If we make an electromagnet by winding 3000 turns on 5-inch-long cast iron bar, and provide $12 \;\text{V}$ and $2 \;\text{A}$ of current, it will produce the magnetic field with certain magnitude.

The question is: If we increase the weight of the objects being lifted by that electromagnet, will it consume more current? If we lift much smaller weight by that electromagnet, will it use lesser current?

The input current and voltage is constant, which means there's no resistor (other than the wire's resistance) between battery and electromagnet.

  • $\begingroup$ you wrote: "The input current and voltage is constant", so if they are fixed they must also be independent of the weight $\endgroup$
    – hyportnex
    Jul 19, 2021 at 15:25

3 Answers 3


It is not possible that both the input voltage and current are given as constant. Therefore this answer assumes that:

  • The 12V power supply is a stable DC power supply such as a car-battery.
  • The current of 2A mentioned in the question is the measured current.

Under these assumptions, the question becomes:

If we increase the weight of the objects being lifted by that electromagnet, will we measure more current?

The answer is no, because:

When we connect 12V to the winding, the current will raise very fast while the magnetic field builds up, and once the magnetic field is stable, the current will only depend on the resistance of the winding (assuming the 12V DC is stable).

When we add some weight to the load of the magnet, there may be a change in the magnetic field and that may cause a very short change in the current, but once the magnetic field is steady again, the measured current will be the same as before because it depends only on the resistance of the winding.

The resistance of the winding will not be affected by the load, it depends only on the material (such as copper), the cross section, the length and the temperature.


As far as I know, there is no simple yes-or-no answer to this question. The theory of electrodynamics uses Maxwell's equations to calculate forces on charges and magnetic moments, it has nothing to do with the motion of bodies or gravity, for which their mass (having weight in some gravitational field) is important. This means, mass will never directly influence some electromagnetic field.

However, if the magnetic force accelerates the body, it's magnetic moments will also be accelerated which could again generate electromagnetic fields influencing the current in the coil. To calculate this effect, though, at least the geometry of the body and the material of which it consists would have to be known.


To start with, this situation is not feasible (compactible with Ohm's Law). If you have 12 V and 2 A you have to have a resistance, more specifically impedance, of exactly 6 $\Omega $.

You can't just say that no I will have way lesser resistance but I have to have 12 V, 2 A. So, if you have 6 $\Omega $ resistance, the power should be given by $I^2 R$.

If you are wondering about how can an electromagnet can lift a heavy as well as a light weight efficiently (as in some cranes), the answer is straight forward, they have a regulator that controls the current that flows through the circuit by changing the impedance (increasing the resistance will decrease the current but that's very inefficient since the power dissipated across that resistance becomes huge). By the way this is exactly the principle on which fans (ceiling /table) works where producing more wind requires more power and so more current (which can be verified if you experiment with your monthly electricity bills).


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