Faraday and Lenz’s Laws: Flux, Signs and Current Direction

1. Find the direction without calculus

Imagine a closed conducting loop lying flat on this page. You are looking straight at the front of the page. “Out of the page” means toward you; “into the page” means away from you. A dot represents the tip of an arrow coming toward you. A cross represents the tail of an arrow going away.

Keep the loop’s area and orientation fixed, with a uniform external field perpendicular to it. The direction results below describe a resistive loop with negligible self-inductance, so its current follows the induced emf. We use conventional current, the direction positive charge would move.

Lenz’s law asks what the induced current’s magnetic field would do to the change in flux. A growing outward field calls for an induced inward field. A weakening outward field calls for an induced outward field: the response resists the loss.

1. External field out of the page, strengthening: induced field into the page; current clockwise.
2. External field out of the page, weakening: induced field out of the page; current counterclockwise.
3. External field into the page, strengthening: induced field out of the page; current counterclockwise.
4. External field into the page, weakening: induced field into the page; current clockwise.

For the right-hand check, point your right thumb in the induced field direction. Your curled fingers show conventional current. Thumb toward you gives counterclockwise current; thumb away gives clockwise current.

Try saying the whole reason: “The inward flux is decreasing, so the induced field points inward to oppose that decrease; from the front, the current is clockwise.” “It opposes the field” would give the wrong result here. The response need not cancel the applied field or stop its change completely.

If the loop, its orientation and the external field all remain unchanged, there is no induced emf. An open loop can still have an induced emf, but it has no sustained circulating conduction current. The current answers in this lesson require a closed circuit.

2. Choose a normal and measure the right angle

A normal is a direction perpendicular to a surface. Choose one side of the loop as positive and draw a unit normal $\hat{\mathbf n}$ pointing that way. For a flat loop in a uniform magnetic field:

$$\Phi_B=BA\cos\theta$$

• $B$ is the external field’s magnitude, in teslas ($\mathrm T$).
• $A$ is the area enclosed by one turn, in square metres ($\mathrm{m^2}$).
• $\theta$ is the angle from the chosen normal to the field, from $0^\circ$ to $180^\circ$.
• $\Phi_B$ is the signed magnetic flux through one turn, in webers: $1\,\mathrm{Wb}=1\,\mathrm T\,\mathrm{m^2}$.

The component along the normal is what counts. A field aligned with the positive normal gives $+BA$; an oppositely directed field gives $-BA$. A field parallel to the plane gives zero flux, even when its magnitude is large. This simple product assumes the field is uniform over the area; a nonuniform field requires adding the contributions from different parts of the surface.

Worked example 1: 30° to the plane is 60° to the normal

A rectangular loop measures $0.20\,\mathrm m$ by $0.15\,\mathrm m$. A uniform $0.40\,\mathrm T$ field makes an angle of $30^\circ$ with its plane and points toward the chosen positive side. Find the flux.

$$A=(0.20)(0.15)=0.030\,\mathrm{m^2}$$

$$\theta=90^\circ-30^\circ=60^\circ$$

$$\Phi_B=(0.40)(0.030)\cos60^\circ$$

$$\Phi_B=0.0060\,\mathrm{Wb}=6.0\,\mathrm{mWb}$$

Equivalently, use $BA\sin30^\circ$ for this geometry. Using $BA\cos30^\circ$ gives about $10.4\,\mathrm{mWb}$, because it selects the wrong component.

Check the limits: if the field lies exactly in the plane, the angle to the normal is $90^\circ$ and the flux is zero. If the field points along the positive normal, the flux is $12\,\mathrm{mWb}$. Our answer lies between those values. A single flux value says nothing yet about an induced emf; we still need to know how the flux changes with time.

Changing a convention is different from turning the loop. If you simply replace $\hat{\mathbf n}$ by $-\hat{\mathbf n}$ for the same physical loop, the signed flux reverses. If you physically rotate the loop in a fixed field, its flux can actually change with time and produce an emf.

3. Turn changing flux into a signed emf

Once the normal is chosen, use the right-hand rule to choose the positive direction around the wire. For a normal pointing toward our front observer, positive circulation is counterclockwise. Positive emf drives current that way in the resistive model; negative emf drives it clockwise.

For a coil of $N$ series-connected turns, each threaded by the same flux $\Phi_B$, Faraday’s law gives the average induced electromotive force over a time interval:

$$\overline{\mathcal E}=-N\frac{\Phi_{B,f}-\Phi_{B,i}}{t_f-t_i}$$

Here $t_i$ and $t_f$ are the initial and final times; $\Phi_{B,i}$ and $\Phi_{B,f}$ are the corresponding initial and final fluxes per turn. The bar marks a time average.

Despite its name, electromotive force (emf) is measured in volts, not newtons. It describes energy transferred per unit charge around the circuit. $N\Phi_B$ is the coil’s flux linkage. If a problem already gives flux linkage, do not multiply by $N$ again.

When a flux-time graph is a straight line on an interval, its slope is constant, and this calculation gives the instantaneous emf everywhere inside that interval too. For a curved graph it gives a time average; a tangent slope is needed for one instant.

• Signed flux rising: positive slope, negative emf.
• Signed flux falling: negative slope, positive emf.
• Signed flux constant: zero slope, zero emf, whether the flux is zero or nonzero.

Keep the sign until you have translated it into a direction. For magnitude alone, take $|\mathcal E|$. The unit check is $\mathrm{Wb/s}=\mathrm V$; a slope in $\mathrm{mWb/s}$ gives millivolts per turn.

When can you use current = emf ÷ resistance?

Our current calculations use a closed, purely resistive circuit of total resistance $R>0$, no other emf sources and negligible self-inductance and capacitive effects:

$$I=\frac{\mathcal E}{R}$$

The signed current uses the same positive circulation as the emf. Self-inductance is a coil’s response to changes in its own current. If that response matters, current can lag behind the external emf or continue briefly after it becomes zero. Then $I=\mathcal E_{\rm external}/R$ is not the complete transient model. The zero-current conclusions below belong to our stated resistive approximation, not to every real coil.

In the following examples, $\Phi_B$ denotes flux from the external field. We neglect the coil’s self-induced contribution when using the resistive approximation.

4. Read a piecewise flux-time graph

Worked example 2: the height gives flux; the slope gives emf

A stationary coil has $80$ turns, area $0.020\,\mathrm{m^2}$ per turn and total resistance $12\,\Omega$. A uniform external field is perpendicular to its plane. Choose the normal out of the page and look from the front. The graph gives signed flux per turn; join the listed points with straight segments.

From 0 to 2 s: positive slope

$$\frac{\Delta\Phi_B}{\Delta t}=\frac{8-2}{2-0}=3\,\mathrm{mWb/s}$$

$$\mathcal E=-80(3\times10^{-3})=-0.240\,\mathrm V$$

$$I=\frac{-0.240}{12}=-0.0200\,\mathrm A$$

The current is $20.0\,\mathrm{mA}$ clockwise. Outward flux is growing, so the induced field points inward.

From 2 to 4 s: nonzero flux, zero slope

The flux stays at $8\,\mathrm{mWb}$. Therefore $\mathcal E=0$ and $I=0$ in our model. The external field is still $B=0.008/0.020=0.40\,\mathrm T$, directed outward. Zero emf does not mean zero field.

From 4 to 5 s: negative slope through zero flux

$$\frac{\Delta\Phi_B}{\Delta t}=\frac{-4-8}{5-4}=-12\,\mathrm{mWb/s}$$

$$\mathcal E=-80(-12\times10^{-3})=+0.960\,\mathrm V$$

$$I=\frac{0.960}{12}=+0.0800\,\mathrm A$$

The current is $80.0\,\mathrm{mA}$ counterclockwise throughout this open interval. Initially the outward field is weakening; after flux passes through zero, an inward field is strengthening. Both changes call for an outward induced field. Passing through zero flux does not make the emf zero.

From 5 to 7 s: negative flux with positive slope

$$\frac{\Delta\Phi_B}{\Delta t}=\frac{0-(-4)}{7-5}=2\,\mathrm{mWb/s}$$

$$\mathcal E=-80(2\times10^{-3})=-0.160\,\mathrm V$$

$$I=-0.0133\,\mathrm A\quad\text{(approximately)}$$

The current is about $13.3\,\mathrm{mA}$ clockwise. The inward external field is weakening, so the induced field points inward. Negative flux alone does not determine the sign of the emf.

At the corners: the ideal graph has different left and right slopes at $t=2,4,5\,\mathrm s$, so it has no unique derivative there. Our answers apply only inside each interval. We do not assign an instantaneous emf at those corners or at the two outer endpoints. A physical transition has a finite time scale that this simplified graph does not specify.

5. Optional calculus: a smooth changing flux

If derivatives are new, you can finish Questions 1–5 first. For this extension, review how to differentiate a polynomial. Replace the straight-segment slope with the instantaneous rate of change:

$$\mathcal E(t)=-N\frac{d\Phi_B}{dt}$$

Worked example 3: zero emf at the top of the flux curve

A fixed $50$-turn coil has area $0.020\,\mathrm{m^2}$ per turn and total resistance $5.0\,\Omega$. The field is uniform and perpendicular to it. Keep outward normal and counterclockwise positive. For $0\le t\le4\,\mathrm s$, the external flux per turn is:

$$\begin{aligned}\Phi_B(t)={}&6\,\mathrm{mWb}\\&+(4\,\mathrm{mWb/s})t\\&-(1\,\mathrm{mWb/s^2})t^2.\end{aligned}$$

The units on each coefficient matter: every term is a flux. Differentiate the constant, linear and quadratic terms separately:

$$\frac{d\Phi_B}{dt}=4\,\mathrm{mWb/s}-(2\,\mathrm{mWb/s^2})t$$

• At $t=1\,\mathrm s$, the slope is $+2\,\mathrm{mWb/s}$. Thus $\mathcal E=-0.100\,\mathrm V$ and $I=-0.0200\,\mathrm A$: $20.0\,\mathrm{mA}$ clockwise.
• At $t=2\,\mathrm s$, the slope is zero, so $\mathcal E=I=0$. But $\Phi_B=10\,\mathrm{mWb}$, and $B=0.010/0.020=0.50\,\mathrm T$ outward.
• At $t=3\,\mathrm s$, the slope is $-2\,\mathrm{mWb/s}$. Thus $\mathcal E=+0.100\,\mathrm V$ and $I=+0.0200\,\mathrm A$: $20.0\,\mathrm{mA}$ counterclockwise.

Over the whole interval from 0 to 4 s, the initial and final fluxes are equal, so the average emf is zero. That does not make the emf zero throughout the interval. The clockwise and counterclockwise contributions cancel in the signed time average.

6. Check where the energy comes from

An induced current heats a resistor at a rate $P=I^2R$. Reversing current does not make that heating negative. The energy must come from somewhere: the source changing the field, mechanical work changing the geometry, or stored energy released by the wider system.

For example, when a magnet moves toward a resistive loop, the induced response resists the approach. Keeping the magnet moving requires mechanical input; if it is free to move, it can lose kinetic energy. Lenz’s sign is consistent with that energy transfer. It does not create an extra supply of energy.

Choose the system before claiming that energy is conserved within it. The loop alone can receive energy from outside, and mechanical energy alone can decrease while thermal energy increases. Use the same system-and-transfer reasoning in work, energy and momentum: choosing the method.

7. Try six graduated questions

Use a closed resistive circuit with no other emf source, negligible self-inductance and negligible capacitive effects whenever a current is requested. Unless a rotation is stated, area and orientation are fixed. All stated fields are uniform over each turn. Try the question before opening the hint or solution.

8. Keep a five-line solution routine

1. Name the observer, choose the normal, and pair it with positive circulation.
2. Identify what changes: field, area, angle, or a combination.
3. Calculate flux per turn; keep the angle to the normal and the units explicit.
4. Take the appropriate slope, multiply by $-N$, and use $I=\mathcal E/R$ only when its assumptions hold.
5. Translate the sign into a physical direction and check it with Lenz’s law.

For adjacent circuit work, practise current arrows and circuit signs. To return to your course guide, choose Dawson 203-SN2-RE, Concordia PHYS 205 or McGill PHYS 102.

If you want to work through a diagram or a difficult sign together, see electricity and magnetism tutoring.

References and course connections

Checked 19 October 2026. The sources below support the physical principles and topic connections.

• OpenStax University Physics, volume 2, §13.1: Faraday’s law and §13.2: Lenz’s law. For the current-response limitation, see §14.4: RL circuits.
• Dawson’s current 203-SN2-RE description, Concordia’s PHYS 205 description and McGill’s 2026–27 PHYS 102 description include electromagnetic induction.