Full chapter
Electromagnetic Induction
All 6 topics and the revision summary on one page.
01
Flux through a coil
Magnetic flux combines a field with the area it passes through. Flux linkage adds the contributions through a coil's turns. Define the area and its orientation before calculating either quantity.
For a uniform magnetic flux density B and an area A perpendicular to the field, magnetic flux is:
B describes the local field and is measured in teslas, T. Flux Φ describes the field through the area and is measured in webers, Wb. A field-line drawing helps represent direction and distribution; counting its drawn lines does not determine a numerical flux.
Use the angle to the area normal
A flat surface's normal is a direction perpendicular to its plane. If θ is the angle between B and that normal, the perpendicular projected area is A cos θ:
When the plane is perpendicular to B, its normal is parallel to B and flux magnitude is greatest. When the plane is parallel to B, its normal is perpendicular to B and flux is zero. If an acute angle is given to the plane instead, use the complementary angle or the corresponding sine projection.
0° between the field and the area normal
The plane is perpendicular to B. Its full area is perpendicular to the field, so the flux magnitude is greatest.
60° between the field and the area normal
The angle to the normal is 60 degrees; the acute angle to the plane is 30 degrees. Use A cos(60 degrees), not A cos(30 degrees).
90° between the field and the area normal
The plane is parallel to B. Its perpendicular projected area, flux and linkage are zero.
Choosing a positive normal makes flux signed. A field in the opposite direction gives negative flux, and reversing the chosen normal reverses the assigned sign. The physical field does not reverse merely because a reference changes. For a nonuniform field, a single point's B cannot generally represent the whole area without an appropriate approximation.
Sum flux through the turns
For N turns, each linking the same flux Φ, define flux linkage Λ by:
Λ = NBA cos θ for the stated normal angle
The equal-flux condition matters. If different turns link different fluxes, add their individual fluxes. Increasing N increases linkage for the same imposed field and per-turn area; it does not itself increase that external B.
Worked orientation comparison
Separate one-turn flux from total linkage
A rough field of 0.2 T through about 5 × 10-3 m2 perpendicular to it gives flux of order 10-3 Wb. A few tens of equally linked turns then give a few hundredths of a weber-turn.
Use B = 0.200 T, A = 5.00 × 10-3 m2 per turn and N = 40. At a 60° angle between B and the chosen normal:
= 5.00 × 10-4 Wb
Λ = 40Φ = 2.00 × 10-2 Wb-turn
The angle between the plane and B is 30°, not 60°. With the normal instead parallel to B, one-turn flux is 1.00 × 10-3 Wb and linkage is 0.0400 Wb-turn. At a 90° normal angle, both are zero.
Check units and notation
= 1 kg m2 s-2 A-1
The turn count N is dimensionless. Writing linkage in Wb-turn makes the counting explicit, but a turn introduces no new physical dimension. Here Λ denotes linkage; it is distinct from Φ for flux through one turn and from symbols used for wavelength or decay elsewhere.
Optional check A 40-turn coil has area 5.00 x 10^-3 m^2 per turn in a uniform 0.200 T field. Every turn links the same flux and its positive normal is at 60 degrees to B. What is the flux linkage?
02
Evidence for induction
Compare changes while keeping the coil, leads and viewpoint fixed. A moving meter needle is an observation; the explanation must identify the changed flux linkage, the induced direction and the circuit that carries current.
Connect a fixed coil to a sensitive centre-zero current meter and move a permanent magnet along its axis. The coil and return leads form a complete conducting loop. In the stated winding, A-to-B traversal is clockwise when viewed from the magnet on the left. Keep A connected to the meter's positive terminal and B to its negative terminal throughout the comparison.
A fixed winding and a complete meter circuit
This connected current-meter setup shows a current pulse during approach. Stopping gives no sustained deflection; withdrawal reverses it. A high-impedance voltmeter is a different measurement that can show induced e.m.f. with negligible current.
A deflection in this closed setup indicates a current and supports the presence of induced e.m.f. They are different quantities: e.m.f. is energy supplied per charge, measured in volts, while current is charge flow per time, measured in amperes. A high-impedance voltage measurement can reveal induced e.m.f. with negligible current. An open path prevents sustained loop current, not necessarily induced voltage.
Use controlled observations to infer a relationship
These are comparisons to make and interpret with appropriate apparatus or supplied traces. Record the actual response; an ideal prediction is not itself a measured result.
| Controlled change | Observation or prediction | Inference |
|---|---|---|
| Hold the magnet still, approach the coil, then stop. | No sustained deflection when stationary; a pulse during approach, returning to zero when motion stops. | Changing linkage produces e.m.f.; a constant field alone does not. |
| Approach and withdraw the same pole on the same path. | Opposite meter deflections. | Reversing the linkage change reverses the induced direction. |
| Pass through the same positions at greater speed. | A larger voltage peak with comparable sampling and instrument response. | A greater rate of linkage change gives a greater e.m.f. |
| Repeat the same per-turn field history with twice the linked turns. | The ideal model predicts twice the e.m.f.; compare voltage records to test it. | Total linkage, including turn count, determines the induced e.m.f. |
Changing turn count can also change coil resistance. Therefore twice the e.m.f. does not automatically mean twice the meter current. Use voltage measurements, or account for the known circuit response, before making that inference.
A greater change of B, a larger equally exposed perpendicular area, more linked turns or a faster orientation change can increase the rate of linkage change. Identify what is varied and what remains fixed. A large constant flux still gives zero induced e.m.f.
Infer the opposing direction
View the coil face from the magnet. An approaching north pole increases flux into that face. The induced current produces an outward field, making the near face north and opposing the approach. Its current is anticlockwise from that viewpoint. In the stated winding this is B to A through the coil, with the external return from A through the meter to B.
North approaches: an induced north face
The approaching N increases flux into the face. An outward induced field opposes that increase. Replacing N by S reverses the current and near pole.
North withdraws: an induced south face
Withdrawing N reduces flux into the face. An inward induced field opposes that decrease. The induced field need not oppose the existing external field.
Withdrawing that north pole makes the induced near face south, attracting the departing pole and opposing the change. The current is clockwise. Replacing north with south reverses the predictions: south approach gives clockwise current; south withdrawal gives anticlockwise current.
The clockwise label depends on the viewing side. Looking from the opposite end reverses its apparent sense. Use the coil field-direction rule and a named viewing side rather than memorising an unlabelled clockwise arrow.
Choose controls and diagnose a limited record
Keep the magnet and pole, coil geometry and turns, relative path, lead polarity and acquisition settings fixed when comparing speed. To compare turns, match the per-turn field-change history instead. Use suitable low-voltage laboratory equipment, secure the coil and record repeated traces without changing the reference.
A short pulse can be missed by a slow measurement, so a near-zero display alone does not establish that nothing changed. Choose a range that contains both signs, adequate response and a short enough sample interval. Check the no-change baseline and leads before interpreting a pulse.
Repetition can reveal a repeatable pattern or random variation. A shifted zero needs a baseline correction, reversed leads need the reference restored, and loose contacts need repair. Preserve the original traces and explain a specific reason before excluding a result; do not remove a reading merely because it disagrees with an expected shape.
Optional check A coil experiences changing magnetic flux linkage. A high-impedance voltmeter connected across its terminals draws negligible current. Which conclusion is justified?
03
Linkage changes and induced direction
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.
Local-rate software exercise
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.
Optional check 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?
04
Alternating generation
Rotating a coil changes the area it presents perpendicular to a magnetic field. The changing flux linkage induces e.m.f.; a complete output circuit allows current to flow. Mechanical input supplies the transferred electrical energy.
For a uniform field and N equally linked turns, Λ = NBA cos θ, where θ is the angle from B to the chosen area normal. Faraday's law gives Eind = -dΛ/dt. Follow that same normal and the same terminal labels throughout a rotation.
Keep each coil terminal on its own slip ring
View the coil along its axle, with B to the right and rotation clockwise. Physical active leg A starts above the axle and B below it. Their far ends join; the two near ends connect to separate complete slip rings. A stationary brush touches each ring, connecting the rotating coil to a resistive load.
Each coil end stays connected to its own ring throughout the rotation. The rings are insulated from each other and from the shaft. They maintain the circuit while the output reverses; they are not the two halves of a split-ring commutator.
Define output as VA - VB at the near terminals. The positive coil traversal runs from B-near to B-far, across the far join to A-far, then to A-near. At the starting orientation its right-hand area normal points right, along B.
Two full slip rings preserve two separate coil connections
A stays attached to Ring A and B to Ring B throughout rotation. Each stationary brush remains on its own full ring. At this quarter-turn, near A is positive and supplies the resistive load. Half a turn later the same connections carry the opposite current. The dashed axle is insulated, not a conducting return.
At a quarter-turn, A is on the right and moves down. The magnetic force on positive charge is towards its near terminal, making A positive relative to B. Conventional current in the external load runs from A's brush to B's brush. Half a turn later A is on the left and moves up, so its terminal polarity and the load current reverse.
Worked rotating-coil model
Find output from the linkage gradient
Use N = 40, B = 0.200 T, area A = 5.00 × 10-3 m2 per turn and rotation frequency f = 25.0 Hz. Choose t = 0 with the area normal along B. Rotation is uniform, so:
T = 1/f = 0.0400 s = 40.0 ms
Λ = 0.0400 cos(50πt) Wb-turn
The cosine's angle is in radians and t is in seconds. Applying the negative gradient:
= (0.0400)(50π) sin(50πt)
= 2π sin(50πt) V
The e.m.f. amplitude is 6.28 V. With negligible winding and lead resistance and negligible self-inductance, the terminal output VA - VB equals this induced e.m.f. Those assumptions are needed for the equality under load.
Initial orientation: 0 ms
A is above the axle and B below. The selected normal points right along B, so linkage is at its positive maximum. Its instantaneous rate of change is zero. This same orientation returns at T = 40.0 ms.
Quarter-turn: 10.0 ms
A is right and moving down; B is left and moving up. The selected normal points down. Positive charge is driven towards near A, so A is positive relative to B. In the connected resistive model, current in A is out towards you and current in B is in.
Half-turn: 20.0 ms
A is below the axle and B above. The selected normal points left, opposite B, giving maximum negative linkage. Its instantaneous rate of change and the output are zero again.
Three-quarter-turn: 30.0 ms
A is left and moving up; B is right and moving down. The selected normal points up. Near A is negative relative to near B; current in A is in away from you and current in B is out. The ring and terminal labels have not been exchanged.
The same rotating coil: signed linkage
One turn takes 0.0400 s. The normal starts along B, turns against it after half a turn and returns after a full turn. This graph and the output graph have different vertical quantities and units, but the same horizontal time scale.
Output follows the negative linkage gradient
At a linkage extremum the output is zero; at a zero crossing of linkage its gradient magnitude and output magnitude are greatest. Peak output is 2 pi V, about 6.28 V, with period 0.0400 s. These are model values, not mains or measured data.
At t = 0 the linkage is maximally positive, but its gradient and the e.m.f. are zero. At 10.0 ms, the linkage crosses zero while decreasing most rapidly: output is +6.28 V. At 20.0 ms, linkage is maximally negative and output is zero. At 30.0 ms, linkage crosses zero while increasing most rapidly: output is -6.28 V.
The coil plane is perpendicular to B at the maximum-linkage orientations, and parallel to B at the zero-linkage orientations. Maximum flux linkage does not mean maximum e.m.f. It is the rate of change that matters.
Account for the mechanical input
When current flows through the load, magnetic forces on the coil oppose the driven rotation. An external turning effect must do work to maintain its speed. In the ideal model this mechanical input becomes electrical output; real devices also dissipate energy. An open output can still have induced voltage, with negligible load current and correspondingly negligible load-related opposition.
Resolve the alternating output
This 25 Hz model illustrates why acquisition settings matter. Samples at 0, 20 and 40 ms all give zero, although the output reaches positive and negative peaks at 10 and 30 ms. Use a shorter sampling interval and adequate instrument bandwidth to resolve the waveform. A faster display alone does not guarantee that a short peak was captured.
If available range choices were ±1 V and ±10 V, the latter would contain the roughly ±6.3 V output; the former would clip it. Check the actual instrument's resolution and response when choosing a range. These are supplied range choices, not specifications for a particular instrument.
A baseline offset, reversed leads, clipping and missed peaks have different causes. Check the zero reference, restore the intended polarity, select a sufficient range, or improve time resolution as appropriate. Repetition alone does not correct all four.
Optional check A coil rotates uniformly in a uniform field with linkage Lambda = 0.0400 cos(50 pi t) Wb-turn. At maximum positive linkage, what is the induced e.m.f.?
05
Magnetic braking and energy
A conductor moving into a field can develop a current whose magnetic force opposes the motion. Trace the changing linkage and the complete current path before predicting a braking force.
A rigid rectangular conducting loop moves right into a bounded uniform field directed out of the page. Initially only its right vertical edge is inside the field. The area of the loop exposed to B increases, so outward flux increases.
Lenz's law requires an induced inward field. Viewed from the front, current is therefore clockwise. On the right vertical edge it flows down; in the outward external field, the force on that current-carrying conductor points left. It opposes entry.
Entering the field increases the linked outward flux
Brown arrows follow the complete induced-current path. The right-edge current is down, so its magnetic force is left. At steady speed the applied pull balances that force. Horizontal-edge forces cancel vertically; Fv = 0.0960 W is transferred to I squared R heating.
The complete loop is now inside the uniform field
The linked area and B are now constant even though the loop moves. This closed rigid-loop control has no sustained induced current or braking force in the stated model. It does not describe every eddy-current path in a solid conductor near a field boundary.
Worked entry and energy account
Derive the e.m.f. from changing area
Use B = 0.400 T, vertical edge length l = 0.200 m, speed v = 3.00 m/s and total loop resistance R = 0.600 Ω. Treat the field as uniform within its boundary, resistance as constant and self-inductance as negligible.
During an interval Δt while the loop is entering, its edge moves vΔt and the exposed area increases by lvΔt. The one-turn flux change therefore gives:
= (0.400)(0.200)(3.00) = 0.240 V
I = |Eind|/R = 0.240/0.600
= 0.400 A, clockwise
The right edge is perpendicular to B, so its opposing force has magnitude:
= 0.0320 N
For steady speed, an applied pull of the same magnitude acts right. Its mechanical input power is:
Pheating = I2R = (0.400)2(0.600)
= 0.0960 W
The matching powers check the energy account. External work maintains the speed while the loop warms. Without that pull, and with no other energy input, the heating comes from a decrease in the loop's kinetic energy.
Identify when the braking stops
Once the entire rigid loop is inside the unchanged uniform field, its area, orientation and B are constant. Its total flux is unchanged even while it translates. In the stated negligible-self-inductance model there is no sustained induced e.m.f. around the loop, no current and no associated braking force.
The cancellation requires the whole-loop condition. During entry, only one vertical edge is in the field; after full entry, both are. Motion by itself is not sufficient evidence of induction. A stationary conductor does not acquire a persistent braking current merely because a static B is nonzero.
Extend the reasoning to a solid conductor
A moving solid conductor can contain closed induced-current paths within its material. These eddy currents can create opposing forces and heating where the field and motion change their linkage. Slots interrupt or restrict some current paths and can reduce the braking effect.
Apply the changing-linkage test to the actual geometry. The entering-loop result does not imply that every solid conductor translating anywhere within a uniform field must experience sustained magnetic braking.
Optional check A rigid closed loop has 0.400 A induced while entering a bounded uniform field. It then moves wholly inside the same unchanging uniform field, with its area and orientation fixed. Neglect self-inductance and other effects. What is its sustained model current there?
06
Transformer voltage and current
A transformer uses a changing magnetic flux to transfer energy between electrically separate windings. The turns ratio sets the ideal voltage ratio; conservation of energy then sets the inverse current ratio.
The primary winding connects to an alternating source. Its changing current establishes changing flux in a closed iron core, which links the secondary turns and induces e.m.f. A connected secondary load completes a separate circuit and draws current.
A changing core flux links two separate circuits
Each winding is insulated from the core and from the other winding. Use the supplied turn counts, not the number of drawn turns. In the ideal rms model, 230 V becomes 12.0 V, with input and output both 24.0 W. The separate real example has 30.0 W input, 24.0 W output and 6.0 W loss; it is not the ideal current-ratio model. The supplied 230 V calculation is not a mains investigation.
The core guides flux linking both windings. Charge does not travel through the iron from the primary to the secondary. A steady direct current provides no continuing flux change and therefore no sustained secondary e.m.f. after the switching transient; it is not an appropriate continuous input for this transformer.
State the ideal conditions before using the ratios
In the ideal model, every turn in each winding links the same changing core flux. Winding resistance, flux leakage and core losses are negligible. Faraday's law then makes each winding's e.m.f. proportional to its number of turns. With negligible winding voltage loss, the terminal voltage magnitudes satisfy:
For the sinusoidal source and resistive load used here, V and I are rms values. The ideal input and output powers are equal:
Ns/Np = Vs/Vp = Ip/Is
A step-up transformer increases voltage and reduces the corresponding current; it does not increase ideal power. A step-down transformer does the reverse. Keep peak and rms values distinct when using the mean-power relationship.
Worked ideal transformer
Find the output before finding input current
First estimate the scale. Roughly 1000 primary turns and 50 secondary turns, with a supplied primary voltage near 200 V, suggest an output near 10 V. A load of a few ohms then draws a few amperes and receives tens of watts, requiring primary current of order 0.1 A.
Now use Np = 1150, Ns = 60, Vp = 230 V rms and a secondary load R = 6.00 Ω:
Is = Vs/R = 12.0/6.00 = 2.00 A rms
Pout = VsIs = 24.0 W
Ip = Pout/Vp = 24.0/230
= 0.104 A rms
The smaller primary current is consistent with the larger primary voltage and equal ideal power. This is a supplied calculation model; the operating principle can be investigated with suitable low-voltage laboratory equipment or supplied observations.
Keep a real-loss account separate
In a real transformer, winding resistance produces heating, while changing core magnetisation and induced eddy currents dissipate energy in the core. Flux leakage reduces the flux common to both windings; leakage itself is not simply another name for heating.
Insulated laminations divide the iron core and restrict eddy-current paths. A suitable core material reduces magnetisation losses. These changes reduce losses but do not make the device create energy.
For a separate supplied nonideal case, input power is 30.0 W and useful output is 24.0 W:
Power dissipated = 30.0 - 24.0 = 6.0 W
Do not insert this loss into a calculation still using the ideal equal-power current ratio. Identify whether the question supplies an ideal transformer or a real input/output account before choosing an equation.
Optional check An ideal transformer has 1150 primary turns and 60 secondary turns. The primary receives 230 V rms; the secondary supplies a 6.00 ohm resistor. What is the primary rms current?
Revision
Electromagnetic induction at a glance
Choose the area normal, identify what changes, then connect the linkage rate to e.m.f. Use the actual circuit to decide whether current flows and where energy is transferred.
Flux, linkage and induced e.m.f.
Φ = BA cos θ for angle θ to the normal
Λ = NΦ for equally linked turns
Mean Eind = -ΔΛ/Δt
Instantaneous Eind = -dΛ/dt
- The normal is perpendicular to the plane. A plane parallel to B has zero flux; a plane perpendicular to B has maximum flux magnitude.
- Match the positive normal to positive circuit traversal using the right-hand orientation. Reversing a reference changes the assigned sign, not the physical effect.
- Lenz's law opposes the change producing the effect. An induced field can reinforce an existing field if that field is decreasing.
- A finite graph chord gives an interval mean. A local tangent gives the instantaneous rate. For measured data, shorter intervals can increase sensitivity to reading noise.
- Induced e.m.f. can exist with negligible current. A sustained loop current needs a complete conducting path; I = Eind/R also requires the stated resistance-only model.
In the signed graph example, rising, constant and falling linkage produce -2.00, zero and +2.00 V. The final induced field remains outward as external B passes through zero, because the signed external field is still decreasing. Revisit the matched graph and current references when checking a direction.
Evidence needs controlled comparisons
The three central inferences are that changing linkage induces e.m.f., the induced direction opposes the producing change, and a greater linkage-change rate increases e.m.f. magnitude. Stationary/moving, approach/withdrawal and matched faster/slower comparisons address different parts of that evidence.
Keep the pole, path, coil, viewing side, leads and acquisition settings appropriate to the comparison. More turns can also change resistance, so a current ratio alone does not establish the e.m.f. ratio. A zero-looking record may have missed a pulse; a clipped record cannot give its true peak. The observation and inference table separates recorded responses from ideal predictions.
Applications and their conditions
| Application | Reasoning to retain |
|---|---|
| Rotating generator | Orientation changes linkage. Separate full slip rings preserve each coil connection. Maximum linkage gives zero e.m.f.; the maximum linkage-change rate gives peak output. Equating terminal voltage with induced e.m.f. requires the stated negligible winding/lead resistance and self-inductance. Mechanical input supplies loaded output. |
| Entering loop | Increasing outward flux gives clockwise current and a leftward force on the right edge. For steady entry, Fv = I2R in the supplied ideal model. A complete rigid loop wholly inside an unchanged uniform field has constant linkage and no sustained model current. |
| Iron-core transformer | Changing common flux links separate windings. Ideal ratios are Ns/Np = Vs/Vp = Ip/Is. Use rms values for the stated sinusoidal resistive-load power account. Steady d.c. gives no sustained secondary e.m.f.; real losses require a separate energy account. |
Quantities and units
| Quantity and symbol | SI unit | Meaning or condition |
|---|---|---|
| Magnetic flux density B | T | The magnetic field at a location; use uniform B for the simple area product. |
| Flux Φ | Wb | 1 Wb = 1 T m2 = 1 V s. |
| Flux linkage Λ = NΦ | Wb-turn | Equal per-turn flux is assumed. A turn adds no dimension. |
| Area A; turn count N | m2; no unit | A is one turn's actual area; its perpendicular projection sets flux. |
| Induced e.m.f. Eind; Vp, Vs | V | Eind is a voltage, not field strength. Transformer voltages here are rms. |
| Current I, Ip, Is | A | Specify reference direction or rms magnitude as appropriate. |
| Resistance R | Ω | The braking calculation uses total loop resistance. |
| Time t; period T | s | Period T is distinct from the unit symbol T for tesla. |
| Frequency f; angular frequency ω | Hz; rad/s | ω = 2πf; use radians in the supplied cosine and sine model. |
| Power P | W | Mechanical input, electrical transfer and heating must share the same account. |
In base units, 1 Wb = 1 kg m2 s-2 A-1. Dividing linkage change by time gives Wb/s = V. These checks help distinguish a flux, a voltage and a current even when their graphs share the same time axis.