Lesson 3 of 6
Linkage changes and induced direction
I can recall and solve problems using Faraday's law and Lenz's law.Syllabus
Syllabus 9478, 18(e). Recall and solve problems using Faraday's law and Lenz's law.
Make a guess. You are not marked.
A light copper ring hangs from a thread. The north pole of a strong magnet is pushed quickly towards it along its axis. What does the ring do?
Show the answer
It swings away from the magnet, because its induced current repels it.
The flux through the ring increases, so a current is induced. By Lenz's law it opposes the change: the ring's near face becomes a north pole and is pushed away.
Faraday's law connects induced e.m.f. to the rate of change of magnetic flux linkage. Lenz's law fixes the opposing direction, represented by a minus sign when the normal and circuit references are matched.
Write linkage as Λ = NΦ and induced e.m.f. as Eind. This Eind is a voltage, not electric field strength or elementary charge. Over a finite interval:
Instantaneous Eind = -dΛ/dt
The instantaneous value is the negative gradient of the linkage-time graph. A straight segment has constant gradient; a curved graph needs a local tangent. A finite chord gives an interval mean, not automatically the instantaneous value at either endpoint or at its midpoint. Since turns are dimensionless, linkage-rate units Wb/s reduce to volts.
Match the normal and circuit sense
For the following coil, choose the positive normal out of the page and positive traversal anticlockwise viewed from the front. They follow the right-hand orientation. Positive Eind acts around that chosen traversal; in a resistive closed loop it drives positive anticlockwise current.
Lenz's law says the induced effect opposes the change producing it. If an outward field decreases, an induced outward field opposes that decrease. The induced field does not always oppose the existing external field. In a closed loop the resulting opposition is consistent with conservation of energy.
Worked signed linkage graph
Use each interval's own endpoints
A coil has N = 50 equally linked turns and perpendicular area A = 4.00 × 10-3 m2. Its uniform normal field rises from +0.100 to +0.400 T over 0-0.030 s, stays there to 0.050 s, then falls linearly to -0.100 T at 0.100 s. Positive field is outward.
A linkage change around 0.06 Wb-turn in about 0.03 s suggests an e.m.f. of a few volts. The precise supplied linkage points are:
| t / s | Λ / Wb-turn |
|---|---|
| 0 | +0.020 |
| 0.030 | +0.080 |
| 0.050 | +0.080 |
| 0.100 | -0.020 |
Eind = -(0.080 - 0.020)/(0.030 - 0)
= -2.00 V
0.030 < t < 0.050 s:
Eind = 0 V
0.050 < t < 0.100 s:
Eind = -(-0.020 - 0.080)/(0.100 - 0.050)
= +2.00 V
The ideal sharp corners have no unique instantaneous gradient; the displayed levels apply within the open intervals. A complete loop with total R = 10.0 Ω, negligible self-inductance and no other source has I = Eind/R: -0.200 A, zero and +0.200 A in those intervals.
The first current is clockwise and produces an inward field, opposing the increase in outward flux. The last is anticlockwise and produces an outward field throughout the falling interval. External B crosses zero at 0.090 s, but its continuing decrease still calls for the same outward induced field. Opening the loop removes sustained current while a changing-linkage e.m.f. can remain.
Signed linkage: follow the chosen normal
Positive flux normal: out of the screen. Positive circuit traversal: anticlockwise. The first gradient is +2.00 Wb/s, the middle is zero and the last is -2.00 Wb/s. The final segment crosses zero linkage at 0.090 s.
Induced e.m.f. is the negative gradient
At a sharp ideal corner, do not assign a unique gradient. With a closed 10.0 ohm resistive loop and negligible self-inductance, the interval currents are -0.200 A, 0 and +0.200 A. An open path removes sustained current, not the changing-flux e.m.f.
A local tangent is different from an interval chord
Solid curve and filled dots: supplied cosine model. Dashed line and hollow rings: true tangent and its guides. At 0.010 s the tangent gradient is -2 pi Wb/s, so the instantaneous e.m.f. is +2 pi V. The nearby endpoint samples differ from the guides by about 0.000206 Wb-turn; finite-window estimates remain approximations.
Paper 4 practice: spreadsheet
Use generated linkage values without calling them instantaneous rates
Consider the supplied smooth model Λ = 0.0400 cos(50πt) Wb-turn, with t in seconds and the angle in radians. This is a mathematical trace. The table contains rounded generated values, not instrument readings; nine displayed decimal places are not a claim about measurement resolution.
Enter A1 = Time / s and B1 = Linkage / Wb-turn, then put these numeric values in A2:B6. The motion-data workflow explains numeric entry, formula copying and XY plotting.
| Row | A: time / s | B: linkage / Wb-turn |
|---|---|---|
| 2 | 0.008 | +0.012360680 |
| 3 | 0.009 | +0.006257379 |
| 4 | 0.010 | 0.000000000 |
| 5 | 0.011 | -0.006257379 |
| 6 | 0.012 | -0.012360680 |
Set C1 = Interval midpoint / s and D1 = Mean induced emf / V. Enter =(A2+A3)/2 in C2 and =-(B3-B2)/(A3-A2) in D2. Fill C2:D2 through row 5 only; the final supplied point in row 6 has no following interval.
Plot numeric linkage B vertically against time A. If comparing interval e.m.f. estimates, plot D against its midpoint times C separately. Label units and use numeric XY axes. Keep the raw values while formatting their display; do not force a straight-line fit across the curved trace.
To compare symmetric windows centred at 0.010 s, label G1 as reference time / s, H1 as 2 ms window estimate / V, and I1 as 4 ms window estimate / V:
| Cell | Formula |
|---|---|
| G2 | =A4 |
| H2 | =-(B5-B3)/(A5-A3) |
| I2 | =-(B6-B2)/(A6-A2) |
Compare interval means, centred estimates and the true local value
The four interval means are about 6.1033, 6.2574, 6.2574 and 6.1033 V, at midpoint times 0.0085, 0.0095, 0.0105 and 0.0115 s. These are finite-interval means from the rounded endpoints, not four exact instantaneous voltages.
Cell H2 gives 6.2574 V and cell I2 gives 6.1803 V. The supplied model's true instantaneous value at 0.010 s is 2π V = 6.283185... V. The narrower window better approximates this smooth-model value.
The centred secant uses two curve points on either side of the reference time. Its slope estimates the derivative there, but the secant line is not the tangent. The true tangent is Λ = -2π(t - 0.010) in the stated units. Its guide points at 0.008 and 0.012 s have linkage about +0.01256637 and -0.01256637 Wb-turn; those are tangent-only values, not extra curve readings.
In real records, smaller time intervals can amplify the effect of reading noise on a difference. Balance local resolution against that limitation rather than assuming the shortest interval is always best. A three-point polynomial fit is not automatically a measured tangent. Use the actual-record procedure when working with measured data, retaining its source, timing and uncertainty information.
Try it step by step
Work it out in steps: Change in flux linkage in Wb, then Average induced e.m.f. in V.
Show the answer
Problem 1.
Change in flux linkage in Wb: Change in linkage = N x change in B x A = 300 x 0.40 x 2.5 x 10-3 = 0.30 Wb
Average induced e.m.f. in V: e.m.f. = 0.30 / 0.20 = 1.5 V
Problem 2.
Change in flux linkage in Wb: Change in linkage = 120 x 0.40 x 5.0 x 10-3 = 0.24 Wb
Average induced e.m.f. in V: e.m.f. = 0.24 / 0.080 = 3.0 V
Problem 3.
Change in flux linkage in Wb: The field changes by 0.40 T, so change in linkage = 400 x 0.40 x 1.5 x 10-3 = 0.24 Wb
Average induced e.m.f. in V: e.m.f. = 0.24 / 0.050 = 4.8 V
Watch out for this
A student says: "The induced e.m.f. in a coil is largest when the flux linkage is largest." What is wrong with this?
Show the answer
The e.m.f. depends on the rate of change of flux linkage, not on its size. A large but steady flux linkage induces no e.m.f. at all.
Drill
Lenz's law: five in a row
Streak: 0 of 5
The north pole of a magnet moves towards a closed coil.
Check your understanding
Positive flux normal is out of the page and positive circuit traversal is anticlockwise. Linkage decreases linearly from +0.080 Wb-turn at 0.050 s to -0.020 Wb-turn at 0.100 s. What is the induced e.m.f. during that final open interval?
Show the answer
+2.00 V.
The linkage gradient is (-0.020 - 0.080)/(0.100 - 0.050) = -2.00 Wb/s. Faraday's minus sign gives +2.00 V around the positive anticlockwise traversal.
Practise an exam question
Write your answer first, then tick the mark points it makes.
Exam-style question
State Faraday's law and Lenz's law of electromagnetic induction. [2]
Saved on this browser only. Anyone using this browser can see it.