Making an Electromagnet Lift Three Times Its Load
A scrap dealer lifts iron scrap with an electromagnet built from a 12 V battery, a switch and a solenoid of 300 turns wound on a soft-iron core. The coil uses 1.0 mm² enamelled copper wire, whose safe continuous current is 4 A; its measured resistance is 6 Ω, so it draws 2 A and lifts 30 kg. He now needs to lift 75 kg. His assistant proposes rewinding the same core with 600 turns of the same wire, arguing that twice the turns must give twice the magnetism. A second worker suggests leaving the coil alone and simply doubling the supply to 24 V. The dealer has a variable 0–30 V supply, a spare reel of the same wire, the same core, and time for one rewind. The load must drop cleanly the instant the switch is opened, and the coil must not exceed the wire's 4 A limit even briefly. Lifting force on this core is proportional to the product of turns and current (the ampere-turns).
1. Test the assistant's proposal quantitatively: 600 turns of the same wire on the same core, still on 12 V. What load will it lift?
Read the worked answer
For the same wire and the same core, resistance is proportional to the number of turns: 300 turns gave 6 Ω, so 600 turns gives about 12 Ω.
Current = V/R = 12/12 = 1 A. Ampere-turns = 600 × 1 = 600 - exactly what the old coil had (300 × 2 = 600). The lift stays at 30 kg, not 60 kg.
The general result is worth stating: at a fixed supply voltage, nI = nV/R = nV/(k·n) = V/k, which does not contain n at all. Adding turns adds resistance in the same proportion, so the extra turns are cancelled by the extra resistance. Winding more turns only helps if the current can be held constant, which here it cannot.
For comparison, the second worker's 24 V on the original 300-turn coil gives 4 A and 1200 ampere-turns, i.e. 60 kg - better, but still short of 75 kg and sitting exactly on the wire's limit.
2. Using the 0–30 V supply and one rewind, find a combination that lifts 75 kg while respecting the 4 A limit. Show the numbers and choose the safer option.
Read the worked answer
Lift is proportional to ampere-turns: 600 A·turns lifts 30 kg, so 75 kg needs 600 × 75/30 = 1500 A·turns.
Option A - 375 turns: R = 6 × 375/300 = 7.5 Ω. To reach 1500 A·turns the coil needs 4 A, so V = 4 × 7.5 = 30 V. It works exactly at the wire's 4 A limit, leaving no margin for a warm coil or a slightly low-resistance batch of wire.
Option B - 500 turns: R = 10 Ω. At 3 A the ampere-turns are 500 × 3 = 1500, and the required voltage is 3 × 10 = 30 V - inside the supply's range, with 1 A of headroom below the wire rating.
Option B is the better engineering choice: same lift, 25% margin on current, and heat produced is I²R = 9 × 10 = 90 W rather than 120 W. Either way the coil should be rated for intermittent duty, since 90 W in a compact winding will still get hot.
3. Explain why the core is soft iron rather than steel, and what would go wrong with a steel core when the switch is opened.
Read the worked answer
Soft iron has high permeability, so a given number of ampere-turns produces a much stronger field than an air core would - that is why the same coil lifts far more with the core in place. It also has low retentivity and low coercivity: it magnetises strongly while current flows and loses almost all its magnetism when the current stops.
That second property is essential here. When the switch is opened the scrap must fall away immediately. With a steel core, high retentivity means the core stays magnetised after switch-off and the load hangs on; the operator would have to reverse the current briefly to demagnetise it, which is slow and unsafe over a lorry or a furnace.
Laminating the core is not needed for steady DC, since there is no changing flux to drive eddy currents, but it would matter if the coil were fed with alternating current.