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.
The complete route is shown at the positive quarter-turn: A connects through its full ring and brush to the load, then through the other brush and full ring to B. The far join completes the coil. Winding and lead resistance and self-inductance are neglected, so the labelled terminal voltage equals the induced e.m.f. Drawn dimensions and turn count are schematic.
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:
ω = 2πf = 50π rad/s 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:
Eind = -dΛ/dt = (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.
The four distinct orientations use the same physical A/B legs, rightward field and clockwise rotation. The initial orientation repeats at T. Linkage and output graphs share time coordinates, with separate vertical quantities and units.
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.?