Copper wire wrapped around an iron core produces current when a strong magnet is moved in its vicinity.

I gather the "magnet movement" part is expressed in units of "weber" (kg⋅m2⋅s−2⋅A−1) or perhaps "tesla" ( kg⋅s−2⋅A−1)

The copper wire of course will be carrying a certain voltage, at some number of amps - dependent on (I think) the number of "turns" of the copper wire, and (I'm guessing) something to do with the speed of the magnet (the unit second-2), and it's strength (the unit kg) and maybe something to do with size (the unit meters2).

Can anyone explain that relationship in a practical (easy to understand in the real world) way?

Like - if I get an N52 neodymium magnet in a 1 cm cube, and I wrap 10 turns of copper around an iron core that's also a 1cm cube - and I move the magnet past the iron (so close it almost touches - so - 0 mm away) at a speed of 10 meters per second... what's my voltmeter and ammeter going to say?

This is not a "homework" question, nothing to do with scams, and I seriously spend ages trying to work this out before asking - please don't down vote my question: if you don't understand something ASK me.

  • \$\begingroup\$ You may think the voltage matters, it really doesn't. If you double the number of turns and halve the wire area, you get twice the voltage, but only half the current capability. So you see it's the power that really matters, the voltage can vary over a wide range as long as the insulation can take it. Your geometries, 1cm3 moving past each other, are inappropriate for motors/generators, far too complicated. Instead, consider a changing field inside a coil. Later on, consider how that field changes when you move magnets within a fixed iron stator assembly. \$\endgroup\$
    – Neil_UK
    Feb 17, 2019 at 11:23
  • 1
    \$\begingroup\$ Remember that the induced voltage is proportional to the rate of change of the magnetic field, and if that voltage is allowed to drive current that current will cause a magnetic field which opposes the change of the outer field. \$\endgroup\$
    – JimmyB
    Feb 17, 2019 at 16:34
  • \$\begingroup\$ Voltage is of course extremely important, because V=IR ... if you need to move a lot of energy a long way, you need to use a lot of volts, or else lose all your power. Also, as all R/C folk know, motor speed is directly proportional to voltage, not current. So the math to work this stuff out in a design situation is critical. \$\endgroup\$ Feb 18, 2019 at 0:17

2 Answers 2


You want the 'weber' unit for magnetic flux, and the significant equation is that the EMF in the external circuit formed by the windings and load is

$$EMF = \int{{\bf E} \centerdot d{\bf L}} = {{d\Phi}\over {dt}}$$ $$ Flux == {\Phi} = \int{\bf B} \centerdot d{\bf A}$$ where the first integral is taken over the whole helical coil path, and Phi, the flux, is cumulative over all the turns whose enclosed area has a magnetic field, B. This is the Maxwell-Faraday law, and it IS practical, but not simple. The development of motors and generators depending on this phenomenon took some decades.

A 'magnet' has some external B-field, and that field, both in magnitude and direction, matters. You cannot complete the calculation without the field B known in all of the region.

That EMF is a voltage equivalent value, and is applied to the whole circuit. If the external circuit is a short, current flows to completely eliminate the flux change (though of course there's always some resistance in common wire). This is because the current in the wire CAUSES flux that opposes the external field which is changing.

The equation holds whether the flux change is the cause of the EMF, or if a voltage is applied to the coil to cause a flux change.

what's my voltmeter and ammeter going to say ?

The voltmeter, the ammeter, and the wires to connect them, will determine that. An ideal ammeter is an electric short circuit, and an ideal voltmeter is an electrical open circuit; connecting both to the coil terminals would be impractical (would show zero volts). There would be current, but only to the meter, not to an external load...

  • \$\begingroup\$ I'm not qualified enough to know if this is better than the other answer - anyone suggest how I know which one to "tick" for my question? \$\endgroup\$ Feb 18, 2019 at 8:51

Biot-Savart Law and Faraday Law of Induction can be combined, into a relation between a FIXED WIRE with a changing current, and that wire having a FIXED DISTANCE to a loop with FIXED AREA; for easy math, the wire is in the same plane as the loop.

The resulting equation is

Vinduce = [MU0 * MUr * LoopArea / (2 * PI * Distance_wire_loop)] * dI/dT

and for MU0 = 4 * PI * 1e-7, and MUr = 1 (air,copper, plastic) the equation becomes

Vinduce = 2e-7 * Area/Distance * dI/dT

This equation is conservative, in that an EXACT answer uses natural_log of distance ratios.

Thus you might examine what you need to vary. And read up on Biot_Savart Law.


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