$	product rule, terms add	Ex 6
C8	Non-linear in time	$B \propto t^{2}$	EMF itself varies in time	Ex 7
C9	Real-world word problem	card / coil swipe	translate words to knobs	Ex 8
C10	Exam twist: limiting value	field switched off in $Δ t \to 0$	spike / average EMF	Ex 9

Example 1 — Changing field, fixed loop `(cell C1)`

Identify which knob turns. The loop doesn't move ( $A$ fixed) and doesn't tilt ( $θ = 0$ fixed). Only $B$ changes. Why this step? Picture the spaghetti strands getting denser while the hoop sits still — more strands cram through the same opening. Naming the one moving knob kills two of the three product-rule terms, leaving only $\frac{d B}{d t} A cos θ$ .
Compute the area. $A = s^{2} = (0.20)^{2} = 0.040 m^{2}$ . Why this step? We need area to turn a field-rate into a flux-rate — the loop's size sets how many extra strands each unit rise in $B$ pushes through.
Compute the field's rate of change. Steady ramp means $\frac{d B}{d t} = \frac{0.80 - 0}{4.0} = 0.20 T/s$ . Why this step? "Steadily" means constant slope, so rise-over-run gives the exact rate; no calculus needed for a straight line.
Apply Faraday's law (magnitude, $cos 0 = 1$ ): $∣ E ∣ = N A \frac{d B}{d t} = 50 \times 0.040 \times 0.20 = 0.40 V .$ Why this step? Each of the $N = 50$ turns links the same flux, so every turn adds its own little EMF and they stack in series.

Example 2 — Sliding rod, changing area `(cell C2)`

See the swept area (mint strip in Figure 1). As the rod moves right by $d x$ , the loop's area grows by $d A = L d x$ . Why this step? Think of the field as a carpet of vertical spaghetti strands standing on the page. As the rod sweeps rightward it rakes in new strands; the mint area is exactly the patch of new strands now caught inside the loop. $B$ and $θ$ never change — only area does — so this is Motional EMF.
Rate of area change. $\frac{d A}{d t} = L \frac{d x}{d t} = Lv = 0.25 \times 4.0 = 1.0 m^{2} / s$ . Why this step? Speed is exactly "how fast $x$ grows," so $d x / d t = v$ ; multiply by the rod length to get area swept per second.
Faraday gives $∣ E ∣ = B \frac{d A}{d t} = B Lv = 0.30 \times 0.25 \times 4.0 = 0.30 V .$ Why this step? Field density $B$ times new-area-rate = new-strand-rate = flux rate. Same number as the parent's $B v L$ force-on-charges derivation — two pictures, one answer.
Current $I = \frac{E}{R} = \frac{0.30}{1.5} = 0.20 A .$ Why this step? The EMF acts like a battery of $0.30 V$ pushing current through the resistor (Ohm's law).

Example 3 — Rotating coil (generator), with sign per quadrant `(cell C3)`

Set the time reference, then write the flux. We chose $t = 0$ at $θ = 0$ , so $θ = ω t$ and $Φ_{B} = B A cos (ω t)$ . Why this step? The phase depends on where you start the clock. Had we started with the loop edge-on ( $θ = 90°$ at $t = 0$ ), we would write $Φ_{B} = B A cos (ω t + 90°) = - B A sin (ω t)$ , and the EMF would come out as a $cos$ instead — the shape is identical, only shifted, and a different starting phase flips which sign the EMF has at $t = 0$ . Stating the reference is what pins the sign.
Differentiate. $E = - N \frac{d}{d t} (B A cos ω t) = N B A ω sin ω t$ . Why this step? The derivative of $cos$ is $- sin$ ; the two minus signs (Faraday's and the derivative's) cancel, leaving a clean $+ sin$ . This sine is why generators give AC — see Electric Generators and AC.
Peak value (when $sin = 1$ ): $E_{0} = N B A ω = 200 \times 0.40 \times 0.015 \times 100 = 120 V .$ Why this step? $sin$ tops out at $1$ ; the peak is the amplitude of the wave.
At $θ = 60°$ : $E = E_{0} sin 60° = 120 \times 0.8660 = 103.9 V .$ Why this step? The instantaneous EMF is the amplitude times $sin θ$ .
Sign in each quadrant. With $θ$ running $0 \to 360°$ :
- $θ \in (0°, 180°)$ : $sin θ > 0$ , so $E > 0$ — flux falling, current one way.
- $θ \in (180°, 360°)$ : $sin θ < 0$ , so $E < 0$ — flux rising, current reverses.
Why this step? This is the sign trap of AC: the same magnitude appears twice per turn but with opposite current direction (Lenz keeps track). At $θ = 240°$ , for instance, $sin 240° = - 0.866$ , so $E = - 103.9 V$ — same size as at $60°$ , opposite sign.

Example 4 — Two degenerate cases `(cells C4 & C5)`

Case (a) — constant field. $\frac{d B}{d t} = 0$ , area fixed, angle fixed. Every term of the product rule is zero. $E = 0 V .$ Why this step? EMF tracks change, never the size of the flux. A giant frozen field induces nothing — this is the parent's headline mistake ("big flux ⇒ big EMF" is false).
Case (b) — edge-on loop. Here $θ = 90°$ , so $cos θ = 0$ . The flux is $Φ_{B} = B A cos 90° = 0$ at every instant, even as $B$ grows. $Φ_{B} (t) = 0 \Rightarrow \frac{d Φ _{B}}{d t} = 0 \Rightarrow E = 0 V .$ Why this step? Field sliding along the loop's plane threads nothing through it, so no matter how fast $B$ ramps, zero strands pass the hoop. The $cos θ$ factor is doing exactly the job the parent described.

Example 5 — The sign / direction of current `(cell C6)`

Magnitude. $∣ E ∣ = \frac{d Φ _{B}}{d t} = 0.60 V .$ Why this step? One turn, single knob — straight application.
Direction from Lenz (see Lenz's Law). The downward flux is growing. The loop opposes the change, so it makes its own magnetic field pointing up inside the loop, to resist the increase. Why this step? The minus sign in Faraday's law is this opposition — energy conservation forbids the loop from helping the change.
Right-hand rule for the current. Curl the right hand so the thumb points up (the loop's induced field); the fingers curl counterclockwise as seen from above. Why this step? The current direction and its magnetic field are locked together by the right-hand rule.

Example 6 — Two knobs turning at once `(cell C7)`

Product rule. With $θ = 0$ (so $cos θ = 1$ and its rate is zero), only two terms survive: $\frac{d Φ _{B}}{d t} = \frac{d B}{d t} A + B \frac{d A}{d t} .$ Why this step? Two knobs turning means two product-rule terms; we cannot use a single-term shortcut.
Plug in. $\frac{d Φ _{B}}{d t} = (0.40) (0.050) + (0.30) (0.010) = 0.020 + 0.003 = 0.023 Wb/s .$ Why this step? Evaluate each term at the given instant, then add — both effects grow the flux, so they reinforce.
EMF: $∣ E ∣ = 0.023 V = 23 mV .$

Example 7 — Non-linear field in time `(cell C8)`

Why a derivative and not a slope. $B$ is a curve now, not a straight ramp, so the rate $\frac{d B}{d t}$ is different at every instant — we must differentiate. Why this step? "Rise over run" only works for straight lines; here the slope changes.
Differentiate. $\frac{d B}{d t} = \frac{d}{d t} (k t^{2}) = 2 k t .$ Why this step? Power rule: the derivative of $t^{2}$ is $2 t$ . This is why the EMF isn't constant — it grows linearly with $t$ .
EMF magnitude. $∣ E ∣ = A \frac{d B}{d t} = A \cdot 2 k t .$ At $t = 3.0 s$ : $∣ E ∣ = 0.020 \times 2 \times 0.50 \times 3.0 = 0.060 V .$

Example 8 — Real-world word problem `(cell C9)`

Translate words to a knob. The coil doesn't tilt or resize; the field it experiences drops. So this is the $\frac{d B}{d t}$ (changing-field) case, cell C1's cousin. Why this step? Word problems are just a knob in disguise; naming it picks the right formula. Here the strands the coil sees suddenly vanish as it clears the strip.
Average field-rate. $\frac{Δ B}{Δ t} = \frac{2.0 \times 1 0 ^{- 3}}{5.0 \times 1 0 ^{- 4}} = 4.0 T/s .$ Why this step? "Average" means use the total change over the total time, not an instantaneous derivative — the field is not given as a smooth formula.
Plug into Faraday's law. $∣ \overset{ˉ}{E} ∣ = N A \frac{Δ B}{Δ t} = 300 \times (1.0 \times 1 0^{- 4}) \times 4.0 = 0.12 V .$ Why this step? Same $N A d B / d t$ machinery as Ex 1 — the coil is fixed, only the field changes, and the $N = 300$ turns stack the EMF so the tiny per-turn signal becomes readable.
Check the magnitude sense. $0.12 V = 120 mV$ — hundreds of millivolts, not microvolts and not volts. Why this step? A quick order-of-magnitude check catches slipped powers of ten; $120 mV$ is a signal a real reader can detect.

Example 9 — The limiting spike `(cell C10)`

Total flux change. $∣Δ Φ_{B} ∣ = A ∣Δ B ∣ = 0.10 \times 0.60 = 0.060 Wb .$ Why this step? The flux drops from $B A$ to $0$ ; the amount of change is fixed no matter how fast we do it — the same bundle of strands must leave the hoop.
Average EMF for (a). $∣ \overset{ˉ}{E} ∣ = \frac{∣Δ Φ _{B} ∣}{Δ t} = \frac{0.060}{0.20} = 0.30 V .$ Why this step? Average EMF is total flux change spread over the switch-off time.
The limit for (b). The numerator $∣Δ Φ_{B} ∣ = 0.060 Wb$ is fixed, so $∣ \overset{ˉ}{E} ∣ = \frac{0.060}{Δ t} Δ t \to 0 \infty.$ Why this step? Dividing a fixed number by an ever-smaller time blows up. In real circuits the self-inductance of the coil resists this sudden change and you get a large voltage spike — the reason opening a switch on a coil can throw a spark.

Coverage check — did we hit every cell?

Recall Every matrix cell, with its example

C1 changing field ::: Ex 1 (and Ex 8's cousin) C2 changing area / motional ::: Ex 2 C3 rotating loop / generator + sign per quadrant ::: Ex 3 C4 degenerate edge-on ( $θ = 90°$ ) ::: Ex 4(b) C5 degenerate constant field ::: Ex 4(a) C6 sign / current direction (Lenz) ::: Ex 5 C7 two knobs at once (product rule) ::: Ex 6 C8 non-linear in time ( $B \propto t^{2}$ ) ::: Ex 7 C9 real-world word problem ::: Ex 8 C10 limiting value / spike ::: Ex 9

Connections

Magnetic Flux — the quantity $Φ_{B} = B A cos θ$ we differentiate throughout
Motional EMF — the physics behind Ex 2's sliding rod
Lenz's Law — the direction rule used in Ex 5
Electric Generators and AC — the sine output of Ex 3
Inductance — why the Ex 9 spike is finite in real coils
Maxwell's Equations — Faraday's law in field form
Hinglish parent note

Worked examples — Faraday's law — EMF = −dΦ - dt