Slug length and pitch

How long is one magnetised chunk of the medium, and how far apart are they?

ℓ = v \cdot T1 = \frac{v \cdot D}{f}, \qquad p = \frac{v}{f}

Method

1. Convert the velocity to metres a second (50 in/s is 1.27 m/s) and the duty to a fraction.
2. The period is 1 ÷ f, and the on-time T₁ is the duty times the period. At 100 Hz and 50 % that is 5 ms.
3. The slug length is the velocity times the on-time: everything that passed through the coil while it was energised. The pitch is the velocity times the period, v ÷ f, and the gap between slugs is the pitch less the slug.
4. The current in a coil switched onto a voltage rises as 1 − e^(−t/τ) with τ = L/R. The medium that passed during the rise, v·τ, is the leading-edge width; the fraction of the final current reached by the end of the pulse is 1 − e^(−T₁/τ).
5. Nothing further happens here. Whether the slug is still a slug when it reaches the pickup coil is the smearing calculation's question.

Assumptions

• Plug flow: every part of the cross-section moves at the mean velocity, so a slug has flat ends. In laminar flow the centre moves at twice the mean and the wall not at all, and the slug is a paraboloid rather than a cylinder from the moment it is written — that is handled downstream, not here.
• The medium magnetises as fast as the field rises. For micron particles in a liquid the magnetic response is effectively instantaneous on a millisecond scale; what lags is the current, and that is the τ modelled.
• The current falls as fast as it rose, so the trailing edge is as graded as the leading one and the tail is a mirror of the front. A free-wheeling diode makes the fall slower than the rise, and a snubber makes it faster.
• Nothing here is Meyer's except the velocity the EPG #1 holding specifies. The frequency, duty and coil time constant are the reader's; the arithmetic is two multiplications and one exponential.