Lenz's law — opposing induced current: Worked examples

Before anything, one symbol contract so we never sneak notation past you:

Definition The symbols we will use (each earned once, here)

$Φ_{B}$ = magnetic flux — think of it as the number of magnetic field lines threading through the loop. Big loop or strong field or lines hitting the loop straight-on → more lines through → bigger $Φ_{B}$ . Full build in Magnetic flux. Formula for a uniform field: $Φ_{B} = B A cos θ$ .
$B$ = strength of the external magnetic field (unit: tesla, T).
$A$ = area of the loop the field passes through (unit: $m^{2}$ ).
$θ$ = the angle between the field $B$ and the loop's normal (an imaginary arrow poking straight out of the loop's face).
$N$ = number of turns (loops of wire stacked in a coil). $N$ identical loops each catch the same flux, so the total is $N$ times bigger — that's why $N$ multiplies the EMF for a coil.
$\frac{d Φ _{B}}{d t}$ = how fast the flux is changing per second. The $\frac{d}{d t}$ means "rate of change with time". Positive = flux growing; negative = flux shrinking; zero = flux frozen.
$ε$ = induced EMF (the electrical "push", in volts) given by $ε = - \frac{d Φ _{B}}{d t}$ (Faraday's law of induction).
$R$ = resistance of the loop (ohms, $Ω$ ); the induced current is $I = \frac{∣ ε ∣}{R}$ .

The scenario matrix

Everything Lenz's law does is decide which way the loop fights. Only three things can change flux $Φ_{B} = B A cos θ$ : the field $B$ , the area $A$ , or the angle $θ$ . Combine that with the sign of the change and the degenerate cases, and here is the complete map:

#	Cell (case class)	What changes	Sign of $d Φ_{B} / d t$	Covered by
C1	Field grows	$B ↑$ , $A, θ$ fixed	$+$	Ex 1
C2	Field shrinks	$B ↓$	$-$	Ex 2
C3	Area grows (moving conductor)	$A ↑$	$+$	Ex 3
C4	Rotation (angle sweeps)	$θ$ changes	oscillating $\pm$	Ex 4
C5a	Degenerate: field constant, no motion	nothing	$0$ (no current!)	Ex 5a
C5b	Degenerate: motion parallel to field	$θ = 9 0^{\circ}$ always	$0$	Ex 5b
C6	Limit: very fast vs very slow motion	$v \to \infty$ , $v \to 0$	scales with $v$	Ex 6
C7	Real-world word problem	falling magnet / brake	$-$ then damping	Ex 7
C8	Exam twist: loop shrinks / sign trap	$A ↓$ or reversed $B$	flips	Ex 8

We now fill every cell.

Cell C1 — Field grows

Worked example Example 1 — Ramping up an electromagnet (cell C1)

A flat circular loop of radius $r = 0.10 m$ lies flat, face-up, inside a uniform field pointing straight up through it ( $θ = 0$ ). The field is turned up from $B_{0} = 0.20 T$ to $B_{1} = 0.80 T$ in $t = 0.30 s$ , steadily. The loop's resistance is $R = 2.0 Ω$ . Find $∣ ε ∣$ , the current, and the current's direction (viewed from above).

Forecast: the field points up and grows. Guess: does the loop's induced current push with or against that upward field? Write your guess before reading.

Area of the loop. $A = π r^{2} = π (0.10)^{2} = 0.0314 m^{2}$ . Why this step? Flux needs area; the field goes straight through so $cos θ = cos 0 = 1$ and $Φ_{B} = B A$ .
Rate of flux change. Only $B$ changes, so $\frac{d Φ _{B}}{d t} = A \frac{d B}{d t} = A \frac{B _{1} - B _{0}}{t} = 0.0314 \times \frac{0.80 - 0.20}{0.30}$ . Why this step? Faraday needs the rate; $A$ is constant so it just multiplies the slope of $B$ . $\frac{d B}{d t} = \frac{0.60}{0.30} = 2.0 T/s$ , so $\frac{d Φ _{B}}{d t} = 0.0628 Wb/s$ .
EMF magnitude. $∣ ε ∣ = 0.0628 V = 62.8 mV$ .
Current. $I = \frac{∣ ε ∣}{R} = \frac{0.0628}{2.0} = 0.0314 A = 31.4 mA$ .
Direction (Lenz). Upward flux is increasing → the loop opposes the increase → its induced field must point downward inside the loop → by the Right-hand rule (curl fingers so thumb points down), the current runs clockwise viewed from above. See the figure below.

Verify: units — $T \cdot m^{2} / s = Wb/s = V$ ✓. If your forecast said "the loop pushes with the field," that would help the increase → free runaway energy → forbidden. The correct answer opposes it. ✓

Figure — Lenz's law — opposing induced current

Figure 1 (cell C1). The blue up-arrows are the growing external field $B$ . The loop fights the increase by making its own induced field (red, pointing down) inside the loop. Curling the right hand so the thumb points down gives the yellow clockwise current, seen from above.

Cell C2 — Field shrinks

Worked example Example 2 — Switching the electromagnet off (cell C2)

Same loop as Ex 1 ( $A = 0.0314 m^{2}$ , $R = 2.0 Ω$ ). Now the field is dropped from $0.80 T$ to $0$ in $0.20 s$ . Find current and direction.

Forecast: field was up, now vanishing. Which way does the loop fight now — the opposite of Ex 1?

Rate. $\frac{d Φ _{B}}{d t} = A \frac{0 - 0.80}{0.20} = 0.0314 \times (- 4.0) = - 0.1256 Wb/s$ . Why this step? Negative because flux is falling. The sign matters for direction.
EMF. $∣ ε ∣ = 0.1256 V$ .
Current. $I = \frac{0.1256}{2.0} = 0.0628 A = 62.8 mA$ .
Direction (Lenz). Upward flux is decreasing → the loop wants to prop it up → induced field points up (same as the dying field) → RHR gives counterclockwise viewed from above — the exact reverse of Ex 1.

Verify: the sign of $d Φ_{B} / d t$ flipped from $+$ (Ex 1) to $-$ (Ex 2), so the current direction flips. This is the parent note's key point: oppose the change, not the field. ✓ Here "opposing the change" means pointing the induced field with $B$ . ✓

Cell C3 — Area grows (moving conductor)

Worked example Example 3 — Sliding rod, full numbers (cell C3)

A rod of length $L = 0.50 m$ slides at $v = 3.0 m/s$ on rails in a field $B = 0.40 T$ pointing into the page. Loop resistance $R = 1.5 Ω$ . Find $∣ ε ∣$ , $I$ , the drag force, the power, and the current direction. (Deep quantitative build lives in Motional EMF and sliding rod.)

Forecast: the enclosed area grows as the rod slides. Guess the current direction (clockwise / counterclockwise) before computing.

Flux. With the rod at position $x$ , area $= Lx$ , so $Φ_{B} = B Lx$ . Why this step? Field is uniform and perpendicular to the loop ( $θ = 0$ ), so $Φ_{B} = B A = B Lx$ .
Rate. $\frac{d Φ _{B}}{d t} = B L \frac{d x}{d t} = B Lv$ . Only $x$ changes. $∣ ε ∣ = B Lv = 0.40 \times 0.50 \times 3.0 = 0.60 V$ .
Current. $I = \frac{∣ ε ∣}{R} = \frac{0.60}{1.5} = 0.40 A$ .
Direction (Lenz). Flux into the page is growing → induced field must point out of the page inside the loop → RHR ⇒ current counterclockwise. See the figure below.
Drag force. A current-carrying rod in the field feels $F = B I L = 0.40 \times 0.40 \times 0.50 = 0.080 N$ , directed against the motion. Why this step? Lenz guarantees the force opposes what you're doing — you must push.
Power balance. Mechanical power in $= F v = 0.080 \times 3.0 = 0.24 W$ . Electrical power dissipated $= I^{2} R = (0.40)^{2} \times 1.5 = 0.24 W$ .

Verify: $F v = I^{2} R = 0.24 W$ — they match exactly. This is Conservation of energy: every joule you push in becomes heat. ✓ Units: $T \cdot m \cdot m/s = V$ ✓.

Figure 2 (cell C3). The blue "×" marks are the field $B$ going into the page. As the yellow rod slides right at speed $v$ (green), the enclosed area — and so the inward flux — grows. Lenz forces the red counterclockwise current (making field out of page inside the loop), and the rod feels the red drag force pointing back against $v$ .

Cell C4 — Rotation (angle sweeps)

Worked example Example 4 — Rotating loop (AC generator) (cell C4)

A coil of area $A = 0.020 m^{2}$ with $N = 50$ turns spins at angular speed $ω = 100 rad/s$ in a field $B = 0.30 T$ . Its normal makes angle $θ = ω t$ with the field. Find the peak EMF and explain the oscillating sign.

Forecast: now nothing is entering or leaving; the coil just turns. Where in the spin is the EMF biggest, and where is it zero? Guess before reading.

Flux vs angle. $Φ_{B} = N B A cos θ = N B A cos (ω t)$ . Why this step? Rotation changes the angle between field and normal, so the $cos θ$ factor does the work here — not $B$ or $A$ . The factor $N$ appears because all $N$ turns catch the same flux.
Rate. $\frac{d Φ _{B}}{d t} = - N B A ω sin (ω t)$ , so $ε = N B A ω sin (ω t)$ . Why this step? The derivative of $cos$ is $- sin$ ; the minus from $- d Φ/ d t$ and the minus from the derivative give a clean $+ sin$ .
Peak EMF. $sin$ maxes at $1$ : $ε_{m a x} = N B A ω = 50 \times 0.30 \times 0.020 \times 100 = 30 V$ .
Where zero, where max? When the coil's face is perpendicular to the field ( $θ = 0$ , flux maximum) the rate $sin θ = 0$ , so $ε = 0$ . When the face is parallel to the field ( $θ = 9 0^{\circ}$ , flux zero) the rate is fastest, so $ε$ is peak. The sign flips every half turn — that's why this is alternating current.

Verify: peak EMF $= 30 V$ . Sanity: units $T \cdot m^{2} \cdot s^{- 1} = V$ ✓. The famous counterintuitive fact — EMF is zero when flux is maximum — falls straight out of " $ε$ tracks the rate, not the value." ✓

Cell C5a and C5b — Degenerate cases (no current at all)

Cell C6 — Limiting behaviour (fast vs slow)

Worked example Example 6 — How drag scales with speed (cell C6, limits)

Take the sliding rod of Ex 3 ( $B = 0.40 T$ , $L = 0.50 m$ , $R = 1.5 Ω$ ). Write the drag force as a function of $v$ , then evaluate at $v = 0$ , $v = 3.0 m/s$ , and $v = 30 m/s$ .

Forecast: does doubling the speed double the drag, or worse? And at $v = 0$ ?

General drag. From Ex 3: $I = \frac{B Lv}{R}$ and $F = B I L = \frac{B ^{2} L ^{2} v}{R}$ . Why this step? Chain the two results — force is linear in $v$ .
Coefficient. $\frac{B ^{2} L ^{2}}{R} = \frac{( 0.40 ) ^{2} ( 0.50 ) ^{2}}{1.5} = \frac{0.16 \times 0.25}{1.5} = 0.02667 N\cdots/m$ .
Evaluate.
- $v = 0$ : $F = 0$ — a rod at rest feels no magnetic drag. (Degenerate limit, matches C5a.)
- $v = 3.0$ : $F = 0.02667 \times 3.0 = 0.080 N$ — matches Ex 3 ✓.
- $v = 30$ : $F = 0.02667 \times 30 = 0.80 N$ — ten times the speed, ten times the drag (linear).

Verify: the parent warned against "faster ⇒ instant stop." Because $F \propto v$ (not $v^{2}$ or worse), the drag rises smoothly and only ever removes energy — it never reverses the motion or creates energy. As $v \to \infty$ the force is large but finite for finite $v$ ; as $v \to 0$ it fades to nothing. ✓

Cell C7 — Real-world word problem

Worked example Example 7 — Magnet falling through a copper pipe (cell C7)

A small magnet is dropped down a vertical copper pipe. Instead of accelerating at $g$ , it drifts down slowly at a nearly constant "terminal" speed. Explain with Lenz's law and estimate the terminal condition.

Forecast: why is a magnet slower in copper than in a plastic pipe of the same size? Guess the mechanism.

Falling = changing flux. As the magnet falls, the flux through each ring-shaped slice of pipe below it grows (magnet approaching) and through each slice above it shrinks (magnet leaving). Both changes drive eddy currents in the copper walls. Why this step? Copper is a conductor, so loops of induced current can circulate — that's what Lenz needs.
Lenz direction. Below the magnet the current makes a face that repels it (fights the approach); above, a face that attracts it (fights the departure). Both act upward — a retarding force opposing the fall.
Terminal condition. The magnet speeds up until the magnetic drag $F_{drag} (v)$ (which grows with $v$ , exactly like Ex 6) balances gravity: $F_{drag} (v_{term}) = m g$ . After that, net force $= 0$ , so speed is constant.

Verify: In a plastic pipe no current can flow ( $R = \infty$ , so $I = 0$ ), no drag — the magnet falls at $g$ . In copper the drag is real, and all the lost gravitational energy becomes heat in the pipe (Conservation of energy). This is the same physics as Eddy currents and magnetic braking used in train brakes and drop-tower rides. ✓

Cell C8 — Exam twist (sign trap)

Worked example Example 8 — Shrinking loop with a reversed-field trap (cell C8)

A square loop, side $s$ , lies in a field $B = 0.50 T$ pointing out of the page. Someone squeezes the loop so its side shrinks from $s_{0} = 0.20 m$ to $s_{1} = 0.10 m$ in $t = 0.40 s$ . $R = 0.80 Ω$ . Find current magnitude and direction. Then answer the twist: if the field instead pointed into the page, would the current direction change?

Forecast: the loop is shrinking — is that an increase or decrease of flux? And which way does the current go? Guess.

Flux. $Φ_{B} = B s^{2}$ (field straight through, $cos 0 = 1$ ).
Rate. $\frac{d Φ _{B}}{d t} = B \frac{d ( s ^{2} )}{d t}$ . Using average slope: $\frac{s _{1}^{2} - s _{0}^{2}}{t} = \frac{0.01 - 0.04}{0.40} = \frac{- 0.03}{0.40} = - 0.075 m^{2} / s$ . So $\frac{d Φ _{B}}{d t} = 0.50 \times (- 0.075) = - 0.0375 Wb/s$ — flux decreasing (smaller loop catches fewer lines). Why this step? Area falls with $s^{2}$ ; shrinking is a decrease even though you're "doing something."
EMF and current. $∣ ε ∣ = 0.0375 V$ , $I = \frac{0.0375}{0.80} = 0.0469 A \approx 46.9 mA$ .
Direction (Lenz). Out-of-page flux is decreasing → loop props it up → induced field points out of page → RHR ⇒ current counterclockwise.
The twist. If the field pointed into the page instead, then flux (into page) is also decreasing → induced field must point into the page to prop it up → RHR ⇒ current clockwise — the opposite direction.

Verify: the current magnitude is identical in both field orientations ( $∣ ε ∣$ only depends on $∣ d Φ_{B} / d t ∣$ ), but the direction reverses when the field reverses. The trap is thinking "shrinking loop = increasing something." No: shrinking area with fixed field = decreasing flux. ✓

Recall Self-test: name the cell before you solve

A coil sits in a field that is constant in size but the coil is slowly tilted from face-on to edge-on. Increase or decrease of flux? Which cell? ::: Decrease ( $cos θ$ falls from $1$ to $0$ ); it's a C4-type (angle change) sliding toward the C5b degenerate zero-flux limit. The sliding rod is suddenly stopped. What is the induced current the instant after? ::: Zero — $v = 0$ so $∣ ε ∣ = B Lv = 0$ (the C6 $v \to 0$ limit). Which quantity flips the current direction: the sign of $B$ , or the sign of $d Φ_{B} / d t$ , or both? ::: Both — reversing either one alone flips the current; reversing both leaves it unchanged (Ex 8).

Connections

Faraday's law of induction — supplies $∣ ε ∣ = ∣ d Φ_{B} / d t ∣$ used in every example.
Magnetic flux — the $Φ_{B} = B A cos θ$ we differentiated throughout.
Right-hand rule — turned each "oppose the change" field into a current direction.
Motional EMF and sliding rod — the full theory behind Ex 3 and Ex 6.
Eddy currents and magnetic braking — the mechanism of Ex 7.
Conservation of energy — the power balance $F v = I^{2} R$ verified in Ex 3.
Self-inductance and back-EMF — Lenz applied to a coil's own changing current.