This question came up when I used our vacuum cleaner (230 V, 50 Hz AC, 1 kW) without pulling out most of the cable, i.e. about 8 meters were still wrapped around a small cable reel (maybe 5-10 cm diameter) inside the device. I got told this should be avoided because the cable might produce an electromagnetic field or heat up strongly when it's wound up like that.

Is this true? Would it be dangerous to operate a device consuming much AC power with a tightly wound up cable? What could happen worst-case and how high power consumption would be needed to expect any effect? Or is that just a myth?

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    $\begingroup$ The coil of wire inside would just be a bad solenoid for which the magnetic field is given by: $$B = \frac{\mu_{o} \ N \ I}{2 \ \pi \ r}$$ where $r$ is the radius of the loop, $N$ is the number of wraps, and $I$ is the current through the wire. Vacuum cleaners tend to use a lot of current (small ones use $\gtrsim$15 amps and larger ones even more). So you can go through and figure out how large $B$ would be. However, I think heat would be more of an issue than the field... $\endgroup$ – honeste_vivere Sep 26 '16 at 12:07
  • $\begingroup$ It is declared to use 1 kW on 230 V, so that would be around 4 A. But isn't the magnetic field actually zero because we have current flowing in both directions inside each cable? $\endgroup$ – Byte Commander Sep 26 '16 at 12:11
  • $\begingroup$ Yes, I agree with Byte Commander. Current flows in both directions in that wire loops thus net field will be zero. $\endgroup$ – soosai steven Sep 26 '16 at 13:05
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    $\begingroup$ @ByteCommander - The current is alternating, thus it is not simultaneously flowing in both directions. The coil of wire would act like an inductor (e.g., an RL-circuit in the simplest case) but I used solenoid as an over simplification for a hand-wavy estimate of the potential field strength to illustrate that it's not an issue. The field would be zero if the coil were wrapped evenly and tightly, but this is almost certainly not true... Regardless, as I said before, heat is your issue not the field... $\endgroup$ – honeste_vivere Sep 26 '16 at 13:54
  • $\begingroup$ @ByteCommander - Oh I see what you mean... I thought you were confusing something else... Yes, there is a two-way current (one out and one return current), but as I said, this would only cancel if the cable were wrapped perfectly. Regardless, the field would be very small and heat would be the issue... $\endgroup$ – honeste_vivere Sep 26 '16 at 13:56

A typical (cheap) extension reel (10 metres, 10 amps) has the following specifications:

Uncoiled 2400W (10 A) 240 V
Coiled 720 W (3 A) 240 V

The reel had 4 sockets and a 13 A fuse in the mains plug.

Measuring the lead resistances I found them to be approximately $0.3 \; \Omega$ and $0.4 \; \Omega$ so a total of $0.7 \Omega$.

Suppose that the maximum current $13$ A hen the power dissipated in the whole cable would be $13^2 \times 0.7 \approx 120$ W.

If the reel is tightly wound and assuming that the electrical insulation on the cables is also a good insulator of heat you have the potential for quite a rise in the temperature of the cable even if some of the $120$ watts is lost through the outer surfaces of the reel.

In the past I have certainly noticed a wound up reel getting hot (rather than just warm) so unless I am using small power devices have always at least partially unwound a reel.


A quick sum shows that the radius of the copper conductor has to be a rather low 0.4 mm to give a resistance of 0.35 $\Omega$ for a 10 m extension reel so I am undertaking a deconstruction job on the extension reel.

Even if the resistance of the copper cable is a factor of $10$ were and the power dissipated is of the order of $10$ watts this still has the possibility of raising the temperature of the reel to a dangerous, melting insulation, level.

Update 2

The cable used had a copper conductor cross sectional area of 2.5 mm$^2$ and PVC insulation rated up to a maximum working temperature of 70 $^\circ$C.
The total calculated resistance of a 10 m reel is 0.13 $\Omega$ and so with 13 A passing through the reel the power dissipated as heat is 22 watts.
Still sufficient over a period of time to raise the temperature inside a tightly coiled reel to a level where the temperature rise will be quite high.

  • $\begingroup$ So winding the cable up does only affect its ability to cool down? Or is there also an electromagnetic effect? $\endgroup$ – Byte Commander Sep 26 '16 at 12:59
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    $\begingroup$ I do not think there is an electromagnetic effect. As one of the other comments points out the currents in the live and neutral are in opposite directions so the net field will be small. $\endgroup$ – Farcher Sep 26 '16 at 13:53

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