Rewinding a Scrap-Yard Lifting Magnet
A JTS trainee at a scrap dealer's yard is asked to rewind the coil of a circular lifting electromagnet used on a 3-tonne crane. The iron core is cast iron with a mean magnetic path length of 0.40 m and a uniform cross-section of 20 cm². When the magnet sits flat on a steel plate the working air gap is 2 mm, and at the working flux density of 1.2 T the relative permeability of the cast iron is about 200. The former carries 1200 turns of copper wire; after rewinding, the coil resistance measures 30 Ω and it is energised from the yard's 110 V DC supply. The dealer complains that the rewound magnet picks up thin plates well but drops thicker, slightly warped plates that the old coil used to hold, and that the coil body becomes noticeably warm after about twenty minutes of continuous lifting. The trainee has to decide whether the winding is at fault, whether the supply is adequate, and how much the air gap matters.
1. Using μ0 = 4π×10⁻⁷ H/m, calculate the mmf needed to establish 1.2 T in the gap and the current the coil must carry. Compare it with the current the 30 Ω coil actually draws at 110 V and state whether the rewind is adequate.
Read the worked answer
Air-gap reluctance (A = 20 cm² = 2×10⁻³ m², lg = 2×10⁻³ m): Rg = lg/(μ0A) = 2×10⁻³/(4π×10⁻⁷ × 2×10⁻³) ≈ 7.96×10⁵ A·t/Wb. Iron reluctance: Ri = l/(μ0μrA) = 0.40/(4π×10⁻⁷ × 200 × 2×10⁻³) ≈ 7.96×10⁵ A·t/Wb. Total R ≈ 1.59×10⁶ A·t/Wb. Required flux Φ = B×A = 1.2 × 2×10⁻³ = 2.4×10⁻³ Wb. mmf = ΦR = 2.4×10⁻³ × 1.59×10⁶ ≈ 3,820 A·t, so I = 3820/1200 ≈ 3.2 A. The coil actually draws 110/30 = 3.67 A, about 15% more than the minimum, so the rewind is adequate - the magnet is not weak because of too few turns or too little current. Note the coil dissipates 3.67² × 30 ≈ 404 W; this is a high-current, intermittent-duty magnet, and twenty minutes of continuous lifting is outside its duty, which explains the warmth. The complaint must therefore be traced to the air gap and to core saturation, not to the winding.
2. On warped plates the effective gap is closer to 4 mm. Recalculate the flux density obtainable from the same coil and estimate how the holding force changes.
Read the worked answer
Doubling the gap doubles the air-gap reluctance to 1.59×10⁶ A·t/Wb, giving a total of 2.39×10⁶ A·t/Wb. The coil can still produce only NI = 1200 × 3.67 = 4,400 A·t, so: Φ = 4400/2.39×10⁶ = 1.84×10⁻³ Wb and B = 1.84×10⁻³/2×10⁻³ = 0.92 T. Holding force F = B²A/(2μ0):
- at 1.2 T: F = 1.44 × 2×10⁻³/(8π×10⁻⁷) ≈ 1,150 N (about 117 kgf);
- at 0.92 T: F = 0.846 × 2×10⁻³/(8π×10⁻⁷) ≈ 675 N (about 69 kgf). The force falls to roughly 59% for a 2 mm change in gap, because force varies as B² and the air gap dominates the magnetic circuit. Practical conclusion: the plate face and the poles must be clean and flat, rust and scale must be removed, and a magnet designed for a 2 mm gap cannot be judged by a test on a warped plate. Check the coil for shorted turns only after the gap has been corrected.
3. Explain why the same magnet design cannot simply be connected to an AC supply, and what changes AC operation would force.
Read the worked answer
On AC the mmf must alternate, so the flux reverses every half cycle. Since force is proportional to B², the pull falls to zero twice per cycle: the magnet chatters, is noisy, and loses its grip on the load - a shaded ring or DC holding coil is needed to keep the force unidirectional. The alternating flux also induces eddy currents in a solid cast-iron core, giving heavy eddy-current loss and heating, so the core would have to be laminated (or made of soft iron with a low-loss magnetic path). Finally, on AC the coil current is limited mainly by inductive reactance (I ≈ V/ωL) rather than by resistance, so the same 1200 turns at the same voltage would not produce the same mmf; the winding must be redesigned for the required flux at 50 Hz. A solid steel plate lying across the poles also behaves like a short-circuited secondary turn, further reducing the flux. DC operation is preferred for lifting magnets for exactly these reasons.